The Humongous Book of Algebra Problems

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Ltd., 24 Sturdee Avenue, Rosebank, Johannesburg 2196, South Africa Penguin Books Ltd., Registered Offices: 80 Strand, London WC2R 0RL, England Copyright © 2008 by W. Michael Kelley All rights reserved. No part of this book shall be reproduced, stored in a retrieval system, or transmitted by any means, electronic, mechanical, photocopying, recording, or otherwise, without written permission from the publisher. No patent liability is assumed with respect to the use of the information contained herein. Although every precaution has been taken in the preparation of this book, the publisher and author assume no responsibility for errors or omissions. Neither is any liability assumed for damages resulting from the use of information contained herein. For information, address Alpha Books, 800 East 96th Street, Indianapolis, IN 46240. International Standard Book Number: 978-1-59257-722-4 Library of Congress Catalog Card Number: 2008920832 10â•… 09â•… 08â•…â•… 8â•… 7â•… 6â•… 5â•… 4â•… 3â•… 2â•… 1 Interpretation of the printing code: The rightmost number of the first series of numbers is the year of the book’s printing; the rightmost number of the second series of numbers is the number of the book’s printing. For example, a printing code of 08-1 shows that the first printing occurred in 2008. Printed in the United States of America Note: This publication contains the opinions and ideas of its author. It is intended to provide helpful and informative material on the subject matter covered. It is sold with the understanding that the author and publisher are not engaged in rendering professional services in the book. If the reader requires personal assistance or advice, a competent professional should be consulted. The author and publisher specifically disclaim any responsibility for any liability, loss, or risk, personal or otherwise, which is incurred as a consequence, directly or indirectly, of the use and application of any of the contents of this book. Most Alpha books are available at special quantity discounts for bulk purchases for sales promotions, premiums, fund-raising, or educational use. Special books, or book excerpts, can also be created to fit specific needs. For details, write: Special Markets, Alpha Books, 375 Hudson Street, New York, NY 10014. Contents Introduction Your one-stop shop for a review of numbers Chapter 1: Arithmetic Fundamentals 1 Numbers fall into different groups Number Classification ................................................................................................. 2 subtract, multiply, and divide positive and negative numbers 5 Expressions Containing Signed Numbers ..Add, ...................................................................... When numbers band together, deal with them first Grouping Symbols ....................................................................................................... 8 Basic assumptions about algebra Algebraic Properties ................................................................................................... 11 Chapter 2: Rational Numbers em Understanding fractions sure beats being afraid of th 17 Proper and improper fractions, decimals, and mixed numbers 18 Rational Number Notation . ....................................................................................... Reducing fractions to lowest terms, like 1/2 instead of 5/10 Simplifying Fractions . ............................................................................................... 23 Add, subtract, multiply, and divide fractions Combining Fractions ................................................................................................. 26 37 Chapter 3: Basic Algebraic Expressions Time for x to make its stunning debut The alchemy of turning words into math Translating Expressions . ........................................................................................... 38 Rules for simplifying expressions that contain powers Exponential Expressions ............................................................................................ 40 Multiply one thing by a bunch of things in parentheses Distributive Property . ................................................................................................ 45 My dear Aunt Sally is eternally excused Order of Operations . ................................................................................................. 48 Replace variables with numbers Evaluating Expressions ............................................................................................. 51 Chapter 4: Linear Equations in One Variable How to solve basic equations 55 Add to/subtract from both sides Adding and Subtracting to Solve an Equation .............................................................. 56 Multiply/divide both sides Multiplying and Dividing to Solve an Equation . .......................................................... 59 Nothing new here, just more steps Solving Equations Using Multiple Steps ....................................................................... 61 Most of them have two solutions Absolute Value Equations . ......................................................................................... 70 Equations with TWO variables (like x and y) or more 73 Equations Containing Multiple Variables . ................................................................... The Humongous Book of Algebra Problems iii Table of Contents Identify the points that make an e quat ion true Chapter 5: Graphing Linear Equations in Two Variables 77 Which should you use to graph? Number Lines and the Coordinate Plane ...................................................................... 78 Plug in some x’s, plot some points, call it a day Graphing with a Table of Values ................................................................................. 83 The easiest way to plot two points on a line quickly Graphing Using Intercepts ......................................................................................... 90 Figure out how slanty a line is Calculating Slope of a Line ........................................................................................ 93 Don’t miss the point in these graphs (Get it?) Graphing Absolute Value Equations ...........................................................................100 Chapter 6: Linear Equations in Two Variables Generating equations of lines 105 Point +Â€slopeÂ€=Â€equation Point-Slope Form of a Linear Equation .......................................................................106 Lines that look likeÂ€yÂ€=Â€mxÂ€+Â€b Slope-Intercept Form of a Linear Equation ...................................................................110 Graphing equations that are solved for y Graphing Lines in Slope-Intercept Form ......................................................................113 Write equations of lines in a uniform way Standard Form of a Linear Equation ..........................................................................118 Practice all the skills from this chapter Creating Linear Equations ........................................................................................121 Chapter 7: Linear Inequalities 127 They’re like equations without the equal sign Dust off your equation-solving skills from Chapter 4 Inequalities in One Variable ......................................................................................128 Shoot arrows into number lines Graphing Inequalities in One Variable . ......................................................................132 Two inequalities for the price of one Compound Inequalities . ...........................................................................................135 Break these into two inequalities Absolute Value Inequalities . ......................................................................................137 A fancy way to write solutions Set Notation ............................................................................................................140 Lines that give off shade in the coordinate plane Graphing Inequalities in Two Variables ......................................................................142 W ork wit h more than one equation at a time 147 Chapter 8: Systems of Linear Equations and Inequalities Graph two lines at once Graphing Linear Systems ..........................................................................................148 Solve one equation for a variable and plug it into the other The Substitution Method . .........................................................................................153 Make one variable disappear and solve for the other one Variable Elimination ................................................................................................162 The answer is where the shading overlaps Systems of Inequalities ..............................................................................................168 Use the sharp points at the edge of a shaded region Linear Programming ................................................................................................173 181 Chapter 9: Matrix Operations and Calculations Numbers in rows and columns The order of a matrix and identifying elements Anatomy of a Matrix . ..............................................................................................182 Combine numbers in matching positions Adding and Subtracting Matrices ..............................................................................183 Not as easy as adding or subtracting them Multiplying Matrices . ..............................................................................................188 Values defined for square matrices only Calculating Determinants .........................................................................................192 Double-decker matrices that solve systems Cramer’s Rule .........................................................................................................200 iv The Humongous Book of Algebra Problems Table of Contents Chapter 10: Applications of Matrix Algebra Advanced matrix stuff 207 Extra columns and lots of 0s and 1s Augmented and Identity Matrices . .............................................................................208 Swap rows, add rows, or multiply by a number Matrix Row Operations ............................................................................................211 More matrices full of 0s with a diagonal of 1s Row and Reduced-Row Echelon Form .........................................................................216 Matrices that cancel other matrices out Inverse Matrices ......................................................................................................228 Chapter 11: Polynomials 237 Clumps of numbers and variables raised to powers Labeling them based on the exponent and total terms Classifying Polynomials ............................................................................................238 Only works for like terms Adding and Subtracting Polynomials .........................................................................239 FOIL and beyond Multiplying Polynomials . .........................................................................................244 A lot like long dividing integers Long Division of Polynomials ....................................................................................246 Divide using only the coefficients Synthetic Division of Polynomials ...............................................................................251 Chapter 12: Factoring Polynomials The opposite of multiplying polynomials 257 Largest factor that divides into everything evenly Greatest Common Factors ..........................................................................................258 You can factor out binomials, too Factoring by Grouping ..............................................................................................265 Difference of perfect squares/cubes, sum of perfect cubes Common Factor Patterns ...........................................................................................267 Turn one trinomial into two binomials Factoring Quadratic Trinomials . ...............................................................................270 Square r oots, c Chapter 13: Radical Expressions and Equations ube roots, and fractional exponents 275 Moving things out from under the radical Simplifying Radical Expressions . ...............................................................................276 Fractional powers are radicals in disguise Rational Exponents . ................................................................................................281 Add, subtract, multiply, and divide roots Radical Operations ..................................................................................................283 Use exponents to cancel out radicals Solving Radical Equations ........................................................................................288 Numbers that contain i, which equals √— –1 Complex Numbers.....................................................................................................290 Solve equations containing x2 295 Chapter 14: Quadratic Equations and Inequalities Use techniques from Chapter 12 to solve equations Solving Quadratics by Factoring ................................................................................296 Make a trinomial into a perfect square Completing the Square ..............................................................................................300 Use an equation’s coefficients to calculate the solution Quadratic Formula ..................................................................................................305 What b2Â€–Â€4ac tells you about an equation Applying the Discriminant ........................................................................................312 Inequalities that contain x2 One-Variable Quadratic Inequalities ...........................................................................316 The Humongous Book of Algebra Problems ˘ Table of Contents Chapter 15: Functions 323 Named expressions that give one output per input What makes a function a function? Relations and Functions ...........................................................................................324 +, –, ·, and ÷ functions Operations on Functions ...........................................................................................326 Plug one function into another Composition of Functions . ........................................................................................330 Functions that cancel each other out Inverse Functions . ...................................................................................................335 Function rules that change based on the x-input Piecewise-Defined Functions . .....................................................................................343 Chapter 16: Graphing Functions 347 Drawing graphs that aren’t lines Plug in a bunch of things for x Graphing with a Table of Values ................................................................................348 What can you plug in? What comes out? Domain and Range of a Function ..............................................................................354 Pieces of a graph are reflections of each other Symmetry ................................................................................................................360 The graphs you need to understand most Fundamental Function Graphs...................................................................................365 Move, stretch, squish, and flip graphs Graphing Functions Using Transformations ................................................................369 These graphs might have sharp points Absolute Value Functions...........................................................................................374 Chapter 17: Calculating Roots of Functions 379 RootsÂ€=Â€solutionsÂ€=Â€x-intercepts Factoring polynomials given a head start Identifying Rational Roots ........................................................................................380 The ends of a function describe the ends of its graph Leading Coefficient Test ............................................................................................384 Sign changes help enumerate real roots Descartes’ Rule of Signs ............................................................................................388 Find possible roots given nothing but a function Rational Root Test ...................................................................................................390 Factoring big polynomials from the ground up Synthesizing Root Identification Strategies ...................................................................394 Chapter 18: Logarithmic Functions 399 Contains enough logs to build yourself a cabin Given logaÂ€bÂ€=Â€c, find a, b, or c Evaluating Logarithmic Expressions . .........................................................................400 All log functions have the same basic shape Graphs of Logarithmic Functions ...............................................................................402 What the bases equal when no bases are written Common and Natural Logarithms .............................................................................406 Calculate log values that have weird bases Change of Base Formula ...........................................................................................409 Expanding, contracting, and simplifying log expressions Logarithmic Properties ..............................................................................................412 Functions with a variable in the exponent 417 Graphs that start close to yÂ€=Â€0 and climb fast Graphing Exponential Functions ...............................................................................418 Chapter 19: Exponential Functions They cancel each other out Composing Exponential and Logarithmic Functions .....................................................423 Cancel logs with exponentials and vice versa Exponential and Logarithmic Equations .....................................................................426 Use f(t)Â€=Â€Nekt to measure things like population Exponential Growth and Decay . ................................................................................433 vi The Humongous Book of Algebra Problems Table of Contents Chapter 20: Rational Expressions 439 Fractions with lots of variables in them Reducing fractions by factoring Simplifying Rational Expressions ...............................................................................440 Use common denominators Adding and Subtracting Rational Expressions .............................................................444 denominators not necessary Multiplying and Dividing Rational Expressions .Common .........................................................452 Reduce fractions that contains fractions Simplifying Complex Fractions ...................................................................................457 Rational functions have asymptotes Graphing Rational Functions . ..................................................................................459 r 20 pte a h Chapter 21: Rational Equations and Inequalities Solve equations using the skills from C 465 When two fractions are equal, “X” marks the solution Proportions and Cross Multiplication .........................................................................466 Ditch the fractions or cross multiply to solve Solving Rational Equations ......................................................................................470 Turn a word problem into a rational equation Direct and Indirect Variation .....................................................................................475 Critical numbers, test points, and shading Solving Rational Inequalities ....................................................................................479 Chapter 22: Conic Sections Parabolas, Circles, Ellipses, and Hyperbolas 487 Vertex, axis of symmetry, focus, and directrix Parabolas ...............................................................................................................488 Center, radius, and diameter Circles ....................................................................................................................494 Major and minor axes, center, foci, and eccentricity Ellipses ...................................................................................................................499 Transverse and conjugate axes, foci, vertices, and asymptotes Hyperbolas ..............................................................................................................506 st is earned? intere h c u 515 Chapter 23: Word Problems If two trains leave the station full of consecutive integers, how m Integer and age problems Determining Unknown Values ...................................................................................516 Simple, compound, and continuously compounding Calculating Interest . ................................................................................................521 Area, volume, perimeter, and so on Geometric Formulas . ................................................................................................525 Distance equals rate times time Speed and Distance ..................................................................................................529 Measuring ingredients in a mixture Mixture and Combination ........................................................................................534 How much time does it save to work together? Work ......................................................................................................................538 Appendix A: Algebraic Properties 545 Appendix B: Important Graphs and Graph Transformations 547 Appendix C: Key Algebra Formulas 551 Index 555 The Humongous Book of Algebra Problems vii Introduction Are you in an algebra class? Yes? Then you NEED this book. Here’s why: Fact #1: The best way to learn algebra is by working out algebra problems. There’s no denying it. If you could figure this class out just by reading the textbook or taking good notes in class, everybody would pass with flying colors. Unfortunately, the harsh truth is that you have to buckle down and work problems out until your fingers are numb. Fact #2: Most textbooks only tell you WHAT the answers to their practice problems are, but not HOW to do them! Sure, your textbook may have 175 problems for every topic, but most of them only give you the answers. That means if you don’t get the answer right you’re totally out of luck! Knowing you’re wrong is no help at all if you don’t know why you’re wrong. Math textbooks sit on a huge throne like the Great and Terrible Oz and say, “Nope, try again,” and we do. Over and over. And we keep getting the problem wrong. What a delightful way to learn! (Let’s not even get into why they only tell you the answers to the odd problems. Does that mean the book’s actual author didn’t even feel like working out the even ones?) Fact #3: Even when math books try to show you the steps for a problem, they do a lousy job. Math people love to skip steps. You’ll be following along fine with an explanation and then all of a sudden BAM, you’re lost. You’ll think to yourself, “How did they do that?” or “Where the heck did that 42 come from? It wasn’t there in the last step!” Why do almost all of these books assume that in order to work out a problem on page 200, you’d better know pages 1 through 199 like the back of your hand? You don’t want to spend the rest of your life on homework! You just want to know why you keep getting a negative number when you’re calculating the minimum cost of building a pool whose length is four times the sum of its depth plus the rate at which the water is leaking out of a train that left Chicago at 4:00 a.m. traveling due west at the same speed carbon decays. viii The Humongous Book of Algebra Problems Introduction Fact #4: Reading lists of facts is fun for a while, but then it gets old. Let’s cut to the chase. Just about every single kind of algebra problem you could possibly run into is in here—after all, this book is HUMONGOUS! If a thousand problems aren’t enough, then you’ve got some kind of crazy math hunger, my friend, and I’d seek professional help. This practice book was good at first, but to make it great, I went through and worked out all the problems and took notes in the margins when I thought something was confusing or needed a little more explanation. I also drew little skulls next to the hardest problems, so you’d know not to freak out if they were too challenging. After all, if you’re working on a problem and you’re totally stumped, isn’t it better to know that the problem is SUPPOSED to be hard? It’s reassuring, at least for me. All of my notes are off to the side like this and point to the parts of the book I’m trying to explain. I think you’ll be pleasantly surprised by how detailed the answer explanations are, and I hope you’ll find my little notes helpful along the way. Call me crazy, but I think that people who want to learn algebra and are willing to spend the time drilling their way through practice problems should actually be able to figure the problems out and learn as they go, but that’s just my two cents. Good luck and make sure to come visit my website at www.calculus-help.com. If you feel so inclined, drop me an email and give me your two cents. (Not literally, though—real pennies clog up the Internet pipes.) Acknowledgments Special thanks to the technical reviewer, Paula Perry, an expert who doublechecked the accuracy of what you’ll learn here. I met Paula when she was a student teacher (and I had only a year or two under my belt at the time). She is an extremely talented educator, and it’s almost a waste of her impressive skill set to merely proofread this book, but I am appreciative nonetheless. The Humongous Book of Algebra Problems ix Trademarks All terms mentioned in this book that are known to be or are suspected of being trademarks or service marks have been appropriately capitalized. Alpha Books and Penguin Group (USA) Inc. cannot attest to the accuracy of this information. Use of a term in this book should not be regarded as affecting the validity of any trademark or service mark. Dedication For my son Nick, the all-American boy who loves soccer, Legos, superheroes, the Legend of Zelda, and pretending that he knows karate. You summarized it best, kiddo, when you said, “You know why I love you so much, Dad? Because we’re the same.” For my little girls, Erin (who likes to hold my hand during dinner) and Sara (who loves it when I tickle her until she can’t breathe). In a strange way, I am proud that at three years old, you’ve both mastered the way to say “Daaadeeeee...,” which suggests that I both amuse and greatly embarrass you at the same time. Most of all, for my wife, Lisa, who cheers me on, pulls me through, picks me up, and makes coming home the only reason I need to make it through every day. ˘ The Humongous Book of Algebra Problems Chapter 1 Algebraic Fundamentals Your one-stop shop for a review of numbers Algebra, at its core, is a compendium of mathematical concepts, axioms, theorems, and algorithms rooted in abstraction. Mathematics is most powerful when it is not fettered by the limitations of the concrete, and the first step toward shedding those restrictions is the introduction of the variable, a structure into which any number of values may be substituted. However, algebra students must first possess considerable knowledge of numbers before they can make the next logical step, representing concrete values with abstract notation. This chapter ensures that you are thoroughly familiar with the most common classifications used to describe numbers, provides an opportunity to manipulate signed numbers arithmetically, and investigates the foundational mathematical principles that govern algebra. You might be anxious to dive into the nuts and bolts of algebra, but don’t skip over the stuff in this chap ter. It’s full of key vocabulary words , such as “rational number” and “comm utative property.” You also learn th ings like the difference between real an d complex numbers and whether 0 is odd or even. Some of the problems might be easy, but you might be surprised to learn something new. Chapter One — Algebraic Fundamentals Number Classification The natural numbers are also called the “counting numbers,” because when you read them, it sounds like you’re counting: 1, 2, 3, 4, 5, and so on. Most people don’t start counting with 0. Numbers fall into different groups 1.1 Describe the difference between the whole numbers and the natural numbers. Number theory dictates that the set of whole numbers and the set of natural numbers contain nearly all of the same members: {1, 2, 3, 4, 5, 6, …}. The characteristic difference between the two is that the whole numbers also include the number 0. Therefore, the set of natural numbers is equivalent to the set of positive integers {1, 2, 3, 4, 5, …}, whereas the set of whole numbers is equivalent to the set of nonnegative integers {0, 1, 2, 3, 4, 5, …}. 1.2 What set of numbers consists of integers that are not natural numbers? What mathematical term best describes that set? The integers are numbers that contain no explicit fraction or decimal. Therefore, numbers such as 5, 0, and –6 are integers but 4.3 and So you get the whole numbers by taking the natural numbers and sticking 0 in there. are not. Thus, all integers belong to the set {…, –3, –2, –1, 0, 1, 2, 3, …}. According to Problem 1.1, the set of natural numbers is {1, 2, 3, 4, 5, …}. Remove the natural numbers from the set of integers to create the set described in this problem: {…, –4, –3, –2, –1, 0}. This set, which contains all of the negative integers and the number 0, is described as the “nonpositive numbers.” 1.3 Is the number 0 even or odd? Positive or negative? Justify your answers. By definition, a number is even if there is no remainder when you divide it by 2. To determine whether 0 is an even number, divide it by 2: . (Note that 0 divided by any real number—except for 0—is equal to 0.) The result, 0, has no remainder, so 0 is an even number. However, 0 is neither positive nor negative. Positive numbers are defined as the real numbers greater than (but not equal to) 0, and negative numbers are defined as real numbers less than (but not equal to) 0, so 0 can be classified only as “nonpositive” or “nonnegative.” 1.4 Numbers, like 8, that aren’t prim because they are e divisible by too m a things, are called ny “composite numbe rs.” Identify the smallest positive prime number and justify your answer. A number is described as “prime” if it cannot be evenly divided by any number other than the number itself and 1. According to this definition, the number 8 is not prime, because the numbers 2 and 4 both divide evenly into 8. However, the numbers 2, 3, 5, 7, and 11 are prime, because none of those numbers is evenly divisible by a value other than the number itself and 1. Note that the number 1 is conspicuously absent from this list and is not a prime number. By definition, a prime number must be divisible by exactly two unique values, the number itself and the number 1. In the case of 1, those two values are equal and, therefore, not unique. Although this might seem a technicality, it excludes 1 from the set of prime numbers, so the smallest positive prime number is 2. The Humongous Book of Algebra Problems Chapter One — Algebraic Fundamentals 1.5 List the two characteristics most commonly associated with a rational number. The fundamental characteristic of a rational number is that it can be expressed as a fraction, a quotient of two integers. Therefore, and are examples of rational numbers. Rational numbers expressed in decimal form feature either a terminating decimal (a finite, rather than infinite, number of values after the decimal point) or a repeating decimal (a pattern of digits that repeats infinitely). Consider the following decimal representations of rational numbers to better understand the concepts of terminating and repeating decimals. 615384615384 1.6 615384 The irrational mathematical constant p is sometimes approximated with the fraction Little bars like this are used to indicate which digits of a repeating decimal actually repeat. Sometimes, a few digits in front won’t repeat, but the number is still rational. For example, is a rational number. . Explain why that approximation cannot be the exact value of p. When expanded to millions, billions, and even trillions of decimal places, the digits in the decimal representation of p do not repeat in a discernable pattern. Because p is equal to a nonterminating, nonrepeating decimal, p is an irrational number, and irrational numbers cannot be expressed as fractions. 1.7 Which is larger, the set of real numbers or the set of complex numbers? Explain your answer. Combining the set of rational numbers together with the set of irrational numbers produces the set of real numbers. In other words, every real number must be either rational or irrational. The set of complex numbers is far larger than the set of real numbers, and the reasoning is simple: All real numbers are complex numbers as well. The set of complex numbers is larger than the set of real numbers in the same way that the set of human beings on Earth is larger than the set of men on Earth. All men are humans, but not all humans are necessarily men. Similarly, all real numbers are complex, but not all complex numbers are real. Complex numbers are discussed in more detail late r in the book, in Problems 13.3713.44. The Humongous Book of Algebra Problems ˘ Chapter One — Algebraic Fundamentals According to Problem 1.1, the single element that the whole numbers contain and the natural numbers exclude is the number 0. Any infinitely long decimal that has no pattern of repeating digits represents an irrational number. On the other hand, rational decimals either have to repeat or terminate. Because there are a lot more ways to write irrational numbers as decimals than there are to write rational numbers as decimals, there are a lot more irrational numbers than rational numbers. 1.8 List the following sets of numbers in order from smallest to largest: complex numbers, integers, irrational numbers, natural numbers, rational numbers, real numbers, and whole numbers. Although each of these sets is infinitely large, they are not the same size. The smallest set is the natural numbers, followed by the whole numbers, which is exactly one element larger than the natural numbers. Appending the negative integers to the whole numbers results in the next largest set, the integers. The set of rational numbers is significantly larger than the integers, and the set of irrational numbers is significantly larger than the set of rational numbers. The real numbers must be larger than the irrational numbers, because all irrational numbers are real numbers. The complex numbers are larger than the real numbers, as explained in Problem 1.7. Therefore, this is the correct order: natural numbers, whole numbers, integers, rational numbers, irrational numbers, real numbers, and complex numbers. 1.9 Describe the number 13 by identifying the number sets to which it belongs. Because 13 has no explicit decimal or fraction, it is an integer. All positive integers are also natural numbers and whole numbers. It is not evenly divisible by 2, so 13 is an odd number. In fact, 13 is not evenly divisible by any number other than 1 and 13, so it is a prime number. You can express 13 as a fraction , so 13 is a rational number. It follows, therefore, that 13 is also a real number and a complex number. In conclusion, 13 is odd, prime, a natural number, a whole number, an integer, a rational number, a real number, and a complex number. 1.10 Describe the number Because by identifying the number sets to which it belongs. is less than 0 (i.e., to the left of 0 on a number line), it is a negative number. It is a fraction, so by definition it is a rational number and, therefore, it is a real number and a complex number as well. Any number divided by itself is 1, so . ˘ The Humongous Book of Algebra Problems Chapter One — Algebraic Fundamentals Expressions Containing Signed Numbers Add, subtract, multiply, and divide positive and negative numbers 1.11 Simplify the expression: 16 + (–9). This expression contains adjacent or “double” signs, two signs next to one another. To simplify this expression, you must convert the double sign into a single sign. The method is simple: If the two signs in question are different, replace them with a single negative sign; if the signs are the same (whether both positive or both negative), replace them with a single positive sign. Some algebra books write positive and negative signs higher and smaller, like this: 16 + –9. I’m sorry, but that’s just weird. It’s perfectly fine to turn that teeny floating sign into a regular sign: 16 + –9. In this problem, the adjacent signs are different, “+ –,” so you must replace them with a single negative sign: –. 1.12 Simplify the expression: –5 – (+6). This expression contains the adjacent signs “– +.” As explained in Problem 1.11, the double sign must be rewritten as a single sign. Because the adjacent signs are different, they must be replaced with a single negative sign. –5 – (+6) = –5 – 6 To simplify the expression –5 – 6, or in fact any expression that contains signed numbers, think in terms of payments and debts. Every negative number represents money you owe, and every positive number represents money you’ve earned. In this analogy, –5 – 6 would be interpreted as a debt of $5 followed by a debt of $6, as both numbers are negative. Therefore, –5 – 6 = –11, a total debt of $11. Think of it this way. If the two signs agree w it each other (if th h ey’re both positive or bo negative), then th th a good thing, a POS t’s a IT thing. On the othe IV E r hand, when two signs ca n’t agree with each other (one’s positive and one’ negative), then th s at’s no good. That’s NEGATIV E. 1.13 Simplify the expression: 4 – (–5) – (+10). This expression contains two sets of adjacent or “double” signs: “– –” between the numbers 4 and 5 and “– +” between the numbers 5 and 10. Replace like signs with a single + and unlike signs with a single –. 4 – (–5) – (+10) = 4 + 5 – 10 Simplify the expression from left to right, beginning with 4 + 5 = 9. 4 + 5 – 10 = 9 – 10 There’s one other technique you can use to add and subtract signed numbers. If two numbers have different signs (like 9 and –10), then subtract them (10 – 9 = 1) and use the sign from the bigger number (10 > 9, so use the negative sign attached to the 10 to get –1 instead of 1). If the signs on the numbers are the same, then add the numbers together and use the shared sign. In other words, to simplify –12 – 4, add 12 and 4 to get 16 and then stick the shared negative sign out front: –16. The Humongous Book of Algebra Problems ˘ Chapter One — Algebraic Fundamentals To simplify 9 – 10 using the payments and debts analogy from Problem 1.12, 9 represents $9 in cash and –10 represents $10 in debts. The net result would be a debt of $1, so 9 – 10 = –1. 1.14 Simplify the expression: . Choosing the sign to use when you multiply and divide numbers works very similarly to the method described in Problem 1.11 to eliminate double signs. When two numbers of the same sign are multiplied, the result is always positive. If, however, you multiply two numbers with different signs, the result is always negative. In this case, you are asked to multiply the numbers 6 and –3. Because one is positive and one is negative (that is, their signs are different), the result must be negative. You could also write , but you don’t HAVE to write a + sign in front of a positive number. If a number has no sign in front of it, that means it’s positive. There’s no multiplication sign written between (3) and (–3), so how did you know to multiply them together? It’s an “unwritten rule” of algebra. When two quantities are written next to one another and no sign separates them, multiplication is implied. That means things like 4(9), 10y, and xy are all multiplication problems. ˘ 1.15 Simplify the expression: . When signed numbers are divided, the sign of the result once again depends upon the signs of the numbers involved. If the numbers have the same sign, the result will be positive, and if the numbers have different signs, the result will be negative. In this case, both of the numbers in the expression, –16 and –2, have the same sign, so the result is positive: . 1.16 Simplify the expression: (3)( –3)(4)(–4). Multiply the signed numbers in this expression together working from left to right. In this way, because you are multiplying only two numbers at a time, you can apply the technique described in Problem 1.14 to determine the sign of each result. The leftmost two numbers are 3 and –3; they have different signs, so multiplying them together results in a negative number: (3)( –3) = –9. (3)( –3)(4)( –4) = (–9)(4)( –4) Again multiply the two leftmost numbers. The signs of –9 and 4 are different, so the result is negative: (–9)(4) = –36. (–9)(4)( –4) = (–36)( –4) The remaining signed numbers are both negative; because the signs match, multiplying them together results in a positive number. The Humongous Book of Algebra Problems (–36)( –4) = 144 Chapter One — Algebraic Fundamentals 1.17 Simplify the expression: . The straight lines surrounding 9 in this expression represent an absolute value. Evaluating the absolute value of a signed number is a trivial matter—simply make the signed number within the absolute value bars positive and then remove the bars from the expression. In this case, the number within the absolute value notation is already positive, so it remains unchanged. You are left with two signed numbers to combine: +4 and –9. According to the technique described in Problem 1.11, combining $4 in assets with $9 in debt has a net result of $5 in debt: 4 – 9 = –5. 1.18 Simplify the expression: . The absolute value of a negative number, in this case –10, is the opposite of the negative number: . 1.19 Simplify the expression: . If this problem had no absolute value bars and used parentheses instead, your approach would be entirely different. The expression –(5) – (–5) has the double sign “– –,” which should be eliminated using the technique described in Problems 1.11–1.13. However, absolute value bars are treated differently than parentheses, so this expression technically does not contain double signs. Begin by evaluating the absolute values: and . See? There’s the double sign. When turned in (+5), the negative to si in front of the a gn bsolute values didn’t go awa In the next step, y. you eliminate the dou ble sign “– +” to get –5 – 5. Absolute value bars are the antidepressants of the mathematical world. They make everything inside positive. To say that more precisely, they take away the negative of the number inside. That means . However, the moodaltering lines have no effect on positive numbers: . Absolute values are simple when there’s only one number inside. If the number inside is negative, make it positive and drop the absolute value bars. If the number’s already positive, leave it alone and just drop the bars. Well, it doesn’t contain double signs YET. It will in just a moment. The Humongous Book of Algebra Problems ˘ Chapter One — Algebraic Fundamentals 1.20 Simplify the expression: . Do not eliminate double signs in this expression until you have first addressed the absolute values. Combine the signed numbers two at a time, working from left to right. Begin with 2 – 7 = –5. –5 + 5 = 0 1.21 Simplify the expression: . This problem contains the absolute value of an entire expression, not just a single number. In these cases, you cannot simply remove the negative signs from each term of the expression, but rather simplify the expression first and then take the absolute value of the result. To simplify the expression 3 + (–16) – (–9), you must eliminate the double signs and them combine the numbers one at a time, from left to right. Grouping Symbols When numbers band together, deal with them first For now, the parentheses and other grouping symbols will tell you what pieces of a problem to simplify first. When parentheses aren’t there to help, you have to apply something called the “order of operations,” which is covered in Problems 3.30– 3.39. ˘ 1.22 Simplify the expression: . When portions of an expression are contained within grouping symbols—like parentheses (), brackets [], and braces {}—simplify those portions of the expression first, no matter where in the expression it occurs. In this expression, is contained within parentheses, so multiply those numbers: . 1.23 Simplify the expression: . The only difference between this expression and Problem 1.22 is the placement of the parentheses. This time, the expression 7 + 10 is surrounded by grouping symbols and must be simplified first. The Humongous Book of Algebra Problems Chapter One — Algebraic Fundamentals By comparing this solution to the solution for Problem 1.22, it is clear that the placement of the parentheses in the expression had a significant impact on the solution. 1.24 Simplify the expression: . Although this expression contains parentheses and brackets, the brackets are technically the only grouping symbols present; the parentheses surrounding –11 are there for notation purposes only. Simplify the expression inside the brackets first. 1.25 Simplify the expression: . This expression contains two sets of nested grouping symbols, brackets and parentheses. When one grouped expression is contained inside another, always simplify the innermost expression first and work outward from there. should be simplified first. In this case, the parenthetical expression A grouped expression still remains in the expression, so it must be simplified next. 1.26 Simplify the expression: . Grouping symbols are not limited to parentheses, brackets, and braces. Though it contains none of the aforementioned elements, this fraction consists of two grouped expressions. Treat the numerator (6 + 10) and the denominator (14 – 8) as individual expressions and simplify them separately. If you’re not sure how turned into , you divide the numbers in the top and bottom and . That process of the fraction by 2: is called “simplifying” or “reducing” the fraction and is explained in Problems 2.11–2.17. Double signs, like in the expression 19 + (–11), are ugly enough, but it’s just too ugly to write the signs right next to each other like this: 19 + – 11. If you look back at Problems 1.11–1.13, you’ll notice that the second signed number is always encased in parentheses if leaving them out would mean two signs are touching. “Nested” means that one expression is inside the other one. In this case, is nested inside the bracketed expression because the expression inside parentheses is also inside the brackets. Nested expressions are like those egg-shaped Russian nesting dolls. You know the ones? When you open one of the dolls, there’s another, smaller one inside? The Humongous Book of Algebra Problems ˘ Chapter One — Algebraic Fundamentals 1.27 Simplify the expression: The “numerator” is the top part of the fraction and the “denominator” is the bottom part. . Like Problem 1.26, this fractional expression has, by definition, two implicit groups, the numerator and the denominator. However, it contains a second grouping symbol as well, absolute value bars. The absolute value expression is nested within the denominator, so simplify the innermost expression, first. Now simplify the numerator and denominator separately. Any number divided by itself equals 1, so = 1, but note that the numerator is negative. According to Problem 1.15, when numbers with different signs are divided, the result is negative. 1.28 Simplify the expression: According to the end of Proble m 1. 27, when you d ivide a number and it s opposite (like 7 a nd –7), you get –1. . This expression consists of two separate absolute value expressions that are subtracted. The left fractional expression requires the most attention, so begin by simplifying it. Now that the fraction is in a more manageable form, determine both of the absolute values in the expression. 1.29 Simplify the expression: Three if you don’t count as a group (because it has only one number inside). Four if you do count it. 10 . This problem contains numerous nested expressions—braces that contain brackets that, in turn, contain parentheses that include an absolute value. Begin with the innermost of these, the absolute value expression. The innermost expression surrounded by grouping symbols is now (3 + 1), so simplify it next. The Humongous Book of Algebra Problems Chapter One — Algebraic Fundamentals The bracketed expression is now the innermost group; simplify it next. 1.30 Simplify the expression: . The numerator and denominator both contain double signs within their innermost nested expressions. Begin simplifying there, and carefully work your way outward. For the moment, ignore the absolute value signs surrounding the entire fraction. Evaluate and Leave the big, outside absolute value bars until the very end, after you have a single number on the top and bottom of the fraction. to continue simplifying. Now that the numerator and denominator each contain a single real number value, take the absolute value of the fraction that remains. Algebraic Properties Basic assumptions about algebra 1.31 Simplify the expressions on each side of the following equation to verify that the sides of the equation are, in fact, equal. (3 + 9) + 10 = 3 + (9 + 10) Each side of the equation contains a pair of terms added within grouping symbols. According to Problem 1.22, those expressions should be simplified first. Both sides of the equation have a value of 22 and are, therefore, equal. The Humongous Book of Algebra Problems 11 Chapter One — Algebraic Fundamentals 1.32 What algebraic property guarantees that the equation (3 + 9) + 10 = 3 + (9 + 10) from Problem 1.31 is true? There’s also an associ-ative property for multiplication, which says that you can regroup numbers that are multiplied together and it won’t change the answer. Here’s an example: The only difference between the sides of the equation is the placement of the parentheses. According to the associative property of addition, if a set of numbers is added together, the manner in which they are grouped will not affect the total sum. 1.33 Simplify the expressions on each side of the following equation to verify that the sides of the equation are, in fact, equal: There are no grouping symbols present to indicate the order in which you should multiply the numbers on each side of the equation. Therefore, you should multiply the numbers from left to right, starting with on the left side of the equation and on the right. The rule stating that you should multiply a of numbers from string le right is part of th ft to e order of operations. Prob lem 3.30 –3.39 cover th s is in more detail. 1.34 What algebraic property guarantees that the equation in Problem 1.31 is true? The sides of the equation in Problem 1.33 contain the same values; however, they are listed in a different order. The commutative property of multiplication states that re-ordering a set of real numbers multiplied together will not affect the product. “Product” is a fancy word fo r “what you get w hen you multiply thin gs,” like “sum” is a fancy wa say “what you ge y to t when you add things.” 12 The Humongous Book of Algebra Problems Just like the associative property, the commutative property works for both addition and multiplication. If you’ve got a big list of numbers added together, you can add them in any order you want, and you’ll get the same thing. In case you’d like to see visual evidence, here’s Exhibit A: Chapter One — Algebraic Fundamentals 1.35 According to the associative properties of addition and multiplication, the manner in which values are grouped does not affect the value of the expression. However, Problems 1.22 and 1.23, which contain only addition and multiplication, prove that . How is it possible that grouping the expressions differently changed their values, despite the guarantees of the associative properties? The associative properties of addition and multiplication are separate and cannot be combined. In other words, you can apply the associative property of addition only when addition is the sole operation present, and you can apply the associative property of multiplication only when the numbers involved are multiplied. Neither associative property can be applied to the expression because it contains both addition and multiplication. 1.36 Describe the identity properties of addition and multiplication, including the role of the additive and multiplicative identities. According to the identity property of addition, adding 0 (the additive identity) to any real number will not change the value of that number. Similarly, the multiplicative identity states that multiplying a real number by 1 (the multiplicative identity) doesn’t change the value either. 1.37 Complete the following statement and explain your answer: According to the ______________ property, if a = b, then b = a. The symmetric property guarantees that two equal quantities are still equal if written on opposite sides of the equal sign. In other words, if x = 5, then it is equally correct to state that 5 = x. 1.38 According to the distributive property, if a, b, and c are real numbers, then expression 3(2 – 7). . Apply the distributive property to simplify the The distributive property applies to expressions within grouping symbols that are multiplied by another term. Here, the entire expression (2 – 7) is multiplied by 3. The distributive property allows you to multiply each term within the parentheses by 3. Don’t over think this one—it ’s nothing you don’t already know. If yo multiply a numbe u r by 1 or add 0 to it, the number’s IDENTIT Y doesn’t change: 5 + 0 = 5 and . So if you’re as old as I am, then I am as old as you are. Hmmmm. Not very shocking or particularly groundbreaking. Multiply 3(2) and 3(–7) before adding the terms together. According to the algebraic order of operations, multiplication within an expression should be completed before addition. For a more thorough investigation of this topic, see Problems 3.30–3.39. The Humongous Book of Algebra Problems 13 Chapter One — Algebraic Fundamentals 1.39 Simplify the expression 3(2 – 7) (from Problem 1.38) once again, this time calculating the sum within the grouping symbols first. Verify that the result matches the answer produced by the distributive property in Problem 1.38. Although the distributive property applies to this expression (as explained in Problem 1.38), simplifying the expression inside the grouping symbols first makes the problem significantly easier. Problem 1.36 explains why the additive identity is 0 and the multiplicative identity is 1. The expressions in Problems 1.38 and 1.39, though simplified differently, have the same value. 1.40 Explain the additive inverse property and demonstrate it mathematically using a real number. The reciprocal of an integer like 2 equals the fraction 1 divided by that number: . (So the reciprocal of –4 is and the reciprocal of 7 is .) The reciprocal of a fraction is the fraction you get by reversing the numerator and denominator. (So the reciprocal of is and the reciprocal of is , or just –10.) The additive inverse property states that adding any real number to its opposite results in the additive identity, 0. Consider the number 6; the sum of 6 and its opposite, –6, is 0: 6 + (–6) = 0. The property applies to negative numbers as well. Adding –3 to its opposite, +3, also results in 0: –3 + 3 = 0. 1.41 Explain the multiplicative inverse property and demonstrate it mathematically using a real number. The multiplicative inverse property states that multiplying a number by its reciprocal results in the multiplicative identity, 1. For instance, if you multiply 2 by its reciprocal 1.42 Complete the following statement and explain your answer: According to the transitive property, if a = b and b = c, then _______________. The transitive property describes the relative equality of three quantities. Here, the quantity a is equal to the quantity b. In turn, b is equal to a third quantity, c. If b is equal to both a and c, it follows logically that a and c must also be equal. Therefore, the equation a = c correctly completes the statement. If multiplying or simplifying fractions makes you nauseated, don’t worry. You’ll find lots of practice in Problems 2.33– 2.37. 14 , the product is 1. The Humongous Book of Algebra Problems Chapter One — Algebraic Fundamentals 1.43 Identify the mathematical property that justifies the following statement and explain your answer: Both sides of the equation contain the same numbers in the same order. The only difference between the sides of the equation is the way the numbers are grouped by the parentheses. Therefore, this statement is true by the associative property of multiplication, which states that the way a set of real numbers is grouped does not affect its product. 1.44 Identify the mathematical properties that justify the following statement and explain your answer: If 3 = x and x = y, then y = 3. According to the transitive property, if 3 = x and x = y, then 3 = y. To rewrite 3 = y as y = 3 to match the given statement, you must apply the symmetric property. If the numbers had been in a different order on either side of the equal sign, the commutative property would have come into play. The commutative property says “order doesn’t matter,” and the associative property says “it doesn’t matter where you stick the parentheses.” See Problem 1.42 for more information about the transitive property and Problem 1.37 for more info about the symmetric property. The Humongous Book of Algebra Problems 15 Chapter 2 Rational Numbers eing afraid of them t s b a e b e r u s s n o i t c f a r g n i d n a t s r e Und Chapter 1 introduced the concept of rational numbers, real numbers that can be expressed as a fraction, a terminating decimal, or a repeating decimal. Rational numbers are truly Gestalt, which is to say they are greater than the sum of their parts. By nature they are merely quotients of integers, but their complexity requires a unique set of concepts (such as the least common denominator) and procedures (such as reducing to lowest terms and transforming rational numbers from fractions to decimals and vice versa). Through the study of rational numbers, and the careful, often rigorous, techniques that surround them, unique and otherwise obfuscated properties of integers are uncovered. When you divide two integers, you ge t a fraction. The fancy name for “fraction” is “rational number,” an d that’s what you deal with in this chapter. Fractions aren’t as ha rd to handle as most people think— you just need to know that when fra ctions are in the mix, you need to follow specific rules. For instance , you can only add and subtract fractions if they have the same de nominator. Simple enough. However, if fractions have DIFFERENT denom inators, you CAN multiply and divide them. Spend some time getting familiar wit h fractions in this chapter. When you finish, you’ll be able to change decimals into fractions; change fractions into decimals; simplify fra ctions; identify a least common denominator; and add, subtract, mu ltiply, and divide fractions to your heart’s content. Chapter Two — Rational Numbers Rational Number Notation Proper and improper fractions, decimals, and mixed numbers 2.1 You can put as many zeroes as you want at the beginning of a decimal: 0.25, 00.25, 00 0.25, and 00 00 00 00 00.25 all mean the same thing. You can also add zeroes at the end of a decimal: 1.5, 1.50, 1.500, and so on. Express 0.013 as a percentage. Decimal numbers, like percentages, express a value in terms of a whole. This whole value can be expressed in decimal form, as 1 (or 1.0), and as a percentage, 100%. Transforming a decimal into a percent is as simple as moving the decimal point exactly two digits to the right. Here, 0.013 = 1.3%. 2.2 Express 0.25% as a decimal. According to Problem 2.1, converting from a decimal to a percentage requires you to move the decimal point two digits to the right. It comes as no surprise, then, that performing the opposite conversion, from percentage to decimal, requires you to move the decimal exactly two digits to the left. In this problem, only one digit, 0, appears to the left of the decimal. The second, unwritten, digit is also 0. Therefore, 0.25% = 00.25% = 0.0025%. Note: Problems 2.3–2.4 refer to the rational number . The number you’re dividing BY is called the “divisor” and the number you’re dividing INTO is called the “dividend.” The answer you get once you’re done dividing is called the “quotient.” 2.3 You can add as many zeroes as you want, and you can do it at any time during the problem. Here’s your goal: you want the answer to have either terminated or begun to repeat. If it hasn’t done either, pop some more zeroes up there and keep going. 18 Express the fraction as a decimal. The rational number represents the quotient . To express the fraction as a decimal, use long division to divide 4 by 1. Set up the long division problem, writing an additional zero at the end of the dividend. Copy the decimal point above the division symbol. For the moment, ignore the decimal point within the dividend and imagine that 1.0 is equal to 10. Because 4 divides into 10 two times, place a 2 above the rightmost digit of 10. Because 4 does not divide evenly into 10, a remainder will exist. Multiply 2 in the quotient by the divisor (4) and write the result below the dividend (10). Draw a horizontal line beneath 8. The Humongous Book of Algebra Problems Chapter Two — Rational Numbers Subtract 8 from 10 and write the result below the horizontal line. The difference (10 – 8 = 2) is not 0, so the quotient has not yet terminated. Place another 0 on the end of the dividend and on the end of the number below the horizontal line. This time, dividing the bottommost number by the divisor produces no remainder; 4 divides evenly into 20. Write the result above the division symbol next to the 2 already there. Multiply the newest digit in the quotient (5) by the divisor (4) and write the result beneath 20. Subtract the bottom two numbers. Because the remainder is 0, the division problem is complete: . Rational numbers either repeat or terminate. When you get a remainder of 0, you stop dividing, and the decimal terminates. The Humongous Book of Algebra Problems 19 Chapter Two — Rational Numbers Note: Problems 2.3–2.4 refer to the rational number . 2.4 Express the fraction as a percentage. According to Problem 2.3, . To transform a decimal into a percentage, move the decimal point two digits to the right: 0.25 = 25.0%, or simply 25%. In Problem 2.3, you had to add a zero because 4 divides into 10 but it rea lly doesn’t divide into very well. In Proble 1 m 2.5, however, 6 doe s divide into 11, so you don’t need to write 11 as 110. When you’re long dividing, what you’re dividing INTO has to be bigger than what you’re dividing BY. If it’s not, add a zero to the dividend and the bottom number. Note: Problems 2.5–2.7 refer to the rational number 2.5 . Express the fraction as a decimal. Use the method described in Problem 2.3 to rewrite the fraction as a long division problem. Here, however, there is no immediate need to place a 0 in the dividend. No decimal point is written explicitly, so write one at the end of the dividend and copy it above the division symbol. Six divides into eleven one time, so write 1 above the rightmost digit of 11. Multiply the divisor by the digit just written above the dividend the result below the dividend and subtract. , write The divisor (6) cannot divide into the result (5) a whole number of times. As Problem 2.3 directed, change 5 into 50 and also add a zero to the end of the dividend. Six divides into 50 eight times, so place 8 at the right end of the quotient. Multiply the new digit by the divisor (6), write the result below 50, and subtract. 20 The Humongous Book of Algebra Problems Chapter Two — Rational Numbers Six does not divide into 2 a whole number of times, so once again insert zeroes after 2 and 11.0. Six divides into 20 three times. Take the appropriate actions in the long division problem. Put a 3 above the division symbol, multiply it by the divisor (6), write that multiplication result (18) below 20, and then subtract it from 20. Once again, 2 is the bottommost number in the long division problem. If you place zeros after it and after the dividend, it produces another 3 in the quotient. Once again, 2 is the bottommost number. Repeating this process is futile—each time a zero is added, it produces another 3 in the quotient and a difference of 2, the same number with which you started. Because the division problem has turned into an infinite loop producing the same pattern of digits, you can conclude that . If the decimal form of a rational numbe r doesn’t terminate , one or more digit sw repeat infinitely. ill Wri the decimal wit te ha little bar over th e 3 to indicate th at it’s an infinitely repeating digit in the decimal. The Humongous Book of Algebra Problems 21 Chapter Two — Rational Numbers Note: Problems 2.5–2.7 refer to the rational number 2.6 . Express the fraction as a percentage. According to problem 2.5, . Move the decimal two places to the right to convert the decimal into a percentage: Improper doesn’t mean unacceptable. It’s fine to have fractions with bigger numbers on top than on bottom, and most teachers would rather you leave fractions in improper form, rather than write them as mixed numbers. Note: Problems 2.5–2.7 refer to the rational number 2.7 22 . Express the fraction as a mixed number. The fraction is considered improper because its numerator is greater than its denominator. To express it as a mixed number, divide 11 by 6. There is no need to use long division—it is sufficient to conclude that 6 divides into 11 one time with a remainder of 5. Thus . The whole number portion of the mixed number is the number of times the divisor divides into the dividend; the remainder is the numerator of the fraction. The denominator of the fraction matches the denominator of the original fraction. In other words, given the improper fraction times with a remainder of r, then 2.8 Here’s a quick way to figure this out. Either long divid e or use a calculator. You ge t 16.25. The numbe r of the decimal (1 left 6) is the nonfraction pa of the mixed nu rt mber. To figure out the remainder, multi ply that whole numbe r by the denomina tor and subtract what yo u get from the numerator: 65 – 64 = 1. . Express the improper fraction , if x divides into y a total of w . as a mixed number. Four divides into 65 a total of 16 times with a remainder of 1. Therefore . The fractional part of the mixed number consists of the remainder divided by the original denominator. 2.9 Express as an improper fraction. To convert a mixed number into a fraction, multiply the denominator and the whole number and then add the numerator. Divide by the denominator of the mixed number: 2.10 Express . as an improper fraction. According to Problem 2.9, the formula. The Humongous Book of Algebra Problems . Substitute a = 4, b = 5, and c = 12 into Chapter Two — Rational Numbers Simplifying Fractions Reducing fractions to lowest terms, like instead of 2.11 Explain the process used to reduce a fraction to its lowest terms. A fraction is in lowest terms only when its numerator and denominator no longer share any common factors. In other words, the fraction is reduced if no number divides evenly into both the numerator and denominator. One effective way to reduce a fraction is to identify the greatest common factor (GCF) of both numbers and then divide each by that GCF. Note: Problems 2.12–2.13 refer to the rational number . Except 1, because 1 can divide evenly into anything. 2.12 Identify the greatest common factor of the numerator and denominator. List all the factors of the numerator (numbers that divide evenly into 8) and the denominator (numbers that divide evenly into 24). The largest factor common to both lists is 8, so the greatest common factor of 8 and 24 is 8. Note: Problems 2.12–2.13 refer to the rational number . If you have two dozen eggs, most people would say, “I have one-th of the eggs” rath ird er than “I have 8 of the 24 eggs.” Both are correct, but the first is easier to visualize. 2.13 Reduce the fraction to lowest terms. According to Problem 2.12, the greatest common factor of 8 and 24 is 8, so divide both the numerator and the denominator by 8 to reduce the fraction. Though and are different fractions, they have equivalent values: . Note: Problems 2.14–2.15 refer to the rational number . 2.14 Identify the greatest common factor of the numerator and denominator. List the factors of the numerator and denominator. The largest factor common to both lists is 9, so the greatest common factor of 27 and 63 is 9. There’s no magic trick for generating the list of factors. Start by trying to divide the number by 2, then by 3, and so on, until you identify all the numbers that divide in evenly. Here’s one tip: All the numbers in the list are paired up. For instance, after you figure out that 3 is a factor of 63, then also has to be in the list of factors. The Humongous Book of Algebra Problems 23 Chapter Two — Rational Numbers Note: Problems 2.14–2.15 refer to the rational number . 2.15 Express the fraction in lowest terms. According to Problem 2.14, the greatest common factor of 27 and 63 is 9, so divide the numerator and denominator by 9 to reduce the fraction. 2.16 Reduce the fraction to lowest terms. List the factors of the numerator and denominator. Although this fraction is negative, the technique you use to reduce it to lowest terms remains unchanged. Divide the numerator and denominator by 4, the greatest common factor. 2.17 Reduce the fraction to lowest terms and identify the greatest common factor of 2,024 and 8,448. The technique demonstrated in Problems 2.11–2.16 is not convenient when the numbers in the numerator and denominator are very large. Because 2,024 and 8,448 each have many factors, rather than list all of them, identify one common factor (it does not matter which) and use it to reduce the fraction. In this case, both numbers are even, so you can divide each by 2. If it would take too long to write all the factors, find something that divides evenly into the top and the bottom of the fraction and reduce it. Keep doing that until you can’t find any common factors. 24 The fraction is not yet reduced to lowest terms. Notice that the numerator and denominator of the resulting fraction are again even, so divide both by 2. Once again the fraction consists of even numbers. Divide by 2. The Humongous Book of Algebra Problems Chapter Two — Rational Numbers Continue to look for common factors. Notice that 253 and 1,056 are evenly divisible by 11. Now that the fraction is reduced to lowest terms, calculate the greatest common factor by multiplying each of the numbers that were eliminated from the fraction. 2.18 Express 0.45 as a fraction in lowest terms. The decimal 0.45 is read “forty-five hundredths,” because it extends two digits to the right of the decimal point. Forty-five hundredths literally translates into the fraction . Reduce the fraction to lowest terms. 2.19 Express 1.843 as an improper fraction. The decimal 1.843 is read “one and eight hundred forty-three thousandths.” Therefore, you should divide 843 by one thousand to convert the decimal into a mixed number: When you’re looking for factors, don’t stop checking at 10, or you’ll miss this one. You don’t have to try every number in the world, though—it depends on the smaller of the two numbers in the numerator and denominator. If that number is less than 225, you can stop looking for factors at 15. If it’s less than 400, you can stop looking for factors at 20. Basically, if the smaller number is less than a number N, then stop checking at . . The number left of the decimal, 1, becomes the whole part of the mixed number. That fraction cannot be reduced, because the greatest common factor of 843 and 1,000 is 1. Use the method described in Problem 2.7 to convert the mixed number into an improper fraction. 0.5 is five tenths, 0.05 is five hundredths, 0.005 is five thousandths, 0.0005 is five ten thousandths, etc. 2.20 Express 0.5 as a fraction. The decimal 0.5 repeats infinitely (0.5 = 0.5555555...) ; therefore, you cannot use the method described in Problems 2.18–2.19 to convert this number into a fraction. If the digits of a repeating decimal begin repeating immediately after the decimal point (that is, the repeated string begins in the tenths place of the decimal), then you can apply a shortcut to rewrite the decimal as a fraction: Divide the repeated string by as many 9s as there are digits in the repeated string. In Problems 2.18 decimal told you –2 .19, the number of digits after w th decimal never en hat power of 10 to divide by. How e ever, this ds, so you can’t ge t the power of 10 you need. Each digit right of the decimal point is a power of 10. Because 0.45 ends two digits right of the decimal, you divide it by 10 2. To make 0.12986 into a fraction, you divide it by 10 5, because it has five digits left of the decimal. The Humongous Book of Algebra Problems 25 Chapter Two — Rational Numbers If the repeated string is two digits long, divide by 99. If it’s three digits long, divide by 999. In this case, the repeated string is consists of one digit (5). To convert fraction, divide the repeated string by 9. into a The rationale behind this shortcut is omitted here, as it is based on skills not discussed until Chapter 4. This technique, in its more rigorous form, is explained in greater detail in Problems 4.26–4.28. 2.21 Express as a fraction. The repeated string of 0.727272 … consists of two digits, so divide the repeated string (72) by two nines (99) and reduce the fraction to lowest terms. Remember that this technique applies only when the repeated string of digits begins immediately to the left of the decimal point. If the repeated string begins farther left in the decimal, you should apply the technique described in Problems 4.26–4.28. Combining Fractions Add, subtract, multiply, and divide fractions 2.22 Explain what is meant by a least common denominator. Equivalent fractions might have different denominators. For instance, Problem 2.13 demonstrated that and have the same value, as is expressed in lowest terms. It is often useful to rewrite one or more fractions so that their denominators are equal. Usually, there are numerous options from which you can choose a common denominator, and the least common denominator is the smallest of those options. In other words, has no remainder. Note: Problems 2.23–2.25 refer to the fractions and . 2.23 Identify the least common denominator of the fractions. Begin by identifying the largest of the given denominators; here, the largest denominator is 10. Because the other denominator (2) is a factor of 10, then 10 is the least common denominator (LCD). The LCD is never smaller than the largest denominator. 26 The Humongous Book of Algebra Problems Chapter Two — Rational Numbers Note: Problems 2.23–2.25 refer to the fractions and . 2.24 Generate equivalent fractions using the least common denominator. To rewrite using the least common denominator, divide the LCD by the current denominator: . Multiply the numerator and denominator of by that result. Because already contains the least common denominator, it does not need to be rewritten. Note: Problems 2.23–2.25 refer to the fractions and . 2.25 Calculate the sum of the fractions. Multiplying the top and bottom of the fraction by 5 is like multiplying the entire fraction by . You’re allowed to do that because , and multiplying any number by 1 doesn’t change it, according to the multiplicative identity property in Problem 1.36. To calculate the sum or difference of fractions, those fractions must have a common denominator. According to Problem 2.24, . Add the numerators of the fractions, but not the denominators. In other words, if you want to ADD or SUBTRACT fractions …. Unless otherwise directed, you should always reduce answers to lowest terms. Note: Problems 2.26–2.27 refer to the fractions , , and . 2.26 Identify the least common denominator. The largest denominator of the three fractions is 9. However, both of the remaining denominators are not factors of 9, so 9 is not the LCD. To identify another potential LCD candidate, multiply the largest denominator by 2: . All the denominators (3, 6, and 9) are factors of 18, so it is the LCD. If 18 didn’t work, you’d test to see if were the LCD. If 27 didn’t work, you’d multiply 9 by 4, then 5, then 6, and so on, until finally all the denominators divided in evenly. The Humongous Book of Algebra Problems 27 Chapter Two — Rational Numbers Note: Problems 2.26–2.27 refer to the fractions 2.27 Simplify the expression: Don’t simplify these fractions, or you’ll end up w ith what you started with and destroy the common denominators. , , and . . To rewrite the fraction using the least common denominator, divide each denominator into 18: , , and . Multiply the numerator and denominator of each fraction by the corresponding result. In other words, multiply the first fraction by third fraction by , the second fraction by , and the . Combine the numerators into a single numerator divided by the least common denominator and simplify. The result is already in lowest terms: 2.28 Simplify the expression: . . The largest denominator (16) is not evenly divisible by both of the other denominators (3 and 12) so multiply it by 2: . However, 32 is not divisible by all of the denominators either, so multiply the largest denominator by 3: . Because 48 is divisible by 3, 12, and 16, it is the least common denominator. Rewrite the expression using the method described by Problem 2.27: Divide each denominator into the LCD and multiply the numerator and denominator of each fraction by the corresponding result. 28 The Humongous Book of Algebra Problems Chapter Two — Rational Numbers 2.29 Simplify the expression: . The largest denominator (12) is not divisible by both of the remaining denominators, and neither is . However, is divisible by 4, 9, and 12, so 36 is the least common denominator. Rewrite the expression using equivalent fractions with the LCD. Divide the numerator and denominator by 2 to reduce the fraction. Note: Problems 2.30–2.31 refer to the fractions and . 2.30 Identify the least common denominator. The technique described in Problem 2.23, calculating the least common denominator using consecutive multiples of the largest denominator, are not well suited to these fractions; 64 is not divisibly by 30, and neither are , , , or . Rather than continue to follow this pattern, you can use the prime factorizations of the denominators to determine the LCD. The first step, appropriately, is to determine the prime factorization of each denominator, 30 and 64; to accomplish this, write each number as a product. There is no single correct product, so it is equally valid to express 30 as , , or . Similarly, you could write 64 as , , or . Factor any composite numbers in the products. For instance, 6 can be expressed as and 8 can be expressed as . Continue factoring until only prime numbers remain. The only remaining composite number is 4, so express 4 as Every number unravels into a unique product of prime numbers, according to something called the Fundamental Theorem of Algebra. A prime factorization is like a number’s unique fingerprint. In other words, leave all prime numbers (like 5) alo However, if you ca ne. n rewrite a numbe r as a multiplication prob le containing somet m hing besides 1, you ne ed to do that. The Humongous Book of Algebra Problems 29 Chapter Two — Rational Numbers Exponents are covered more in depth in Problems 3.8–3.18. For now it’s enough to know that Count up how many 2s are multiplied together and turn that number into a power. Rewrite the factorizations by listing the factors in order from least to greatest, using exponents when possible. . Now that the prime factorizations have been identified, you can begin to construct the least common denominator. Start by listing each unique factor that appears in either factorization. For instance, 30 has prime factors 2, 3, and 5; 64 has a single prime factor, 2, which is already included in the prime factorization of 30, so it isn’t included a second time. Each factor in the LCD should be raised to the highest power of that factor in either factorization. Both 3 and 5 only appear in the prime factorization of 30 and are raised to the first power, so 3 and 5 are also raised to the first power in the LCD. However, 2 appears in both factorizations, once to the first power and once to the sixth power. Choose the higher of the two powers, 26, for the LCD. Multiply the factors of the LCD together. The least common denominator is 960. Note: Problems 2.30–2.31 refer to the fractions and . 2.31 Simplify the expression: According to problem 2.30, the least common denominator is 960. Multiply the numerator and denominator of the first fraction by 32 (because ). Similarly, the numbers in the second fraction should be multiplied by . These operations rewrite the expression using equivalent fractions with the least common denominator. 67 is a prime number, so you can only reduce the fraction if 960 is divisible by 67, and it’s not. 30 The Humongous Book of Algebra Problems Chapter Two — Rational Numbers 2.32 Use the prime factorization technique described in Problem 2.30 to simplify the expression: . Identify the prime factorizations of 945 and 1,575. Both factorizations include the same factors (3, 5, and 7), so the LCD includes those factors as well. The highest power of 3 is 33 (from the factorization of 945), and the highest power of 5 is 52 (from the factorization of 1,575), so apply those powers to the factors in the LCD. The remaining factor, 7, is raised to the same power in both factorizations (1). Both numbers are divisible by 5: 945 = (5)(189) and 1,575 = (5)(315). Additionally, 189 and 315 are divisible by 9: 945 = (5)(9)(21) and 1,575 = (5)(9)(35). Finally, 21 and 35 are both divisible by 7. Divide the LCD by each denominator to identify the value by which you should multiply to generate the equivalent fractions: and . 2.33 Simplify the expression: . Multiplying and dividing fractions does not require a common denominator. Multiply the numerators of the fractions together and divide by the product of the denominators. To reduce the fraction to lowest terms, divide the numerator and denominator by 2. 6÷2 40 ÷ 2 The Humongous Book of Algebra Problems 31 Chapter Two — Rational Numbers Note: Problems 2.34–2.35 refer to the expression . 2.34 Simplify the expression using the method described in Problem 2.33. Multiply the numerators and divide by the product of the denominators. Note that the product of numbers with different signs is negative. Divide 126 and 504 by their greatest common factor (126) to reduce the fraction. This is a great trick because yo ur answers won’t ne ed to be reduced a t th end of the proble e m. In other words, leave it as a big string of numbers multiplied together. Note: Problems 2.34–2.35 refer to the expression . 2.35 Reduce the numerators and denominators of the expression before calculating the product. If a number in any numerator shares a common factor with a number in any denominator, it is beneficial to express the numbers as products that include the common factor. Here, the left numerator and the right denominator share the common factor 7; the left denominator and the right numerator share the common factor 18. Multiply the fractions together but do not calculate the individual products. Eliminate the matching pairs of factors in the numerator and denominator. If you cross out a 7 in the top of the fraction, you can cross out a 7 in the denominator. It’s a one-for-one exchange—one number on the top for one matching number on the bottom. 32 No factors remain in the numerator other than the (usually) unwritten factor 1. Notice that this answer and the answer to Problem 2.34 are equal. The method you use to multiply the fraction has no affect on the product. 2.36 Simplify the expression: . Write the product as a single fraction. The Humongous Book of Algebra Problems Chapter Two — Rational Numbers Identify common factors of the numerator and denominator: 6 and 15 have common factor 3; 14 and 35 have common factor 7. Rewrite the fraction so that the common factors are explicitly identified. Eliminate pairs of common factors from the numerator and denominator to reduce the fraction. 2.37 Simplify the expression: . Integers are rational numbers, as any number divided by 1 is equal to itself: . Multiply the fractions. Note that 6 and 9 have common factor 3. 2.38 Describe the relationship between multiplying and dividing fractions. Dividing by a fraction is equivalent to multiplying by the reciprocal of that fraction. Some students call this the KFC rule (it has nothing to do with chicken), which stands for keep, flip, change. When you divide something by a fraction, KEEP the first fraction the same, FLI P the fraction you’re dividing by (take the reciprocal), and CHANGE the division sign to a multiplication sign. The Humongous Book of Algebra Problems 33 Chapter Two — Rational Numbers So dividing by 2 is the same as multiplying by . That makes sense— dividing a number by 2 and taking half of the number mean the same thing. 2.39 Simplify the expression: . Express the integer as a rational number and rewrite the quotient as a product, using the method described in Problem 2.38. Multiply the fractions and reduce the product to lowest terms. 2.40 Simplify the expression: If you don’t like looking for common factors and canceling them out, you don’t have to. You can always wait until the very last step and divide by the greatest common factor. However, it’s usually easier to spot common factors when the numbers are smaller. 34 . Rewrite the quotient as a product, multiply the fractions, and write the result in lowest terms. Recall that the quotient (and product) of two negative numbers is positive. The Humongous Book of Algebra Problems Chapter Two — Rational Numbers 2.41 Simplify the complex fraction: . A complex fraction contains a fraction in its numerator, denominator, or both. This fraction is complex because its denominator is the fraction . All fractions, including complex fractions, can be rewritten as a quotient. Convert the quotient into a product, calculate the product, and reduce the result to lowest terms. 2.42 Simplify the complex fraction: . Use the method described in Problem 2.41: Rewrite the complex fraction as a quotient and then as a product. Calculate the product and ensure the result is reduced to lowest terms. The Humongous Book of Algebra Problems 35 Chapter Two — Rational Numbers 2.43 Simplify the complex fraction: . Rewrite the complex fraction as a quotient and then as a product. Calculate the product and ensure the result is reduced to lowest terms. Notice that 20 and 16 have common factor 4; 27 and 9 have common factor 9. 36 The Humongous Book of Algebra Problems Chapter 3 Basic algebraic expressions Time for x to make its stunning debut This chapter further develops the preceding chapters’ rigorous discussion of numbers by introducing a new element, the variable. Variables provide entirely new avenues of exploration, and the study of algebraic expressions containing these unknown values immediately supersedes all discussion of numeric expressions. Variables aren’t drastically differen t than numbers. After all, the entire purpose of a variable is to represent a number, either one that’s unknown or one that could ch ange values. In this chapter, you learn how to do many of the same things you were doing in the last tw o chapters, but now x’s and y’s will be in the mix. However, this chapter’s not just a one-trick pony. It has a fe w other things up its sleeves, like an investigation of exponents, the ord er of operations, and scientific notation. Chapter Three — Basic Algebraic Expressions Translating Expressions The alchemy of turning words into math 3.1 See Problem 1.34 for more information. If the phrase had been “Eight IS more than a number,” you’d translate that into 8 > x. The difference? The verb “is.” This chapter deals with math PHRASES , not math SENTENCES that include verbs—those are covered in Chapter 4. Because subtraction is not commutative, as Problem 3.1 indicated. When it comes to scientific notation (like in Problems 3.19–3.22), × notation is better because you’re dealing with decimals. It’s easy to confuse multiplication dots and decimal points if you’re writing fast. 38 Translate the following phrases into algebraic expressions and evaluate the expressions: “the sum of three and six” and “the difference of three and six.” Add two numbers to calculate the sum (3 + 6 = 9) and subtract two numbers to calculate the difference (3 – 6 = –3). Because addition is commutative, the order in which you add is inconsequential: 3 + 6 = 6 + 3 = 9. However, subtraction is not commutative. Ensure that you subtract in the order stated. In this case, the difference of three and six translates to the expression 3 – 6, not 6 – 3. 3.2 Translate into an algebraic expression: eight more than a number. In this instance, “more than” indicates addition; if one number is 8 more than another, then adding 8 to the smaller number produces the larger number. Unlike Problem 3.1, neither number is stated explicitly. Rather than adding known values like 3 and 6, you are asked to add 8 to an unknown number. Use a variable to represent the unknown number: x + 8. Although it is common to represent the unknown value with x, any variable might be used. Thus, y + 8, k + 8, and u + 8 are equally valid answers. 3.3 Explain the difference between the following expressions: “10 less a number” and “10 less than a number.” To understand the subtle difference in meaning, replace “a number” with a real number value, such as 7. “10 less 7” represents the subtraction problem 10 – 7, whereas “10 less than 7” represents “7 – 10.” Though both phrases indicate subtraction, the order in which the operation is performed differs, and the result differs as well. Algebraically, “10 less a number” is translated as 10 – x, and “10 less than a number” is interpreted as “x – 10.” 3.4 Translate into an algebraic expression: the product of a number and 7. The word “product” indicates multiplication. Thus, the product of an unknown number x and seven is x · 7 or 7x. There are two important things of which to take note. First, the notation 7x is preferred to x7; numbers are typically written before variables in a product and are called coefficients. In this case, x has a coefficient of 7. Second, dot notation ·(rather than the traditional multiplication operator ×) is preferred to avoid confusing the operator × with the variable x. For that reason, and with rare exception, the dot notation is used to represent multiplication for the remainder of this book. The Humongous Book of Algebra Problems Chapter Three — Basic Algebraic Expressions 3.5 Translate into an algebraic expression: the quotient of a number and five fewer than that number. If x represents the unknown number, then x – 5 is five fewer than x. To calculate the quotient of two numbers, divide them. Note that, like subtraction, division is not commutative—divide the first value stated by the second value stated. 3.6 Translate into an algebraic expression and simplify the expression: one-third of a number less half of the same number. The word “of,” when used alone, usually indicates multiplication. Specifically, it usually indicates multiplication by a rational number. Here, one-third of a number is written ber should be written or . Similarly, one-half of the same unknown numor You could also write x as a . The word “less” indicates subtraction. Because that operation is not commutative, ensure that the values are written in the correct order: Because both terms of the expression contain the same variable (x), they are like terms and can be combined. To combine them, however, the fractions must have a common denominator. Use the technique outlined in Problem 2.23 (or the more advanced technique in Problem 2.30) to identify a least common denominator and combine the coefficients of the terms. fraction . and multiply: . When you have like terms, combine the coefficients (the numbers) and attach the matc hing variable. For exa mple, to add 9y and 3y , 9 + 3 and attach add the y that both term s have in common: 12y. Pr oblem 3.6 is a little roug he because you have r to add fractions, bu t it’s the same principle. The Humongous Book of Algebra Problems 39 Chapter Three — Basic Algebraic Expressions 3.7 You can also simplify using common denominators and write John’s age as a single fraction: John is exactly 4 years less than half of Annie’s age. Express John’s age in terms of a, Annie’s age. Annie’s age, although unknown, is defined as a. Writing John’s age in terms of Annie’s age simply writing John’s age using the variable a. Express “half of Annie’s age” as a product years less than that: or as a quotient or . John’s age is 4 . Exponential Expressions Rules for simplifying expressions that contain powers 3.8 Simplify the expression and verify the result: (23)(24). To calculate the product of two exponential expressions with the same base, raise the common base to the sum of the powers: (xa)(xb) = xa + b. When you raise a number to a power, that number is called the base. In this problem, 23 and 24 have the same base: 2. To multiply things with the same base, add the powers (3 + 4 = 7) and raise the base to that power: 27. (23)(24) = 23 + 4 = 27 To verify the result, expand the expressions 23 and 24. An exponential expression indicates that the given base is multiplied by itself repeatedly, according to the value of the exponent. To calculate 23, you must multiply three 2s together, and to calculate 24, multiply four 2s. Substitute 23 = 8 and 24 = 16 into the original expression and verify that the product equals 27. 3.9 Simplify the expression: y 3 · y 6 · y13 · y. The rightmost factor in this expression has no explicit exponent value, so its implicit exponent is 1. Rewrite the expression so that each factor has a corresponding exponent. If no power is written, the power is 1. 40 y 3 · y 6 · y13 · y = y3 · y 6 · y13 · y1 All the factors have common base y, so add the powers to calculate the product (as explained in Problem 3.8). The Humongous Book of Algebra Problems Chapter Three — Basic Algebraic Expressions 3.10 Simplify the expression: (x 3y 5)(xy 9). Multiplication is commutative, so the order in which the factors are written has no impact upon the product. Rewrite the expression by grouping the common bases and express xy 9 as x 1y 9 so that each factor has an explicitly stated exponent. To calculate the product of exponential expressions with a common base, raise the common base to the sum of the exponents. In this case, x 3 and x 1 share a common base, as do y 5 and y 9. 3.11 Simplify the expression: . To determine the quotient of two exponential expressions with a common base, calculate the difference of the exponents and raise the common base to that power: . 3.12 Simplify the expression: Even though there are two groups defined by the parentheses, everything’s just getting multiplied together in a big pile, so arrange the pile any way you want. The best way is to stick the x’s and y’s together. . Let’s say you have two things with the same ba such as w 6 and w 2 se, . To multiply them, a dd the exponents: (w 6)(w 2 )= w 6 + 2 = w 8. To div ide them, subtract the exponents: w 6 ÷ w2 = w 6 – 2 = 4 w. As explained in Problem 3.11, the quotient of two exponential expressions with a common base is the common base raised to the difference of the exponents. 3.13 Simplify the expression: . Assume that w ≠ 0. Rewrite the product as a single fraction by multiplying the numerators together and dividing by the product of the denominators. State the exponents of x and z in the denominator explicitly: x = x 1 and z = z1. Apply the exponential property described in Problem 3.11: The quotient of exponential expressions with the same base is equal to the common base raised to the difference of the powers. All the factors in the top of the fraction can be moved around, as can the factors in the bottom, if the things on top stay on top and the things in the bottom stay down there. The Humongous Book of Algebra Problems 41 Chapter Three — Basic Algebraic Expressions Any nonzero number raised to the 0 power is equal to 1. The problem states that w ≠ 0, so w 0 = 1. 3.14 Simplify the expression: . The entire expression 5xy 2 is squared (raised to the second power), so square each of the factors individually. Express x as x1 to state its exponent explicitly. When two common bases are multiplied, add the powers: x2(x8) = x10. When something raised to a power is raised to another power, multiply the powers: (x2) 8 = x2 · 8 = x16. When an exponential expression is, itself, raised to a power, the result is the original base raised to the product of the powers: . 3.15 Simplify the expression: 2w(6w 2z3)7. The entire expression 6w 2z3 is raised to the seventh power, so raise each individual factor to that exponent. Note that 2w is not raised to the seventh power. Raise exponential expressions to a power by multiplying the exponents. Note that . Simplify the product. (w1)(w14) = w1+14 = w15 42 The Humongous Book of Algebra Problems Chapter Three — Basic Algebraic Expressions 3.16 Simplify the expression: (2x)3(8x 2y)2. Raise each factor of 2x to the third power and raise each factor of 8x 2y to the second power. Multiply the integers together, as well as the exponential factors with a common base (x 3 and x 4). 3.17 Simplify the expression: . To simplify x 3(x 4), don’t multiply th e exponents, add th em: x 3(x 4) = x7. Multipl y the exponents w he ONE base has a n po raised to a power wer : (x3) 4 = x12. The entire fraction is raised to the –3 power, so raise the numerator and denominator (and all the factors therein) to the –3 power. To raise an exponential expression to a power (even a negative power), multiply the exponents. Move factors with a negative exponent to the opposite side of the fraction bar. In other words, 3 –3 and x –3 should be moved from the numerator to the denominator; conversely, y –6 should be moved from the denominator to the numerator. After the factors are relocated, remove the negative sign from the exponent. When an entire fraction is raised to a negative power (like in Problem 3.17), you can take the reciprocal before you do anything else to eliminate the negative exponent right away: Unless otherwise directed, a final answer should never contain a negative exponent—such answers are not considered fully simplified. The Humongous Book of Algebra Problems 43 Chapter Three — Basic Algebraic Expressions 3.18 Simplify the expression: . Raise each factor of 8x –2y 5 to the second power and raise each factor of 4x 3y –1 to the –3 power. Make sure to change the powers to positive after you move the factors. Divide the entire product by 1 to express it as a fraction. Move each factor that is raised to a negative power to the denominator of the fraction. To multiply exponential expressions with a common base, add the exponents. Divide the numerator and denominator by 64 (the greatest common factor of the coefficients) to reduce the fraction to lowest terms. 3.19 Express the number 23,900,000 using scientific notation. Scientific notation allows you to express very large and very small numbers easily. To express large numbers using scientific notation, move the decimal point (whether it is explicitly stated or implicitly understood to follow the number) to the left until only one nonzero digit remains left of the decimal point. The number 23,900,000 has no explicit decimal point, so assume it follows the number. 23,900,000 = 23,900,000.0 Move the decimal point seven digits to the left, leaving the single nonzero digit 2 left of the decimal point. 2.39000000 = 2.39 44 The Humongous Book of Algebra Problems Chapter Three — Basic Algebraic Expressions Multiply the resulting decimal (2.39) by 10n , where n is the number of digits you moved the decimal to the left. 23,900,000 = 2.39 × 10 7 3.20 Express the number 958,000,000,000,000,000,000 using scientific notation. Assume the decimal point is located at the end of the number, to the right of the 18th zero. Move it 20 digits to the left so that only the nonzero digit 9 remains left of the decimal point. Multiply the resulting decimal (9.58) by 10n , where n is the number of digits the decimal point moved left. The “×” multiplication sym is used in scientifi bol c notation, so that the multiplication dot and the decima l point don’t get confused. 958,000,000,000,000,000,000 = 9.58 × 1020 3.21 Express the number 0.000049 using scientific notation. To write small numbers using scientific notation, count the number of times you must move the decimal point to the right until exactly one nonzero digit appears left of the decimal point. In this case, moving the decimal five digits to the right produces the decimal 4.9. Multiply that decimal by 10 –n , where n is the number of digits the decimal moved right. 0.000049 = 4.9 × 10 –5 3.22 Express the number –0.00000000412 in scientific notation. Apply the technique described in Problem 3.21—scientific notation does not require different procedures for positive and negative numbers. Move the decimal point nine places to the right so that a single nonzero digit appears left of the decimal point. When you move the decimal point RIGHT, the 10 in scientific notation has a negative exponent. When you move the decimal point LEFT, 10 has a positive exponent. Remember that, because in math, right usually means positive and left usually means negative. –0.00000000412 = –4.12 × 10 –9 Distributive Property Multiply one thing by a bunch of things in parentheses 3.23 Simplify the expression: –2(4x + 3y). To multiply the entire parenthetical quantity (4x + 3y) by –2, apply the distributive property: a(b + c) = ab + ac. –2(4x + 3y) = (–2)(4x) + (–2)(3y) In other words, every term in the parentheses gets multiplied by the number outside the parentheses, one at a time. Multiply the coefficient of each term by –2. Multiply the numbers together and then stick the variable on at the end. The Humongous Book of Algebra Problems 45 Chapter Three — Basic Algebraic Expressions 3.24 Simplify the expression: 8x(7x 2 – 3). Multiply each term of the quantity 7x 2 – 3 by 8x. 8x(7x 2 – 3) = (8x)(7x 2) + (8x)(–3) To calculate the product of 8x and 7x 2, multiply the coefficients and then apply the technique described in Problem 3.8 to multiply the x-terms. Both exponential terms have the same base, so the product is the common base raised to the sum of the individual powers. If a term doesn’t have a coefficient—the variable doesn’t have a number in front of it—then you should assume the coefficient is 1. (Just like the power of x is 1 if there’s no power written: x = x1.) That means you can write –y3 as –1y3 and y6 as 1y6. 3.25 Simplify the expression: –y3(–12x 2 + y 6). Apply the distributive property, multiplying each term of the expression –12x 2 + y 6 by –y 3. Writing the coefficients of –y 3 and y 6 explicitly might facilitate the calculation of the products. Because y 3 and y 6 have the same base, calculate their product by adding the powers. Note that x 2 and y3 have different bases, so those powers should not be summed. 3.26 Simplify the expression: x(9x 3 – 6x 2 + 2x + 1). You don’t HAVE to do this. It’s sort of annoying to writ as x1, so when you e x get the problems can right without those 1s, I’d stop writing them . 46 Distribute x to each term in the parenthetical quantity. Express the implicit exponents explicitly. The Humongous Book of Algebra Problems Chapter Three — Basic Algebraic Expressions 3.27 Simplify the expression: 5xy(x 2 – 4xy + 25y 2). Distribute 5xy to each term in the parenthetical quantity. 5xy(x 2 – 4xy + 25y 2) = (5x1y1)(x 2) + (5x1y1)(–4x 1y1) + (5x1y1)(25y 2) Multiply the coefficients of each term, followed by the variables. 3.28 Simplify the expression: 6x(2x – 3y) – y(11x – 5y). Distribute 6x to the terms of the expression (2x – 3y) and distribute –y to the terms of the expression (11x – 5y). Because the terms –18xy and –11xy contain the same variable expression (xy), they are like terms. Combine like terms by combining their coefficients and multiplying the result by the common variable expression: –18xy – 11xy = (–18 – 11)xy = –29xy. Instead of distributing –y, you could distribute a positive y 2 to get –(11xy – 5y ) and THEN distribute the negative sign. Either way, you’ll end up with –11xy + 5y2 . 12x 2 – 18xy – 11xy + 5y 2 = 12x 2 – 29xy + 5y 2 3.29 Apply the distributive property to the expression and eliminate negative exponents: 10x 3y 2z –1(4x 2y + 8xy –5z2 – 3y12z –4. Do not combine the resulting fractions. Apply the distributive property. This explanation of like terms is not very satisfying, but if you’re confused, don’t worry. The whole concept of terms (including like terms) is explained much more in depth in Problems 11.9–11.16. Move factors with negative exponents across the fraction bar into the denominator, a process explained in Problem 3.18. The Humongous Book of Algebra Problems 47 Chapter Three — Basic Algebraic Expressions Order of Operations Most people use the expression “Please Excuse My Dear Aunt Sally” to remember the order of operations. Take the first letter of each word and you get PEMDAS, which stands for Parentheses (grouping symbols); Exponents; Multiplication and Division; and Addition and Subtraction. My dear Aunt Sally is eternally excused 3.30 In what order should arithmetic operations be performed when simplifying an expression? Any expressions within grouping symbols should be performed first, followed by exponential expressions. Next, any multiplication and division in the expression should be performed from left to right. Multiplication is not performed before division, nor is the opposite true. Rather, both operations are performed in the same step. Finally, any addition and subtraction in the expression should be performed from left to right. 3.31 Simplify the expression: 8 – 2 · 7. According to the order of operations presented in Problem 3.30, multiplication should be completed before subtraction. 8 – 2 · 7 = 8 – 14 The only remaining operation is subtraction; calculate the difference of 8 and 14. 8 – 14 = –6 If you get 42 , you’re not doing the operations in the right order. Even thou gh subtraction is wri tt first, it’s last in th en e order of operation s. Problem 3.30.) Mul (See tiply first, then subtra ct. Notice that –10 2 = –100, not 10 0. The negative sign isn’t raised to the second power: –10 2 = –(10 2) = –1 00. If there were pa rentheses around –1 0, however, that ch ang every-thing: (–10 2 es ) = (–10)(–10) = 10 0. 48 3.32 Simplify the expression: (8 – 2) · 7. This expression is very similar to the expression in Problem 3.31, but in this case, parentheses are present. Perform operations within grouping symbols first. Here, the presence of grouping symbols alters the order in which the expression is simplified, and thus produces a different result than Problem 3.31. 3.33 Simplify the expression: 6 – 102 ÷ 4. This expression includes subtraction, an exponential power, and division. According to the order of operations presented in Problem 3.30, the exponent should be addressed first. 6 – 102 ÷ 4 = 6 – 100 ÷ 4 Two operations remain: subtraction and division. Division should be performed first. The Humongous Book of Algebra Problems Chapter Three — Basic Algebraic Expressions 3.34 Simplify the expression: (6 – 10)2 ÷ 4. Though similar to Problem 3.33, this expression contains grouping symbols. The expression within parentheses must be simplified first. (6 – 10)2 ÷ 4 = (–4)2 ÷ 4 The exponent must be addressed before division can be performed. 3.35 Is the following statement true or false? –52 = –(–5)2 Simplify the expressions on both sides of the equal sign separately and compare the results. Because both expressions have the same value, the statement –52 = –(–5)2 is true. See my note on Problem 3.33 if you don’t understand why –5 2 = –25 and not +25. 3.36 Simplify the expression: 10 ÷ 2 – 6 + 8 · 3 – 1. This expression contains addition, subtraction, multiplication, and division. Multiplication and division should be performed first, in the same step, from left to right. Calculate the quotient 10 ÷ 2 = 5 first, followed by the product 8 · 3 = 24. Addition and subtraction remain in the expression; perform the operations from left to right. 3.37 Simplify the expression: 10 · 4 – [3 + 6(–2)] ÷ 5. Simplify the expression within grouping symbols first. That expression, 3 + 6(–2), must be simplified according to the order of operations as well, so multiply before adding. The Humongous Book of Algebra Problems 49 Chapter Three — Basic Algebraic Expressions Technically, –[–9] is the multiplication problem –1[–9] = 9. Multiply and divide from left to right. Add the remaining values using the least common denominator. Start by writing 40 as a fraction: . Flip back to Problems 2.23 –2.25 if you need help calculating the least common denominator or adding the fractions after you figure out the LCD. 3.38 Simplify the expression: {10 – [8(3 ÷ 2)]} + 6(–1). This expression contains three nested sets of grouping symbols: parentheses within brackets within braces. According to Problem 1.25, you must start with the innermost group and work outward. Therefore, begin with the expression inside parentheses, then simplify the expression in brackets, and finally simplify the expression in braces. Of the remaining operations, multiplication and addition, multiplication should be completed first. 50 The Humongous Book of Algebra Problems Chapter Three — Basic Algebraic Expressions 3.39 Simplify the expression: 2 ÷ 5 + [–10 ⋅ (–5 + 11) –1]. According to the order of operations, the grouped expression should be simplified first. The operations in that bracketed expression should be completed in the following order: Simplify the inner group, apply the exponential power, and then multiply. The bracketed expression has a nested group inside that’s surrounded by parentheses: (–5 + 11)–1. Of the operations that remain, division and subtraction, division should be completed first. Calculate the difference of the fractions using the least common denominator. Raising something to the –1 power moves it a cross the fraction ba r. W you raise an inte hen ger or fraction to the –1 power, it’s the same thin ga taking the recipr s ocal. The reciprocal of is Evaluating Expressions Replace variables with numbers 3.40 Evaluate the expression 14x + (–x)2 if x = –2. Substitute x = –2 into the expression. 14x + (–x)2 = 14(–2) + (–[–2])2 Before you address the exponent, simplify the expression within the parentheses: –[–2] = –1 ⋅ [–2] = 2. . The LCD is 15. The larger of the two denominators is 5. The smaller denominator doesn’t divide evenly into 5 ⋅ 1 = 5 or 5 ⋅ 2 = 10, but it does divide evenly into 5 ⋅ 3 = 15. In other words, change all the x’s to “–2’s.” The Humongous Book of Algebra Problems 51 Chapter Three — Basic Algebraic Expressions 3.41 Calculate the value of m in the formula A negative times a negative is positive, so both negative signs vanish in the next step. if A = 7 and B = –28. Substitute A and B into the formula. Divide the numerator and denominator by 7 to reduce the fraction to lowest terms. 7÷7 28÷7 3.42 Evaluate the expression Follow the order of operations— multiply before you subtract in the numerator and denominator. if x = 1 and y = –4. Substitute the given values of x and y into the expression, simplify the numerator and denominator separately, and reduce the fraction to lowest terms. 3.43 Evaluate the expression b 2 – 4ac if a = –2, b = –3, and c = 5. Substitute the given values of a, b, and c into the expression. b 2 – 4ac = (–3)2 – 4(–2)(5) According to the order of operations, you should begin by simplifying the exponential expression: (–3)2 = (–3)(–3) = 9. (–3)2 – 4(–2)(5) = 9 – 4(–2)(5) 52 The Humongous Book of Algebra Problems Chapter Three — Basic Algebraic Expressions Multiplication precedes subtraction in the order of operations, and this expression contains two instances of multiplication. Work from left to right, beginning with –4(–2) = 8. 3.44 Evaluate the expression (a + b)(a 2 – ab + b 2) if and . Substitute the given values of a and b into the expression. The right factor contains exponential expressions, which should be simplified next: and The square of any real number is positive. . According to the order of operations, multiplication should be completed before addition and subtraction: . Combine the fractions in each set of parentheses independently. The least common denominator of the left group is 4, and the LCD of the right group is 16. The Humongous Book of Algebra Problems 53 Chapter 4 Linear Equations in One Variable How to solve basic equations This chapter explores basic linear equations—equations with degree one— that are expressed in terms of a single, unique variable. Unlike the expressions of Chapter 3, equations are complete algebraic sentences, statements that express the equality of two expressions that are separated by an equal sign. The difference between an expre ssion and an equation can be expressed in two words: equal sign. Expressions don’t have them but equations do. That one little symbol makes a huge difference. For one thing, your overall goal is diffe rent. You spent most of Chapter 3 simplifying equations, writing the sa me expression in a different, usuall y shorter, way. Equations are easier to deal with than expressions, because you actually get a tangible numeric answer, one that you can go back and check to make sur e you got it right. In this chapter, you start out with basic equation–solving techniques, and they’ll slowly evolve, becoming more and more complicated. By the end, you should be able to solve just about any equation they throw at you, if there’s only one uniqu e variable is in the equation (like x) and that variable is only rai sed to the first power. Chapter Four — Linear Equations in One Variable Adding and Subtracting to Solve an Equation Add to/subtract from both sides Note: Problems 4.1–4.3 refer to this statement: “Five more than a number is equal to thirteen.” 4.1 Express the statement as an algebraic equation in terms of x. According to Problem 3.2, the statement “more than” in this context indicates a sum, so “five more than a number” is expressed as x + 5. The phrase “is equal to” indicates the presence of an equal sign, so the algebraic equivalent of “five more than a number is equal to thirteen” is x + 5 = 13. Or 5 + x. That works, too. Note: Problems 4.1–4.3 refer to this statement: “Five more than a number is equal to thirteen” “Isolating” a variable means that it, alone, is on one side of the equal sign and everything else is on the other side. After you isolate x, you end up with “x =” or “= x,” an equation with only x on either the left or right side. If you add some-thing to (or subtract something from) one side of an equation, do the same thing to the other side. The two sides only stay equal if you treat them exactly the same way. 4.2 Solve the equation generated in Problem 4.1 for x. To solve an equation for a variable, you must isolate that variable on one side of the equal sign, usually the left side. The equation x + 5 = 13 has two things left of the equal sign, x and the number 5. If you subtract 5 from the left side of the equation to eliminate it, you must subtract 5 from the right side of the equation as well to maintain the equality of the statement. When five is subtracted from both sides of the equation, isolating x left of the equal sign, you are left with x = 8, the solution to the equation. Note: Problems 4.1–4.3 refer to this statement: “Five more than a number is equal to thirteen.” 4.3 Verify that the solution to Problem 4.2 is correct. To verify that x = 8 is the solution to the equation x + 5 = 13, substitute 8 into the equation for x. Because substituting x = 8 into the equation produces a true statement (13 = 13), x = 8 is the correct solution. There’s only one correct sol ution. When equations have one unique variable and the highest power of that varia ble is 1, you only get one solution. 56 The Humongous Book of Algebra Problems Chapter Four — Linear Equations in One Variable 4.4 Express the following statement as an equation and solve it: –6 equals the difference of a number and 19. The word “difference” indicates subtraction, so the difference of a number and 19 is expressed as x – 19. According to the statement, –6 is equal to that difference, so the corresponding equation is –6 = x – 19. To solve the equation, you must isolate x. In this case, it is simpler to isolate x right of the equal sign by adding 19 to both sides of the equation. You want to get x by itself, but x has a –19 next to it. Get rid of that negative 19 with a positive 19. Add 19 to both sides. The solution to the equation is 13 = x, or x = 13. According to the symmetric property described in Problem 1.37, those solutions are equivalent. Note: Problems 4.5–4.6 refer to the equation 2x + 1 = x – 3. 4.5 Solve the equation for x. To isolate x, all the variable terms (the terms containing x) should be moved to the left side of the equation and all the constant terms (the terms with no variables) should be moved to the right side of the equation. In other words, move x on the right side of the equation by subtracting x from both sides, and move 1 on the left side of the equation by subtracting 1 from both sides. The solution to the equation is x = –4. Note: Problems 4.5–4.6 refer to the equation 2x + 1 = x – 3. 4.6 Verify the solution generated in Problem 4.5. Substitute x = –4 into the equation 2x + 1 = x – 3. Ensure that you replace the x’s on both sides of the equation with –4. When you subtract things, write them under their like terms. In other words, when you subtract x from the right side, write –x underneath 2x, and when you subtract 1 from the left side, write –1 underneath –3. 2x and –x are like terms, but 2x does NOT equal – x 2! don’t cancel out You the variables. Remem ber, –x is the same a s –1 so 2x – 1x = 1x. T x, wo of something (2x) m inus one of those thin gs or x) leaves behi (1x nd one thing (1x or x). Because substituting x = –4 into the equation produces a true statement (–7 = –7), x = –4 is the correct solution. The Humongous Book of Algebra Problems 57 Chapter Four — Linear Equations in One Variable Check out Problem 3.23 if you need to review the distributive property. 4.7 Solve the equation 5 + 6(x – 1) = 11 + 5x for x and verify the solution. Before you attempt to isolate x, simplify the left side of the equation using the distributive property. “Moving” terms is a by– product of getting rid of terms you don’t want. You don’t want –1 on the left side, so you zap it by adding 1. However, you have to do the same thing to both sides of the equation, so that banished 1 moves over to the right side of the equation, where it’s combined with the constant term 11 that’s already there. Further simplify the left side of the equation by combining like terms 5 and –6: 6x + 5 – 6 = 11 + 5x 6x – 1 = 11 + 5x Isolate x on the left side of the equation by moving the variable terms left of the equal sign and moving the constant terms right of the equal sign. The solution to the equation is x = 12. Verify the solution by substituting it into the original equation. Because substituting x = 12 into the equation produces a true statement (71 = 71), x = 12 is the correct solution. 4.8 Solve the equation and verify the solution: 4(x + 7) – 18 = 3(x + 10) + 6. Apply the distributive property to simplify both sides of the equation. To remove 3x from the right side of the equation, subtract it from both sides. 58 The Humongous Book of Algebra Problems Chapter Four — Linear Equations in One Variable Subtract 10 from both sides of the equation to isolate x. The solution to the equation is x = 26. Verify the solution by substituting it into the original equation. Because substituting x = 26 into the equation produces a true statement (114 = 114), x = 26 is the correct solution. Multiplying and Dividing to Solve an Equation Multiply/divide both sides Note: Problems 4.9–4.10 refer to the statement “the product of three and a number is 42.” 4.9 Express the statement as an algebraic equation in terms of x. The word “product” indicates multiplication—the product of 3 and a number x is 3 · x, or 3x. According to the problem, that product is equal to 42: 3x = 42. Note: Problems 4.9–4.10 refer to the statement “the product of three and a number is 42.” 4.10 Solve the equation generated in Problem 4.9 for x. To solve an equation, you must isolate x on one side of the equal sign. In this case, isolating x requires you to eliminate the coefficient 3; to do so, divide both sides of the equation by 3. Reduce the fraction to lowest terms. 3 and x are multiplied, so to get rid of 3, you have to do the opposite of multiplying: dividing. The Humongous Book of Algebra Problems 59 Chapter Four — Linear Equations in One Variable 4.11 Solve the equation 10x = 16 for x. To reduce this fraction, divide the top and bottom by 2. Multiplying by is the same as dividing by . To isolate x on the left side of the equation, you must eliminate its coefficient. Apply the technique described in Problem 4.10; divide both sides of the equation by 10. 4.12 Solve the equation for x. Eliminate the coefficient of x to isolate the variable on the left side of the equation. To eliminate the coefficient its reciprocal, , multiply both sides of the equation by . 4.13 Solve the equation for x. Rewrite the fraction as the product of x and a fractional coefficient: . If x is alone in the top of a fraction, you ca n pu it out of the fra ll ctio and leave a 1 in n its place. 60 Multiply both sides of the equation by the reciprocal of The Humongous Book of Algebra Problems to isolate x. Chapter Four — Linear Equations in One Variable 4.14 Solve the equation for x. Before you can isolate x, it must appear in the numerator of the fraction. Take the reciprocal of both sides of the equation to accomplish this—it will have no impact on the equality statement. Write the fraction using an explicit coefficient: To eliminate the coefficient . , and thereby isolate x, multiply by its reciprocal: . The statement is true, and it stays true if you flip both fractions upside . After down: you learn cross multiplication in Problems 21.1–21.8, it won’t matter if the x is in the numerator or denominator. For now, though, you’re solving for x, not , so that x needs to be up top. Solving Equations Using Multiple Steps Nothing new here, just more steps 4.15 Solve the equation 4x + 3 = 27 for x. Subtract 3 from both sides of the equation so that only the variable term 4x is left of the equal sign and the constants are right of the equal sign. To eliminate the coefficient of 4x, and thereby solve for x, divide both sides of the equation by the coefficient. Keep anything with an x in it on the left side of th equation (becaus e e you’re solving for x). Any term without an x (suc ha in this equation) s 3 gets moved to the right side. The Humongous Book of Algebra Problems 61 Chapter Four — Linear Equations in One Variable Note: Problems 4.16–4.17 refer to the equation 19 – (x – 5) = 4. Pretend the negative sign outside (x – 5) is a –1 and multi ply x and –5 by –1. 4.16 Solve the equation by isolating x on the right side of the equal sign. Apply the distributive property to simplify the left side of the equation. You are directed to isolate x right of the equal sign, so add x to both sides of the equation to eliminate variable terms on the left side. 24 = 4 + x Subtract 4 from both sides of the equation to solve for x. Note: Problems 4.16–4.17 refer to the equation 19 – (x – 5) = 4. 4.17 Solve the equation by isolating x on the left side of the equal sign and verify that the solutions to Problems 4.16 and 4.17 are equal. Simplify the left side of the equation. You are instructed to isolate x on the left side of the equation, so eliminate the constant left of the equal sign by subtracting it from both sides of the equation. You could also divide both sides by –1. The equation is not solved for x, because x has a coefficient of –1. Eliminate the coefficient by multiplying both sides of the equation by –1. Problems 4.16 and 4.17 have the same solution: x = 20. 4.18 Solve the equation 6x – 3 = 9x + 2 for x. Group the x–terms left of the equal sign by subtracting 9x from both sides of the equation. Group the constant terms right of the equal sign by adding 3 to both sides of the equation. 62 The Humongous Book of Algebra Problems Chapter Four — Linear Equations in One Variable Don’t forget that you can always check your answer by plugging it back into the original equation for x: Divide both sides of the equation by the coefficient of x. 4.19 Solve the equation for x. Move the x-terms to the numerators of the fractions by taking the reciprocal of both sides of the equation. Write each fraction as the product of a rational number and x. Move the variable terms left of the equal sign by subtracting sides of the equation. from both Use the least common denominator 15 to combine the like terms. Both of these terms have the same variable (x), so they’re like terms. Combine their coefficients and multiply the result by the common variable x. Multiply both sides of the equation by the reciprocal of . Although x = 0 appears to be a valid solution, it is not. Substituting x = 0 into the original equation produces undefined statements, therefore invalidating the solution. The Humongous Book of Algebra Problems 63 Chapter Four — Linear Equations in One Variable You’re not allowed to divide by 0. It’s a fundamental math law. Because substituting x = 0 into the equation produces a 0 in at least one of the denominators—it actually makes both of the denominators 0—it breaks the law, and 0 isn’t a valid answer. The equation has no real number solutions. 4.20 Solve the equation 4x + 2(5 – x) = –7 for x. Use the distributive property to simplify the left side of the equation. Move 10 left of the equal sign by subtracting it from both sides of the equation. Divide both sides of the equation by 2 to isolate x and, therefore, solve the equation. Note: Problems 4.21–4.23 refer to the equation . 4.21 Solve the equation for x. Expand the right side of the equation by applying the distributive property. Separate the variable and constant terms on opposite sides of the equal sign by subtracting 64 The Humongous Book of Algebra Problems from, and adding 8 to, both sides of the equation. Chapter Four — Linear Equations in One Variable These two fractions are like terms becaus they have the sa e m variable (x). They e even have a com mon denominator. Eliminate the coefficient of x. Note: Problems 4.21–4.23 refer to the equation . 4.22 Verify the solution generated in Problem 4.21. Substitute statement. into the equation and verify that the result is a true Use common denominators to combine these pairs of fractions. The Humongous Book of Algebra Problems 65 Chapter Four — Linear Equations in One Variable Note: Problems 4.21–4.23 refer to the equation This gets a little tricky when the least common denominator contains variables, and that’s discussed in Problems 20.9– 20.19. . 4.23 Create an equivalent equation by eliminating the fractions and verify that its solution matches the solution generated in Problem 4.21 and verified in Problem 4.22. Eliminate the fractions in an equation by multiplying every term by the least common denominator. The fractions in this equation both have the same denominator, 3, the LCD by default. (x + 6) is already multiplied by , so when you multiply by 3, you multiply everything inside (x +6) by 3 as well. All the fractions have been eliminated from the equation. Solve the resulting equation for x. Note: Problems 4.24–4.25 refer to the equation . 4.24 Create an equivalent equation by eliminating the fraction and solve that equation for x. Use the method described in Problem 4.23 to eliminate the fraction: Multiply each term of the equation by 5, the least common denominator. (x + 3) is already multiplied by , so multiplying by 5 automatically m ulti everything inside plies (x + 3) by 5. 66 The Humongous Book of Algebra Problems Chapter Four — Linear Equations in One Variable Note: Problems 4.24–4.25 refer to the equation . 4.25 Solve the equation for x without eliminating the fractions in the original equation. Verify that the solution is equal to the solution generated in Problem 4.24. Apply the distributive property to simplify the left side of the equation and then isolate x on that side of the equal sign. Combine like terms on both sides of the equal sign using the least common denominator 5. 4.26 Express the rational number as a fraction. If a decimal contains a repeating string of digits that begins immediately to the right of the decimal point, in the tenths place, apply the technique described in Problems 2.20–2.21 to convert the decimal into a fraction. The decimal , however, requires a slightly more rigorous approach. Set x equal to the repeating decimal. x = 0.9444444444… Multiply both sides of this equation by 10n , where n is the number of digits in the repeating string. Here, the single digit 4 repeats infinitely, so n = 1. The Humongous Book of Algebra Problems 67 Chapter Four — Linear Equations in One Variable When you subtract, assum e that all the infini te many 4s on top ca ly nc out the infinitely el many 4s below. T ha means the equa t tion boils down to this subtraction prob lem: 9. 4 – 0.9 = 8.5. Subtract the original equation (x = 0.9444444444…) from the modified equation (10x = 9.444444444…). Express 8.5 as an improper fraction, using the technique described in Problem 2.19. Eliminate the coefficient of x to solve for x and thereby express x = 0.9444444444… as a fraction. The equation contains a fraction, so rather than divide both sides by 9, multiply both sides by the reciprocal of 9. 4.27 Express the rational number as a fraction. Apply the technique described in Problem 4.26. The repeating string in this decimal consists of one digit (3), so subtract the equation x = 0.5833333333… from 101(x) = 101(0.5833333333…). Express 5.25 as an improper fraction and solve the equation for x. 68 The Humongous Book of Algebra Problems Chapter Four — Linear Equations in One Variable Express the fraction in lowest terns. 4.28 Express the rational number as an improper fraction. Apply the technique described in Problem 4.26. The decimal 6.13939393939… contains a two–digit repeating string, so subtract the equation x = 6.13939393939 from 102(x) = 102(6.1393939393939). Express 607.8 as an improper fraction and solve the equation for x. You multiply by 10 to the SECOND power because TWO digits (3 and 9) repeat infinitely. Express the fraction in lowest terms. The Humongous Book of Algebra Problems 69 Chapter Four — Linear Equations in One Variable Use the word “or” to separate possible solutions, because you get a true statement by plugging either x = 6 OR x = –6 into the equation. Absolute Value Equations Most of them have two solutions 4.29 What values of x satisfy the equation This absolute value equation has two valid solutions, the value on the left side of the equation and its opposite: x = 6 or x = –6. Substituting either for x results in a true statement. The solutions to the equation r and –r. So whatever’s inside the absolute value bars either equals the left side of the equation or the opposite of the left side. ? 4.30 Solve the equation (where r is a positive real number) are for x and verify the solutions. Isolate the absolute value expression left of the equal sign by adding 3 to both sides of the equation. According to Problem 4.29, x = –4 or x = 4. To verify the solutions, substitute each into the original equation and verify that the results are true statements. After you isolate the absolute value on the left side of the equation, set whatever’s inside the bars (in this case x) equal to the right side of the equation (4) and the opposite of that number (–4). Those are the solutions. 70 4.31 Explain why the equation has no real solutions. The expression left of the equal sign, , is entirely enclosed by absolute value bars. Therefore, the left side of the equation must be positive no matter what value of x is substituted into it. The right side of the equation contains a negative constant, –8. If the left side of the equation must be positive for all real x and the right side of the equation is always negative, no x–value can be substituted into the equation that results in a true statement. The Humongous Book of Algebra Problems Chapter Four — Linear Equations in One Variable 4.32 Solve the equation for x. The absolute value expression is isolated on one side of the equation, so create two new equations—one that sets the contents of the absolute value expression equal to 12 and one that sets the same expression equal to –12. The solution is x = –16 or x = 8. 4.33 Solve the equation The first step is to get the absolute values on one side of the equal sign and everything else on the other side. In this problem, the absolute value bars are already all alone on the left side. for x. The absolute value expression is isolated on one side of the equation, so create two new equations—one that sets the contents of the absolute value expression equal to 15 and one that sets the same expression equal to –15. The solution is x = –6 or x = 9. 4.34 Solve the equation for x. Isolate the absolute value expression left of the equal sign by subtracting 1 from both sides of the equation. Create two equations by setting the expression within the absolute value symbols equal to 8 and its opposite, –8. The solution is x = –3 or x = 13. The Humongous Book of Algebra Problems 71 Chapter Four — Linear Equations in One Variable 4.35 Solve the equation This works just like a regular, nonabsolute value, equation. To solve 3x = 27 for x, you’d divide both sides by 3. If you’re isolating something (either a variable or an absolute value expression), its coefficient’s got to go. for x. To isolate the absolute value expression left of the equal sign, add 9 to both sides of the equation. The absolute value expression is not yet isolated because of its coefficient. Divide both sides of the equation by 3. Rewrite the statement as two equations that do not contain absolute values and solve them. It’s the same thing you’ve been doing since Problem 4.30. Set whatever’s inside the absolute value bars equal to the number on the right side of the equation. Then set it equal to the opposite of the number on the right. Move the constant to the left side (by adding 12) before you eliminate the absolute value coefficient . The solution is . 4.36 Solve the equation for x. Isolate the absolute value expression on the left side of the equation. Create two new equations using the technique described in Problem 4.32. The solution is 72 or The Humongous Book of Algebra Problems or . Chapter Four — Linear Equations in One Variable Equations Containing Multiple Variables Equations with TWO variables (like x and y) or more 4.37 Solve So, r is multiplied by three things— two numbers (2 and ) and a variable (h). Either divide both sides by or multiply both , the formula for the lateral surface a right circular cylinder, for r. Whether an equation contains one variable or many, the same technique is employed to solve for (or isolate) a variable. To isolate r on the left side of the equation, eliminate the values by which r is multiplied. to sides by move them away from r. 4.38 Solve , the formula that converts degrees Celsius to degrees Fahrenheit, for C. Isolate C left of the equal sign by subtracting 32 from both sides of the equation and then multiplying every term by The equivalent equation The reciprocal of C’s coefficient. . , obtained by applying the distributive property, represents another acceptable answer. Note: Problems 4.39–4.40 refer to the equation 5x – 3y = 30. 4.39 Solve the equation for x. Move the non x–term to the other side of the equation before you deal with x’s coefficient. To isolate x, add 3y to both sides of the equation and then divide each term by 5, the coefficient of x. It’s a good idea to write va before number terms. You ca riable terms n pull that y out in front of its fraction because . The Humongous Book of Algebra Problems 73 Chapter Four — Linear Equations in One Variable Note: Problems 4.39–4.40 refer to the equation 5x – 3y = 30. 4.40 Solve the equation for y. To isolate y, subtract 5x from both sides of the equation and then divide each term by –3, the coefficient of y. Note: Problems 4.41–4.42 refer to the equation a – (b + 4) = 2a + 3b – 6. 4.41 Solve the equation for a. Apply the distributive property to the left side of the equation. Think of –(b + 4) as (–1)(b + 4): a – b – 4 = 2a + 3b – 6 Move all of the a–terms, the terms that contain the variable for which you are solving, left of the equal sign by subtracting 2a from both sides of the equation and combine like terms. In this case, a – 2a = 1a – 2a = –1a = –a. Move all terms not containing a to the right side of the equation and simplify the result. Multiply each term of the equation by –1 to isolate a left of the equal sign and thereby solve the equation for a. 74 The Humongous Book of Algebra Problems Chapter Four — Linear Equations in One Variable Note: Problems 4.41–4.42 refer to the equation a – (b + 4) = 2a + 3b – 6. 4.42 Solve the equation for b. Use the technique described in Problem 4.41, this time isolating b left of the equal sign. 4.43 Solve the equation xy + 2 = –7y(x + 1) for x. (Assume y ≠ 0.) Apply the distributive property. Move all terms containing x, the variable for which you are solving, to the left side of the equation. Move all other terms right of the equal sign. To isolate (and, therefore, solve the equation for) x, divide each term in the equation by 8y. xy and 7xy are like terms because their variables match— they both conta in xy. Combining like te rms means combining their coefficients : xy + 7xy = 1xy + 7xy = 8xy. You can’t write as . You can only pull y out of the fraction when it’s in the numerator, and this y is in the denominator. The Humongous Book of Algebra Problems 75 Chapter 5 Graphing Linear Equations in Two Variables equation true Identify the points that make an After you solve equations, it is useful to represent those solutions graphically. Unlike the equations in Chapter 4, some linear equations have more than one solution. In fact, most linear equations in two variables have an infinite number of solutions, and the most effective way to present those solutions is visually, by means of a graph. In this chapter, you learn how to graph the solutions of equations that contain only one variable (usually x), equations that contain two variables (usually x and y), and absolute value equations in two variables. Aside from these graphing techniques, the problems also introduce the concept of slope, the key component necessary to create linear equations, a task undertaken in Chapter 6. Graphs allow you to visualize the sol utions of an equation. There’s really no need to visualize a solution when there’s only one (like x = 2) but when there are an infinite numb er of values that make an equatio n true, listing them all is impossible. Gr aphing them is the next best thing. You’ll learn two ways to graph lines: using a table of values and using intercepts. You’ll also learn how to ca lculate the slope of a line and how to graph an equation containin g absolute values. Chapter Five — Graphing Linear Equations in Two Variables Number Lines and the Coordinate Plane A line is one-dimensional because it just allows horizontal travel, left and right along the line. On a coordinate plane, however, you can travel horizontally and vertically, in two dimensions. Think of an ant crawling along a stick as opposed to crawling around on a piece of paper. Which should you use to graph? 5.1 What characteristic of an equation dictates whether its solution should be graphed on a number line or a coordinate plane? A number line is, like any line, one-dimensional, whereas a coordinate plane is two-dimensional. Equations with one variable are one-dimensional, so their solutions must be graphed on a number line. Equations with two different variables have a two-dimensional graph. 5.2 Graph the values w = 6, x = –5, y = Figure 5-1. , and z = on the number line in Figure 5-1: Plot w, x, y, and z on this number line. To plot w, count six units to the right of 0 and mark the value with a solid dot. Because x is negative, it should be five units to the left of 0. Plot y two-thirds of a unit to the right of 0 ( of the distance from 0 to 1 on the number line). It is easier to plot z if you first convert it from am improper fraction into a mixed If you’re not sure how to do this, check out Problems 2.7–2.8. number: . It is located 2.25 units to the left of 0, two full units and then one-fourth of a unit beyond that. All four values, w, x, y, and z, are illustrated in Figure 5-2. Figure 5-2: Negative values, like x and z, are located left of the number 0, and positive values, like w and y, are located to the right of 0. 5.3 Solve the equation 2x – 3(x + 1) – 5 = –6 and graph the solution. Use the method described in Problems 4.1–4.36 to isolate x and solve the equation. Begin by distributing –3 through the parentheses, then combine like terms, isolate x on the left side of the equation and eliminate the coefficient of x. 78 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables To plot the solution, mark the value –2 on a number line, as illustrated by Figure 5-3. Figure 5-3: Only one real number, x = –2, satisfies the equation 2x – 3(x + 1) – 5 = –6. 5.4 Identify lines k and m and point A on the coordinate plane in Figure 5-4. Figure 5-4: Lines k and m intersect at point A = (0,0). A coordinate plane is a two-dimensional plane created by two perpendicular lines. Typically, the horizontal line (m in Figure 5-4) is called the x-axis and has equation y = 0. The vertical line (k in Figure 5-4) is called the y-axis and has equation x = 0. The axes intersect at a point called the origin, which has coordinates (0,0), point A in Figure 5-4. A point on the plane is represented by coordinates, sort of like an address for each point. It’s made up of two numbers in parentheses and looks like B = (2,–1). The first number in the parentheses gives the horizontal location of the point (+2 means two units right of the origin), and the second number represents its vertical location (–1 means one unit below the origin). You’ll practice plotting points in Problem 5.5. The Humongous Book of Algebra Problems 79 Chapter Five — Graphing Linear Equations in Two Variables Note: Problems 5.5–5.6 refer to the points A = (–1,5); B = (4,4); C = (6,–2); D = E = (0,1); and F = (–6,0). 5.5 ; Plot the points on the coordinate plane in Figure 5-5. Figure 5-5: Use the coordinates of points A, B, C, D, E, and F to plot the points on this coordinate plane. Each coordinate pair has the form (x,y). The left value, x, indicates a signed horizontal distance from the y-axis. In other words, positive values of x correspond with points to the right of the y-axis, and negative x-values produce points left of the y-axis. Similarly, the right value of each coordinate pair, y, indicates a signed vertical distance from the x-axis. Coordinates with positive values of y are located above the x-axis, and negative y-values indicate points below the x-axis. With this in mind, plot each of the points; the results are graphed in Figure 5-6. 80 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables The x- and y-axes split the plane into four rectangular regions, called quadrants. They’re numbered in a standard way, but that standard way is kind of strange. Check out Problem 5.6 for more information. Figure 5-6: One point appears in each of the four quadrants of the graph and the remaining points fall on the axes of the coordinate plane. Note: Problems 5.5–5.6 refer to the points A = (–1,5); B = (4,4); C = (6,–2); D = E = (0,1); and F = (–6,0). 5.6 ; Identify the quadrants in which each of the points A, B, C, and D are located. The upper-right quadrant of the coordinate plane is quadrant I, or the first quadrant. The remaining quadrants are numbered counterclockwise from there, as illustrated in Figure 5-7. For some reason, the quadrants are usually labeled with Roman numerals, so quadrants one, two, three and four are written I, II, III, and IV, as shown in Figure 5-7. Figure 5-7: R eversing quadrants I and II, thereby mistakenly identifying the top-left quadrant as quadrant I, is a common error. According to Figure 5-6, B is in quadrant I, A is in quadrant II, D is in quadrant III, and C is in quadrant IV. The Humongous Book of Algebra Problems 81 Chapter Five — Graphing Linear Equations in Two Variables Because the num ber in the equati on is –3. If the equa ti were x = 5, then on th graph would be a e vertical line five un right of the y-axi its s. 5.7 Graph the line x = –3 on a coordinate plane and identify two points on its graph. Normally, you would graph x = –3 on a number line, much like Problem 5.3 graphed the solution x = –2. However, this problem states that you are to graph a line, not a point. You can only graph a line on a coordinate plane—not on a number line. Lines of the form “x = number” are graphed as vertical lines on the coordinate plane. According to Problem 5.4, the y-axis—another vertical line in the coordinate plane—has an equation of this form: x = 0. The vertical line with equation x = –3 should be drawn three units left of the y-axis, as illustrated by Figure 5-8. After all, the equa-tion is x = 3, so that’s the only rule for these points. The y-value doesn’t m atter at all. Each poin t needs a y-value, of course, but any real number will do. Figure 5-8: Vertical lines in the coordinate plane have form “x = n,” where n is the signed distance from the y-axis. A positive value of n indicates a line right of the y-axis, and a negative value of n indicates a vertical line left of the y-axis. All the points on the graph in Figure 5-8 have an x-coordinate of –3. If the x-coordinate of a point is –3, its y-value need only be a real number to ensure that the graph of x = –3 passes through the point. Therefore, the points (–3,0), (–3,–9.6), 5.8 , and all belong to the graph. Graph the line y = 2 on a coordinate plane and identify the coordinate at which it intersects the graph of the line x = –3 (from Problem 5.7). The graph of y = 2, illustrated in Figure 5-9, is a horizontal line two units above the x-axis. All horizontal lines have equation “y = n,” where n is the signed distance from the x-axis. A positive value of n indicates a line above the x-axis, and a negative value of n indicates a horizontal line below the x-axis. The lines y = 2 and x = –3 intersect at (x,y) = (–3,2). 82 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables Figure 5-9: The graph of y = 2 is a horizontal line two units above the x-axis. Graphing with a Table of Values Plug in some x’s, plot some points, call it a day Note: Problems 5.9–5.11 refer to the linear equation 2x – y = 4. 5.9 How many coordinates are required to draw the graph of the equation? Geometric principles dictate that two points on the same plane are required to draw the line that contains those points. If, however, you use coordinates to draw a linear graph, identifying and plotting at least one additional point is advised. The third point serves as a quick way to check your work. If it lies on the same line as the other two points, you’re far less likely to have made an error in your calculations. I call the third point a “check point.” If you make a mistake finding either of the other two points, the check point won’t be on the graph. That’s a red flag to go back and check your arithmetic for all three coordinates. Note: Problems 5.9–5.11 refer to the linear equation 2x – y = 4. 5.10 Use a table of values to identify three points on the graph of the line. A table of values is a brute force arithmetic technique that generates lists of coordinate pairs satisfying the given equation. After a sufficient number of points has been identified and plotted, all that remains is to “connect the dots” on the coordinate plane to graph the equation. The number of points necessary to create an accurate graph varies based on the complexity of the equation. The Humongous Book of Algebra Problems 83 Chapter Five — Graphing Linear Equations in Two Variables The absolute value graphs at the end of the chapter are a little trickier. You could get away with using only two points to make the graph, but you’d be showing off. Better to use more points as the equations get more complicated. See Problems 4.37– 4.43 if you need help with this step. Whenever possible, choose simple, small numbers like x = –1, x = 0, and x = 1. Because 2x – y = 4 is a linear equation in two variables, a minimum of two points is required to graph it. However, as explained in Problem 5.9, three coordinates should be plotted to better ensure your calculations are correct. Begin by solving the equation for y. Construct a table with columns labeled, from left to right, “x,” “y = 2x – 4,” and “y.” The outside columns, x and y, will eventually house the coordinates you will plot on the graph. The inner column contains the equation just solved for y. Choose three values of x to plug into the equation y = 2x – 4. Select x-values that do not produce large or unnecessarily complicated results. Write the three xvalues you chose in the left column of the table. Substitute each x-value into the equation to determine the corresponding values of y and record those in the right column. Combine the x- and y-values in each row to conclude that the graph contains the points (–1,–6), (0,–4), and (1,–2). 84 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables Note: Problems 5.9–5.11 refer to the linear equation 2x – y = 4. 5.11 Graph the linear equation using the table of values generated by Problem 5.10. Plot the points identified in Problem 5.10 and connect them to create the linear graph illustrated by Figure 5-10. –4 Figure 5-9: The graph of 2x – y = 4 passes through the points (–1,–6), (0,–4), and (1,–2). Note: Problems 5.12–5.13 refer to the linear equation x + 3y = –2. 5.12 Use a table of values to identify three points on the graph of the equation. Solve the equation for y. Create a table of values based on three simple values of x, as demonstrated in Problem 5.10. This time the table of values does not contain x = –1 a nd x = 0. If you plug in 1, however, the fr x = a turns into an inte ction ger. That’s why the ot he two values, x = –2 r and 4, are used as w ell— they eliminate th e fraction. The Humongous Book of Algebra Problems 85 Chapter Five — Graphing Linear Equations in Two Variables –(–2) is a double sign and should be rewritten as 2, according to Problem 1.11. You can also interpret –(–2) as a multiplication problem; imagine that the outer “–” equals –1: –(–2) = (–1)(–2) = 2. According to the table of values, points (–2,0), (1,–1) and (4,–2) belong to the graph of x + 3y = –2. Note: Problems 5.12–5.13 refer to the linear equation x + 3y = –2. 5.13 Graph the linear equation using the points generated by Problem 5.12. Plot the points (–2,0), (1,–1) and (4,–2) on the coordinate plane and connect them to graph the linear equation. Figure 5-11: The graph of x + 3y = –2 passes through the points (–2,0), (1,–1) and (4,–2). 86 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables 5.14 Use a table of values to graph the linear equation 2x – (y – 4x) = y + 2. Begin by solving the equation for y. Substitute simple values of x into the equation to generate coordinates and graph the linear equation, as illustrated by Figure 5-12. Figure 5-12: A table of values containing x = –1, x = 0, and x = 1 produces three coordinates: (–1,–4), (0,–1), and (1,2). Those three points are sufficient to graph 2x – (y – 4x) = y + 2. 5.15 Through which of the following points, if any, does the graph of pass? A = (–12,1); B = (3,–5); C = (24,9) Design a table of values using the x-coordinates of each point: x = –12, x = 3, and x = 24. Substitute each into the equation to determine whether the resulting y-values satisfy the equation. The Humongous Book of Algebra Problems 87 Chapter Five — Graphing Linear Equations in Two Variables A linear graph that’s not vertical only has one y-value that goes with each x-valu e. In this case, the ta ble of values says when x = –12, y = –15. Therefore, the gr aph can’t contain the point (–12 ,1) as well. According to the table of values, the points (3,–5) and (24,9) belong to the graph but (–12,–1) does not. 5.16 Through which of the following points, if any, does the graph of 3x – 5y = 2 pass? A = (–10,–16); B = (4,–2); C = (39,23) Use the method outlined in Problem 5.15—solve the equation for y, create a table of values containing x = –10, x = 4, and x = 39, and compare the results from the table with the y-coordinates in the given points. 88 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables Because the coordinate pair C = (39,23) matches the values generated in the last row of the table, point C belongs to the graph of 3x – 5y = 2. 5.17 If the graph of the line x – 2y = 10 passes through the points (8,a) and (b,3), what is the value of a + b? If the line x – 2y = 10 passes through the point (8,a), then substituting x = 8 and y = a into the equation produces a true statement. Similarly, substituting x = b and y = 3 into the equation also produces a true statement. Solve the left equation for a and the right equation for b. Don’t try and get creative and look for underlying meaning in the expression a + b. Just figure out what a and b are by plugging them into the equation and then add them together at the end. If a = –1 and b = 16, then a + b = –1 + 16 = 15. The Humongous Book of Algebra Problems 89 Chapter Five — Graphing Linear Equations in Two Variables Graphing Using Intercepts In other words, if a graph has an x-intercept of –4, that means it passes through the xaxis four units left of the origin. The easiest way to plot two points on a line quickly 5.18 What are the x- and y-intercepts of a graph? An x-intercept is a value at which a graph intersects the x-axis. It is usually reported as a signed value, indicating the location of the intersection point as though the x-axis were a number line. Similarly, the y-intercept reports the y-value at which a graph intersects the y-axis. 5.19 If a graph has x-intercept –1 and y-intercept 9, what are the coordinates of those intercepts? All x-intercepts are located on the x-axis, which has equation y = 0 (according to Problem 5.4). Therefore, the coordinate of an x-intercept must always have a y-value of 0. Similarly, all y-intercepts are located on the y-axis, which has the equation x = 0. Therefore, the coordinate of a y-intercept must always have an x-value of 0. The graph in this problem must pass through the points (–1,0) and (0,9). Note: Problems 5.20–5.21 refer to the equation x – 3y = 12. 5.20 Identify the x- and y-intercepts of the linear graph. All lines (excluding horizontal and vertical lines) have exactly on e x-intercept and one y-intercept, so aft er you’ve found one xintercept for this lin you’ve found all e, of them. According to Problem 5.19, an x-intercept has a corresponding y-value of 0. Substitute y = 0 into the equation and solve for x to identify the x-intercept. The point at which a graph intersects the y-axis has an x-value of 0. Substitute x = 0 into the equation to determine the y-intercept. The equation x – 3y = 12 has x-intercept 12 and y-intercept –4. Note: Problems 5.20–5.21 refer to the equation x – 3y = 12. 5.21 Graph the equation using the intercepts calculated in Problem 5.20. According to Problem 5.20, the intercepts are x = 12 and y = –4, so the graph passes through the points (12,0) and (0,–4). Plot those points on the coordinate plane and connect them to graph the line, as illustrated by Figure 5-13. 90 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables Figure 5-13: The graph of x – 3y = 12 has x-intercept 12 and y-intercept –4. Note: Problems 5.22–5.23 refer to the equation 2x – y = 0. 5.22 Identify the x- and y-intercepts of the linear graph. Use the technique described in Problem 5.20; substitute y = 0 into the equation to calculate the x-intercept and substitute x = 0 to calculate the y-intercept. The x- and y-intercepts of the graph are both 0. Note: Problems 5.22–5.23 refer to the equation 2x – y = 0. 5.23 Explain why you cannot graph the equation using intercepts alone. Graph the equation using a different technique. According to Problem 5.22, the x- and y-intercepts are both 0, which means the graph passes through the origin. Unfortunately, when a linear graph passes through the origin, its x- and y-intercepts are represented by the same point: (0,0). However, two distinct points are required to graph a line. To overcome this obstacle, substitute a nonzero x- or y-value into the equation and solve for the remaining variable to identify a second point on the line. For instance, substitute x = 2 into the equation and solve for y. The problem isn’t that the intercepts are both equal— that’s normally fine. The problem is that they’re both 0. If the intercepts were both 3, for example, then you’d know the graph passed through the points (3,0) and (0,3). Those are two different points. The Humongous Book of Algebra Problems 91 Chapter Five — Graphing Linear Equations in Two Variables Because y = 4 when x = 2, the point (2,4) also belongs to the graph. Connect this point and (0,0) to graph the line, as illustrated by Figure 5-14. Figure 5-14: The graph of 2x – y = 0 passes through the points (0,0) and (2,4). Note: Problems 5.24–5.25 refer to the equation 5x = 3y + 16. 5.24 Identify the x- and y-intercepts of the linear graph. Use the technique described in Problem 5.20; substitute y = 0 into the equation to calculate the x-intercept and substitute x = 0 to calculate the y-intercept. The equation has x-intercept 92 The Humongous Book of Algebra Problems and y-intercept . Chapter Five — Graphing Linear Equations in Two Variables Note: Problems 5.24–5.25 refer to the equation 5x = 3y + 16. 5.25 Graph the equation using the intercepts calculated in Problem 5.24. Calculate the mixed number equivalents of the intercepts identified in Problem 5.24 and , plot the points, and connect them to draw the graph, which is illustrated in Figure 5-15. Figure 5-15: The graph of 5x = 3y + 16 passes through points and Calculating Slope of a Line Figure out how slanty a line is 5.26 The slope of a line is defined as . Explain how to calculate the slope of a line that passes through points (x 1,y1) and (x 2,y 2). The slope of a line is the quotient of the vertical change of a line (the change in the y direction) and the horizontal change of a line (the change in the x direction). To calculate the slope of a line, identify two points on the line, subtract their y-values, and divide by the difference of the corresponding xvalues. Here, a line passes through points (x1,y1) and (x 2,y 2), so the slope of the line is . . The little triangles are the Greek letter delta, which means “change in” when it comes to math. So the slope is equal to the change in y divided by the change in x. The letter m is usually used to represent slope, even though the word “slope” doesn’t have an m in it. Here’s another one that boggles my mind: The y-intercept of a line is usually represented by the variable b. Go figure. The Humongous Book of Algebra Problems 93 Chapter Five — Graphing Linear Equations in Two Variables 5.27 Describe the difference between a linear graph with a positive slope and a graph with a negative slope. As illustrated by Figure 5-16, lines with positive slopes rise from left to right. As the x-values increase from negative to positive (as you travel right along the x-axis), the y-values increase as well (the graph climbs vertically). On the other hand, lines with negative slopes decline from left to right in the coordinate plane. Figure 5-16: L ine k has a positive slope, so it increases from left to right in the coordinate plane. Conversely, line l has a negative slope and decreases. In other words, x1 = 1, y1 = 4, x2 = 6, and y = 2. subscripts ma2tch, If the make sure the x- and y-values come from the sa me coordinate pair. 5.28 Calculate the slope of the line that passes through points (1,4) and (6,2). To apply the formula from Problem 5.26 , set (1,4) = (x1,y1) and (6,2) = (x 2,y 2). Substitute the values into the slope formula to determine the slope of the line. 5.29 Calculate the slope of the line that passes through points (–5,9) and (1,–9). Substitute x 1 = –5, y1 = 9, x 2 = 1, and y 2 = –9 into the slope formula. 94 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables 5.30 Calculate the slope of the line that passes through points and . Substituting these values into the slope formula produces a complex fraction. To simplify the numerator and denominator of the slope, ensure that the fractions you combine have common denominators. For more information about simplifying complex fractions, see Problems 2.41– 2.43. Once the numerator and denominator are rational numbers, rewrite the fraction as a quotient and simplify. 5.31 Calculate the slope of the horizontal line y = 2. To calculate slope using the formula m = , you need two points, (x 1,y1) and (x 2,y 2), on the line. Every point on the line y = 2 has a y-value of 2; no matter what real number is used for the x-value, the point (x,2) belongs to the horizontal line. For instance, set x = 0 and x = 5 to get points (0,2) and (5,2) on the graph of y = 2. Apply the slope formula. A fraction over a fraction is just a division problem, and divid is the same thing ing as multiplying by a reciprocal. That’ s how you change into . The slope of the line y = 2 is 0. Every horizontal lin the line have th e has slope 0. All the points on e same y-value, and when you su equal y-values in btract th 0. Zero divided e numerator of the slope formul a, you get by any real num ber except zero will equal 0. The Humongous Book of Algebra Problems 95 Chapter Five — Graphing Linear Equations in Two Variables If a and b are equal, you get 0 divided by 0, which is a topic for another day. For now, let’s just say a can’t equal b. If you don’t like using the abstract yvalues a and b, you don’t have to. You can pick any real numbers, say for example y = 1 and y = 4, instead. The slope of the line through (c,1) and (c,4) is also undefined. Remember, the x-value of a y-intercept is 0, just like the y-value of an x-intercept is 0. A proportion equation an is with one fraction on each side. You usually use cross multiplication to solve proportions, and that process is explained in Problems 21.1–21.8. 96 5.32 Calculate the slope of the vertical line x = c. Assume that c is a real number. All coordinate pairs on the line x = c have an x-value of c. Any value of y can be used to complete the coordinate pairs. If a and b are distinct real numbers, then (c,a) and (c,b) belong to the graph of x = c. Apply the slope formula to calculate the slope. Division by zero produces an undefined result. The slope of line x = c, like the slope of any vertical line, is undefined. It is equally correct to say that the line has “no slope.” However, it is incorrect to state that x = c has zero slope, because the number zero is defined. Horizontal lines have zero slope, and vertical lines have no slope (or an undefined slope). 5.33 If line k in the coordinate plane has slope and line l is parallel to line k, what is the slope of l? If two lines are parallel, the slopes of those lines are equal. Therefore, line l also has slope . 5.34 Assume s and t are parallel lines. If line s passes through points (0,–2) and (–5,12) and line t has x-intercept 3, what is the y-intercept of line t? If s and t are parallel lines, their slopes are equal. Use the given points to calculate ms , the slope of line s. The slope of line s (and, therefore, the slope of line t) is . If line t has x-intercept 3, then it passes through the point (3,0). Let (0,y) be the y-intercept of line t. In this instance, the slope is already known . Apply the slope formula again, substituting x 1 = 3, y1 = 0, x 2 = 0, and y 2 = y. Solve the resulting proportion for y. You should multiply both sides of this equation by –1 to cancel out the negative signs; that’s why they’re gone in the next step. The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables 5.35 If line k has slope and line l is perpendicular to line k, what is the slope of l? The slopes of perpendicular lines are opposite reciprocals. In other words, the slopes are reciprocals and have opposite signs. If the slope of line k is the slope of line l must be , then . 5.36 Assume s and t are perpendicular lines. Line s passes through point (–6,–2) and line t passes through point (x,8); the lines intersect at (–5,1). Find x. Lines s and t intersect at (–5,1), so both lines pass through that point. Therefore, line s contains the points (–6,–2) and (–5,1). Calculate ms , the slope of line s. Because s and t are perpendicular lines, their slopes are opposite reciprocals, so if ms = , then mt = . Recall that line t passes through points (–5,1) and (x,8). Apply the slope formula. Cross multiply and solve the proportion for x. 5.37 Calculate the x- and y-intercepts of the linear equation Ax + By = C, and use them to generate a formula for the slope of a line written in that form. Assume that A, B, and C are real numbers and . Calculate the x-intercept by substituting y = 0 into the equation and calculate the y-intercept by substituting x = 0 into the equation. - - If A, B, and C aren’t fractions and A is positive, then a line in the form Ax + By = C is in “standard form.” More on this in Problems 6.29– 6.36. The Humongous Book of Algebra Problems 97 Chapter Five — Graphing Linear Equations in Two Variables The graph of Ax + By = C intersects the x-axis at at and intersects the y-axis . Apply the slope formula to calculate the slope of the line containing those points. Rewrite the complex fraction as a quotient and simplify. This is a handy shortcut, as you’ll see in the next problem. The slope of line Ax + By = C is . 5.38 Use the formula generated in Problem 5.37 to determine the slope of the line with equation 3x = 6y + 5. Subtract 6y from both sides of the equation 3x = 6y + 5 so that both of its variable terms appear on the left side of the equation and the constant term is isolated on the right side. 3x – 6y = 5 The equation is now in the form Ax + By = C, and you can calculate the slope of the line using the shortcut formula. 98 The Humongous Book of Algebra Problems Chapter Five — Graphing Linear Equations in Two Variables 5.39 Graph the line that passes through (–3,1) and has slope . The slope of a line indicates the vertical and horizontal change between points on the line. The numerator indicates vertical distance, and the denominator indicates horizontal distance. Here, the slope is negative, so place the negative sign either in the numerator or the denominator—your choice will not affect the outcome. Plot the given point (–3,1) on the coordinate plane. According to the slope , if you travel down five units and right two units from that point, you arrive at another point on the line. Because the slope is equivalent, you can also travel up five units and left two units to reach another point on the line, as illustrated by Figure 5-17. Connect the points to graph the line. A positive numerator means “up,” and a negative numerator means “down.” A positive denominator means “right,” and a negative denominator means “left.” Figure 5-17: The point (–1,–4) is five units below and two units right of the starting point (–3,1). The point (–5,6) is five units above and two units left of (–3,1). The Humongous Book of Algebra Problems 99 Chapter Five — Graphing Linear Equations in Two Variables Graphing Absolute Value Equations When you toss absolute values into a linear equa tion, you change the graph as well. Instead of a straight line gra ph, yo end up with somet u hing that’s pointy and V-shaped. Don’t miss the point in these graphs (Get it?) 5.40 Graph the equation using a table of values. Substitute values of x that are less than 1 as well as values that are greater than 1, or you will omit an important change in the graph at x = 1. Figure 5-18: T he point at which the direction of a linear absolute value graph changes is called its vertex. The vertex of Note: Problems 5.41–5.42 refer to the equation The vertex is the sharp point of the V-shaped graph. is (1,–3). . 5.41 Without using a table of values, identify the vertex of the graph. To identify the vertex of a linear absolute value graph written in terms of x, set the expression within the absolute value bars equal to 0 and solve for x. The resulting value is the x-coordinate of the vertex. The equation is in terms of x because it’s solved for y. Substitute into the equation and solve for y. The vertex of the absolute value graph is ngous Book of Algebra Problems 100 The Humo . Chapter Five — Graphing Linear Equations in Two Variables Note: Problems 5.41–5.42 refer to the equation 5.42 Graph the equation. According to Problem 5.41, the vertex of the graph is , or . The direction of the graph will change at the vertex, so you must plot at least one point left of the vertex and one point right of the vertex to draw the graph. Substitute one x-value less than (such as –4) and one value greater than (such as 0) into the equation and calculate the corresponding y-values. Therefore, the points (–4,7) and (0,9) belong to the graph. As illustrated in Figure 5-19, the graph consists of two rays, one that extends from through the point (–4,7) and one that extends from (0,9). Figure 5-19: The graph of has vertex through the point . The Humongous Book of Algebra Problems 101 Chapter Five — Graphing Linear Equations in Two Variables To get the x-coordinate of the vertex, set whatever’s inside the absolute value bars equal to 0. Plug your answer back into the original equation for x to figure out the y-coordinate of the vertex. 5.43 Graph the equation and identify the vertex of the graph. Use the method described in Problem 5.42 to identify the vertex. The vertex of the graph is (0,–5). Substitute one x-value less than 0 (such as x = –4) and one x-value greater than 0 (such as x = 4) into the equation to identify two additional points on the graph. The graph consists of two rays, one that extends from (0,–5) through (–4,–3) and one that extends from (0,–5) through (4,–3), as illustrated by Figure 5-20. Figure 5-20: The graph of ngous Book of Algebra Problems 102 The Humo . Chapter Five — Graphing Linear Equations in Two Variables 5.44 Graph the equation and identify the vertex of the graph. Use the method described in Problem 5.42 to identify the vertex. The vertex of the graph is , or . Identify two additional points on the graph by substituting an x-value less than greater than (such as x = 5) into the equation and solve for y. One ray of the graph extends from from (such as x = 3) and an x-value through (3,6), and the other extends through (5,4), as illustrated by Figure 5-21. Figure 5-21: The graph of . The Humongous Book of Algebra Problems 103 Chapter 6 Linear Equations in Two Variables Generating equations of lines Chapter 5 introduced the concept of linear slope, a constant that describes the steepness of a line in the coordinate plane. This chapter explores constructing linear equations using the point-slope and slope-intercept formulas, expressing equations in standard form, and graphing linear equations that are expressed in point-slope form. You need only two things to create the equation of a line: the slope of that line and one of the points on the line. Now that you know how to calculate slope (if you don’t, stop rig ht now and work through Problems 5.26–5.39), you can use one of two for mulas to create a linear equation: the point-slope formula or the slopeintercept formula. After you generate the equation of a line, you can sketch a quick graph of it (say goodbye to tables of values) and then (if required) write it in standard form. Chapter Six — Linear Equations in Two Variables Point-Slope Form of a Linear Equation Point + slope = equation 6.1 Identify the values of the constants in the point-slope form of a line. The point-slope form of a linear equation states that a line with slope m that passes through point (x 1,y1) has equation y – y1 = m(x – x 1). Any point (x,y) that satisfies the equation lies on the graph of the line. x1 and y1 are only needed at the beginning of the problem, whe n you create the equa ti The equation you on. co up with will not co me ntain x1 and y1, but it w ill contain x and y. 6.2 Describe the difference between the subscripted variables in the point-slope form of a line (x1 and y1) and the variables without subscripts (x and y). The point-slope form of a line is used to create the equation of a line if the slope of that line and one of the points on that line are known. The coordinates (x1,y1) represent the x- and y-values of the point through which the line is known to pass. The point-slope formula also contains the variables x and y, which correspond to any point (x,y) through which the line passes. Note: Problems 6.3–6.4 refer to line j, which has slope –4 and passes through the point (–1,7) on the coordinate plane. 6.3 Use the point-slope formula to write an equation representing line j. Line j has slope m = –4 and passes through the point (–1,7), so x 1 = –1 and y1 = 7. Substitute the values of m, x 1, and y1 into the point-slope formula. Note: Problems 6.3–6.4 refer to line j, which has slope –4 and passes through the point (–1,7) on the coordinate plane. Substitute into th e equation of line j from Problem 6.3. 6.4 If line j also passes through the point (a,3), what is the value of a? If line j passes through the point (a,3), then substituting x = a and y = 3 into the equation y – 7 = –4(x + 1) must result in a true statement. ngous Book of Algebra Problems 106 The Humo Chapter Six — Linear Equations in Two Variables Solve the equation for a. Because a = 0, line j passes through the point (a,3) = (0,3). Note: Problems 6.5–6.6 refer to line k, which has slope (–6,–5) on the coordinate plane. 6.5 and passes through the point Use the point-slope formula to create the equation of line k. Expand the resulting equation and solve it for y. Substitute , x 1 = –6, and y1 = –5 into the point-slope formula. Treat the negatives outside these parentheses as –1s: (–1)(–5) = 5 and –1(–6) = 6. A negative times a negative equals a positive. Expand the right side of the equation and solve for y. Note: Problems 6.5–6.6 refer to line k, which has slope (–6,–5) on the coordinate plane. 6.6 If line k also passes through the point Substitute and passes through the point , what is the value of c? and y = c into the equation generated by Problem 6.5. Use the least common denominator to combine these numbers: The Humongous Book of Algebra Problems 107 Chapter Six — Linear Equations in Two Variables 6.7 For more info on x- and yintercepts, check out Problems 5.18–5.25. Use the point-slope formula to create the equation of line l, which has slope and x-intercept 1. If line l has x-intercept 1, it passes through the point (1,0). Substitute x 1 = 1, and y1 = 0 into the point-slope formula. Line l has equation . Applying the distributive property produces another valid representation of line l: 6.8 Apply the ive ribut dist property and simplify. . Use the point-slope formula to identify the equation of the line that passes through points (3,–1) and (–7,–9). Expand the resulting equation, simplify it, and solve for y. Apply the technique described in Problems 5.28–5.30 to calculate the slope of the line; substitute x1 = 3, y1 = –1, x 2 = –7, and y 2 = –9 into the slope formula. Substitute You could substitute the coordinates of the other point into the formula instead: x1 = –7 and y1 = –9. Either way, you’ll get the same final answer. , , x1 = 3, and y1 = –1 into the point-slope formula. Expand and simplify the right side of the equation. Solve for y and use a common denominator to combine the constant terms. ngous Book of Algebra Problems 108 The Humo Chapter Six — Linear Equations in Two Variables 6.9 Use the point-slope formula to identify the equation of the line with xintercept 1 and y-intercept –9. Expand the resulting equation, simplify it, and solve for y. If the line has x-intercept 1, then it passes through the point (1,0). Similarly, because the line has y-intercept –9, it passes through the point (0,–9). Calculate the slope of the line. x1 = 1, y1 = 0, x2 = 0, and y2 = –9. Substitute m = 9, x 1 = 1, and y1 = 0 into the point-slope formula. 6.10 Use the point-slope formula to identify the equation of the line that passes through the points (a,b) and (c,d). Assume that a, b, c, and d are real numbers and that a ≠ c. Apply the technique described in Problems 6.8–6.9. Begin by substituting x1 = a, y1 = b, x 2 = c, and y 2 = d into the slope formula. Substitute m, x1, and y1 into the point-slope formula. Substituting (x 1,y1) = (c,d) into the point-slope formula, rather than using the coordinate (a,b), produces the equally valid equation If you set x1 = c, y1 = d, x2 = a, and y2 = b, you get a differentlooking (but equal) slope. You can use that slope to come up with two different (and equivalent) equations that are also acceptable answers: . and . The Humongous Book of Algebra Problems 109 Chapter Six — Linear Equations in Two Variables Slope-Intercept Form of a Linear Equation The coefficient of the xterm is the slope of the line, and the constant marks the point where the line crosses the y-axis. Lines that look like y = mx + b 6.11 Identify the values of the constants in the slope-intercept form of a line. The slope-intercept form of a line is y = mx + b, where m is the slope of the line and b is the y-intercept. Note that an equation in slope-intercept form is solved for y (that is, y is isolated on the left side of the equal sign). The right side of the equation is the sum of an x-term and a constant. Note: Problems 6.12–6.13 refer to line s, which has slope –8 and y-intercept 3. 6.12 Write the equation of line s in slope-intercept form. The slope of the line is –8 so set m = –8. The y-intercept of the line is 3, so set b = 3. Substitute m and b into the slope-intercept formula. Note: Problems 6.12–6.13 refer to line s, which has slope –8 and y-intercept 3. The equation created in Problem 6.12 6.13 If line s passes through point (13,v), what is the value of v? If line s passes through (13,v), then substituting x = 13 and y = v into the equation y = –8x + 3 results in a true statement. Note: Problems 6.14–6.15 refer to line t, which has slope and y-intercept . 6.14 Write the equation of line t in slope-intercept form. Substitute and into the slope-intercept formula. Note: Problems 6.14–6.15 refer to line t, which has slope 6.15 If line t passes through point Substitute x = w and and solve for w. 110 The Humongous Book of Algebra Problems and y-intercept . , what is the value of w? into the linear equation generated by Problem 6.14 Chapter Six — Linear Equations in Two Variables 6.16 What is the x-intercept of the line with y-intercept –2 and slope Substitute ? and b = –2 into the slope-intercept formula to generate the equation of the line. Substitute y = 0 into the equation and solve for x to determine the x-intercept. 6.17 Use the slope-intercept formula to determine the slope of the line with x-intercept –4 and y-intercept . Substitute the given y-intercept into the slope-intercept formula: . The x-intercept is located on the xa All points on the xis. x-axis have a y-value of 0, so set y = 0. In other words, don’t use the slope formula from Problems 6.8 and 6.9 to calculate the slope of the line through (–4,0) and . Use the The x-intercept of the line is (–4,0), so substituting x = –4 and y = 0 into the equation produces a true statement. slope-intercept formula instead. The Humongous Book of Algebra Problems 111 Chapter Six — Linear Equations in Two Variables The equation has fractions in it so don’t divide by , 4 to get rid of the xcoefficient. Instea d, multiply by , th e reciprocal of 4. Solve the equation for m to determine the slope of the line. 6.18 Problem 6.5 describes line k, which has slope Problem 6.5 asked you to expand the equation and solve for y after you plugged everything into pointslope form. By solving for y, you were actually writing the equation in slope-intercept form! So it doesn’t matter whether you use the pointslope or slopeintercept formula to create a linear equation. Both give you the same result when you put them into the same form, in this case solving for y to put them both in slopeintercept form. 112 and passes through the point (–6,–5). Use the slope-intercept formula to identify the equation of line k and verify the solution to Problem 6.5. The slope is given, but the y-intercept is not. Substitute and the known x- and y-values (x = –6 and y = –5) into the slope-intercept formula and solve for b. The slope of the line is , and the y-intercept is b = –1. Substitute these values into the slope-intercept formula to identify the equation of the line. This answer is identical to the equation generated in Problem 6.5. 6.19 Use the slope-intercept formula to identify the equation of the line that passes through points (10,–4) and (9,8). Calculate the slope of the line. The Humongous Book of Algebra Problems Chapter Six — Linear Equations in Two Variables Substitute m = –12 and the x- and y-values from one of the given points into the slope-intercept formula, for instance, x = 10 and y = –4. (It does not matter which point you select, as long as you substitute x- and y-values from the same coordinate.) Here’s what the work looks like if you plug in x = 9 and y = 8 instead: Substitute m and b into the slope-intercept formula to generate the equation of the line. 6.20 Apply the point-slope formula to identify the equation of the line with slope m and y-intercept b. If a line has y-intercept b, then it passes through the point (0,b). Substitute the slope m, x1 = 0, and y 1 = b into the point-slope formula. This is interesting because it shows you where the slope-intercept formula comes from. Solve the equation for y. Graphing Lines in Slope-Intercept Form Graphing equations that are solved for y 6.21 Graph the linear equation . The equation is in slope-intercept form, y = mx + b. Therefore, the coefficient of the x-term is the slope of the line and the constant is the y-intercept of the line (b = –2). To graph the line, plot the y-intercept (0,–2) and use the slope to identify another point on the line. The slope is , so the point one unit above and three units right of the y-intercept is also on the line. Connect those points to graph the line, as illustrated by Figure 6-1. The numerator of the slope tells you how far to up or down. (Positive means go up, negative means go down.) The denominator of the slope tells you how far to go left or right. (Positive means go right, negative means go left.) Here, the top is 1 (go up one) and the bottom is 3 (go right 3). The Humongous Book of Algebra Problems 113 Chapter Six — Linear Equations in Two Variables Figure 6-1: The graph of passes through the point (3,–1), which is one unit above and three units right of the y-intercept (0,–2). 6.22 Graph the linear equation . The equation is in slope-intercept form, so the slope is and the y-intercept is b = 1. Plot the y-intercept on the coordinate plane. The negative sign of the slope can be placed in either the numerator or the denominator, for the purposes of plotting the second point on the line: The y-intercept is located at point (0,1). If you express the slope as . , count three units down and five units right of the y-intercept to plot the point (5,–2). If you, instead, express the slope as , travel up three units and left five units to plot the point (–5,4). All three points—(0,1), (5,–2), and (–5,4)—belong to the same line, as illustrated by Figure 6-2. Figure 6-2: The graph of has y-intercept 1. According to the slope, you c an travel up 3 and left 5 units, or down 3 and right 5 units, to reach another point on the line. 114 The Humongous Book of Algebra Problems Chapter Six — Linear Equations in Two Variables Note: Problems 6.23–6.24 refer to the equation y – 11 = –3(x + 5). 6.23 Express the linear equation in slope-intercept form. The equation’s in slope-intercept form when it looks like y = mx + b. In other words, y’s on one side, and on the other side, you have an xterm and a constant term, in that order. Expand the right side of the equation and solve for y. Note: Problems 6.23–6.24 refer to the equation y – 11 = –3(x + 5). 6.24 Graph the linear equation. According to Problem 6.23, the slope-intercept form of the equation is y = –3x – 4, so the y-intercept is –4, and the graph passes through the point (0,–4). The slope of the line is –3. Express the slope as a fraction and use it to identify a second point on the line: . Connect the two points to graph the line, as illustrated in Figure 6-3. You can go down three and right one or up three and left on e from the y-intercept (0,– 4) to reach another po int on the line. Figure 6-3: The graph of y – 11 = –3(x + 5) is equivalent to the graph of y = –3x – 4, which has slope –3 and y-intercept –4. Note: Problems 6.25–6.26 refer to the equation 5x – 2y = 4. 6.25 Express the linear equation in slope-intercept form. Solve the equation for y to express it in slope-intercept form. The Humongous Book of Algebra Problems 115 Chapter Six — Linear Equations in Two Variables Note: Problems 6.25–6.26 refer to the equation 5x – 2y = 4. 6.26 Graph the linear equation. According to Problem 6.25, the slope-intercept form of the line is . Plot the y-intercept, point (0,–2) on the coordinate plane. The slope dictates that the point (2,3), which is 5 units above and two units right of the y-intercept, also belongs to the line. Graph the line by connecting those two points, as illustrated by Figure 6-4. Figure 6-4: T he graph of 5x – 2y = 4 or, in slope-intercept form, 116 The Humongous Book of Algebra Problems . Chapter Six — Linear Equations in Two Variables Note: Problems 6.27–6.28 refer to the points A = (3,11), B = (–6,–10), and C = . 6.27 Graph A, B, and C to demonstrate that the points are collinear. Points are collinear if they lie on the same line. It’s easier to graph if you write the coordinates as mixed numbers: . Figure 6-5: A, B, and C are collinear because one line on the coordinate plane passes through all three points. Note: Problems 6.27–6.28 refer to the points A = (3,11), B = (–6,–10), and C = . 6.28 Prove that the points are collinear. Calculate the slope of the line containing points A and B. Two points define a line, so only one line exists in the coordinate plane that passes through those two points. Now calculate the slope of the line that contains points A and C. Again, only one line exists in the coordinate plane that passes through those two points. This time, use algebra instead of a graph to prove that the points are all on the same line. You could calculate the slope of the line through points B and C instead. Use the least-common denominator 4 to combine the terms in the numerator and denominator. The Humongous Book of Algebra Problems 117 Chapter Six — Linear Equations in Two Variables Multiply the numerator and denominator by the reciprocal of the denominator. To find the equation of a line, you need a point and a slope. Both of these slopes are equal and both lines pass through B, so according to the point-slope formula, both would have the same equation. The line connecting A and C has the same slope as the line connecting A and B; it also shares a common point, A. Therefore, the lines are not unique. Instead, points A, B, and C all belong to the same line. Standard Form of a Linear Equation Write equations of lines in a uniform way 6.29 What are the characteristics of a linear equation in standard form? If all the coefficients have a common factor , divide all of the terms by that number. Fo example, if all of r th coefficients are e even, divide everything by 2. The standard form of a linear equation is Ax + By = C. In standard form, the variable terms are located left of the equal sign, and the constant is located right of the equal sign. There are additional restrictions on the constants. For one thing, A, B, and C cannot share common factors (other than 1). The coefficients must all be integers, which means that no fractions are present in the equation. Finally, A, the coefficient of the x-term, must be nonnegative. If A is negative, multiply each term of the equation by –1. 6.30 According to Problem 5.37, the slope of Ax + By = C, a linear equation in standard form, is . Use this shortcut formula to calculate the slope of x + 4y = –3, and verify the answer by writing the equation in slope-intercept form. When no coefficient is written, you should assume the coefficient is 1, so A = 1. The equation x + 4y = –3 is in standard form, so the coefficients of x and y are A and B, respectively: A = 1 and B = 4. Substitute these values into the slope shortcut formula. Solve the equation x + 4y = –3 for y to express it in slope-intercept form. The slope of an equation in slope-intercept form is the coefficient of the x-term. The slope of this line is formula. 118 The Humongous Book of Algebra Problems , which verifies the slope generated by the shortcut Chapter Six — Linear Equations in Two Variables 6.31 Calculate the slope of the line 4x – 9y = 14 using the slope shortcut formula and verify the answer by writing the equation in slope-intercept form. Substitute A = 4 and B = –9 into the slope shortcut formula. The formula from Problem 6.30 Solve the equation 4x – 9y = 14 for y. The slope of an equation in slope-intercept form is the coefficient of its x-term. The slope of this line is formula. , which verifies the slope generated by the shortcut 6.32 What linear equations can be expressed in standard form but cannot be expressed in slope-intercept form? Equations of vertical lines cannot be written in slope-intercept form. Vertical lines have form x = c, where c is the distance between the vertical line and the y-axis. Because the equations do not contain y, they cannot be solved for y—one major characteristic of lines in point-slope form. Furthermore, vertical lines have an undefined slope and are, therefore, missing yet another key component of slope-intercept form. 6.33 Express the equation y – 3 = –7(x – 1) in standard form. To convert this equation from point-slope form to standard form, begin by expanding the right side of the equation. Move –7x left of the equal sign and –3 right of the equal sign by adding those quantities to both sides of the equation. This equation is in standard form Ax + By = C because the x- and y-terms are left of the equal sign, the coefficients and constant share no common factors, no fractions are present in the equation, and the coefficient of x is positive. “Undefined slope” means the same as “no slope.” Horizontal lines have a slope of zero, which is a real number. Vertical lines don’t have a slope at all. You can’t write an equation in slopeintercept form if there’s no slope! The x- and y-terms also have to be in th e right order; x ha s to come before y, so the equation y + 7x = 10 —thou gh equivalent—is no t in standard form. The Humongous Book of Algebra Problems 119 Chapter Six — Linear Equations in Two Variables 6.34 Express the equation in standard form. Begin by expanding the right side of the equation. An equation in standard form cannot contain constants or coefficients that are fractions. To eliminate fractions from the equation, multiply each term by 2, the least common denominator. Move the x-term left of the equal sign by subtracting 3x from both sides of the equation. Similarly, move the constant 8 right of the equal sign by subtracting it from both sides. It doesn’t matter if the y-term or the constant is negative, but an equation in standard form has to start with a positive x-term. This equation is not yet in standard form because the coefficient of the x-term is negative. Multiply each term by –1. 6.35 Express the equation in standard form. Use the technique described in Problem 6.34 to convert this slope-intercept equation into standard form: eliminate the fraction, move the x-term left of the equal sign, and multiply all the terms by –1 to make the x-term positive. ngous Book of Algebra Problems 120 The Humo Chapter Six — Linear Equations in Two Variables 6.36 Express the linear equation 6y = 5x + 2(x + 10) – 16 in standard form. Expand and simplify the right side of the equation. Subtract 7x from both sides of the equation and multiply all the terms by –1 so that the equation has form Ax + By = C, where A > 0. You could subtract 6y and 4 from both sides of the equation instead to get –4 = 7x – 6y. Then you’d swap the sides of the equation to get 7x – 6y = –4. Creating Linear Equations Practice all the skills from this chapter Note:Problems 6.37–6.39 refer to line j, which passes through point (3,–1) and is parallel to line p, which has equation . 6.37 Apply the point-slope formula to determine the equation of line j. Line p is in slope-intercept form, so the x-coefficient is the slope of line p. The slopes of parallel lines are equal, so the slope of line j must be Substitute as well. and the coordinates of the point through which j passes (x1 = 3 and y1 = –1) into the point-slope formula. This property of parallel lines was first introduced back in Problem 5.33. The Humongous Book of Algebra Problems 121 Chapter Six — Linear Equations in Two Variables Note: Problems 6.37–6.39 refer to line j, which passes through point (3,–1) and is parallel to line p, which has equation . 6.38 Express line j in slope-intercept form. Expand the right side of the point-slope equation generated in Problem 6.37 and solve for y. Note: Problems 6.37–6.39 refer to line j, which passes through point (3,–1) and is parallel to line p, which has equation . 6.39 Express line j in standard form. Multiply each term of the equation in slope-intercept form (from Problem 6.38) by 3 to eliminate the fraction. Subtract 5x from both sides of the equation and multiply both sides by –1 to rewrite the equation in form Ax + By = C, such that A > 0. ngous Book of Algebra Problems 122 The Humo Chapter Six — Linear Equations in Two Variables Note: Problems 6.40–6.42 refer to line k, which passes through point (–8,7) and is perpendicular to the line l, which has equation 2x – 3y = 12. 6.40 Calculate the slope of line k. Calculate the slope of line l by applying the slope shortcut formula from Problem 6.30. Line k is perpendicular to line l, so the slope of k is of line l’s slope. , the opposite reciprocal Note: Problems 6.40–6.42 refer to line k, which passes through point (–8,7) and is perpendicular to the line l, which has equation 2x – 3y = 12. 6.41 Use the slope-intercept formula to create the equation of line k. Line k passes through point (–8,7) and, according to Problem 6.40, has slope . Substitute The slopes of perpendicular lines are opposites—if one is positive then the other is negative— and reciprocals of each other. If you feel like you heard that somewhere before, you might remember it from Problem 5.35. , x = –8, and y = 7 into the slope-intercept formula in order to calculate b. Apply the slope-intercept formula to generate the equation of line k. Note: Problems 6.40–6.42 refer to line k, which passes through point (–8,7) and is perpendicular to the line l, which has equation 2x – 3y = 12. 6.42 Express line k in standard form. According to Problem 6.41, the equation of line k in slope-intercept form is . Multiply all the terms by the least common denominator (2) and add the x-term to both sides of the equation. The resulting equation has form Ax + By = C such that A > 0. The Humongous Book of Algebra Problems 123 Chapter Six — Linear Equations in Two Variables 6.43 Write the equation of the line graphed in Figure 6-6 in slope-intercept form. Figure 6-6: Y ou can determine the equation of this line using either point-slope or slopeintercept form, but write your final answer in slope-intercept form. The line in Figure 6-6 passes through points (–5,–1) and (4,3). Use the slope formula to calculate the slope of the line. Substitute the slope and the x- and y-values from one of the points through which the line passes into the point-slope formula. Subtract 1 from both sides of the equation to isolate y in this step. Instead of “–1,” you can write . It’s the same value ( ), but you need a denominator in common with so that you can add those numbers together. Expand the right side of the equation and solve for y to express the equation in slope-intercept form. ngous Book of Algebra Problems 124 The Humo Chapter Six — Linear Equations in Two Variables 6.44 Write the equation of the line with x-intercept and y-intercept in standard form. Calculate the slope of the line that passes through points The slope of the line is and the y-intercept is stated: and . . Substitute these values into the slope-intercept formula to create the equation of the line. To get rid of this complex fraction, multiply the num era and denominato tor r by 2, the least com mon denominator. Multiply each of the terms by the least common denominator (14) to eliminate the fractions in the equation. Subtract 10x from both sides of the equation and multiply each term by –1 so that the equation has form Ax + By = C and A > 0. The Humongous Book of Algebra Problems 125 Chapter 7 Linear Inequalities e They’re like equations without th equal sign Replacing an equation’s equal sign with an inequality sign might, at first glance, seem a trivial change, but the implications are significant. For instance, whereas a linear equation in one variable has a single solution, a typical linear inequality in one variable has infinitely many solutions. The differences are not limited to one-dimensional equations. As discussed in Chapter 5, the graphs of linear equations are lines in the coordinate plane, but the graphs of linear inequalities are regions of the plane. Linear inequalities LOOK a lot like equations, but they’re very different. The good news: Linear ine qualities and linear equations act a lot alike. If you can solve the equa tions in Chapter 4, and graph the equations in Chapter 5, linear inequ alities will be a piece of cake. The bad news: Absolute value inequalitie s are harder than absolute value equations. Chapter Seven — Linear Inequalities Inequalities in One Variable Don’t say, “I am not the same height as my cousin,” when you can say, “I am shorter than my cousin.” The second statement doesn’t just state that you have different heights— it also explains why. There’s no way you can meet BO T conditions. No nu H mber is both equal to itself AND less th an itself. If –4 ≥ –4 is true, then either the left side is bigger than the right side (it’s not) or both sides are equal (they are). Think of a number line. If you plot –12 and –15, –15 is farther to the left on the numbe r line. The farther left you go, the “less” th number. The fart e he right you go, the r “greater” the number. Dust off your equation-solving skills from Chapter 4 7.1 Identify the five most commonly used inequality symbols. The five most used algebraic inequality symbols are: <, less than; >, greater than; ≤, less than or equal to; ≥, greater than or equal to; and ≠, not equal to. When possible, use <, ≤, >, and ≥ instead of ≠, because they communicate more information. 7.2 Of the inequality symbols <, ≤, >, and ≥, which correctly complete the following statement? 2 ______ 7 Two is fewer than seven, so the less than symbol correctly completes the statement: 2 < 7. Two is also less than or equal to seven. For the statement 2 ≤ 7 to be true, exactly one of the following conditions must be met: two must either be less than seven (it is), or it must be equal to seven (it is not). 7.3 Of the inequality symbols <, ≤, >, and ≥, which correctly complete the following statement? –4 ______ –4 As stated in Problem 7.2, a statement containing the ≤ symbol is true in one of two cases: if the left quantity is less than the right quantity or both quantities are equal. Here, the left and right sides of the inequality are equal, so the statement –4 ≤ –4 is true. Similarly, the statement –4 ≥ –4 is true. 7.4 Is the following statement true or false? Explain your answer. –15 > –12 The statement –15 > –12 is false. The more negative a number, the less that number is considered. Therefore, –15 is less than –12 because –15 is more negative than –12. 7.5 Solve the inequality x – 3 > 11 for x. To solve a linear inequality in one variable, isolate that variable left of the inequality sign, much like you would solve a linear equation by isolating the variable left of the equal sign. Here, isolate x by adding 3 to both sides of the inequality. ngous Book of Algebra Problems 128 The Humo Chapter Seven — Linear Inequalities The solution to the inequality is x > 14, so substituting any real number greater than 14 into the inequality x – 3 > 11 produces a true statement. 7.6 Not including 14 Solve the inequality 4x + 1 ≤ –11 for x. Isolate the x-term on the left side of the inequality by subtracting 1 from both sides of the equation. Divide both sides of the inequality by 4 to solve for x. 7.7 Solve the inequality 7x – 2(x + 1) ≤ 18. Expand and simplify the left side of the inequality. Isolate 5x on the left side of the inequality and then divide both sides by 5 to eliminate the coefficient. 7.8 Identify the four most common reasons an inequality sign must be reversed. The four most common reasons an inequality symbol must be reversed are: multiplying both sides of an inequality by a negative number, dividing both sides of an inequality by a negative number, exchanging the sides of an inequality, and taking the reciprocal of both sides of an inequality. Reversing an inequality symbol means changing the direction it points. That means < becomes > (and vice versa) and ≤ becomes ≥ (and vice versa). If you swap the sides of an inequality, you have to reverse the inequality symbol. For example, 3 > x becomes x < 3. The Humongous Book of Algebra Problems 129 Chapter Seven — Linear Inequalities 7.9 If x is on the left side of the inequality, then it’s slightly easier to graph—the symbol points in the same direction the arrow will on the graph. Solve the inequality for x. To isolate x, multiply both sides of the inequality by coefficient. , the reciprocal of its It is customary to position x left of the inequality symbol. Exchange the sides of the inequality and, as noted in Problem 7.8, reverse the inequality symbol. 7.10 Solve the inequality Subtract for x. from, and add 1 to, both sides of the inequality to separate the x-terms and constants on opposite sides of the inequality symbol. RU LE OF TH UMB: Why do you reverse the sign when you multiply or divide by a negative number? Here’s an example: –1 < 5, a true inequality statement. If you multiply both sides by –1, you get 1 < –5, which is false—a positive number can’t be less than a negative number. Reversing the inequality symbol fixes the problem. Combine like terms, using the least common denominator 10 to add the fractions. To eliminate the coefficient of x, multiply by . As indicated in Problem 7.8, multiplying both sides of an inequality by a negative number requires the reversal of the inequality sign. ngous Book of Algebra Problems 130 The Humo Chapter Seven — Linear Inequalities 7.11 Solve the inequality 2x + 3 ≥ 8x – (x – 1) for x. Expand the right side of the inequality and simplify it. Subtract the terms 7x and 3 from both sides of the inequality to separate the xterms and constants on opposite sides of the inequality symbol. Divide both sides of the inequality by –5 to solve for x. Recall that dividing by a negative number requires you to reverse the inequality symbol. 7.12 Solve the inequality You can’t solve for x until it’s in th e numerator, wheth er it’s an inequality like this one or an equation like in Problem 4.14. for x. Take the reciprocal of both sides of the inequality and reverse the inequality sign (as directed by Problem 7.8). Write the left side of the inequality as the product of x and a coefficient. Multiply both sides of the inequality by 9, the reciprocal of the x-coefficient. Why reverse the inequality sign? Here’s an example: , but if you take the reciprocal of both sides, you get 4 < 2, which is false. You have to reverse the sign to fix the problem: 4 > 2. Divide the numerator and denominator by 3, the greatest common factor of 12 and 63, to reduce the fraction to lowest terms. The Humongous Book of Algebra Problems 131 Chapter Seven — Linear Inequalities “Open” points are not included in the graph and look like hollow dots. “Closed” points are included on the graph and look like solid dots. Graphing Inequalities in One Variable Shoot arrows into number lines 7.13 What characteristics of an inequality statement determine whether the point or points plotted on its graph are open or closed? Points on the graph of a linear inequality are classified according to the adjacent inequality symbol. If that symbol allows for the possibility of equality (that is, the symbol is ≤ or ≥), plot the value using a closed point. Alternatively, if the symbol indicates strict inequality (the symbol is < or >), an open point should be used to indicate that the value is not a solution to the inequality. 7.14 Compare and contrast the graphs of linear equations and linear inequalities in one variable. If you say, “I can’t run faster than 9 miles an hour,” and you mean that 9 mph is an impossible goal, your speeds are x ≤ 9, and the graph would have an open dot. If you mean you CAN run 9 mph but no more, then use a closed dot to graph x ≤ 9. Because they each contain one variable, both are plotted on a number line (rather than a coordinate plane, which is used when statements are written in terms of two variables). The graphs of linear equations consist of a single point on the number line, whereas the solutions of linear inequalities are intervals consisting of infinitely many values. 7.15 Graph the inequality: x > 3. Plot the value x = 3 on a number line using an open point, as explained in Problem 7.13. Darken the portion of the number line that is right of x = 3, as illustrated by Figure 7.1, as any value in that interval makes the inequality true. Figure 7-1: The graph of x > 3 has an open point at x = 3 because 3 is not a valid Intervals are segments of the number line, so the solution to a linear inequality is something like x > –4 (any real number greater than –4 is a solution), and the solution to a linear equation would be a single number like x = –4. 132 solution to the inequality. 7.16 Graph the inequality x ≤ –1. Plot the value x = –1 on a number line using a closed point, as x = –1 is one of the solutions to the inequality. Darken the portion of the number line that is left of –1 to identify the other solutions of the inequality, as illustrated in Figure 7-2. Figure 7-2: The graph of x ≤ –1 consists of –1 and all of the real numbers less than –1. The inequality > points right, so shade everything right of x = 3 on the number line. This shortcut works onl y when x is on the left side of the inequality. The Humongous Book of Algebra Problems Chapter Seven — Linear Inequalities 7.17 Solve the inequality 3x + 20 ≥ 8 for x and graph the solution. If the inequality symbol is either < or >, use an open dot on the graph. If it’s ≤ or ≥, use a closed dot. Solve the inequality by isolating x left of the inequality sign. Plot x = –4 on the number line using a closed point and darken the portion of the number line greater than –4, as illustrated by Figure 7-3. Figure 7-3: The graph of x ≥ –4, the solution to the inequality 3x + 20 ≥ 8. 7.18 Solve the inequality for x and graph the solution. Multiply both sides of the inequality by 4 to eliminate the fraction and then isolate x left of the inequality symbol. Plot on a number line using an open point and darken the portion of the number line that is less than Figure 7-4: The graph of x < , as illustrated by Figure 7-4. , the solution to the inequality It’s easier to graph this fract ion if you write it as a mixed number. Us e the formula from Prob le 2.7: 5 divides into m 11 twice with a rem ainder of 1, so . (5x – 3) < 2. The Humongous Book of Algebra Problems 133 Chapter Seven — Linear Inequalities 7.19 Solve the inequality –(2x + 12) > 9 + 5x for x and graph the solution. You’re dividing both sides of the inequality by a negative number, so make sure to reverse this inequality sign. Solve the inequality by isolating x left of the inequality symbol. Plot x = –3 on a number line using an open point and darken the portion of the number line that is less than –3, as illustrated by Figure 7-5. Figure 7-5: The graph of x < –3, the solution to the inequality –(2x + 12) > 9 + 5x. 7.20 Solve the inequality 4(x + 2) – 1 ≥ 7x + 5 for x and graph the solution. Here’s another inequality sign th at’ reversed becaus s e yo divide by a nega u tive number. Solve the inequality by isolating x left of the inequality symbol. Plot x = on a number line using an closed point and darken the portion of the number line that is less than Figure 7-6: The graph of ngous Book of Algebra Problems 134 The Humo , as illustrated by Figure 7-6. , the solution to the inequality 4(x + 2) – 1 ≥ 7x + 5. Chapter Seven — Linear Inequalities Compound Inequalities Two inequalities for the price of one Note: Problems 7.21–7.22 refer to the compound inequality –4 < x ≤ 5. 7.21 Express the compound inequality as two inequalities in terms of x. The compound inequality –4 < x ≤ 5 consists of the inequalities x > –4 and x ≤ 5. Therefore, every real number x that satisfies the inequality –4 < x ≤ 5 is between –4 and 5, including x = 5 but excluding x = –4. Note: Problems 7.21–7.22 refer to the compound inequality –4 < x ≤ 5. 7.22 Graph the inequality. A compound inequality describes a segment of the number line defined by two endpoints, in this case x = –4 and x = 5. Plot the endpoints as open or closed points depending upon the adjacent inequality symbol. The < symbol adjacent to the boundary –4 prohibits equality, so plot it using an open point. The ≤ symbol next to x = 5, on the other hand, indicates that 5 is a solution to the inequality, so plot it using a closed point. Complete the graph by shading the portion of the number line lying between the endpoints, as illustrated by Figure 7-7. Figure 7-7: The graph of the inequality –4 < x ≤ 5 includes x = 5 but excludes x = –4. Note: Problems 7.23–7.24 refer to the compound inequality –2 < x – 4 < 1. 7.23 Solve the inequality for x. Just like x is WRITTEN between –4 and 5 in the inequality –4 < x ≤ 5, all the SOLUTIONS are between those boundaries as well. The right boundary is a solution because of the ≤ symbol, but because of the < symbol, the left boundary is not. These rules about when dots are open and whe they’re closed are n consistent with Pr oblem 7.13. Basically, if the symbol includes “o r eq to,” use a solid dot ual . Otherwise, use a hollow one. Isolate x between the inequality symbols by adding 4 to each expression of the inequality. Note: Problems 7.23–7.24 refer to the compound inequality –2 < x – 4 < 1. 7.24 Graph the inequality. According to Problem 7.23, the solution to the inequality is 2 < x < 5. Use open points to plot the endpoints x = 2 and x = 5 and darken the portion of the number line between the boundaries to complete the graph, as illustrated by Figure 7-8. Figure 7-8: The graph of 2 < x < 5, the solution to the inequality –2 < x – 4 < 1. The Humongous Book of Algebra Problems 135 Chapter Seven — Linear Inequalities Note: Problems 7.25–7.26 refer to the compound inequality . 7.25 Solve the inequality for x. Eliminate by multiplying each of the expressions by 2. Isolate x between the inequality symbols. Note: Problems 7.25–7.26 refer to the compound inequality Converting into the mixed number makes it easier to graph. . 7.26 Graph the inequality. Plot the boundary symbol is ≤) and plot using a closed point (because the adjacent inequality using an open point (because the adjacent inequality symbol is <). Darken the portion of the number line bounded by those x-values, as illustrated by Figure 7-9. Figure 7-9: The graph of , the solution to the inequality . ngous Book of Algebra Problems 136 The Humo Chapter Seven — Linear Inequalities Absolute Value Inequalities Break these into two inequalities Note: Problems 7.27–7.28 refer to the inequality . 7.27 Express the absolute value inequality as a compound inequality that does not contain an absolute value expression and solve it for x. Rewrite the absolute value inequality –b ≤ x + a ≤ b. as the compound inequality Isolate x between the inequality symbols. Note: Problems 7.27–7.28 refer to the inequality Take the opposite of the number that’s right of the inequality symbol, follow it up with a copy of that symbol, and after that write the original inequality without the absolute value bars. . 7.28 Graph the solution to the inequality. According to Problem 7.27, the solution to the inequality is –3 ≤ x ≤ –1. Plot both boundaries using closed points, because x = –3 and x = –1 are solutions to the inequality, and darken the portion of the number line between the endpoints. Figure 7-10: The graph of –3 ≤ x ≤ –1, the solution to the inequality 7.29 Solve the inequality . for x and graph the solution. Express as a compound inequality that does not include an absolute value expression, using the method described in Problem 7.27. –5 ≤ 3x + 7 ≤ 5 Isolate x between the inequality symbols. You can only do this “inequality into a compound inequality” switcheroo if two things are true: (1) it’s an absolute value inequality; and (2) the sign is either < or ≤. If the sign is > or ≥, use the technique described in Problems 7.31– 7.33. Graph the solution to the inequality using closed points to mark the boundaries, as illustrated in Figure 7-11. The Humongous Book of Algebra Problems 137 Chapter Seven — Linear Inequalities Figure 7-11: The graph of , the solution to the inequality 7.30 Solve the inequality . for x and graph the solution. Before you express the absolute value inequality as a compound inequality, isolate the absolute value expression left of the equal sign. Important note here: The absolute values always need to be left of the inequality for this method to work. If they’re not, flipflop the sides of the inequality to move it left where it belongs. When you do, don’t forget to reverse the inequality sign. Express the absolute value inequality as a compound inequality and solve. Isolating x requires you to divide by a negative number, so reverse the inequality symbols. The compound inequalities 3 > x > 0 and 0 < x < 3 are equivalent; the graph is illustrated in Figure 7-12. Figure 7-12: The graph of 0 < x < 3, the solution to the inequality Note: Problems 7.31–7.32 refer to the inequality . 7.31 Express the absolute value inequality as two inequalities that do not include To get the first inequality, drop th e absolute value ba rs. To get the second , drop the bars, re verse the inequality sy m and take the op bol, posite of the constant on the right side of the original inequalit y. absolute value expressions and solve both for x. Whereas absolute value expressions that contain either the < or ≤ symbol can be rewritten as compound inequality statements, absolute value expressions containing either the > or ≥ symbol cannot. Instead, express the inequality as “x + a > b or x + a < –b.” x – 4 > 2 or x – 4 < –2 Solve the inequalities. ngous Book of Algebra Problems 138 The Humo . Chapter Seven — Linear Inequalities The solution to the inequality is x < 2 or x > 6. Therefore, any real number less than (but not including) 2 or greater than (but not including) 6 makes the inequality true. Note: Problems 7.31–7.32 refer to the inequality . 7.32 Graph the inequality. According to Problem 7.31, the solution to the inequality is x < 2 or x > 6. The graph of the solution consists of the graphs of x < 2 and x > 6 on the same number line. Plot x = 2 and x = 6 using open points, as illustrated in Figure 7-13; they are not valid solutions to the inequality. Figure 7-13: The graph of x < 2 or x > 6, the solution to the inequality 7.33 Solve the inequality So the graph is two arrows shooting different directions, usually with some space between them— much different than the graphs of absolute value inequalities that contain < or ≤, which look like single segments with two endpoints. . for x and graph the solution. Isolate the absolute value expression left of the inequality symbol. Use the method described in Problem 7.32 to express the absolute value statement as two distinct inequalities that do not include absolute value expressions and solve them. The solution to the inequality, or Figure 7-14: The solution to the inequality The word “or” is important. There’s no number that’s both greater than 6 AND less than 2, so a solution of “x < 2 AND x > 6” doesn’t make sense. , is graphed in Figure 7-14. is or . The Humongous Book of Algebra Problems 139 Chapter Seven — Linear Inequalities 7.34 Solve the inequality for x and graph the solution. Isolate the absolute value expression by subtracting 1 from both sides of the inequality. When you flip-flop the sides of an inequality, you have to reverse the inequality symbol. To solve an absolute value inequality using the methods described in the preceding examples, the absolute value expression must be left of the inequality symbol. Apply the technique described in Problem 7.31 to solve the inequality. The solution to the inequality is x < –10 or x > –4 and is graphed in Figure 7-15. This is an absolute value EQUATION, not an inequality. If you’re not sure how to solve it, flip back to Problem 4.33, which is very similar. Figure 7-15: The graph of x < –10 or x > –4, the solution to the inequality . Set Notation A fancy way to write solutions 7.35 Express the solution to the equation in set notation. Express the absolute value equation as two distinct equations that do not contain absolute values and solve them for x. This is not a coordinate. It’s a list, or set, of all the solutions to the equation. When there’s a fixed number of answers, expressing them in set notation just means listing them inside braces. The solution set of the inequality is ngous Book of Algebra Problems 140 The Humo . Chapter Seven — Linear Inequalities 7.36 Express the solution to the inequality in set notation. Express the absolute value inequality as a compound inequality that does not contain an absolute value expression, as explained in Problems 7.27 and 7.29, and solve the inequality. The solution set of the inequality is written either as {x : –23 < x < –1} or . The “:” in the first set and the “ ” in the second are both read “such that” and serve the same purpose. The symbols are interchangeable, and both solution sets are valid. 7.37 Express the solution to the inequality The solution set basically says, “Any real number x is a solution if that number x is greater than –23 and less than –1.” in set notation. Begin by isolating the absolute value expression left of the inequality symbol. Express the absolute value inequality as two inequalities that do not include absolute value expressions, as explained in Problems 7.31 and 7.33. x ≥ 13 or x ≤ –13 The solution set of the inequality is {x : x ≤ –13 or x ≥ 13}. Like in Problem 7.36, you can replace the colon in the solution set with a short vertical bar to get the equally valid solution set . Drop the bars to get one inequality; then drop the bars, reverse the symbol, and take the opposite of 13 to get the other. The Humongous Book of Algebra Problems 141 Chapter Seven — Linear Inequalities 7.38 Express the solution to the inequality If you plug in x = 6, you get in set notation. Multiply both sides of the inequality by –4 to isolate the absolute value expression. Recall that multiplying by a negative number reverses the inequality symbol. Any other x-valu . e produces a positi ve number. The left side of the inequality consists of an absolute value statement, so no matter what value of x is substituted into , its value is nonnegative. The null set symbol is NOT a zero, because zero is a real number. The solution set for the equation x + 5 = 5 is {0}, because x = 0 makes that equation true and is, therefore, a solution. A null solution set means there are NO solutions. The inequality stipulates that the absolute value expression is less than or equal to –20, but a nonnegative number cannot be less than a negative number, so there are no real number solutions to the inequality. Because the set of solutions is empty, the solution set is the empty, or null, set: . 7.39 Express the solution to the inequality in set notation. Like Problem 7.38, the absolute value expression left of the inequality symbol is nonnegative for any real number x, so all real numbers are solutions to the inequality. In set notation, the solution is {x : x is a real number}; in other words, if x is a real number, then x is a valid solution to the inequality. Graphing Inequalities in Two Variables Lines that give off shade in the coordinate plane 7.40 Graph the inequality y ≤ –3x – 1. If the left side is always greater than or equal to 0, then it’s always larger than the right side, which is –1. The linear inequality is written in terms of two variables, x and y, so it must be graphed on a system with two axes, the coordinate plane. Begin by graphing y = –3x – 1, an equation in slope-intercept form, using the technique described in Problems 6.21–6.26. However, use a dotted line rather than a solid line because the points on the line are not solutions to the inequality. The dotted line separates the coordinate plane into two distinct regions, labeled A and B in Figure 7-16. RU LE OF TH UMB: When you on a see < or > and you’re graphing When ts. do n ope number line, you use ng on phi gra you see < or > and you’re d line. tte a coordinate plane, use a do d dots on Similarly, ≤ and ≥ indicate soli s on the the number line and solid line coordinate plane. 142 The Humongous Book of Algebra Problems Chapter Seven — Linear Inequalities Figure 7-16: T he line y = –3x – 1 has y-intercept (0,–1) and slope –3 but does not represent the graph of the inequality y ≤ –3x – 1. In fact, because the line is dotted, it is not part of the graph at all. Rather, it splits the coordinate plane into two regions, here labeled A and B. Either region A or region B represents the solution to the inequality. Choose one “test” point, from either of the regions, and substitute it into the inequality. If that test point makes the inequality true, then the region to which it belongs is the solution. If it makes the inequality false, then the region not containing the test point is the solution. For instance, the point (–2,0) falls within region A in Figure 7-16. Substituting x = –2 and y = 0 into the inequality y ≤ –3x – 1 produces a true statement. You can pick any test point you want, as long as it’s clearly in one of the two regions. Choosing test points along the line (whether it’s solid or dotted) doesn’t help at all. Because (–2,0) makes the inequality true, region A is the solution to the inequality. Lightly shade that region on the coordinate plane to complete the graph, as illustrated by Figure 7-17. The Humongous Book of Algebra Problems 143 Chapter Seven — Linear Inequalities (0,0) is a great test point unless the line passes through th e origin. Substitutin g x = 0 and y = 0 in to this inequality re su in 0 ≥ 4, which is lts FALSE, so you shou ld shade the region that DOESN’T contain the origin. Figure 7-17: All of the points in the shaded region are solutions to the inequality y ≤ –3x – 1. 7.41 Graph the solution to the inequality . Use the method described in Problem 7.40: Draw the graph of the line , choose a test point to substitute into the inequality, and identify the solution region based upon the results of that test point. Unlike Problem 7.41, the line is part of the solution region, so use a solid line to graph it, as illustrated in Figure 7-18. Figure 7-18: The graph of line ngous Book of Algebra Problems 144 The Humo consists of the shaded region and the . Chapter Seven — Linear Inequalities 7.42 Graph the inequality x ≤ –3. Graph x = –3, a vertical line three units left of the y-axis, and choose a test point to substitute into the inequality. The origin, (0,0), belongs to the region right of the line, and substituting it into the inequality results in the false statement 0 ≤ –3. Therefore, the region left of the line (and the vertical line x = –3 itself) constitute the solution, as illustrated by the graph in Figure 7-19. Figure 7-19: The graph of the inequality x ≤ –3. 7.43 Graph the inequality 4x – 3y > 15. Solve the equation for y to express it in slope-intercept form. Graph in the coordinate plane using a dotted line. Substituting x = 0 and y = 0 into the inequality produces the false statement 0 < –5, so the solution is the region of the plane that does not include the origin, as illustrated by Figure 7-20. The Humongous Book of Algebra Problems 145 Chapter Seven — Linear Inequalities Figure 7-20: The graph of 4x – 3y > 15. ngous Book of Algebra Problems 146 The Humo Chapter 8 Systems of Linear Equations and Inequalities n Work with more than one equatio at a time The four preceding chapters investigate, in some depth, the solutions and graphs of linear equations and inequalities in one and two variables. Though the concepts and techniques vary widely, one universal characteristic unites them: each considers only one equality or inequality statement at a time. This chapter introduces systems, the solutions of which satisfy multiple statements rather than single equations or inequalities. A system of equations is a group of (usually two) equations. Your goal is to find one x-value and one y-valu e that make BOTH of the equatio ns true when you substitute them in. That means the solution of a system of equations is usually a point (x,y). There are a few special cases when there’s no answer or more than one answer, and you’ll deal with those as well. Remember the graphs of inequalitie s from the end of Chapter 7? They were shaded regions in the coo rdinate plane. This chapter deals with systems of inequalities, multiple inequalities on the same graph, which means even more shading. Th e last thing you deal with is linear programming, which is a “real life” ap plication of systems of inequalities. Chapter Eight — Systems of Linear Equations and Inequalities Graphing Linear Systems Graph two lines at once For example, the graph of x + y = 4 passes through the point because plugging (3,1), x = 3 and y = 1 in to equation gives yo the u the true statement 3 + 1 = 4. If all the lines intersect at (a,b), then every line passes through (a,b), and (x,y) = (a,b) is a solution of each equation. Hence, it is also a solution to the system. You could also write the solution like this: x = –4, y = 1. 8.1 Explain why the solution to a system of linear equations in two variables is the coordinate at which the graphs of the equations intersect. The graph of a linear equation in two variables is a line in the coordinate plane. Each of the points (x,y) on that line, when substituted into the equation, results in a true statement. The solution to a system of linear equations consists of the coordinate pair (or coordinate pairs) that make all the equations in a system true. Graphing a system of equations in two variables on the same coordinate plane produces multiple lines. If all the lines intersect at a single point (a,b), then x = a and y = b satisfy each of the equations in the system and, therefore, (x,y) = (a,b) is the solution to the system. 8.2 Solve the following system of equations graphically. Graph each equation of the system on the same coordinate plane. The graph of x = –4 is a vertical line four units left of the y-axis, and the graph of y = 1 is a horizontal line one unit above the x-axis, as illustrated by Figure 8-1. Figure 8-1: The graphs of x = –4 and y = 1 intersect at the point (–4,1). Because the graphs in Figure 8-1 intersect at the point (–4,1), the solution to the system is (x,y) = (–4,1). ngous Book of Algebra Problems 148 The Humo Chapter Eight — Systems of Linear Equations and Inequalities 8.3 Solve the below system of equations graphically and verify the result. The y-intercept of the first equation is 3. The slope is Both equations of the system are in slope-intercept form. Graph them using the , so go up 1 and right 1 technique described in Problems 6.21–6.26 and illustrated in Figure 8-2. to reach another point on the line: (1,4). The second equation has y-intercept , so go –6. Its slope is down 2 and right 1 (or up 2 and left 1) to get to another point on that line. Figure 8-2: The graphs of y = x + 3 and y = –2x – 6, the equations of the system. The graphs appear to intersect at the point (x,y) = (–3,0). Substitute x = –3 and y = 0 into both equations to verify that the coordinate pair satisfies each and, therefore, is the solution to the system. Because the coordinate pair (x,y) = (–3,0) is a solution to both equations of the system, it is the solution to the system of equations. This is what stinks about solving a system graphically. It LOOKS like the lines intersect at (–3,0), but looks aren’t everything. You have to be SURE x = –3 and y = 0 make both equations into true statements when you substitute them in. The Humongous Book of Algebra Problems 149 Chapter Eight — Systems of Linear Equations and Inequalities 8.4 Solve the below system of equations graphically and verify the result. Graph the equations of the system, as illustrated in Figure 8-3. Figure 8-3: The graphs of y = 3x – 7 and , the equations of the system. The lines in Figure 8-3 appear to intersect at (2,–1). Verify that (x,y) = (2,–1) is the solution to the system by substituting x = 2 and y = –1 into the equations. Because (x,y) = (2,–1) is a solution to both equations of the system, it is the solution to the system of equations. ngous Book of Algebra Problems 150 The Humo Chapter Eight — Systems of Linear Equations and Inequalities 8.5 Solve the below system of equations graphically and verify the result. Graph the equations by plotting the x- and y-intercepts of each, as explained in Problems 5.20–5.25. The equation x – 2y = 12 has x-intercept 12 and y-intercept –6. The equation 5x + 3y = 8 has x-intercept and y-intercept , as illustrated in Figure 8-4. Figure 8-4: The graphs of x – 2y = 12 and 5x + 3y = 8, the equations of the system. Verify that (x,y) = (4,–4) is the intersection point of the lines and the solution to the system by substituting the coordinate pair into the equations of the system. Plug x = 0 in to see what the y-intercept is and plug y = 0 in to see what the xintercept is. You don’t HAVE to graph this way—you could always solve for y and graph using slope-intercept form—but using intercepts is faster. RU LE OF TH UMB: The intercepts of 5x + 3y = 8 are ugly fractions, which are hard to graph accurately by hand. Inaccurate graphs lead to inaccurate solutions, so solving systems by graphing works best when fractions aren’t involved in the graph or the solution. The Humongous Book of Algebra Problems 151 Chapter Eight — Systems of Linear Equations and Inequalities 8.6 Describe the solution to the below system of equations. Solve the equation 3x + 6y = –6 for y to express it in slope-intercept form. The equations in a dependent system have the same graph. Usua lly that means you can start with one of th equations and tu e rn it into the other on e. In this case, solving the second equation for turns it into the y first equation. Notice that this equation is exactly the same as the other equation in the system. Two lines with the same slope and the same y-intercept also share the same graph. Therefore, the graphs of and 3x + 6y = –6 overlap, intersecting at every point along the shared line and result in an infinitely large solution set that consists of all the points on the line. Systems of equations that have infinitely many solutions are described as “dependent.” 8.7 Describe the solution to the below system of equations. Solve the equation x – 3y = –9 for y to express it in slope-intercept form. Parallel lines have slopes that are equal. Perpendicular lines have slopes that are opposite reciprocals. 152 Both of the equations in the system have slope illustrated by Figure 8-5. The Humongous Book of Algebra Problems , so the lines are parallel, as Chapter Eight — Systems of Linear Equations and Inequalities Figure 8-5: The graphs of and x – 3y = –9 both have slope and are, therefore, parallel lines. By definition, parallel lines do not intersect in the coordinate plane. It is the intersection point of a system of equations that defines its solution, so the absence of an intersection point indicates the absence of a solution to the system. Systems of equations with no solutions are classified as “inconsistent.” The Substitution Method Solve one equation for a variable and plug it into the other 8.8 To prove that there is no solution to a system of linear equations, all you have to do is prove that all the lines have the same slope (or that all the lines have NO slope— vertical lines are parallel as well). Explain how to apply the substitution method to solve a system of two linear equations in two variables. If one equation of a system can easily be solved for one of its variables, you can substitute the resulting expression into the other equation of the system. For instance, if the first equation of a system can be solved for y, do so, and then substitute the result for y in the second equation. The net result is a new equation written in terms of a single variable, in this case x. When solved, it produces the x-value of the solution to the system, which can then be substituted into one of the equations of the system to determine the corresponding y-value. You’ll learn a couple of ways to solve systems of equations. This on e works best if you can easily solve one of equations for x or the y. It’s even easier w hen an equation is AL READY solved for x or y, like Problems 8.9 and in 8.10. The Humongous Book of Algebra Problems 153 Chapter Eight — Systems of Linear Equations and Inequalities Note: Problems 8.9–8.10 refer to the following system of equations. 8.9 Solve the system using substitution. The second equation in the system is solved for y: y = 11x – 16. According to this statement, the expressions y and 11x – 16 have the same value in this system of equations. Therefore, you may substitute 11x – 16 for y in the other equation of the system, x + y = 8. Solve the equation for x. You could substitute it into the other equation, x + y = 8, instead. You’ll get the same answer. However, it’s almost always quickest to plug x into the equation already solved for y (and vice versa). The solution to a system of linear equations in two variables is a coordinate pair (x,y). To determine the y-coordinate, and therefore complete the solution, substitute x = 2 into the equation y = 11x – 16. The solution to the system of equations is (x,y) = (2,6). Note: Problems 8.9–8.10 refer to the following system of equations. 8.10 Verify the solution generated in Problem 8.9. According to Problem 8.9, the solution to the system of equations is (x,y) = (2,6). To verify that the solution is correct, substitute x = 2 and y = 6 into both equations of the system. Substituting (x,y) = (2,6) into the equations of the system produces true statements, so it is the correct solution to the system. ngous Book of Algebra Problems 154 The Humo Chapter Eight — Systems of Linear Equations and Inequalities 8.11 Solve the following system of equations using substitution. According to the first equation, the expressions x and in this system. Substitute have the same value for x in the second equation and solve for y. Rewrite the coefficient of y (which is 1 even though it’s not written explicitly) using a common denominator: Substitute y = 5 into the equation The solution to the system is (x,y) = (–3,5). to complete the solution. The answer (x,y) = (5,–3) is NOT correct. When you write (x,y), you’re saying that the first value is x and the second value is y, and in this problem x = –3 and y = 5, not the other way around. The Humongous Book of Algebra Problems 155 Chapter Eight — Systems of Linear Equations and Inequalities Note: Problems 8.12–8.13 refer to the following system of equations. 8.12 Solve the system using substitution. When a variable has a coefficient of 1, like y in the first equation of this system, solving for that variable is usually quite easy. Neither of the equations is solved for x or y, but solving 6x + y = 1 for y is a simple matter. Subtract 6x from both sides of the equation to get y = –6x + 1. Substitute this expression for y in the other equation of the system and solve for x. Substitute into the equation you solved for y to complete the solution. The solution to the system is . Note: Problems 8.12–8.13 refer to the following system of equations. 8.13 Verify the solution generated in Problem 8.12. According to Problem 8.12, the solution is . Substitute and y = –2 into both equations of the system and verify that the results are true statements. ngous Book of Algebra Problems 156 The Humo Chapter Eight — Systems of Linear Equations and Inequalities 8.14 Solve the following system using substitution. Solve the second equation for x to get x = –4y – 4. Substitute this expression into the first equation and solve for y. Substitute the solution. Just like in Problem 8.12 , the variable with a coefficient of 1 is the easiest to solve for. You can technically solve for ANY of the variables here, and you’ll end up with the same solution, but why work harder than you have to? into either equation of the system and solve for x to complete The book plugs y into the first equa but you get the same x-value whe tion, plug y into the se n you cond equation: The solution to the system is . The Humongous Book of Algebra Problems 157 Chapter Eight — Systems of Linear Equations and Inequalities 8.15 Solve the following system using substitution. Ignore the signs of the coefficients and pick the smallest number. Here, the coefficients are 2, 3, and 7. The smallest of those is 2, from the term 2y in the second equation, so solve the second equation for y. None of the variables in this system has a coefficient of 1, so none distinguishes itself as the preferred variable to isolate and substitute into the other equation of the system. In these cases, it is usually preferable to solve for the variable whose coefficient has the least absolute value. Therefore, you should solve 3x + 2y = 40 for y. Substitute into the first equation of the system. Multiply each of the terms by 2 to eliminate the fraction, and solve for x. Complete the solution by substituting x = 6 into the equation you solved for y. The solution to the system is (x,y) = (6,11). ngous Book of Algebra Problems 158 The Humo Chapter Eight — Systems of Linear Equations and Inequalities 8.16 Solve the below system using substitution. Solve 2x – 15y = 5 for x, as it has the coefficient with the smallest absolute value. Substitute Problem 8.15 gave this advice: When none of the coefficients is 1, then solve for the coefficient that’s smallest when you ignore the signs. Sure –15 < 2, but when you ignore the signs, 2 < 15, so 2 is the smallest coefficient. into the second equation. Multiply each of the terms by 2 to eliminate the fractions, and solve for y. Substitute into The solution to the system is , the equation you solved for x. If you plug it into the FIRST equation, everything cancels out and you get 0 = 0. That’s because you’re actually plugging another version of the first equation into itself. You have to substitute the equation you solved for x into the OTHER equation of the system. . The Humongous Book of Algebra Problems 159 Chapter Eight — Systems of Linear Equations and Inequalities 8.17 Solve the following system using substitution. Substitution isn’t restricted to systems with two equations. W he there are three n or more, start with th equations that ha e ve the fewest varia bles in them, such as 4z + 1 = 0 in this system. The first equation of this system is written in terms of three variables, and the second is written in terms of two variables. Because the third equation is written in terms of a single variable, it should be solved first. Of the two remaining equations in the system, 8y – 4z = 5 has fewer variables. In fact, substituting which you can solve. Now substitute into the equation leaves a single variable, y, for and into the first equation of the system and solve it for x to complete the solution. The solution to the system is Three different variables are in the system, even though they all don’t appear in each of the equations. That means the solution to the system must have all three: (x,y,z). . 8.18 Solve the below system using substitution. Unlike Problem 8.17, none of the equations in this system is written in terms of a single variable. However, all the equations contain one common variable: y. Solve the third equation for y to get y = 6z + 6 and substitute it into the other equations of the system. ngous Book of Algebra Problems 160 The Humo Chapter Eight — Systems of Linear Equations and Inequalities The end result of substituting y into the first two equations of the system is a new system of two linear equations in two variables. Dividing everything by 2 doesn’t change the solution to the equation, and the added benefit is that x now has a coefficient of 1, so you can solve for it easily. Solve this system of equations using substitution: solve the second equation for x to get x = 27z + 29 and substitute that into the first equation. Substitute z = –1 into the equation you solved for x. Finally, substitute z = –1 into the equation you originally solved for y to complete the solution. The solution to the system is (x,y,z) = (2,0,–1). The Humongous Book of Algebra Problems 161 Chapter Eight — Systems of Linear Equations and Inequalities Variable Elimination Make one variable disappear and solve for the other one 8.19 Explain how to solve a system of two linear equations in two variables using the variable elimination technique. To apply the variable elimination method, the coefficients of either the x-terms or the y-terms of the equations in the system must be opposites. Therefore, when the equations are added, a new equation is generated that contains only one variable. In many cases, you must multiply one or both of the equations by a nonzero real number to introduce the coefficient opposites into the system. Note: Problems 8.20–8.21 refer to the following system of equations. 8.20 Solve the system by eliminating y. The equations of the system have opposite y-coefficients—in the first equation, y has a coefficient of 1, and in the second, the y-coefficient is –1. Add the equations of the system together by combining like terms. The result is a linear equation in one variable: 3x = 9. Solve it for x. The book substituted into the first equation of the system, but you get the same thing when you substitute x = 3 into the second equation: Substitute x into either equation of the system to determine the corresponding value of y. The solution to the system is (x,y) = (3,1). 162 The Humongous Book of Algebra Problems Chapter Eight — Systems of Linear Equations and Inequalities Note: Problems 8.20–8.21 refer to the following system of equations. 8.21 Solve the system by eliminating x. To eliminate the x-variable from the system, the x-coefficients of the equations must be opposites. Multiplying the terms of the second equation by –2 achieves this goal. Add the equations of the modified system to eliminate y. Solve the equation 3y = 3 for y. Substitute y = 1 into either equation of the system to determine the corresponding value of x. The solution to the system is (x,y) = (3,1). You could multiply the first equation by instead, making its x-coefficient –1, the opposite of the xcoefficient in the other equation. However, that’s not a great idea, because it turns the y-coefficient and the constant in the first equation into fractions, and nobody likes fractions. This is the same solution as Proble m 8. 20, so it doesn’t matter which va ria you eliminate—yo ble u up with the sam end e final answer. The Humongous Book of Algebra Problems 163 Chapter Eight — Systems of Linear Equations and Inequalities Note: Problems 8.22–8.23 refer to the system of equations below. 8.22 Solve the system by eliminating x. There’s lots of things you COULD do. For instance, you could multiply the first equation by 10 and the second by –2 and the xcoefficients would be opposites: 1 ˙ 10 = 10 and 5 ˙ (–2) = –10. Just make sure you multiply ALL the terms in both equations by the numbers you choose. The coefficients of the x-terms in the system are 1 and 5. The easiest way to transform those numbers into opposites is to multiply the terms of the first equation by –5. Add the equations of the modified system and solve the resulting equation for y. Substitute y = –9 into either equation of the original system to calculate the corresponding value of x. The solution to the system is (x,y) = (–2,–9). Note: Problems 8.22–8.23 refer to the system of equations below. 8.23 Solve the system by eliminating y. Multiplying the first equation by 3 changes its y-coefficient to the opposite of the other y-coefficient in the system. Add the equations together and solve for x. ngous Book of Algebra Problems 164 The Humo Chapter Eight — Systems of Linear Equations and Inequalities Substitute x = –2 into either equation of the system to complete the solution. The solution to the system is (x,y) = (–2,–9). 8.24 Problem 8.14 used the substitution method to solve the following system of equations. Verify its solution using variable elimination. Multiplying the second equation by –2 makes the x-coefficients of the system into opposites. Add the equations of the modified system and solve for y. Substitute You could multiply the second equation by 3 and turn the ycoefficients opposites: –12 and 12. If you’re feeling crazy, you could even multiply the first equation by 5 and the second by –10 to turn the x-coefficients into 2 ˙ 5 = 10 and 1 ˙ (–10) = –10. into either equation of the system to calculate the corresponding value of x. The solution to the system is Problem 8.14. , the same solution calculated in This way was much faster. Substitution usually requires more steps than variable elimination. The Humongous Book of Algebra Problems 165 Chapter Eight — Systems of Linear Equations and Inequalities 8.25 Problem 8.15 used the substitution method to solve the below system of You could do the same with the x-terms: Multiply the first equation by 3 and the second equation by –7. You’re basically multiplying one equation by the xcoefficient of the other equation. equations. Verify its solution using variable elimination. To eliminate the y-terms, multiply the first equation by 2 and the second by 3. Add the equations and solve for x. Substitute x = 6 into either equation of the system to calculate the corresponding y-value. The solution to the system is (x,y) = (6,11), the same solution calculated in Problem 6.15. You could multiply the first equation by 9 and the second by –2 (or the first by –9 and the second by 2) to eliminate the x’s instead. 8.26 Problem 8.16 used the substitution method to solve the below system of equations. Verify its solution using variable elimination. Multiply the second equation by 3 to get 27x + 15y = 24, add the equations of the modified system, and solve for x. ngous Book of Algebra Problems 166 The Humo Chapter Eight — Systems of Linear Equations and Inequalities Substitute x = 1 into either equation of the system to calculate the corresponding value of y. The solution to the system is Problem 8.16. , the same solution calculated in 8.27 Solve the following system using variable elimination. Multiply the first equation by 8 and the second equation by 3 to eliminate the y-terms of the system. Add the equations of the modified system and solve for x. Substitute x = –4 into either equation of the original system to calculate the corresponding y-value. Like in Problem 8.25, you’re basically multiplying the equation by the y-coefficient of the other equation, ignoring the signs of the coefficients because the signs are already opposites and don’t need adjusting. The solution to the system is (x,y) = (–4,2). The Humongous Book of Algebra Problems 167 Chapter Eight — Systems of Linear Equations and Inequalities 8.28 Solve the following system using variable elimination. You could multiply the first equation by 7 and the second by 3 to eliminate the y’s instead. You could even multiply the first equation by 15 and the second by –6 to eliminate the x’s in a different way, but that’ll make the numbers in the equation pretty big. Like most variable elimination problems, there are various ways to eliminate x or y from this system. One method is to multiply the first equation by 5 and the second by –2 so that the x-coefficients are opposites. Add the equations of the modified system and solve for y. Substitute y = 7 into either equation of the original system to calculate the corresponding x-value. The solution to the inequality is . Systems of Inequalities The answer is where the shading overlaps 8.29 According to Problem 7.40, the graph of a linear inequality in two variables is a region of the coordinate plane. Explain how to generate the graph of a system of linear inequalities in two variables. The shaded solution region of a linear inequality graph is a visual representation of the points that, if substituted into the inequality, would make the statement true. The solution to a system of inequalities is the set of points that makes all the inequalities in the system true. ngous Book of Algebra Problems 168 The Humo Chapter Eight — Systems of Linear Equations and Inequalities To graph a system of inequalities, graph the individual inequalities on the same coordinate plane. The solution is the region of the graph on where the individual solutions overlap. Any point in that region is a solution to all of the inequalities in the system. 8.30 Graph the following system of linear inequalities. Graph the inequalities x > 1 and y ≤ 2 using the techniques described in Problems 7.40–7.43, as illustrated in Figure 8-6. Figure 8-6: The lightly shaded regions represent solutions to one of the inequalities in the system, and the darker region represents the common solution to both inequalities—the solution to the system. All the points left of the vertical line x = 1 satisfy the first inequality of the system, and all the points below (and including) the horizontal line y = 2 satisfy the second inequality. The solution to the system is the dark region of the coordinate plane in Figure 8-6, the set of points that is both right of x = 1 and below y = 2. Graph the inequalities of the system, shading each solution lightly. The solution to the system is the darkest shaded region, because it’s where the graphs overlap. The lightly shaded regions aren’t technically part of the graph. They’re just included so you can see what parts of the inequality graph overlap and what parts don’t. It’s okay just to include the dark region as the graph and leave the lightly shaded regions out. The Humongous Book of Algebra Problems 169 Chapter Eight — Systems of Linear Equations and Inequalities 8.31 Graph the following system of linear inequalities. The graph of y < x consists of all the points below and to the right of the line y = x, and the graph of y > –x consists of all the points above and to the right of the line y = –x. The solution to the system, therefore, is the region right of the y-axis that is bounded above by y = x and bounded below by y = –x, as illustrated by Figure 8-7. Figure 8-7: T he dark shaded region of the coordinate plane contains points that satisfy the inequalities y < x and y > –x. Note: Problems 8.32–8.33 refer to the following system of linear inequalities. 8.32 Graph the system. The linear inequalities are written in slope-intercept form. Graph both on the same coordinate plane and indicate the solution to the system by shading it more darkly than the individual inequality solutions, as illustrated by Figure 8-8. 170 The Humongous Book of Algebra Problems Chapter Eight — Systems of Linear Equations and Inequalities Figure 8-8: The dark shaded region of the graph represents the solution to the system. Note: Problems 8.32–8.33 refer to the following system of linear inequalities. 8.33 Choose one point from the solution region graphed in Problem 8.32 and There are an infinite number of points to choose fr The point (0,–4) belongs to the solution graphed in Figure 8-8. To verify that om in th at solution region (x,y) = (0,–4) is a solution to the system, substitute x = 0 and y = –4 into both . (0,– 4) is only one of inequalities and demonstrate that the results are true statements. th em . You can choose a ny point from the darker regi Figure 8-8, where on of the inequality solution s overlap. verify that it satisfies both inequalities of the system. The Humongous Book of Algebra Problems 171 Chapter Eight — Systems of Linear Equations and Inequalities 8.34 Graph the following system of linear inequalities. Graph the first inequality using a solid line because it is part of the graph, and graph the second inequality using a dotted line because it is not. The inequalities and the solution to the system are graphed in Figure 8-9. Figure 8-9: The dark shaded region is the solution to the system of inequalities. 8.35 Graph the following system of linear inequalities. You need to solve for y (not –y), so multiply all the terms by –1. D forget that multi on’t plying an inequality by a negative number means you have to reve rse the inequality sig n from < to >. 172 The graphs of the first and second inequalities of the system are regions bounded by vertical and horizontal lines, respectively. To graph the third inequality, solve it for y. Figure 8-10 is the graph of the solution to the system of inequalities. The Humongous Book of Algebra Problems Chapter Eight — Systems of Linear Equations and Inequalities Figure 8-10: The shaded region represents the solution to the system of inequalities. Note that the individual inequality graphs, and their accompanying lightly shaded regions, are omitted from Figure 8-10. When a system consists of three or more inequalities, the value of the individual graphs is lessened because the graph begins to look cluttered. If the graph of a system is bounded by multiple inequalities, ensure that you clearly indicate which region constitutes the solution. Linear Programming Use the sharp points at the edge of a shaded region Note: Problems 8.36–8.39 refer to the expression P = 6x – 2y and the following system of linear equations. 8.36 Graph the system of inequalities. In Problems 8.36–8.39, you’ll find the highest and lowest possible values of P given the limitations defined by the system of inequalities. In other words, you’ll find out what point from the shaded region of the graph, when plugged into P = 6x – 2y, makes P the biggest and which point makes P the smallest. Apply the method described in Problems 8.29–8.35 to generate the graph, as illustrated by Figure 8-11. The Humongous Book of Algebra Problems 173 Chapter Eight — Systems of Linear Equations and Inequalities The boundaries are also solutions because all the inequalities contain either “≤” or “≥.” Figure 8-11: All the points in the shaded region and the boundaries of the region represent solutions to the system of inequalities. Note: Problems 8.36–8.39 refer to the expression P = 6x – 2y and the system of linear inequalities identified in Problem 8.36. The vertices are the intersection points of the boundary lines along the edge of the region, basically the “corners” of the shaded region. 8.37 Consider the region graphed in Problem 8.36. Identify the points that could generate an optimal value of P. The optimal (maximum and minimum) values of P will occur at one of the vertices of the solution region, which are identified as points A, B, C, and D in Figure 8-12. The shaded region of the graph is sometimes referred to as the set of “feasible solutions” in linear programming problems. Figure 8-12: Points A, B, C, and D are the vertices of the feasible solution set. 174 The Humongous Book of Algebra Problems Chapter Eight — Systems of Linear Equations and Inequalities Point D is the origin: D = (0,0). Point A is the y-intercept of the line . The line is in slope-intercept form, so the y-intercept is 3, the constant in the equation. Therefore, A = (0,3). Point C is the x-intercept of the line Substitute y = 0 into the equation and solve for x to calculate the x-intercept. Both of the equations are solved for y, so substituting Therefore, into . Point B is the intersection of lines and . Solve the system containing those two equations using substitution to identify B. (or vice versa) ends up being the xexpressions set equal to each other. Multiply every term by 4 to eliminate the fractions. Solve for x. Substitute x = 2 into either of the intersecting linear equations to determine the corresponding value of y. The Humongous Book of Algebra Problems 175 Chapter Eight — Systems of Linear Equations and Inequalities The graphs of and intersect at . Note: Problems 8.36–8.39 refer to the expression P = 6x – 2y and the system of linear inequalities identified in Problem 8.36. 8.38 Evaluate P at each of the coordinate pairs identified in Problem 8.37. Substitute A = (0,3), The optimal (maximum or minimum) values of P will always correspond with a vertex of the graph. No other coordinate pair in the shaded region will result in a higher or lower value of P. 176 , , and D = (0,0) into P = 6x – 2y. Note: Problems 8.36–8.39 refer to the expression P = 6x – 2y and the system of linear inequalities identified in Problem 8.36. 8.39 Identify the maximum and minimum values of P given the constraints defined by the system of linear inequalities. According to Problem 8.38, the maximum value of P is , occurring when and y = 0. The minimum value of P is –6, occurring when x = 0 and y = –3. The Humongous Book of Algebra Problems Chapter Eight — Systems of Linear Equations and Inequalities Note: Problems 8.40–8.43 refer to the expression S = x + 5y and the below system of linear equations. 8.40 Graph the system of inequalities. Apply the method described in Problems 8.29–8.35 to generate the graph in Figure 8-13. Figure 8-13: T he shaded region of the coordinate plane is the set of feasible solutions, coordinates that comply with the constraints defined by the system of inequalities. The inequalities of the system are sometimes called the “constraints” in a linear programming problem. The Humongous Book of Algebra Problems 177 Chapter Eight — Systems of Linear Equations and Inequalities Note: Problems 8.40–8.43 refer to the expression S = x + 5y and the system of linear inequalities identified in Problem 8.40. 8.41 Consider the region graphed in Problem 8.40. Identify the vertices of the feasible solution region. The optimal (maximum and minimum) values of S will occur at one of the vertices of the solution region identified in Figure 8-14. Figure 8-14: Points A, B, C, D, and E are vertices of the feasible solution region. Point E is the origin: E = (0,0). Point A is the y-intercept of x – y = –2. Substitute x = 0 into the equation and solve for y. Therefore, A = (0,2). Point D is the x-intercept of x + y = 4. Substituting y = 0 into that equation results in x = 4, so D = (4,0). Multiply the equation x – y = –2 by 3 to get 3x – 3y = –6 and add it to the other equation. 178 Point B is the intersection of lines x – y = –2 and 2x + 3y = 10. Solve the system of equations consisting of those two lines by eliminating x. The Humongous Book of Algebra Problems Chapter Eight — Systems of Linear Equations and Inequalities Substitute into one of the linear equations intersecting at B to calculate the corresponding value of y. Multiply each of the terms by 5 to eliminate the fraction and solve for y. Therefore, . Point C is the intersection of lines 2x + 3y = 10 and x + y = 4. Identify C by solving that system of linear equations. Multiply the second equation by –2 to get –2x – 2y = –8. Add that equation to 2x + 3y = 10 and solve for y. Substitute y = 2 into one of the linear equations intersecting at C to calculate the corresponding value of x. The solution to the system is the intersection point of 2x + 3y = 10 and x + y = 4: C = (2,2). The Humongous Book of Algebra Problems 179 Chapter Eight — Systems of Linear Equations and Inequalities Note: Problems 8.40–8.43 refer to the expression S = x + 5y and the system of linear inequalities identified in Problem 8.40. 8.42 Evaluate S at each coordinate pair identified in Problem 8.41. Substitute A = (0,2), S = x + 5y. , C = (2,2), D = (4,0), and E = (0,0) into Note: Problems 8.40–8.43 refer to the expression S = x + 5y and the system of linear inequalities identified in Problem 8.40. 8.43 Identify the maximum and minimum values of S given the constraints defined by the system of linear inequalities. According to Problem 8.42, the maximum value of S is and ngous Book of Algebra Problems 180 The Humo , occurring when . The minimum value of S is 0, occurring when x = 0 and y = 0. Chapter 9 Matrix Operations and Calculations Numbers in rows and columns Matrices are orderly collections of numbers arranged in a grid of rows and columns. This chapter introduces matrix notation, common matrix terms, matrix operations (such as addition and multiplication), and determinants. The chapter concludes with Cramer’s Rule, a set of matrix formulas used to calculate the solution to a system of equations. Matrices are pretty useful, even tho ugh they’re just big laundry lists of numbers. This chapter introd uces you to matrices—it tells you what the pieces are called, shows you how to combine them, explains what the heck a determinant is, an d demonstrates how to calculate determinants the easy way and the hard way. Oddly, the chapter ends with Crame r’s Rule, a way to solve the systems of equations that were covered in Problems 8.1–8. 28. Even though it’s nothing more than a new way to do something you already know how to do (if you read Chapter 8, that is), it’s still interesting that clumps of numbers can produce the same an swer that substitution and variable elimination did. Chapter Nine — Matrix Operations and Calculations Anatomy of a Matrix The i and j here represent numbers: a and 11 a23 represent spec ifi positions in the ma c trix, as you’ll see in Problems 9. 2–9.3 and 9.5–9.6. The first number always tells you how many rows are in the matrix, and the second value always represents the number of columns. The first number (i) tells you what horizontal row the element is in (1 = top row, 2 = second row from the top, and so on), and the second number (j) tells you what vertical column the number is in (1 = far left column, 2 = second column from the left, and so on). The order of a matrix and identifying elements Note: Problems 9.1–9.3 refer to matrix A defined below and the elements aij of matrix A. 9.1 What is the order of the matrix? Order indicates the dimensions—the number of rows and columns—of a matrix. Matrix A has two horizontal rows and five vertical columns, so it has order 2 × 5. Note that order is always written r × c, where r is the number of rows and c is the number of columns. Note: Problems 9.1–9.3 refer to matrix A defined below and the elements aij of matrix A. 9.2 Identify element a13. The individual numbers within a matrix are called elements and are labeled using the notation aij . Element aij is in the ith row and the jth column of matrix A. Element a13 is in the first row and the third column from the left: a13 = –3. Note: Problems 9.1–9.3 refer to matrix A defined below and the elements aij of matrix A. 9.3 Identify element a 21. As explained in Problem 9.2, the notation a 21 describes the element in the second row (from the top) and first column (from the left): a 21 = 0. Matrices are labeled with capital letters and elements of ma trices are labeled with lowercase versions of the of matrix A look like a and same letters. Elements elements of matrix B 21 look like b21. ngous Book of Algebra Problems 182 The Humo Chapter Nine — Matrix Operations and Calculations Note: Problems 9.4–9.6 refer to matrix B defined below and the elements bij of matrix B. 9.4 What is the order of the matrix? Matrix B has three horizontal rows and three vertical columns, so it has order 3 × 3. Because B is a square matrix (a matrix with an equal number of rows and columns), you can describe its order using a single number: 3. You can only use a single number to describe the order of a matrix if the matrix is square. The order of any other matrix looks like r × c. Note: Problems 9.4–9.6 refer to matrix B defined below and the elements bij of matrix B. 9.5 Identify element b 23. Element b 23 is in the second row (from the top) and the third column (from the left) of matrix B: b 23 = –5. Note: Problems 9.4–9.6 refer to matrix B defined below and the elements bij of matrix B. 9.6 Identify element b 31. Element b 31 is in the third row (from the top) and the first column (from the left) of matrix B: b 31 = 3. Adding and Subtracting Matrices Combine numbers in matching positions 9.7 What condition must be met by matrices A and B if the matrix C = A + B exists? Matrix C is defined as the sum of matrices A and B, and a matrix sum (or difference) is defined only for matrices of the same order. You can only add or subtract matrices if they’re the same size. In other words, they need a matching number of rows and a matching number of columns. You can’t add a 3 × 2 matrix to a 2 × 3 matrix, but you can add two 3 × 2 matrices or two 2 × 3 matrices together. The Humongous Book of Algebra Problems 183 Chapter Nine — Matrix Operations and Calculations A matrix like this one that has only on row is called a “r e ow matrix.” If a ma trix has only one colu mn, it’s called either a “col matrix” or a “col umn umnar matrix.” 9.8 . Both row matrices are the same size (that is, they have the same order) so the matrix sum is defined and has the same order: 1 × 3. To calculate the sum, add elements in corresponding positions. 9.9 It’s easy: add upper-left to uppe rleft (13 + 12), up perright to upper-righ t (–5 + [–3]), lower-le ft lower-left (–7 + 8) to , and lower-right to low erright (2 + [–1]). Calculate the sum of the matrices: Calculate the sum of the matrices: . Both matrices have the same order (2 × 2) so the sum exists and is also a 2 × 2 matrix. To calculate the sum, add elements in corresponding positions. Note: Problems 9.10–9.11 refer to matrices A and B defined below. 9.10 Perform scalar multiplication to calculate 4A and 3B. To generate 4A, multiply each element of A by 4; to generate 3B, multiply each element of B by 3. A scalar is a real number that’s multiplied by every term in a matrix. It’s like the distributive property, but on a bigger scale. ngous Book of Algebra Problems 184 The Humo Chapter Nine — Matrix Operations and Calculations Note: Problems 9.10–9.11 refer to matrices A and B defined below. 9.11 Evaluate the expression: 4A + 3B. According to Problem 9.10, and . Calculate the sum by adding elements in corresponding positions. Note: Problems 9.12–9.13 refer to matrices M and N defined below. 9.12 Evaluate the expression: 2M + 3N. Multiply each element of M by 2, multiply each element of N by 3, and then add the resulting matrices. Textbooks often write fractions with the numerator and denominator next to each other (instead of above and below each other) when they’re inside matrices. Even though they look a little strange, they’re just ordinary fractions. The Humongous Book of Algebra Problems 185 Chapter Nine — Matrix Operations and Calculations Note: Problems 9.12–9.13 refer to matrices M and N defined below. 9.13 Evaluate the expression: Multiply each element of M by matrices. . and each element of N by –1. Add the resulting The expression contains “–N”, whi the same thing a ch is s “–1N .” Note: Problems 9.14–9.15 refer to matrices C, D, and E defined below. 9.14 Evaluate the expression: C + D – E. Subtracting matrix E is the same thing as adding –E. That’s why you spend the first part of this problem figuring out what –E is. Note that C + D – E = C + D + (–E). Perform scalar multiplication to generate matrix –E. Evaluate the expression C + D + (–E). ngous Book of Algebra Problems 186 The Humo Chapter Nine — Matrix Operations and Calculations Note: Problems 9.14–9.15 refer to matrices C, D, and E defined below. 9.15 Evaluate the expression: 3C – 2D + 4E. Perform scalar multiplication to generate matrices 3C, –2D, and 4E. Add the resulting matrices. The Humongous Book of Algebra Problems 187 Chapter Nine — Matrix Operations and Calculations Multiplying Matrices Not as easy as adding or subtracting them 9.16 If matrix A has order c × d, matrix B has order m × n, and the product matrix If there aren’t as many rows in the second matrix as there are columns in the first matrix, you can’t multiply the matrices. P = A ⋅ B exists, what condition must be met by the dimensions of A and B? What is the order of P? If the product A ⋅ B exists, then the number of columns in matrix A must equal the number of rows in matrix B. In this example, d = m. The dimensions of the product matrix are dictated by the dimensions of A and B as well. Matrix P will have the same number of rows as matrix A and the same number of columns as matrix B. Therefore, matrix P is c × n. Note: Problems 9.17–9.19 refer to matrices A and B (as defined below) and matrix P, such that P = A ⋅ B. 9.17 Calculate element p11 of matrix P. To calculate an element pij of the product matrix P, multiply each element in the ith row of A by the corresponding element in the jth column of B and add the products together. You are calculating p11, so you’ll multiply the numbers in the first row of A (starting at the left side) by the numbers in the first column of B (starting at the top), one at a time. Note: Problems 9.17–9.19 refer to matrices A and B (as defined below) and matrix P, such that P = A ⋅ B. 9.18 Calculate element p12 of matrix P. Multiply each element in the first row of A by the corresponding element in the second column of B. ngous Book of Algebra Problems 188 The Humo Chapter Nine — Matrix Operations and Calculations Note: Problems 9.17–9.19 refer to matrices A and B (as defined below) and matrix P, such that P = A ⋅ B. 9.19 Calculate elements p 21 and p 22 to complete the product matrix P. To calculate p 21, multiply each element in the second row of A by the corresponding element in the first column of B. Remember, the little numbers tell you what row and column an element is in, so p21 is in row 2 column 1 and p22 is in row 2 column 2. To calculate p 22, multiply each element in the second row of A by the corresponding element in the second column of B. Create matrix P, placing elements p 11, p12, p 21, and p 22 in the positions indicated. 9.20 Multiply the matrices: . Calculate each element pij of the product matrix P by multiplying each element in the ith row of the left matrix by the corresponding element in the jth column of the right matrix and then summing those products, as explained in Problems 9.17–9.19. The Humongous Book of Algebra Problems 189 Chapter Nine — Matrix Operations and Calculations Note: Problems 9.21–9.22 refer to matrices A and B (as defined below), and matrix P, such that P = A ⋅ B. 9.21 Explain why P exists and calculate element p42. The product P = A ⋅ B exists because the number of columns in matrix A is equal to the number of rows in matrix B. A has two colum ns and B has two ro ws. To calculate element p42 of matrix P, multiply each element in the fourth row of A by the corresponding element in the second column of B and add the results. Note: Problems 9.21–9.22 refer to matrices A and B (as defined below), and matrix P, such that P = A ⋅ B. 9.22 Generate matrix P. You can multiply A ⋅ B, but you can’t multiply B ⋅ A, because the number of columns in B (3) does not equal the number of rows in A (4). The product matrix has the same number of rows as matrix A and the same number of columns as matrix B. Therefore, P has order 4 × 3. Calculate each element of P using the technique described in Problem 9.21. ngous Book of Algebra Problems 190 The Humo Chapter Nine — Matrix Operations and Calculations Note: Problems 9.23–9.24 refer to matrices A and B defined below. Remember, A ⋅ B has as many rows as the left matrix in the product (matrix A has 2 rows) and as many columns as the right matrix in the product (matrix B has two columns). 9.23 Calculate the product: A ⋅ B. The product matrix P = A ⋅ B has order 2 × 2. Note: Problems 9.23–9.24 refer to matrices A and B defined below. 9.24 Calculate the product: B ⋅ A. The product matrix has order 4 × 4 and, when compared to the product calculated by Problem 9.23, demonstrates that matrix multiplication is not commutative. RU LE OF TH UMB: The product in Problem 9.23 is a 2 × 2 matrix, and the product in Problem 9.24 is 4 × 4! The order in which you multiply matrices has a drastic effect on the size of the product matrix, what elements are inside, and whether a product even exists! The Humongous Book of Algebra Problems 191 Chapter Nine — Matrix Operations and Calculations 9.25 Calculate the matrix product below. The product of two 3 × 3 matrices is a 3 × 3 matrix. Calculating Determinants Values defined for square matrices only 9.26 Given matrix A defined below, calculate A square matrix has the same number of rows and columns. Start at the upper-left corner and multiply diagonally down. Then, subtract what you get when you start at the lowerleft corner and multiply diagonally up. . The determinant of matrix A, written “ ” or “det (A),” is a real number value defined when A is a square matrix. The determinant of a 2 × 2 matrix can be calculated using the below shortcut formula. Set a = 3, b = –2, c = 1, and d = 9. ngous Book of Algebra Problems 192 The Humo Chapter Nine — Matrix Operations and Calculations 9.27 Given matrix B defined below, calculate . Apply the shortcut formula from Problem 9.26 to calculate the determinant of the 2 × 2 matrix. 9.28 Given matrix C defined below, calculate . Apply the shortcut formula from Problem 9.26 to calculate the determinant of the 2 × 2 matrix. 9.29 Given matrix D defined below, calculate . Problem 9.26 describes a shortcut used to calculate the determinant of a 2 × 2 matrix. It requires you to subtract a product of elements along an upward diagonal from a product of elements along a downward diagonal. The Humongous Book of Algebra Problems 193 Chapter Nine — Matrix Operations and Calculations Stick copies of the first two columns on the rig side of the 3 × 3 ht matrix to create a 3 × 5 matrix you’ ll use to calculate the determinant. The shortcut technique for calculating the matrix of a 3 × 3 determinant also involves the difference of products along diagonals, but it requires you to construct a 3 × 5 matrix. The first three columns consist of the original matrix, column four is a copy of column one, and column five is a copy of column two. Beginning with element d11 = 2, calculate the products along the diagonal and sum the results. Do the same with elements d12 = –5 and d13 = 1. Now multiply along the upward diagonals that start at d 31 = 0, d 32 = –1, and d 33 = 8 and add the products. Just like a 2 × 2 determinant is equal to the difference of a downward and an upward diagonal. The determinant of the 3 × 3 matrix is the difference of the downward and upward diagonal sums. 9.30 Given matrix E defined here, calculate det(E). Det(E) and both mean the same thing: the determinant of matrix E. 194 Construct a 3 × 5 matrix by duplicating the first two columns of the matrix, as directed by Problem 9.29. The Humongous Book of Algebra Problems Chapter Nine — Matrix Operations and Calculations Calculate the determinant by adding the products of the elements in the downward diagonals beginning at e 11 = 1, e 12 = 3, and e 13 = –6 and then subtracting the products of the elements in the upward diagonals beginning at e 31 = 4, , and e 33 = 5. Note: Problems 9.31–9.35 refer to matrix A defined below. 9.31 Calculate the minor M11 of A. The minor Mij of a matrix is the determinant of the matrix created by removing the ith row and the jth column. This problem directs you to calculate M11, so eliminate the first row and the first column from the matrix and calculate the determinant of the resulting 2 × 2 matrix. Apply the shortcut formula for calculating 2 × 2 matrices presented in Problem 9.26. Drop the first row (which contains 3 4 –1) and the first column (which contains 3 0 1) and you’re left with this. Note: Problems 9.31–9.35 refer to matrix A defined below. 9.32 Calculate the cofactor C11 of A. The cofactor Cij of a matrix is computed according to the following formula: Cij = (–1)i + j ⋅ Mij . Therefore, the cofactor is equal to the corresponding minor multiplied either by +1 or –1, depending upon the value of the expression i + j. According to Problem 9.31, M11 = 84. Substitute M11, i = 1, and j = 1 into the cofactor formula. You’re calculating C11. The first little number next to C is i and the second little number is j. They’re both equal to 1 here, so i = 1 and j = 1. The Humongous Book of Algebra Problems 195 Chapter Nine — Matrix Operations and Calculations Note: Problems 9.31–9.35 refer to matrix A defined below. It might help to temporarily cross row 2 and column 3 out of the matrix to visualize the 2 × 2 matrix you’ll end up with: 9.33 Calculate the cofactor C 23 of A. Begin by calculating M 23, the determinant of the matrix created by removing the second row and the third column of A. Substitute M 23 = –22, i = 2, and j = 3 into the cofactor formula. Note: Problems 9.31–9.35 refer to matrix A defined below. 9.34 Calculate As you’ll see in th e next problem, whe n you expand along a column you will get the same answer. by performing a cofactor expansion along the third row of A. Though you can calculate the determinant of this 3 × 3 matrix using the shortcut described in Problem 9.29, the cofactor expansion technique can be applied to any square matrix. Multiply each of the elements in a single row or column of the matrix by the corresponding cofactor and add the results. This problem directs you to expand along the third row, but the results would be the same if you chose to expand along either of the other rows or any of the columns. ngous Book of Algebra Problems 196 The Humo Chapter Nine — Matrix Operations and Calculations Calculate the minors M 31, M 32, and M 33 using the technique described in Problem 9.31. Next, calculate the corresponding cofactors: C 31, C 32, and C 33. Multiply each element of the third row by the corresponding cofactor. The sum of those products is the determinant of matrix A. Note: Problems 9.31–9.35 refer to matrix A defined below. 9.35 Verify the solution to Problem 9.34 by performing a cofactor expansion along the first column of A. Note that a 21 = 0, so there is no need to calculate C 21 for the cofactor expansion. According to Problem 9.32, C 11 = 84, and according to Problem 9.34, C 31 = 38. No matter what C21 is, during the expansion it gets multiplied by a21 = 0, and the result will be 0. The Humongous Book of Algebra Problems 197 Chapter Nine — Matrix Operations and Calculations 9.36 According to Problem 9.30, the determinant of matrix E (defined below) is The book shows you how to expand along the first row, but you’ll get the same answer with either of the other two rows or if you expanded one of the columns instead. 412. Perform a cofactor expansion to verify the determinant. To perform a cofactor expansion along the first row, calculate M11, M12, and M13. Calculate the corresponding cofactors. Multiply the elements of the first row by the corresponding cofactors and add the results. 9.37 Given matrix B defined below, calculate Problem 9.35 explains that a 0 in the row or column yo u’re expanding means that you don’t have to calcula te the correspondin g cofactor, and th at means a little less work. . Because b 13 = 0, it is most efficient to expand along either the first row or the third column. To expand along the third column of B, calculate M 23, M 33, and M43. Apply the shortcut technique described in Problem 9.29 to calculate the 3 × 3 determinants. ngous Book of Algebra Problems 198 The Humo Chapter Nine — Matrix Operations and Calculations Calculate the corresponding cofactors. Multiply the elements of the third column by the corresponding cofactors and sum the results. The Humongous Book of Algebra Problems 199 Chapter Nine — Matrix Operations and Calculations Cramer’s Rule Double-decker matrices that solve systems 9.38 According to Cramer’s Rule, the solution to a system of two linear equations in two variables is X, and Y. . Given the system below, identify matrices C, Matrix C consists of the system’s coefficients. The x-coefficients comprise the first column of matrix C and the y-coefficients comprise the second. The constants are the numbers with no variables to them—usually next fo on the right side und of the equal sign. To get the X matrix, replace the xcoefficients in C with the numbers from across the equal sign. To get the Y matrix, replace the y-coefficients instead. The remaining matrices, X and Y, are created by replacing individual columns of C with the column of constants: . To generate matrix X, replace the first column of C with the column of constants. To generate matrix Y, replace the second column of C with the column of constants. Note: Problems 9.39–9.41 refer to the system of equations below. 9.39 Use variable elimination to solve the system. Multiply the first equation by 2 and multiply the second equation by –3. To review variable elimination, check out Problems 8.19–8.28. Add the equations of the modified system and solve for x. ngous Book of Algebra Problems 200 The Humo Chapter Nine — Matrix Operations and Calculations Substitute x = 6 into either equation of the original system to calculate the corresponding value of y. The solution to the system is (x,y) = (6,–7). Note: Problems 9.39–9.41 refer to the system of equations below. 9.40 Construct the matrices C, X, and Y that are required to solve the system using Cramer’s Rule. According to Problem 9.38, the first column of matrix C consists of the system’s x-coefficients and the second column consists of the y-coefficients. To generate matrix X, replace the column of x-coefficients with the constants from the system. Similarly, generate Y by replacing the column of y-coefficients with the column of constants. The Humongous Book of Algebra Problems 201 Chapter Nine — Matrix Operations and Calculations Note: Problems 9.39–9.41 refer to the system of equations below. 9.41 Use Cramer’s Rule to verify the solution to the system generated in Problem 9.39. According to problem 9.40, , , and . Calculate the determinant of each matrix. Substitute , , and calculate the solution to the system. into the Cramer’s Rule formula to 9.42 According to Problem 7.12, the solution to the system below is . Use Cramer’s Rule to verify the solution. The constants in this system are 1 and 17. They replace the xcoefficients (6 and 14) in the X matrix and the y-coefficients (1 and –5) in the Y matrix. Construct matrices C, X, and Y, as directed by Problem 9.40. Calculate the determinants of the matrices. Apply the Cramer’s Rule formula to calculate the solution to the system. ngous Book of Algebra Problems 202 The Humo Chapter Nine — Matrix Operations and Calculations 9.43 Solve the system of equations below using Cramer’s Rule. Before constructing the matrices required by Cramer’s Rule, rewrite the second equation of the system in standard form; move the x-term left of the equal sign and multiply each term by 5 to eliminate the fractions. Line up the x’s, y’s, and constants of the system before you create any matrices. The easiest way to get everything ready for Cramer’s Rule is to put the lines in standard form. (See Problems 6.29– 6.36 to review standard form.) The modified system now contains two linear equations in standard form. Construct matrices C, X, and Y required by Cramer’s Rule and calculate the determinant of each. Apply the Cramer’s Rule formula to calculate the solution to the system. The Humongous Book of Algebra Problems 203 Chapter Nine — Matrix Operations and Calculations 9.44 Solve the system below using Cramer’s Rule. Cramer’s Rule is easily modified to solve systems of three linear equations in three variables. Instead of three 2 × 2 matrices, four 3 × 3 matrices are required: C, X, Y, and Z. Matrix C once again serves as the coefficient matrix—the first column contains the x-coefficients of the system, the second column contains the y-coefficients, and the third column contains the z-coefficients. To generate matrices X, Y, and Z, replace the first, second, and third columns of C, respectively, with the column of constants. Calculate the determinants of all four matrices. ngous Book of Algebra Problems 204 The Humo Chapter Nine — Matrix Operations and Calculations Apply the Cramer’s Rule formula to calculate the solution to the system. The Humongous Book of Algebra Problems 205 Chapter 10 Applications of Matrix Algebra Advanced matrix stuff Chapter 9 introduced matrices, explored matrix operations, and culminated with Cramer’s Rule, a method by which to solve systems of linear equations using matrices. This chapter is organized in a similar fashion. It begins with the manipulation of matrices, specifically row operations and row echelon form, and evolves into a set of matrix applications, including a method by which to solve nontrivial systems of equations. The chapter also explores inverse matrices, which are necessary to solve matrix equations. In this chapter, you learn how to put matrices in row echelon form and reduced-row echelon form. It’s sort of like putting linear equations into standard form, but with great er benefits. The biggest benefit? When you put an augmented matri x in reduced-row echelon form, you ’ve automatically solved the correspon ding system of equations! Before that, however, you need to master row operations—things you’re allowed to do to a matrix to get it in the rig ht form. Chapter Ten — Applications of Matrix Algebra Augmented and Identity Matrices Extra columns and lots of 0s and 1s 10.1 What is an augmented matrix? You can also indicate an augmented matr ix using a solid, inst ead of a dotted, line in matrix name [A |B the ]a in the matrix itse nd lf. An augmented matrix is created by joining together matrices that contain the same number of rows. They are most commonly used to represent systems of equations, whereby the coefficient matrix is joined with a column of constants. Consider matrices A and B defined below. The augmented matrix is constructed by including the values of matrix B as an additional column of matrix A. The new column is usually separated from the original columns of A by a dotted line. 10.2 Construct the augmented matrix for the below system of equations, such that C is the coefficient matrix and D is the column of constants. Not sure what a coefficient matrix or a column of constants is? Check out Problem 9.38. When a term has no explicit coefficient writt en (like x in the equa tion x + 9y = 11, the coefficient is 1. When there’s no coeffi cient but the variable is negative (like –y in 3x – y = 5), the coefficient is –1. The coefficient matrix for this system contains the x-coefficients in the first column and the y-coefficients in the second. The column of constants is a 2 × 1 matrix containing the numbers that appear right of the equal sign. Construct the augmented matrix third column of C. ngous Book of Algebra Problems 208 The Humo by including the elements of D as the Chapter Ten — Applications of Matrix Algebra Note: Problems 10.3–10.5 refer to matrix A defined below. 10.3 Construct matrix B that serves as the additive identity for matrix A and verify that A + B = A. As explained in Problem 1.36, the additive identity for real numbers is 0. Adding 0 to any real number x does not change the value of x. x+0=0+x=x To add matrix B to matrix A without changing the elements of A, every element of B must be 0. Therefore, the additive identity for any m × n matrix is the zero matrix 0 m × n . Verify that B is the additive inverse by demonstrating that A + B = A. A zero matrix is exactly what you’d imagine: a matrix that contains nothing but zeros. It has to have the same dimensions of A because you can’t add matrices unless they’re the same size. Note: Problems 10.3–10.5 refer to matrix A defined below. 10.4 Construct matrix C that serves as the multiplicative identity for matrix A. The identity matrix In is a square matrix with n rows and n columns that contains 0s for each of its elements except for those along the diagonal that begins with the element in the first row and the first column. In this problem, matrix A has two columns, so for the product A · C to exist, C must have two rows. Therefore, the correct identity matrix is C = I2. An identity matrix has ones along the diagonal that stretches from the top-left to the bottomright corner. All the other elements in the matrix are zeros. The Humongous Book of Algebra Problems 209 Chapter Ten — Applications of Matrix Algebra Note: Problems 10.3–10.5 refer to matrix A defined below. 10.5 Verify that A · C = A. According to Problem 10.4, the multiplicative identity is . Calculate the product of A and C using the technique described in Problems 9.17–9.19. Note: Problems 10.6–10.7 refer to matrix B defined below. 10.6 Construct matrix D that serves as the multiplicative identity for matrix B. I 3, like the identity matrix I2 from Problem 10.4, has all zero elements except along the diagonal that goes from topleft to bottom-right. The elements in that diagonal are all ones. 210 Matrix B contains three columns, so the identity matrix must contain the same number of rows: D = I3. The Humongous Book of Algebra Problems Chapter Ten — Applications of Matrix Algebra Note: Problems 10.6–10.7 refer to matrix B defined below. 10.7 Verify that B · D = B, assuming that D is the identity matrix identified in Problem 10.6. Calculate the product of the matrices and verify that it is equal to B. Matrix Row Operations Swap rows, add rows, or multiply by a number 10.8 Identify the three elementary matrix row operations. The three elementary row operations are: exchanging the positions of two rows, multiplying a row by a nonzero number, and replacing a row by adding it to a multiple of another row in the matrix. Note: Problems 10.9–10.12 refer to matrix A defined below. 10.9 Perform the row operation: . The notation Ri refers to the ith row of the matrix, so this problem refers to the second row of A (R 2), and the third row of A (R 3). The symbol “ ” indicates that the rows should be switched. To perform the row operation , move the elements of the second row to the third row and vice versa. For example, you could swap the first two rows so that the second row becomes the first and vice versa. This is the trickiest of the three row operati on To see it in actio s. n, look at Problems 10.11 and 10.12. The Humongous Book of Algebra Problems 211 Chapter Ten — Applications of Matrix Algebra Note: Problems 10.9–10.12 refer to matrix A defined below. Go back to the original matrix A, not the matrix you end up with in Problem 10.9. 10.10 Perform the row operation: . Multiply each element of the first row by 4. Note: Problems 10.9–10.12 refer to matrix A defined below. 10.11 Perform the row operation: . Add the corresponding elements of rows one and two together. Replace the second row with those values. Even though you used the numbers in the first row to get the new values for the second row, the first row is unaffected by the row operation when everything’s all said and done. 212 Notice that the row operation first or third row of A. The Humongous Book of Algebra Problems did not affect the elements in the Chapter Ten — Applications of Matrix Algebra Note: Problems 10.9–10.12 refer to matrix A defined below. 10.12 Perform the row operation: . Multiply the elements of the third row by –5 and add the results to the elements of the first row. Replace the first row with these values. Note: In Problems 10.13–10.15, manipulate matrix B, as defined below, using the indicated elementary row operation in order to make b11 = 1. 10.13 Switch two rows of the matrix. Switch the first and second rows of the matrix from position b 21 in the matrix to position b11. In Problems 10.13 –10.15, you have the same goal: adjust the matrix so that the upper-left element is 1. You’ll just reach that goal three different ways, each time using a different row operation. to move element 1 The Humongous Book of Algebra Problems 213 Chapter Ten — Applications of Matrix Algebra Note: In Problems 10.13–10.15, manipulate matrix B, as defined below, using the indicated elementary row operation in order to make b11 = 1. The multiplicative inverse property (see Problem 1.41) says that the product of any real number and its reciprocal is 1. 10.14 Multiply a row by a nonzero number. Multiply each element of the first row by (the reciprocal of b 11 = –3). Note: In Problems 10.13–10.15, manipulate matrix B, as defined below, using the indicated elementary row operation in order to make b11 = 1. 10.15 Replace a row with the sum of two rows. This row operation also works: Notice that the sum of elements b 11 and b 31 is 1. Apply the row operation . . Basically the whole point of Problems 10.13–10.15 is to show you that there are a lot of different ways to change numbers in a matrix. The matrices you end up with, in each case, look different. 214 The Humongous Book of Algebra Problems Chapter Ten — Applications of Matrix Algebra Note: Problems 10.16–10.17 refer to matrix C defined below. 10.16 Perform an elementary row operation on the first row of the matrix to make c11 = 1. Apply the row operation . Usually, the most efficient way to transform a matrix element into 1 is to multiply its row by the reciprocal of the element. Note: Problems 10.16–10.17 refer to matrix C defined in Problem 10.16. 10.17 Apply a row operation to the matrix generated in Problem 10.16 to make c21 = 0. There is good reason to make c 11 = 1, a step completed in Problem 10.16. Multiplying the first row by the opposite of c21 and adding it to the second row ensures that c21 = 0. In other words, apply the row operation . The last four problems focused on changing the element in the upper-left corner of the matrix into a 1. That’s because it’s the first step when you’re putting a matrix in row echelon or reduced-row echelon form (as you’ll see in Problems 10.18– 10.33). Most of the time, you’ll perform a bunch of row operations on a matrix, one at a time. Each opera tion is applied to the m atr you got from the ix la step, rather tha st n going back to the orig inal matrix and start ing over. The Humongous Book of Algebra Problems 215 Chapter Ten — Applications of Matrix Algebra Row and Reduced-Row Echelon Form More matrices full of 0s with a diagonal of 1s 10.18 What are the characteristics of a matrix in row echelon form? What additional requirements are required for reduced-row echelon form? In other words, the diagonal that starts in the upper-left corner and goes to the lower-right corner of the matrix. Only rows of zeros at the bottom of a matrix are exempt from the “1s in the diagonals and 0s below that” rule. So when you’re getting the diago nal numbers to be 1 and the numbers abo ve and below the d iagonal to be 0, only worry about the numbe rs left of the dotte d line in the augm ented matrix. Both of the forms discussed in this problem are typically applied to augmented matrices (as defined in Problem 10.1). If the augmented matrix is in row echelon form, then each of the elements along the diagonal a11, a 22, a 33, … is 1. Furthermore, all the elements below the diagonal must be 0. Finally, if the matrix contains a row of zeros, that row must be placed at the bottom of the matrix. Matrices and below fulfill these requirements and are, therefore, in row echelon form. Notice that includes a row of zeroes at the bottom of the matrix, which is not required to contain a 1 along the diagonal. Matrices in reduced-row echelon form satisfy one requirement beyond those in row echelon form—not only must the elements below the diagonal be zeros, the elements above the diagonal must be zeros as well. Matrices and are expressed in reduced-row echelon form. Note that the strict requirements of row and reduced-row echelon form only apply to the left matrix in the augmented pair. There are no requirements, for example, on column matrix S in the matrix above. Note: Problems 10.19–10.22 present the steps necessary to solve the system of equations using an augmented matrix. 10.19 Construct augmented matrix to represent the system and perform a row operation to make a11 = 1. A is the 2 × 2 coefficient matrix and C is the 2 × 1 column of constants, just like in Problem 10.2. To change element a11 = 2 into the required value of 1, multiply the first row by the reciprocal of a11: 216 The Humongous Book of Algebra Problems . Chapter Ten — Applications of Matrix Algebra Note: Problems 10.19–10.22 present the steps necessary to solve the system of equations defined in Problem 10.19 using an augmented matrix. 10.20 Perform a row operation on the matrix generated by Problem 10.19 to make a 21 = 0. When a11 = 1 (as a result of Problem 10.19), you can multiply the first row by the opposite of a 21 = 4 and add the rows together to generate new values for the second row: . Note: Problems 10.19–10.22 present the steps necessary to solve the system of equations defined in Problem 10.19 using an augmented matrix. 10.21 Express the matrix generated by Problem 10.20 in row-echelon form. Putting matrices in row echelon form is also called “Gaussian elimination.” To express the matrix in row echelon form, the elements along the diagonal (a11 = 1 and a 22 = 5) must be 1. To change the value of a 22 = 5 into the required value 1, multiply each element of the row by the reciprocal of a 22 = 5. In other words, apply the row operation . Thanks to the work already done in Problem 10.20. The other requirement of row echelon form, that the element below the diagonal of ones must be equal to 0, is already met by the system: a 21 = 0. The Humongous Book of Algebra Problems 217 Chapter Ten — Applications of Matrix Algebra Note: Problems 10.19–10.22 present the steps necessary to solve the system of equations defined in Problem 10.19 using an augmented matrix. 10.22 Express the matrix generated by Problem 10.21 in reduced-row echelon form, and identify the solution to the system of equations. Verify that the solution is correct. Putting matrices in reduced-row echelon form is also called “Gauss-Jordan elimination.” In matrix A, the columns (from left to right) represent x- then y-coefficients. That means the solutions are (from top to bottom) x then y right of the dotted line. Multiply the row containing the newly modified element a 22 = 1 by the opposite of , add the rows together, and record the results in the first row: . The solutions to the system of equations appears right of the dotted line in the augmented matrix: To verify the solution, plug . and y = –3 into both equations of the system (2x – 3y = 10 and 4x – y = 5) to verify that each produces in a true statement. Note: Problems 10.23–10.24 refer to the system of equations below. 10.23 Express the system as an augmented matrix in row echelon form. Construct an augmented matrix the system. 218 The Humongous Book of Algebra Problems from the coefficients and constants of Chapter Ten — Applications of Matrix Algebra Perform the row operation to make a11 = 1. STEP ONE: (to put a matrix in reduced-row echelon form): Multiply the first row by the reciprocal of the number in the upperleft corner. Perform the row operation Perform the row operation to make a 21 = 0. STEP TW O: Make a new second row that equals the current second row plus the first row multiplied by the opposite of a21. . Because the elements in the diagonal (a11 and a 22) are 1 and the element below the diagonal (a 21) is 0, the matrix is in row echelon form. STEP THREE: Divide the second row by a22 (unless a22 already equals 1). The matrix is now in row echelon form. Note: Problems 10.23–10.24 refer to the system of equations defined in Problem 10.23. 10.24 Express the matrix generated by Problem 10.23 in reduced-row echelon form to solve the system of equations. Perform the row operation so that a12 = 0. STEP FO UR: Multiply row two by the opposite of a , 12 add the rows, and replace the first row with the results. The Humongous Book of Algebra Problems 219 Chapter Ten — Applications of Matrix Algebra See Problem 10.4 if you’re not sure what an identity matrix is or what I 2 means. When a matrix is in reduced row-echelon form, the matrix left of the dotted line is an identity matrix, except that it might have one or more rows of zeros on the bottom of it. Because the 2 × 2 matrix left of the dotted line is the identity matrix I2, the augmented matrix is in reduced-row echelon form, and the solutions appear in the column matrix right of the dotted line: (x,y) = (–2,–9). 10.25 Solve the system of equations below by constructing an augmented matrix and expressing it in reduced-row echelon form. Construct the augmented matrix operation and apply the row . Make a 21 = 0 by applying the row operation Apply the row operation Reduce this fraction by dividing the top and bottom by 14. ngous Book of Algebra Problems 220 The Humo to make a 22 = 1. . Chapter Ten — Applications of Matrix Algebra Apply the row operation to make a12 = 0. The solution to the system of equations is . Note: Problems 10.26–10.31 present the steps necessary to solve the system of equations defined below using an augmented matrix. 10.26 Construct augmented matrix to represent the system and perform a row operation to make m11 = 1. Matrix M contains the coefficients of the system and matrix N contains the constants right of the equal sign in each equation. The most efficient way to make m11 = 1 is to switch the first and third rows of the matrix: . The Humongous Book of Algebra Problems 221 Chapter Ten — Applications of Matrix Algebra When you get the upper-left element to equal 1 in a 3 × 3 matr ix focus on the seco , nd row. The left elem ent needs to equal 0 and the middle elemen t needs to be 1. RU LE OF TH UMB: Whenever you’re trying to make something equal 0, you multiply a row by the opposite of the number that’s going to disappear and then add rows together. When you’re trying to make something equal 1, you divide a row by the number that needs to go. Note: Problems 10.26–10.31 present the steps necessary to solve the system of equations defined in Problem 10.26 using an augmented matrix. 10.27 Perform row operations on the matrix generated by Problem 10.26 to make m 21 = 0 and m 22 = 1. Perform the row operation to make m 21 = 0. When m11 = 1 (the goal of Problem 10.26), you should multiply the first row by the opposite of m 21 = 2, add it to the second row, and replace the elements in the second row with the results. Multiply the second row by m 22 = 1: , the reciprocal of m 22 = –7, to make . Note: Problems 10.26–10.31 present the steps necessary to solve the system of equations defined in Problem 10.26 using an augmented matrix. 10.28 Express the matrix generated by Problem 10.27 in row echelon form. Perform the row operation to make m 31 = 0. Use the first row to transform m 31 because it has 1 in the same column as m 31. You’re trying to make all the elements in the diagonal (m11, m22 , and m33) into ones. In this step, you’ve got to make m33 = 1, but to do that, both numbers LEFT of m33 have to be 0. Make m 32 = 0 using the row operation . Note that the second row is used (instead of the first row, which was used to change m 31) because it contains 1 in the same column as m 32 and has zeros left of that element. ngous Book of Algebra Problems 222 The Humo Chapter Ten — Applications of Matrix Algebra Apply the row operation to make m 33 = 1 and thus express the augmented matrix in row echelon form. Note: Problems 10.26–10.31 present the steps necessary to solve the system of equations defined in Problem 10.26 using an augmented matrix. 10.29 Perform a row operation on the matrix generated by Problem 10.28 to make m 23 = 0. Multiply the third row by the opposite of m 23, add rows two and three, and replace row two with the results: Now it’s time to start making the elements ABOV E the diagonal into 0s. . Note that the third row is used to change m 23 because it contains 1 in the same column as m 23 and all the elements left of 1 are zeros. The Humongous Book of Algebra Problems 223 Chapter Ten — Applications of Matrix Algebra Note: Problems 10.26–10.31 present the steps necessary to solve the system of equations defined in Problem 10.26 using an augmented matrix. 10.30 Perform a row operation on the matrix generated by Problem 10.29 to make m 13 = 0. Use the method described in Problem 10.29 to change an element in the third column into 0; multiply the third row by the opposite of the number to be eliminated (m13 = –2, so multiply by 2), add the first and third rows, and replace the elements of the first row with the results: . Note: Problems 10.26–10.31 present the steps necessary to solve the system of equations defined in Problem 10.26 using an augmented matrix. 10.31 Perform a row operation on the matrix generated by Problem 10.30 to make To get rid of m12, multiply the second row by its opposite. Why use the second row? It has 1 in the same column as m12 and zeros everywhere else (except across the dotted line, and that doesn’t matter). m 12 = 0 and therefore express the augmented matrix in reduced-row echelon form. Identify the solution to the system of equations. Perform the row operation to make m 12 = 0. The column of constants contains the solution to the system: (x,y,z) = (0,3,–4). ngous Book of Algebra Problems 224 The Humo Chapter Ten — Applications of Matrix Algebra Note: Problems 10.32–10.33 refer to the system of equations below. 10.32 Construct an augmented matrix that represents the system and express it in row echelon form. Matrix A contains the coefficients of the system and B contains the constants right of the equal sign. Perform the row operation to make a11 = 1. Make a 21 = 0 by applying the row operation Apply the row operation diagonal, a 22. There’s no xterm in the third equation (3y + 8z = 5) so set a31 = 0, because it represents the xcoefficient of the third equation. . to place a 1 in the second element of the The Humongous Book of Algebra Problems 225 Chapter Ten — Applications of Matrix Algebra Note that a 31 already equals 0, but a 32 must equal 0 as well, so apply the row operation . This is getting messy. To add 8 in a33 and to add 5 in b31, you need to us e common denomin ators, so multiply the to p and bottom of th ese fractions by 13. To complete row echelon form, use the row operation a 33 = 1. ngous Book of Algebra Problems 226 The Humo to make Chapter Ten — Applications of Matrix Algebra Note: Problems 10.32–10.33 refer to the system of equations defined in Problem 10.32. 10.33 Express the matrix generated by Problem 10.32 in reduced-row echelon form and identify the solution to the system of equations. Apply the row operation Apply the row operation to make a 23 = 0. to make a13 = 0. Apply the row operation to make a12 = 0 and, therefore, express the matrix in reduced-row echelon form. The solution to the system is (x,y,z) = (–1,7,–2). The Humongous Book of Algebra Problems 227 Chapter Ten — Applications of Matrix Algebra Inverse Matrices Matrices that cancel other matrices out Note: Problems 10.34–10.35 refer to matrix A defined below. The notation A–1 means “the inverse of matrix A,” NOT “matrix A raised to the –1 power.” If you’re not sure what I 2 is, check out Problem 10.4. 10.34 Calculate the inverse matrix A –1 of A. Only square matrices have inverses. If matrix M has order n × n, the first step toward identifying its inverse is to augment M with the identity matrix In . In this problem, A has order 2 × 2 so augment it with the identity matrix I 2: . Use the technique presented in Problems 10.19–10.22 to express A in reducedrow echelon form. Begin by applying the row operation Apply the row operation to make a 21 = 0. Everything you do to the 2 × 2 matrix on the left also trickles over to the 2 × 2 matrix on the right, so don’t forget to apply the row operations to the entire row. Apply the row operation Apply the row operation ngous Book of Algebra Problems 228 The Humo to make a 22 = 1. to make a12 = 0. . Chapter Ten — Applications of Matrix Algebra The 2 × 2 matrix left of the dotted line is A –1, the inverse matrix of A. Note: Problems 10.34–10.35 refer to matrix A defined below. 10.35 Verify that A · A –1 = A –1 · A = I2. Calculate the product A · A –1. If two matrices are inverses, then multiplying them in any order produces an identity matrix of the same size. Verify that the product A –1 · A is equal to I2 as well. The Humongous Book of Algebra Problems 229 Chapter Ten — Applications of Matrix Algebra 10.36 Given matrix B defined below, calculate the inverse matrix B –1. Construct the augmented matrix . Express B in reduced-row echelon form, beginning with the row operation to make b 11 = 1. Now apply the row operation You didn’t have to worry about making b12 = 0 to complete reduced-row echelon form because it’s already 0! to make b 21 = 0. The matrix right of the dotted line is the inverse matrix of B. 10.37 If C and D, as defined below, are inverse matrices, calculate the value of v + w. According to Problem 10.35, the product of two n × n inverse matrices is the identity matrix In . Calculate the product of the matrices. ngous Book of Algebra Problems 230 The Humo Chapter Ten — Applications of Matrix Algebra Because the matrices in this equation are equal, each of the corresponding pairs of elements must be equal. Of particular interest is the pair of elements in the lower-left corners of the matrices. 6w = 0 Solve the equation for w. Substitute w = 0 into the matrix equation. Note that the elements in the first row and second column of each matrix are equal as well. Finally, evaluate the expression v + w, as directed by the problem. The Humongous Book of Algebra Problems 231 Chapter Ten — Applications of Matrix Algebra 10.38 Given matrix N defined below, calculate the inverse matrix N –1. Construct the augmented matrix Look at Problems 10.26– 10.31 if you need to review transforming 3 × 3 matrices into reduced-row echelon form. . Use row operations to express N in reduced-row echelon form, beginning with to make n11 = 1. Apply the row operation ngous Book of Algebra Problems 232 The Humo to make n 21 = 0. Chapter Ten — Applications of Matrix Algebra Apply the row operation Apply the row operation Apply the row operation Apply the row operation to make n 22 = 1 to make n 32 = 0. to make n 33 = 1. to make n 23 = 0. The Humongous Book of Algebra Problems 233 Chapter Ten — Applications of Matrix Algebra Apply the row operation to make n12 = 0 and, therefore, express N in reduced-row echelon form. The inverse matrix N –1 appears right of the dotted line. ngous Book of Algebra Problems 234 The Humo Chapter Ten — Applications of Matrix Algebra Note: Problems 10.39–10.41 refer to the system of equations defined below. 10.39 Express the system as a matrix equation of the form . Construct matrix A using the coefficients of the system and use the constants of the system to generate matrix B. Substitute A and B into the equation . Note: Problems 10.39–10.41 refer to matrix A, matrix B, and the system of equations defined in Problem 10.39. 10.40 Calculate A –1, the inverse of matrix A. Construct the augmented matrix echelon form. Begin with the row operation Apply the row operation Apply the row operation and express matrix A in reduced-row to make a11 = 1. to make a 21 = 0. This matrix equation is just another way to write the system of equations. You could use the row operation instead, but that’s a lot more work than just swapping the rows. Make sure to swap the rows in the augmented part of the matrix across the dotted line as well. to make a12 = 0. The Humongous Book of Algebra Problems 235 Chapter Ten — Applications of Matrix Algebra Notice that matrix A and matrix A–1 are the same! Some matrices are their own inverses. The inverse matrix, A –1, appears right of the dotted line in the augmented matrix. Note: Problems 10.39–10.41 refer to matrix A, matrix B, and the system of equations defined in Problem 10.39. 10.41 Solve the matrix equation generated in Problem 10.39. According to Problem 10.39, the system of equations can be expressed as the following matrix equation. To solve this equation, isolate the column matrix To eliminate matrix , multiply both sides of the equation by the inverse matrix calculated in Problem 10.40: Multiplying a matrix by an identity matrix (like I2) is basically the same as multiplying a real number by 1. You’d write 1(3) as 3 and just drop the 1 because there’s no need for it. Same thing goes with the identity matrix here. ngous Book of Algebra Problems 236 The Humo left of the equal sign. . Chapter 11 Polynomials aised to powers Clumps of numbers and variables r Though a polynomial is no more than a sum of a finite number of terms, each term usually the product of a real number and a variable raised to some integer, the importance of polynomials cannot be overstated. This chapter presents the most basic polynomial skills: classifying polynomials and calculating polynomial sums, differences, products, and quotients. Polynomials are like chains. Small clu mps of numbers and variables multiplied together called terms ar e strung together with addition and subtraction signs. You’ve been dealing with polynomials for a while now, even though you may not have known it. For instance, 2x + 1 is a polynomial made up of two terms, 2x and 1. This chapter teaches you how to describe polynomials and how to add them, subtract them, multiply them, and divide them. There are polynomials out there that have an infinite nu of terms, but you don’t deal mber with those in algebra. Chapter Eleven — Polynomials It doesn’t have degree one because there’s only one variable. It has degree one because the only variable there is raised to the first POWER. Classifying Polynomials Labeling them based on the exponent and total terms 11.1 Classify the polynomial 7x + 10 according to its degree and the number of terms it contains. The degree of a polynomial is the highest exponent to which the variable in that polynomial is raised. The only variable present in this expression is x, so the polynomial has degree one. Such polynomials are described as linear. The polynomial contains two terms, 7x and 10, so it is a linear binomial. 11.2 Classify the polynomial 6x 2 according to its degree and the number of terms it contains. It’s a BInomial because it has TWO terms (bi = two). A MONOmial has ONE term, and a TRInomial has THREE terms. RU LE OF TH UMB: The degree is the highest power of the variable. Here’s how polynomial classifications break down by degree: Degree 1: Linear Degree 2: Quadratic Degree 3: Cubic Degree 4: Quartic Degree 5: Quintic There are others, but these are the most common. The polynomial 6x 2 consists of a single term, so it is a monomial. That term contains a variable raised to the second power, which classifies the polynomial as a quadratic monomial. 11.3 Classify the polynomial –3x 2 + 5x according to its degree and the number of terms it contains. The polynomial –3x 2 contains two terms: –3x 2 and 5x. Of those terms, the highest power of x is 2, so the polynomial has degree two. A polynomial with two terms and degree two is called a quadratic binomial. 11.4 Classify the polynomial 12x 3 according to its degree and the number of terms it contains. This polynomial consists of a single term, so it is a monomial. Furthermore, the polynomial has degree three, which means it is cubic. In conclusion, 12x 3 is a cubic monomial. Note: Problems 11.5–11.6 refer to the polynomial 4x2 – 7x + 6. 11.5 Classify the polynomial according to its degree and the number of terms it contains. This polynomial contains three terms (4x 3, –7x, and 6) and has degree two (because the highest power of x is two). Therefore, 4x 2 – 7x + 6 is a quadratic trinomial. ngous Book of Algebra Problems 238 The Humo Chapter Eleven — Polynomials Note: Problems 11.5–11.6 refer to the polynomial 4x2 – 7x + 6. 11.6 Identify the leading coefficient and the constant of the polynomial. According to Problem 11.5, the polynomial has degree two because x 2 represents the highest power of x. The coefficient of x 2 is the leading coefficient of the polynomial: 4. A constant is a term with no explicitly stated variable. Of the terms in 4x 2 – 7x + 6, the rightmost term has no variable, so 6 is the constant. Note: Problems 11.7–11.8 refer to the polynomial ax2 – bx3 – c. Assume that a, b, and c are nonzero integers. 11.7 Classify the polynomial according to its degree and the number of terms it contains. The polynomial consists of three terms: ax 2, –bx 3, and –c. The highest power of x among the terms is 3, so the polynomial has degree three. Therefore, ax 2 – bx 3 – c is a cubic trinomial. Note: Problems 11.7–11.8 refer to the polynomial ax2 – bx3 – c. Assume that a, b, and c are nonzero integers. 11.8 Identify the leading coefficient and the constant of the polynomial. According to Problem 11.7, the term containing the highest variable power is –bx 3. Its coefficient, –b, is the leading coefficient of the polynomial. The term –c is a constant because it contains no variables. It’s called the “leading coefficient” beca us you normally write e a polynomial in or der from the highest power of the variable to the lowest. That puts the term with the bi ggest exponent out fron t, its coefficient en and ds up at the beginn ing the polynomial, in of the “lead” spot. It’s called a constant because with no variables inside it, its value never varies, remaining constant no matter what value of x gets plugged into the other terms. Adding and Subtracting Polynomials Only works for like terms 11.9 Simplify the expression: (5x + 2) + (8x – 3). Terms that contain the same variable expression are described as “like terms” and can be combined via addition or subtraction. In this expression, 5x and 8x are like terms because they both contain x. The constants 2 and –3 are like terms as well, as they share the same (lack of) variables. Rewrite the expression by grouping the like terms. 8y and –5y would be like terms (because they both have y). However, 2x2 and 9x are NOT like terms. They both have an x, but those x’s are raised to different powers. The variables of like terms must match exactly. You’re allowed to move the terms around when you’re adding real numbers, according to the commutat ive Problem 1.34 for more inform property. (See ation.) The Humongous Book of Algebra Problems 239 Chapter Eleven — Polynomials (5x + 8x) + (2 – 3) To combine like terms, add the coefficients and multiply the result by the shared variable. For example, to add 5x and 8x, add the coefficients (5 + 8 = 13) and multiply the result by the shared variable (x): 5x + 8x = 13x and 2 – 3 = –1. (5x + 8x) + (2 – 3) = 13x – 1 11.10 Simplify the expression: (2x2 + 3x + 1) + (9x 2 – 6x – 13). This expression contains three pairs of like terms: 2x 2 and 9x 2 ; 3x and –6x; 1 and –13. Rewrite the expression by grouping the like terms. (2x 2 + 9x 2) + (3x – 6x) + (1 – 13) Combine like terms to simplify the expression. 11x 2 – 3x – 12 11.11 Simplify the expression: . Before you combine the terms of the expression, distribute expressions in the adjacent parentheses. and 3 to the The distributive property says tha t you can multiply a number outside parenthe ses to an addition or su btraction problem inside. Tha t 3(x – 2) = 3(x) + 3( means –2) = 3x – 6. Rewrite the expression by grouping like terms. Combine like terms, using a common denominator to add the x-terms. Therefore, ngous Book of Algebra Problems 240 The Humo . Chapter Eleven — Polynomials 11.12 Simplify the expression: 6x – (8x + 4). Distribute –1 through the expression 8x + 4. When an expression in parentheses has a negative in front of it, think of that negative sign as a –1 and distribute it through the parentheses. Combine like terms 6x and –8x. (6x – 8x) – 4 = –2x – 4 11.13 Simplify the expression: (6x 2 + 5x + 2) – 2x 2(7x – 1). Apply the distributive property to the expression –2x 2(7x – 1). Rewrite the expression by grouping like terms. –14x 3 + (6x 2 + 2x 2) + 5x + 2 Combine like terms. –14x 3 + 8x 2 + 5x + 2 11.14 Simplify the expression: (x 2 + 6y 2 + 2x – 3y) + 2x(5x + y – 10). Apply the distributive property. When you multiply x2 and x, you get x2(x) = x2(x1) = x2 + 1 = X 3. If you need to review this exponential rule, look at Problems 3.25–3.29; there are a handful of distributive property practice problems there that contain variables. Group like terms and combine them to simplify the expression. 11.15 Simplify the expression: (x 2 + 4) – 6x(3x – 9) – (2x 2 – 5x). Apply the distributive property. Group and combine the like terms. You don’t HAVE to rewrite the expression so tha t the like terms are ne xt each other befo to re combine them, bu you t helps to make su it re you don’t leave anyth ing out. (x 2 – 18x 2 – 2x 2) + (54x + 5x) + 4 = –19x 2 + 59x + 4 The Humongous Book of Algebra Problems 241 Chapter Eleven — Polynomials 11.16 Simplify the expression: 3y(2x 2 + 9xy – 7y 2) + 2x(x 2 – 2xy – 4y2). Apply the distributive property. Group and combine like terms. 11.17 Given the following equation, evaluate the expression a + b. The word “quantity” is used here to describe something in parentheses. If you say “ 3 times the quantity x + 2,” you’re saying that 3 is not just multiplied by x but by 2 as well: 3(x + 2). 4(x 2 + ax + 1) – 3(x 2 + b) = x 2 – 8x + 79 Distribute 4 and –3 through the respective quantities. Group and combine like terms. Note that x is the only variable in this equation. Although the real number values of a and b are not yet known, they are fixed— and therefore constant—values nonetheless. Subtract x 2 from both sides of the equation. Treat a and b like numbers, not like variables. That means 4 and –3b are like terms because neither of them contains x. Both sides of this equation contain an x-term and a constant. The x-term left of the equal sign has a coefficient of 4a and the coefficient of the x-term right of the equal sign is –8. For the equation to be true, the x-terms must be equivalent, so the coefficients must be equal. The constants of the polynomials (4 – 3b on the left side of the equal sign and 79 on the right) must be equal as well. ngous Book of Algebra Problems 242 The Humo Chapter Eleven — Polynomials Evaluate the expression a + b. 11.18 Given the following equation, evaluate the expression a + 3b + 2c. 2(x 2 – 4x + a) – 3(–2x 2 + bx – 1) + 5(cx 2 + 5x + 6) = 23x 2 + 17x – 5 Apply the distributive property to the left side of the equation. Group and combine like terms. Calculate a by setting the constants on both sides of the equation equal. The right side of the equation has th re terms and the le e ft side has three gr oups. The group 2a + 33 represents a cons ta because it has no nt x’ Set 2a + 33 equa s. l to the constant tha t’s right of the equa l sign: –5. Set the x-terms on both sides of the polynomial equation equal to calculate b. For the moment, assume that x ≠ 0. Set the x 2-terms on both sides of the polynomial equation equal to calculate c. Again, assume for the moment that x ≠ 0. The Humongous Book of Algebra Problems 243 Chapter Eleven — Polynomials Evaluate the expression a + 3b + 2c. Multiplying Polynomials Don’t spend too much energy memorizing what FOIL stands for. Instead, realize what’s going on when you multiply polynomials. Basically, you take the terms of the left polynomial one at a time and distribute them through the terms in the right polynomial. FOIL and beyond 11.19 Explain how to calculate the binomial product (a + b)(c + d) using the FOIL method. FOIL is an acronym for “First Outside Inside Last” and refers to pairs of terms in the product. To multiply the binomials (a + b) and (c + d), you add the products of the first terms in each binomial (a and c), the outside terms (a and d), the inside terms (b and c), and the last terms of each binomial (b and d). (a + b)(c + d) = ac + ad + bc + bd In practice, the FOIL technique distributes a, the first term of the first binomial, through the second binomial: a(c + d) = ac + ad. It then distributes b, the remaining term of the first binomial: b(c + d) = bc + bd. This interpretation of polynomial multiplication is more valuable because it can be generalized to calculate products of any two finite polynomials. 11.20 Calculate the product and simplify: (x + 2)(x + 6). The left polynomial in the product is x + 2. Distribute x, its first term, through the polynomial on the right (x + 6). x(x + 6) = x 2 + 6x Now distribute 2, the second term of the left polynomial, through the terms of the right polynomial. 2(x + 6) = 2x + 12 The product (x + 2)(x + 6) is the sum of the terms calculated above. ngous Book of Algebra Problems 244 The Humo Chapter Eleven — Polynomials 11.21 Calculate the product and simplify: (y – 1)(y + 3). Distribute the terms of the left binomial, y and –1, through the right binomial, (y + 3). 11.22 Calculate the product and simplify: (3x – 4)2. Rewrite the expression as a product: (3x – 4)2 = (3x – 4)(3x – 4). Calculate the product using the technique described in Problems 11.19–11.21. Squaring something means multiplying it times itself. 11.23 Calculate the product and simplify: (a + 3b)(2a – 7b). Distribute each term of the left binomial through the right binomial. 11.24 Calculate the product and simplify: (x – 3y)(4x 2 – y 2). Distribute each term of the left binomial through the right binomial. 11.25 Calculate the product and simplify: (x + 2)(x2 – 4x + 6). Distribute each term of the left binomial through the right trinomial. The righthand polynomial in this product has three terms, not two, so you can’t use the FOIL method. No big deal—just distribute the terms of the left-hand polynomial (x and 2) through x2 – 4x + 6 and add up the results. The Humongous Book of Algebra Problems 245 Chapter Eleven — Polynomials 11.26 Calculate the product and simplify: (x – y)(x 2 + 6xy – 9y 2). Distribute each term of the left binomial through the right trinomial. Even though both polynomials have three terms, it doesn’t change the way you multiply. Distribute x, then 3y, and then –1 through the right-hand polynomial and add everything together: x(2x – y + 7) + 3y(2x – y + 7) – 1(2x – y + 7). Before you try to perform long division on polynomials, make sure you remember how the process works with whole numbers—flip back to Problems 2.3 and 2.5. The two techniques are basically the same, except with whole numbers you divide one DIGIT at a time, and with polynomials you divide one TERM at a time. 11.27 Calculate the product and simplify: (x + 3y – 1)(2x – y + 7). Distribute each term of the left trinomial through the right trinomial. 11.28 Calculate the product and simplify: (a – 3b + 2c)(4a – b + 5c). Distribute each term of the left trinomial through the right trinomial. Long Division of Polynomials A lot like long dividing integers 11.29 Calculate (x + 4) ÷ x using long division. Rewrite the expression as a long division problem, with the divisor outside the division symbol and the dividend within it. The divisor is what you’re dividing BY and the dividend is what you’re dividing INTO. ngous Book of Algebra Problems 246 The Humo Chapter Eleven — Polynomials Divide the leftmost term of the dividend by the divisor: x ÷ x = 1. Write this number above its like term in the dividend. Multiply the newly placed number by the divisor: 1 · x = x. Write the opposite of the result beneath the dividend, once again lining up like terms. Combine like terms: x – x = 0. The number 4 in the dividend x + 4 has the same variable as 1 (because neither of them have any variables). That means 1 and 4 are like terms. When you write numbers above the division symbol, line them up with their like terms. Add the next term of the dividend (+4). Whenever you multiply a number on top of the division symbol by the number out front, write the OPPOSITE of that number (in this case –x instead of x) beneath the dividend. Combine like terms: 0 + 4 = 4. The degree of the divisor x is greater than the degree of the polynomial below the horizontal line, so the process of long division is complete. The number above the division symbol (1) is the quotient and the number below the horizontal line (4) is the remainder. Write the solution using the format below. x = x1, so x has degree one. 4 is a constant and doe have any variabl sn’t es, so it has degree zero. Because 1 > 0, yo u’re done dividing. The Humongous Book of Algebra Problems 247 Chapter Eleven — Polynomials Note: Problems 11.30–11.31 refer to the quotient (x2 + 6x – 9) ÷ (x – 2). 11.30 Calculate the quotient and remainder using long division. In other words, x (from the divis or) times what equa 2 ls (from the dividen x d)? The answer’s x: x . x = x2. If you wri the polynomials fr te om highest to lowest powers of x, you’ll always be dealing with the terms on the left: “The le term of the divis ft or times what equa ls the left term of the dividend?” Express the quotient as a long division problem. Divide the leftmost term of the dividend by the leftmost term of the divisor: x 2 ÷ x = x. Write the answer above the like term 6x in the dividend. Multiply the newly placed x by each term of the divisor and write the opposites of the products beneath the dividend. Combine like terms (6x + 2x = 8x) and add the next term of the dividend (–9). Divide the leftmost term below the horizontal line by the leftmost term of the divisor: 8x ÷ x = 8. Write the answer above the division symbol, multiply both terms of the divisor by 8, and write the opposites of the products below 8x – 9. Therefore, ngous Book of Algebra Problems 248 The Humo . Chapter Eleven — Polynomials Note: Problems 11.30–11.31 refer to the quotient (x2 + 6x – 9) ÷ (x – 2). 11.31 Verify the answer generated by Problem 11.30. To verify the answer to a long division problem, multiply the divisor by the quotient and then add the remainder. The result should be equal to the dividend. (divisor)(quotient) + remainder = (x – 2)(x + 8) + 7 Multiply the binomials by distributing each term of x – 2 through the quantity (x + 8), as explained in Problems 11.19–11.24. The remainder in Problem 11.30 is just 7, even though you write the remainder as part of the fraction . Because the result, x 2 + 6x – 9, is equal to the dividend of Problem 11.30, the quotient x + 8 and the remainder 7 are correct. 11.32 Calculate (2x 2 – x – 15) ÷ (x 2 + x – 12) using long division. Rewrite the expression as a long division problem. Divide 2x 2 by x 2 to get 2 and write the answer above the like term –15 in the dividend. Multiply each term of the divisor x 2 + x – 12 by the newly placed 2 and write the opposite of the results below the corresponding like terms of the dividend. The polynomial below the horizontal line has degree 1, which is less than the degree of the divisor, so the long division process is complete: You’re always asking, “The divisor term with the highest degree times WHAT equals the dividend term with the highest degree?” In later steps it changes to, “The divisor term with the highest degree times WHAT equals the term below the horizontal line with the highest degree?” . The Humongous Book of Algebra Problems 249 Chapter Eleven — Polynomials 11.33 Calculate (x 3 + 4x 2 – 7x + 9) ÷ (x + 1) using long division. Apply the technique described in Problems 11.29–11.32. Therefore, When a divisor “divides evenly” into a dividend, the final answer has no remainder. That also means x + 5 is a FACTOR of x3 + x2 – 32x – 60. More on that in Chapter 12 when factoring is covered in depth. . 11.34 Use long division to demonstrate that x + 5 divides evenly into x 3 + x 2 – 32x – 60. Divide x 3 + x 2 – 32x – 60 by x + 5. Because the remainder is 0, x + 5 divides evenly into x 3 + x 2 – 32x – 60. 11.35 Calculate (2x 3 – 4x + 2) ÷ (x – 6) using long division. Note that the dividend does not contain an x 2-term. To preserve the polynomial long division technique, insert the placeholder term 0x 2 into the dividend. If the highest power in the dividend is x3, then you need all the smaller powers of x in there as well: x2 , x, and a constant. If any of those things are missing, you need to put them in with a 0 coefficient to make sure things line up properly. ngous Book of Algebra Problems 250 The Humo Chapter Eleven — Polynomials Therefore, . 11.36 Calculate (5x 3 + 6x 2 – 3) ÷ (x 2 – 1) using long division. Both the dividend and the divisor are missing powers of x. Neither contains an x 1-term, so insert the placeholder 0x into each one before dividing. Therefore, . Synthetic Division of Polynomials Divide using only the coefficients 11.37 Explain why synthetic division cannot be used to calculate (x3 + 1) ÷ (x 2 + 1). Synthetic division, a technique used to divide polynomials that is simpler and more compact than long division, can only be applied when the divisor is a linear binomial. In this problem, the divisor x 2 + 1 is a quadratic binomial, so long division must be used to calculate the quotient. 11.38 Calculate (x 2 – 4x – 12) ÷ (x – 6) using synthetic division. It’s a quadratic because the highest power of x is 2. It’s a binomial because there are two separate terms. List the coefficients of the dividend in a horizontal row. Note that x 2 does not have an explicitly stated coefficient, so the implied coefficient is 1. Beneath the coefficients, draw a horizontal line. To the left of the numbers, list the opposite of the constant in the divisor. Separate it from the coefficients using a box, as demonstrated below. Drop the leftmost coefficient, 1, below the horizontal line. The constant in the divisor x – 6 is –6. Take the opposite of that number and put it here. The Humongous Book of Algebra Problems 251 Chapter Eleven — Polynomials Multiply the boxed number (6) by the number below the horizontal line (1) and write the result (1 · 6 = 6) between the next coefficient (–4) and the horizontal line. Add –4 and 6 (–4 + 6 = 2) and write the sum in the same column as those values, below the horizontal line. Multiply the boxed number (6) by the newly placed value beneath the horizontal line (2) and write the result (6 · 2 = 12) between the final coefficient (–12) and the horizontal line. Combine the numbers in the rightmost column (–12 + 12 = 0) and write the result in the same column, below the horizontal line. The dividend is x2 – 4x – 12, which has degree two (because the highest power of x is two). The degree of the quotient is one less, so its first term has degree one: x1. The numbers below the horizontal line represent the coefficients of the quotient and the remainder. Note that the degree of the quotient is always one less than the degree of the dividend, so this quotient has degree one: 1x + 2, or x + 2. The rightmost number below the horizontal line represents the remainder, which is 0 for this problem. Therefore, (x 2 – 4x – 12) ÷ (x – 6) = x + 2. 11.39 Calculate (3x 2 + 5x – 1) ÷ (x + 4) using synthetic division. The divisor is x + 4, so write –4 (the opposite of the constant) in a box left of the coefficients. Place the coefficients of the dividend in a horizontal row and the opposite of the divisor’s constant to the left of those values, separated by a box. Beneath the row of numbers, draw a horizontal line. Drop the first coefficient (3) below the horizontal line, multiply it by the boxed number (–4), and write the result (–4 · 3 = –12) between the next coefficient (5) and the horizontal line. ngous Book of Algebra Problems 252 The Humo Chapter Eleven — Polynomials Add the values in the second column of coefficients (5 – 12 = –7), and write the result in the same column below the horizontal line. Multiply the newly placed value (–7) by the boxed number (–4), write the product (–4 · –7 = 28) between the final coefficient value (–1) and the horizontal line, and add the numbers in that column (–1 + 28 = 27). The degree of the quotient is one (exactly one less than the degree of the dividend, which is two), so the numbers below the horizontal line, from left to right, represent the coefficient of x 1 (3), a constant (–7), and the remainder (27). 11.40 Calculate (2x – 9x – 51x – 40) ÷ (x – 8) using synthetic division. 3 2 Apply the synthetic division technique described in Problems 11.38–11.39. The degree of the quotient is two, which is one less than the degree of the dividend. Therefore, the numbers below the horizontal line represent, from left to right, the coefficient of x 2, the coefficient of x, a constant, and the remainder. Therefore, (2x 3 – 9x 2 – 51x – 40) ÷ (x – 8) = 2x 2 + 7x + 5. 11.41 Calculate (5x 4 + 9x 3 – 3x 2 – x – 10) ÷ (x + 3) using synthetic division. RU LE OF TH UMB: Multiply the numbers below the horizontal line by the number in the box and write the result above the horizontal line in the next column. Add the numbers in that column, write the answer below the horizontal line, and repeat. Rewrite the expression as a synthetic division problem to calculate the values of the quotient and remainder. The Humongous Book of Algebra Problems 253 Chapter Eleven — Polynomials The degree of the quotient is 3, one less than the degree of the dividend. Divide the remainder 128 by the divisor, x + 3. 11.42 According to Problem 11.34, (x 3 + x 2 – 32x – 60) ÷ (x + 5) = x 2 – 4x – 12. Verify the quotient using synthetic division. Rewrite the expression as a synthetic division problem to calculate the values of the quotient and remainder. Therefore, (x 3 + x 2 – 32x – 60) ÷ (x + 5) = x 2 – 4x – 12. 11.43 According to Problem 11.35, . Verify the solution using synthetic division. The dividend is missing an x2-ter m, so just like you ha ve to include 0x2 to long divide in Problem 11 you have to includ .35, e the 0 coefficien t to divide synthetica lly. Rewrite the expression as a synthetic division problem. Therefore, . 11.44 Calculate the value of c in the expression (x 2 – 9x + c) ÷ (x + 2) if the remainder is 26. Use synthetic division to calculate (x 2 – 9x + c) ÷ (x + 2). Set the remainder calculated above (c – 14) equal to the stated remainder (26) and solve for c. ngous Book of Algebra Problems 254 The Humo Chapter Eleven — Polynomials 11.45 Calculate the value of a in the expression x 3 + ax 2 – 61x + 14 if x + 7 divides evenly into the cubic polynomial. Apply synthetic division to calculate (x 3 + ax 2 – 61x + 14) ÷ (x + 7). If x + 7 divides evenly into x 3 + ax 2 – 61x + 14, then the remainder (49a + 98) is equal to 0. Express this using an equation and solve it for a. When you multiply a – 7 by –7 (the number in the box), you get –7(a – 7) = –7a – 7(–7) = –7a + 49. Add that to –61 by combining the constants: –61 + (–7a + 49) = –7a + (–61 + 49) = –7a – 12 . The Humongous Book of Algebra Problems 255 Chapter 12 Factoring Polynomials The opposite of multiplying polynomials This chapter explores factorization, first of integers and then of polynomials. The process of factorization is the reduction of a quantity into a series of prime components, the product of which is the original quantity. Factoring is basically “unmultiplying.” In other words, you can multiply 3 ˙ 5 to get 15, and you can factor 15 to get 3 ˙ 5. You start with numbers, identifying prime factoriza tions of integers—the unique string of prime number factors that exist s for any integers. You’ll then move on to greatest common factors of two monomials. After that, you’ll factor binomials an d trinomials, using procedures called “factoring by grouping” and “factoring by decomposition.” There ’s also a tiny bit of memorization involv ed, formulas that help you factor differences of perfect squares an d the sums and differences of perfect cubes. Chapter Twelve — Factoring Polynomials Natural numbers are positive numbers with no fractions or decimals: 1, 2, 3, 4, 5, 6, and so on. Prime numbers aren’t divisible by any numbers except themselves and 1. (And 1 divides into everything, so that’s no big deal.) 2, 3, 5, 7, and 11 are prime numbers but 4 is not (because it’s divisible by 2) and 100 isn’t prime (because it’s divisible by 2, 4, 5, 10, and a bunch of other numbers as well). In Problem 12 .2, those numbers happened to be prime, but they don’t have to be in the first step. In fact, they usually won’t be. You’ll just factor the ones that aren’t until they’re all prime. 4 is not prime because you can divide it by 2: 4 ÷ 2 = 2. That means 2 ˙ 2 = 4. Write (2 ˙ 2) where 4 was in the factorization of 20. Greatest Common Factors Largest factor that divides into everything evenly 12.1 What does the fundamental theorem of arithmetic guarantee? According to the fundamental theorem of arithmetic, all natural numbers greater than 1 can be expressed as a unique product of prime numbers—no two natural numbers have the same set of prime factors. 12.2 Generate the prime factorization of 21. Only two natural numbers have a product of 21: 3 and 7. Therefore, the prime factorization of 21 is 3 · 7. Note: Problems 12.3–12.4 demonstrate that a number has the same prime factorization regardless of the initial pair of products chosen. 12.3 Generate the prime factorization of 20, given 20 = 4 · 5. To generate a prime factorization, begin by expressing the number to be factored as a product of two natural numbers. This problem directs you to begin with the product 4 · 5. 20 = 4 · 5 Your goal is to express 20 as a product of prime numbers. Whereas 5 is a prime number, 4 is not, so you should express 4 as a product of natural numbers, much like 20 was expressed as the product 4 · 5 one step earlier. 20 = (2 · 2) · 5 Because 2 and 5 are prime numbers, 2 · 2 · 5 is the prime factorization of 20. Complete the problem by expressing 2 · 2 in exponential notation (2 · 2 = 22). 20 = 22 · 5 Note: Problems 12.3–12.4 demonstrate that a number has the same prime factorization regardless of the initial pair of natural numbers chosen. 12.4 Generate the prime factorization of 20, given 20 = 2 · 10. Of the given factors, 2 is prime but 10 is not. Note that 10 is divisible by 5 (10 ÷ 5 = 2). Express 10 as a product of natural numbers. ngous Book of Algebra Problems 258 The Humo Chapter Twelve — Factoring Polynomials Factors 2 and 5 are prime, so the prime factorization of 20 is 2 · 2 · 5. As directed in Problem 12.3, write the repeated factor using exponential notation. 20 = 22 · 5 This prime factorization exactly matches the prime factorization generated by Problem 12.3. The initial natural numbers chosen to factor 20 (whether 4 · 5 or 2 · 10) had no effect on the final answer. Note: Problems 12.5–12.6 demonstrate that a number has the same prime factorization regardless of the initial pair of natural numbers chosen. The number 2 appears twice in the factorization, so write it once but raise it to the second power. 12.5 Generate the prime factorization of 48 given 6 · 8 = 48. Neither of the initial factors (neither 6 nor 8) are prime, so both should be expressed as the product of natural numbers. There is only one way to factor each into a pair of natural numbers: 2 · 3 = 6 and 2 · 4 = 8. Of the three factors (2, 3, and 4), only 4 is not prime. Express it as a product: 2 · 2 = 4. Well, you could write 6 as 3 2, but that’s no˙t re ally different. It’s just the same numbers in a different order. Both of the remaining factors (2 and 3) are prime. Use exponential notation to indicate that the factor 2 appears four times. 48 = 24 · 3 Note: Problems 12.5–12.6 demonstrate that a number has the same prime factorization regardless of the initial pair of natural numbers chosen. 12.6 Generate the prime factorization of 48 given 4 · 12 = 48. Neither 4 nor 12 is prime. Express each as a product of natural numbers. Note that 12 can be expressed either as 2 · 6 or 3 · 4, but both will result in the same prime factorization. The Humongous Book of Algebra Problems 259 Chapter Twelve — Factoring Polynomials “Composite” is the opposite of prime. The composite factor in 2 ˙ 2 ˙ 3 ˙ 4 is 4. One composite factor remains. Express it as a product of natural numbers. This prime factorization exactly matches the factorization generated by Problem 12.5, despite the fact that the problems began with different natural number factors. Note: Problems 12.7–12.9 illustrate the steps used to calculate the greatest common factor of 64 and 112. 12.7 Generate the prime factorization of 64. Because 64 is an even number, it is divisible by 2: 64 ÷ 2 = 32. Therefore, 64 = 2 · 32. 64 = 2 · 32 The number 32 is composite, so express it as a product. 64 = 2 · (4 · 8) Neither 4 nor 8 is prime, so factor each: 4 = 2 · 2 and 8 = 2 · 4. Only one composite factor remains: 4. Note: Problems 12.7–12.9 illustrate the steps used to calculate the greatest common factor of 64 and 112. Here, “factor” means “write 56 as two natural numbers multiplied together.” You could use 8 ˙ 7 instead of the product used by the book, 4 ˙ 14. 12.8 Generate the prime factorization of 112. Notice that 112 is an even number, so it is divisible by 2: 112 ÷ 2 = 56. 112 = 2 · 56 Factor the composite number 56. ngous Book of Algebra Problems 260 The Humo 112 = 2 · (4 · 14) Chapter Twelve — Factoring Polynomials Factor the newly introduced composite numbers, 4 and 14. Note: Problems 12.7–12.9 illustrate the steps used to calculate the greatest common factor of 64 and 112. 12.9 Identify the greatest common factor of 64 and 112. According to Problems 12.7 and 12.8, 64 = 26 and 112 = 24 · 7. To construct the greatest common factor identify the common factors shared by 64 and 112. The number 112 has two different prime factors, 2 and 7, but 64 has only a single prime factor, 2. Thus the only factor 64 and 112 have in common is 2. In the factorization of 64, 2 is raised to the sixth power, and in the factorization of 112, 2 is raised to the fourth power. The greatest common factor is 2 raised to the lower of those powers: 24. Therefore, the greatest common factor of 64 and 112 is 24 = 16. Note: Problems 12.10–12.11 illustrate the steps used to calculate the greatest common factor of 36 and 90. 12.10 Generate the prime factorizations of 36 and 90. Write each number as a product of natural numbers and factor all of the resulting composite numbers. The greatest co m factor has to div mon ide into both numbers evenly, so power means tha choosing the lower ta out if you divide ll of the 2s will cancel both numbers by 4 2: The Humongous Book of Algebra Problems 261 Chapter Twelve — Factoring Polynomials Both factorizations contain 3 2, so technically there is no “lower power” of 3; they’re both equal. When this happens, use the matching power: 32 . Note: Problems 12.10–12.11 illustrate the steps used to calculate the greatest common factor of 36 and 90. 12.11 Identify the greatest common factor of 36 and 90. According to Problem 12.10, 36 and 90 have two factors in common, 2 and 3. The prime factorization of 36 contains 22 and 32; the prime factorization of 90 contains 21 and 32. Choose the lower power of each to include in the greatest common factor (GCF). GCF of 36 and 90: 21 · 32 = 2 · 9 = 18 12.12 Identify the greatest common factor of 9x 2 and 15x 3. Generate the prime factorizations of coefficients 9 and 15. As you factor the numbers, do not alter the variable expressions. Instead of 32 9x 2 and 15x 3 have two factors in common: 3 and x. The greatest common factor (GCF) will include the lower power of both factors, 31 and x 2 . GCF of 9x 2 and 15x 3 : 31 · x 2 = 3x 2. Instead of x3 Note: Problems 12.13–12.14 present the steps used to factor the expression 18xy4 + 27x3y2. 12.13 Identify the greatest common factor of 18xy4 and 27x 3y 2. Generate the prime factorizations of coefficients 18 and 27. 18xy4 and 27x 3y 2 have three factors in common: 3, x, and y. Choose the lower power of each from the factorizations to calculate the greatest common factor (GCF). GCF of 18xy 4 and 27x 3y 2 : 32 · x1 · y 2 = 9xy 2 Note: Problems 12.13–12.14 present the steps used to factor the expression 18xy4 + 27x3y2. 12.14 Factor the greatest common factor out of the expression and verify your answer. According to Problem 12.13, the greatest common factor of 18xy4 and 27x 3y2 is 9xy 2. Divide each term of the expression by 9xy 2. ngous Book of Algebra Problems 262 The Humo Chapter Twelve — Factoring Polynomials Write each term as a product of the greatest common factor and the quotients calculated above. Rewrite the expression as the greatest common factor 9xy 2 times the sum of the remaining factors of each term. 18xy4 + 27x 3y2 = 9xy 2(2y 2 + 3x 2) Verify that 9xy 2(2y 2 + 3x 2) is correct by applying the distributive property. A number raised to the zero power equals one, so x0 = y0 = 1. In other words, pull 9xy2 out of ea term. You’re left ch w 2y2 + 3x2. Put th ith ose leftovers in pare nthe and multiply by th ses e greatest common factor: 9xy2(2y2 + 3x2). The result is equal to the original expression, so the factored form of the expression is correct. Note: Problems 12.15–12.16 present the steps used to factor the expression 40x3y5z7 + 12x2y3z3 + 36x4yz6. 12.15 Identify the greatest common factor of 40x 3y 5z7, 12x 2y 3z 3, and 36x 4yz 6. Generate the prime factorizations of each term. All three terms have four factors in common: 2, x, y, and z. Raise each common factor to its lowest power in all three factorizations to calculate the greatest common factor (GCF). GCF: 22 · x 2 · y1 · z3 = 4x 2yz 3 12x2y3z3 and 36x4yz6 both contain the prime factor 3, but 40x3y5z7 does not, so 3 doesn’t appear in the greatest common factor. Here are all the powers of the common factors, with the lowest of each circled: The Humongous Book of Algebra Problems 263 Chapter Twelve — Factoring Polynomials Note: Problems 12.15–12.16 present the steps used to factor the expression 40x3y5z7 + 12x2y3z3 + 36x4 yz6. 12.16 Factor the greatest common factor out of the expression. According to Problem 12.15, the greatest common factor is 4x 2yz 3. Divide this quantity into each term of the expression. Factor 4x 2yz 3 out of the expression by multiplying it by the sum of the quotients listed above. 40x 3y 5z7 + 12x 2y 3z 3 + 36x 4yz 6 = 4x 2yz 3(10xy4z 4 + 3y 2 + 9x 2z 3) 12.17 Factor the expression: 16x 2y 3 – 12xy 2. Use the technique described in Problems 12.1–12.12 to calculate the greatest common factor of the expression: 4xy 2. Divide each term of the expression by 4xy 2. Distribute 4xy2 to check your answer: 4xy2(4xy) + 4xy2(–3) = 16x2y3 – 12xy2 Factor the expression by writing it as the product of the greatest common factor and the quotients above. 16x 2y 3 – 12xy 2 = 4xy 2(4xy – 3) 12.18 Factor the expression: –4x 7y3 – 20x 5y11. The greatest common factor is negative because both of the terms are negative—both are multiplied by –1, so you can factor that out as well. Divide each term of the expression by the greatest common factor: –4x 5y 3. Factor the expression by writing it as the product of the greatest common factor and the preceding quotients. –4x 7y 3 – 20x 5y11 = –4x 5y3(x 2 + 5y 8) ngous Book of Algebra Problems 264 The Humo Chapter Twelve — Factoring Polynomials 12.19 Factor the expression: 2(x + 1) + 3y(x + 1). This expression consists of two terms, 2(x + 1) and 3y(x + 1). Both terms have one common factor, the binomial (x + 1), so the greatest common factor of the expression is x + 1. Divide both terms by that quantity. Factor the expression by writing it as the product of the greatest common factor and the preceding quotients. 2(x + 1) + 3y(x + 1) = (x + 1)(2 + 3y) Factoring by Grouping You can factor out binomials, too If the expression was 2w + 3yw, it’d be easy to tell that w was the greatest common factor; just plug in (x + 1) where w is and you get Problem 12.19. Common factors don’t have to be monomials. They can be binomials like this (or trinomials, or fractions, or square roots… the list goes on and on). 12.20 Factor the expression: 9x(2y – 1) + 8y – 4. This expression consists of three terms: 9x(2y – 1) is the first term, 8y is the second, and –4 is the third. Notice that the first term is expressed as a product. The greatest common factor of the second and third terms (8y and –4) is 4. Factor it out of the expression 8y – 4 to get 4(2y – 1). Rewrite the expression using the newly factored form of the second and third terms. The expression contains two terms with a binomial greatest common factor: 2y – 1. Factor the expression using the method described in Problem 12.19. 9x(2y – 1) + 4(2y – 1) = (2y – 1)(9x + 4) 12.21 Factor the expression by grouping: 3x 3 + 15x 2 + 2x + 10. To factor by grouping, group together the polynomial terms that have common factors. In this problem, 3x 3 and 15x 2 have common factors, and 2x and 10 share a common factor as well. (3x 3 + 15x 2) + (2x + 10) The greatest common factor of the left group of terms is 3x 2 ; the greatest common factor of the right group is 2. Write each group in factored form. When you factor (2y – 1) out of the expression, you’re left with 9x in the first term and 4 in the second. Add those values inside parentheses (9x + 4) and multiply by the greatest common factor: (9x + 4)(2y – 1). The terms you end up grouping are usually next to each other. 3x 2(x + 5) + 2(x + 5) The Humongous Book of Algebra Problems 265 Chapter Twelve — Factoring Polynomials The expression now contains two terms with a common factor: x + 5. Factor it out. (x + 5)(3x 2 + 2) Therefore, the factored form of 3x 3 + 15x 2 + 2x + 10 is (x + 5)(3x 2 + 2). 12.22 Factor by grouping: 14xy + 21x + 8y + 12. The first two terms share the common factor 7x; the last two terms share the common factor 4. (14xy + 21x) + (8y + 12) Factor each group. 7x(2y + 3) + 4(2y + 3) Factor the binomial 2y + 3 out of both terms. (2y + 3)(7x + 4) The factored form of 14xy + 21x + 8y + 12 is (2y + 3)(7x + 4). 12.23 Factor by grouping: 3x 3 – 12x 2 + x – 4. The first two terms have a greatest common factor of 3x 2; group them together. (3x 3 – 12x 2) + x – 4 Factor the grouped expression. 3x 2(x – 4) + x – 4 The only factor common to the last two terms of the expression (x and –4) is 1. Factor it out. 3x 2(x – 4) + 1(x – 4) Factor the binomial x – 4 out of both terms. (x – 4)(3x 2 + 1) Therefore, the factored form of 3x 3 – 12x 2 + x – 4 is (x – 4)(3x 2 + 1). 12.24 Factor the expression by grouping: 4x 5 – 8x 3 – 5x 2 + 10. You can tell the binomials are oppo si because the sign tes s of the correspondin g terms are opposites: x2 and –x2 are opposites, as are –2 and +2 The first two terms share common factor 4x 3 and the last two terms share common factor 5. Unlike Problems 12.20–12.23, once factored, these groups do not contain a common binomial term. In fact, the binomials x 2 – 2 and –x 2 + 2 are opposites. To apply the factoring by grouping technique, the binomials must be equal. Satisfy this requirement by factoring –1 out of the second group. ngous Book of Algebra Problems 266 The Humo Chapter Twelve — Factoring Polynomials Factor the binomial x 2 – 2 out of the expression. (x 2 – 2)(4x 3 – 5) Therefore, the factored form of 4x 5 – 8x 3 – 5x 2 + 10 is (x 2 – 2)(4x 3 – 5). 12.25 Factor the expression: 96x 4 + 36x 3 – 160x – 60. In Problem 12.23, factoring out a 1 didn’t change the binomial. When you factor out a –1 in this problem, it changes every term in the parentheses into its opposite. Before you factor by grouping, note that all of the terms share the common factor 4. Factor it out of the expression. 4(24x 4 + 9x 3 – 40x – 15) The first two terms of the quartic polynomial share common factor 3x 3 ; the last two terms share common factor –5. Factor the common binomial out of the expression. “Quartic” means the highest power of x in the polynomial is 4. Leave this 4 (that got factored out in the previous st ep) in front of the expression. Common Factor Patterns Difference of perfect squares/cubes, sum of perfect cubes 12.26 Factor the expression: a 2 – b 2. The expression a 2 – b 2 is a difference of perfect squares. In other words, the expression consists of one squared quantity (b2 ) subtracted from another (a2 ). All differences of perfect squares are factored using the formula (a 2 – b 2) = (a + b)(a – b). 12.27 Factor the expression: w 2 – 36. Notice that w 2 and 36 are perfect squares, as each is equal to some quantity times itself: w · w = w 2 and 6 · 6 = 36. w 2 – 36 = w 2 – 62 According to Problem 12.26, a polynomial of the form a 2 – b 2 has factors (a + b)(a – b). Here, a = w and b = 6. A perfect square is a number you get by multiplying something times itself. That means 4 is a perfect square because 2 times itself equals 4: 2 ˙ 2 = 22 = 4. A DIFFERENCE of perfect squares just means subtraction is involved. The Humongous Book of Algebra Problems 267 Chapter Twelve — Factoring Polynomials The answer (w – 6)(w + 6) is also correct. The order of the factors doesn’t matter. “a” is the thing you multiply times itself to get the left term. In this case a = 2x because (2x) 2 = 4x2 . “b” is the thing you multiply times itself to get the term that’s subtracted. Therefore, the factored form of w 2 – 36 is (w + 6)(w – 6). 12.28 Factor the expression: 4x 2 – 81. This binomial is the difference of perfect squares: (2x)(2x) = 4x 2 and (9)(9) = 81. Therefore, it should be factored using the formula a 2 – b 2 = (a + b)(a – b). If a 2 – b 2 = 4x 2 – 81, then a = 2x and b = 9. 12.29 Factor the expression: 98x 2 – 200. Both terms are divisible by 2, so factor that value out of the expression. 2(49x 2 – 100) Always check to see if there is a greatest common factor before yo u try to factor an expression any ot he way. In this proble r m if you don’t fact , or the 2 out first, yo u don’t end up with a difference of perfect squares. The resulting binomial, 49x 2 – 100, is a difference of perfect squares: (7x)2 = 49x 2 and 102 = 100. 2(7x + 10)(7x – 10) 12.30 Factor the expression: 4x 4 – 64. The terms of this expression have a greatest common factor of 4, so factor 4 out of the polynomial. 4(x 4 – 16) The binomial x 4 – 16 is a difference of perfect squares: (x 2)2 = x 4 and 42 = 16. 4(x 2 + 4)(x 2 – 4) The expression is not yet fully factored, because x 2 – 4 is, itself, a difference of perfect squares: (x 2 – 4) = (x + 2)(x – 2). Substitute the factored form of x 2 – 4 into the polynomial. 4x 4 – 64 = 4(x 2 + 4)(x + 2)(x – 2) After factorin difference of pe g the it turns out one of rfect squares, the factors is ALSO a differenc e of perfect squares. You can’t factor 2 x + 4, because there’s no formula for the perfect squares. SUM of ngous Book of Algebra Problems 268 The Humo Chapter Twelve — Factoring Polynomials 12.31 Identify the formulas used to factor a 3 + b 3 (the sum of perfect cubes) and a 3 – b 3 (the difference of perfect cubes). The formulas for the sum and difference of perfect cubes are nearly identical; two signs in the factors constitute the only differences. a 3 + b 3 = (a + b)(a 2 – ab + b 2) a 3 – b 3 = (a – b)(a 2 + ab + b 2) Both expressions factor into the product of a binomial and a trinomial. The sign of b in the binomial matches the sign of b 3 in the unfactored expression, and the sign of ab in the trinomial factor is always its opposite. 12.32 Factor the expression: x 3 + 64. This polynomial is the sum of perfect cubes because each term can be generated by raising some quantity to the third power (that is, cubing it): (x)3 = x 3 and 4 · 4 · 4 = 43 = 64. Apply the formula from Problem 12.31, setting a = x and b = 4. 12.33 Factor the expression: 24x 3 – 375. Both terms share the common factor 3, so factor it out of the polynomial. 3(8x 3 – 125) The resulting cubic binomial is a difference of perfect cubes: (2x)3 = 8x 3 and 53 = 125. Apply the formula from Problem 12.31, setting a = 2x and b = 5. Substitute the factored form of 8x 3 – 125 into 3(8x 3 – 125). Don’t forget that the first step of this problem was to factor 3 out of 24x3 – 375, which produced the difference of perfect cubes 8x3 – 125. That 3 has to be a factor in the final answer. 3(2x – 5)(4x 2 + 10x + 25) The Humongous Book of Algebra Problems 269 Chapter Twelve — Factoring Polynomials 12.34 Factor the expression: x 6 – 1. Express the terms of the polynomials as perfect squares: (x 3)2 = x 6 and 12 = 1. Apply the difference of perfect squares formula. You could factor x 6 – 1 using the differe nc of perfect cube e s formula, becaus e (x2) 3 = x 6 and 13 = 1. You’ll end up with (x + 1)(x – 1)(x 4 + 2 x + 1). There’s no easy way to factor the tr inomial in there, though. T difference of pe he rfect squares formula give a better answer s with more factors. The resulting factors are a sum of perfect cubes (x 3 + 1) and a difference of perfect cubes (x 3 – 1). Apply the factoring formulas for each, setting a = x and b = 1. Factoring Quadratic Trinomials Turn one trinomial into two binomials 12.35 Factor the expression: x 2 + 3x + 2. In other words, when you add the numbers, you get the coefficient of x in the trinomial, and when you multiply them, you get the constant. This works only when the coefficient of x2 is 1 (like in this problem). The quadratic binomial x 2 + ax + b, once factored, has the form below. Each of the empty boxes represents a number. Those missing values have sum a and product b. In this problem, a = 3 and b = 2, so the two missing values must have a sum of 3 and a product of 2. The correct values are 1 and 2, because 1 + 2 = 3 and 1 · 2 = 2. Substitute 1 and 2 into the boxes of the formula. x 2 + 3x + 2 = (x + 1)(x + 2) The order of the factors is irrelevant, so the answer (x + 2)(x + 1) is equally correct. 12.36 Factor the expression: x 2 + 8x + 12. To factor this quadratic binomial, you must identify two numbers with sum 8 and product 12. Begin by subtracting 1 from the required sum (8 – 1 = 7) to identify two numbers with the correct sum: 1 + 7 = 8. Unfortunately, 1 and 7 do not have the required product (1 · 7 ≠ 12). ngous Book of Algebra Problems 270 The Humo Chapter Twelve — Factoring Polynomials To continue the search for the correct values, subtract two from the required sum (8 – 2 = 6) to identify another possible pair of numbers: 2 + 6 = 8. The numbers 2 and 6 satisfy both requirements, the correct sum and the required product: 2 + 6 = 8 and 2(6) = 12. Substitute 2 and 6 into the boxes of the factoring formula. 12.37 Factor the expression: x 2 – 9x + 20. If that didn’t work, you’d subtract 3, then 4, and so on. From start to finish, you’d try 7 + 1, 6 + 2, 5 + 3, and 4 + 4 until you find the pair of numbers that multiplies to give you 12. The two numbers needed to factor this expression have a sum of –9 and a product of 20. Those values are only possible if both numbers are negative; two negative numbers have a negative sum and a positive product. Test a series of negative numbers that add up to –9 until you find a pair of values that equals 20 when multiplied: –8 – 1, –7 – 2, –6 – 3, and –5 – 4. Notice that –5 and –4 have the required sum and product: –5 + (–4) = –9 and –5(–4) = 20. Use those values to factor the polynomial. x 2 – 9x + 20 = (x – 5)(x – 4) 12.38 Factor the expression: x 2 – 16x + 64. Like Problem 12.37, the numbers required to factor the expression have a negative sum and a positive product, so both of the numbers must be negative. Test a series of sums that equal –16 until you find values that have a product of 64: –15 – 1, –14 – 2, –13 – 3, –12 – 4, –11 – 5, –10 – 6, –9 – 7, and –8 – 8. Because –8 + (–8) = –16 and –8(–8) = 64, –8 and –8 are the necessary values. Factor the expression. x 2 – 16x + 64 = (x – 8)(x – 8) Because the factor (x – 8) appears twice, you should use exponential notation. You don’t HAVE to write out this list, but it gives you a place to start if you can’t figure out what the two numbers are right away. x 2 – 16x + 64 = (x – 8)2 12.39 Factor the expression: x + 3x – 18. 2 To factor, you need to identify two numbers with a sum of 3 and a product of –18. Because the product is negative, the numbers must have different signs; the products of two positive numbers and the products of two negative numbers are positive. Because the sum is positive, the positive value must be larger than the negative value. Generate pairs of numbers with opposite signs that have a sum of 3 until you identify a pair with a product of –18: 4 – 1, 5 – 2, and 6 – 3. Because 6 – 3 = 3 and 6(–3) = –18, the values required to factor the expression are –3 and 6. x 2 + 3x – 18 = (x – 3)(x + 6) In other words, if you ignore the sign fo r moment, the posi a tive number will be gr ea than the negati ter ve. For example, +9 and –6 have a positive because the posi sum tive value 9 is greate r than the negative va lue 6. However, –9 and +6 have a negative sum because 6 < 9. The Humongous Book of Algebra Problems 271 Chapter Twelve — Factoring Polynomials So when you find the numbers, write them as coefficients of y in the factors. 12.40 Factor the expression: x 2 – 5xy – 36y2. Much like the polynomial x 2 + ax + b has factors , the polynomial x 2 + axy + by 2 has factors . To factor, you must identify two numbers that have a sum of –5 and a product of –36. Because the product is negative, the numbers have different signs. The sum is negative, so the larger of the two numbers is negative. Generate a list of number pairs that have a sum of –5 until you encounter numbers with a product of –36: –6 + 1, –7 + 2, –8 + 3, and –9 + 4. Notice that –9 + 4 = –5 and –9(4) = –36, so factor the polynomial by placing –9 and 4 into the boxes of the formula . x 2 – 5xy – 36y 2 = (x – 9y)(x + 4y) 12.41 Factor the expression: x 4 + 6x 2 + 9. Identify two numbers with sum 6 and product 9 and place them into the boxes of the formula to factor the polynomial x 4 + 6x 2 + 9. The sum and the product of the numbers are positive, so the numbers themselves are also positive. Generate a list of number pairs that sum to 6 until you identify a pair that has a product of 9: 5 + 1 = 6, 4 + 2 = 6, and 3 + 3 = 6. Notice that 3 + 3 = 6 and 3(3) = 9, so substitute 3 and 3 into the boxes of the factoring formula stated above. x 4 + 6x 2 + 9 = (x 2 + 3)(x 2 + 3) Use an exponent to indicate the presence of a repeated factor. x 4 + 6x 2 + 9 = (x 2 + 3)2 The numbers still add up to the x-coefficient like they did befo but now they don re, ’t multiply to give yo u the constant; their pr od is equal to the pr uct oduct of the constant a nd the coefficient of x 2. 12.42 Factor the expression: 2x 2 + 17x + 36. When the coefficient of x 2 is not 1, the factoring technique described in Problems 12.35–12.41 will not work. Instead, you should factor by decomposition. Like the method presented in the preceding problems, you must identify two numbers with specific characteristics. Given the polynomial ax 2 + bx + c, the two required numbers have sum b and product a · c. In this problem, the two required numbers have a sum of 17 and a product of 2 · 36 = 72. Both the sum and product are positive, so the individual numbers are positive as well. List pairs of numbers that have a sum of 17 until one pair has a product of 72 as well: 16 + 1, 15 + 2, 14 + 3, 13 + 4, 12 + 5, 11 + 6, 10 + 7, and 9 + 8. Notice that 9 + 8 = 17 and 9(8) = 72. ngous Book of Algebra Problems 272 The Humo Chapter Twelve — Factoring Polynomials When factoring by decomposition, you do not immediately place the required numbers into factors. Instead, rewrite the x-coefficient as a sum. According to the commutative property, the order in which you multiply does not affect the product. Therefore, (8 + 9)x and x(8 + 9) are equivalent. Apply the distributive property: x(8 + 9) = 8x + 9x. 2x 2 + 8x + 9x + 36 Factor by grouping, as explained in Problems 12.20–12.25. Therefore, the factored form of 2x 2 + 17x + 36 is (x + 4)(2x + 9). RU LE OF TH UMB: 2 When the x coefficient is 1, finding the two “mystery” numbers means you’re done factoring. However, 2 when the x coefficeint is NOT 1 (in this problem, it’s 2), you replace the xcoefficient with the sum of the mystery numbers and then factor by grouping. 12.43 Factor the expression: 8x 2 – 2xy – 3y 2. Identify two numbers with a sum of –2 and a product of –24. List pairs of numbers with a sum of –2 until one pair also has a product of –24: –3 + 1, –4 + 2, –5 + 3, and –6 + 4. Notice that –6 + 4 = –2 and –6(4) = –24. Replace the x-coefficient with the sum of –6 and 4. Apply the distributive property: (–6 + 4)xy = –6xy + 4xy. 8x 2 – 6xy + 4xy – 3y 2 Factor by grouping. When you multipl the “mystery” y numbers, you shou get the same re ld as multiplying th sult2 e coefficient (8) a x nd the constant (–3). This sign is always positive, no matter what’s in parentheses. Therefore, the factored form of 8x 2 – 2xy – 3y 2 is (4x – 3y)(2x + y). The Humongous Book of Algebra Problems 273 Chapter Twelve — Factoring Polynomials The product is positive, so both numbers have to have the same sign. You also know that the sum is negative, which means the numbers you’re looking for must both be negative. 12.44 Factor the expression: 6x 2 – 41x + 30. In order to get matching (x – 6) binomials, you have to factor –5 out of (–5x + 30) instead of +5. If you don’t understand why, look at Problem 12 .24. 274 To factor by decomposition, list pairs of numbers that have a sum of –41 until one pair also has a product of 6 · 30 = 180: –40 – 1, –39 – 2, –38 – 3, –37 – 4, and –36 – 5. Notice that –36(–5) = 180, so replace the x-coefficient with the sum of –36 and –5. Apply the distributive property and factor by grouping. You can check this (or any other factoring answer) by multiplying the factors to make sure you get the original polynomial back: The Humongous Book of Algebra Problems Chapter 13 Radical Expressions and Equations tional exponents Square roots, cube roots, and frac One of the most important characteristics of the polynomials discussed in Chapters 11 and 12 is the exponential powers to which variables in the terms are raised. This chapter explores roots, written as both rational exponential powers and radical expressions. It begins with the simplification of roots, progresses to operations on (and equations involving) roots, and culminates with a discussion of complex and imaginary numbers and their relationship to negative radicands. You’ve probably heard of square roo ts, which look like this: . A square root answers the question, “W hat times itself gives you the number inside the radical sign?” (Th e radical sign is the symbol that looks like a check mark.) It’s basicall y the opposite of squaring somethin g. Squaring 3 gives you 9: 32 = 9, so the square ROOT of 9 is 3: . Square roots aren’t the only kind of roots out there, as you see in this chapter. After you figure out how to simplify radicals, it’s time to start using them in expressions and equations, which means you’ll learn how to add, subtract, multiply, and divide them. Finally, you’ll look at complex numbers, which answer the question , “What times itself could possibly equal a negative number? ” Chapter Thirteen — Radical Expressions and Equations Simplifying Radical Expressions Moving things out from under the radical Note: Problems 13.1–13.2 refer to the radical expression . 13.1 Identify the radicand and index of the expression. The index of a radical expression is the small number outside and to the left of the radical symbol; the index of is 3. The value inside the radical symbol is called the radicand; the radicand of is 64. Radicals with index 3 are called “cube roots,” just like raising something to the third power is called “cubing” it. Radicals with index 2 are called “square roots,” just like raising something to the second power is called “squaring” it. The 4 inside the radical symbol has exponent 3, and the index of the radical (that tiny number floating out front) is also 3. When the power and the index match, they sort of cancel each other out. The 3 exponent goes away, the index goes away, and the radical sign disappears, too. Note: Problems 13.1–13.2 refer to the radical expression . 13.2 Simplify the expression and verify your answer. As explained in Problem 13.1, the index of is 3. To simplify the cube root, try to identify factors of the radicand that are perfect cubes. It is helpful to memorize the first six perfect cubes, listed here. Notice that 64 is a perfect cube, as 43 = 64. Rewrite the radical expression, identifying the perfect cube explicitly. Any factor of the radicand that is raised to an exponent equal to the index of the radical can be removed from the radical. Therefore, . To verify this solution, raise the answer (4) to an exponent equal to the index of the original radical (3). The result should be the original radicand (64). 43 = 64 Note: Problems 13.3–13.4 refer to the radical expression . 13.3 Identify the index and radicand of the expression. The radicand of is the number inside the radical symbol: 36. No index is written explicitly, which indicates an implied index of 2. Square roots are rarely written with an explicit index, so the notation is preferred to . ngous Book of Algebra Problems 276 The Humo When there’s no little number nestled in the chec k mark outside the radical, that me ans you’re dealing with a squar e root, and you can assume the ind ex is 2. Chapter Thirteen — Radical Expressions and Equations Note: Problems 13.3–13.4 refer to the radical expression . 13.4 Simplify the expression and verify your answer. According to Problem 13.3, the index of is 2. To simplify the square root, factor the radicand using perfect squares. It is helpful to memorize the first 15 perfect squares, listed here. Notice that 36 is a perfect square, as 62 = 36. Rewrite the radical expression, identifying the perfect square explicitly. As stated in Problem 13.2, any factor in the radicand that is raised to a power equal to the index of the radical can be simplified. The index of the radical is 2, and the exponent of 6 is 2. Those 2’ s cancel out and ta ke the radical sign away with them. All that’s left is 6. To verify that , raise the answer (6) to a power that equals the index of the original radical (2). The answer must match the original radicand (36). 62 = 36 13.5 Simplify the radical expression: . The index of is not written explicitly, so according to Problem 13.3, the index is 2 and the radical expression is a square root. Consider the perfect squares listed in Problem 13.4 and determine whether any of those values divide evenly into the radicand. Because dividing 50 by 25 produces no remainder (50 ÷ 25 = 2), 25 is a factor of 50. Rewrite the radicand in factored form, explicitly identifying 25 as a perfect square. The root of a product is equal to the product of the roots. The exponent of 5 and the index of the radical expression are equal, so simplify the radical expression: . Therefore, . Don’t even bother with 12 = 1. Sure 1 is a perfect square, but it divides into every number evenly and is not all that useful for simplifying radicals. In other words, when two things are multiplied inside a root (a radical sy m you can break th bol), em into two separate roots that are multipl ied: . does NOT necess This ari work when two th ly in inside radicals a gs re added or subtra cted: . The Humongous Book of Algebra Problems 277 Chapter Thirteen — Radical Expressions and Equations 13.6 Simplify the radical expression: Don’t bother flipping back to Problem 13.2. Here’s the list: 23 = 8, 33 = 27, 4 3 = 64, 5 3 = 125, 63 = 216. . The index of the radical expression is 3, so to simplify the expression, you must identify factors that are perfect cubes. Consider the abbreviated list provided in Problem 13.2. Notice that 64 divides evenly into 256: 256 ÷ 64 = 4. Factor 256 and identify the perfect cube explicitly. The root of a product is equal to the product of the roots. Simplify , noting that the exponent of the radicand is equal to the index of the radical: . 13.7 Simplify the radical expression: There’s no way that 3 5 = 24 3 could be a factor of 96 because 24 3 is bi gg than 96. The sa er me goes for any num ber larger than 3 ra ised to the fifth power , so don’t even bother checking it. . The index of the radical is 5, so simplifying the expression requires you to identify factors with an exponent of 5. Notice that 25 = 32, which divides evenly into the radicand: 96 ÷ 32 = 3. Rewrite the expression, explicitly identifying the factor that is raised to the fifth power. 13.8 Simplify the radical expression: . The index of the radical is 3, so to simplify the expression, express x 6 as a quantity raised to the third power. To accomplish this, divide the exponent of x (6) by the index of the radical (3): 6 ÷ 3 = 2, so (x 2)3 = x 6. If you have a Now that the index of the radical and the exponent of the radicand are equal, hard time turning x6 into simplify the expression. (x2) 3 to get the exponent of 3 you need, break x6 into as many x3’s as you can: x6 = x3 · x3 (since x3 · x3 = x3 + 3 = x6). Now simplify: ngous Book of Algebra Problems 278 The Humo Chapter Thirteen — Radical Expressions and Equations 13.9 Simplify the expression: . To rewrite the radicand to include perfect cubes, divide each of the variables’ powers by 3, noting the resulting quotients and remainders. Begin with x 17: 17 ÷ 3 = 5 with remainder 2. Therefore, x 17 = (x 5)3 · x 2. Apply the same procedure to y4 : 4 ÷ 3 = 1 with remainder 1, so y4 = (y 3)1 · y1. Finally, express z 8 using perfect cubes: 8 ÷ 3 = 2 with remainder 2, so z 8 = (z 2)3 · z 2. Rewrite the radicand using the above factorizations. Separate the product into two radicals, one that contains factors raised to the third power and one that contains factors raised to other powers. To rewrite x17 as something to the third power, you divide 17 by 3. It goes in 5 times evenly: (x5) 3 = x15. You still have a remainder of 2, so multiply by x raised to the remainder power: (x5) 3 · x2 = x17. Simplify the radical containing the perfect cubes. 13.10 Simplify the expression: . The exponent of the radicand (2) is equal to the index of the radical (2) so the expression can be simplified: . Note the presence of absolute values, which were not required in the preceding problems of this chapter. If n is an even number, then . In other words, if a factor in the radicand is raised to an even power that is equal to the index of the radicand, when simplified, that variable expression must appear in absolute values. According to mathematical convention, a root with an even index must produce a positive number. This holds true for natural numbers as well. Whereas this can easily be ensured for natural numbers, absolute values are used to ensure that roots of variables are positive. 13.11 Simplify the expression: . Express the coefficient as a product that includes a perfect square: . Express x 9 using a perfect square by dividing its exponent by the index of the radical: 9 ÷ 2 = 4 with remainder 1, so x 9 = (x 4)2 · x 1. Apply the same procedure to y 3 : 3 ÷ 2 = 1 with remainder 1, so y 3 = (y1)2 · y1. RULE OF THUMB: If a radical has an even index, then simplifying it has to result in positive values. Even though 22 and (–2) 2 both equal 4, is correct, and . You don’t know if x is positive or negative in Problem 13.10, so use absolute values to make sure that the answer’s positive. Separate the radicand into two radicals, one that contains all of the perfect squares and another that contains the remaining factors. The Humongous Book of Algebra Problems 279 Chapter Thirteen — Radical Expressions and Equations Wondering why x 4 suddenly jumped out of those absolute value ba rs? x is raised to an even power, so x 4 will a lwa be positive, no ma ys tter what x equals. However, y is not raised to an even power, so it has to stay inside the absolute values. Simplify the radical containing the perfect squares. Note that and 4 because x and y are raised to an even power that matches the index of the radical. 13.12 Simplify the expression: . To simplify the coefficient, identify factors that are raised to the fourth power: . Divide the exponents of the variables by the index of the radical to express the variables as factors raised to the fourth power: 7 ÷ 4 = 1 with remainder 3, so x 7 = (x 1)4 · x 3 ; and 12 ÷ 4 = 3, so y12 = (y3)4. The factor z 3 cannot be rewritten because its power is less than the index. Try dividing 40 5 by a few natural numbers raised to the fourth power: 24 = 16, 34 = 81, 4 4 = 256, 5 4 = 625. Of those, only 81 divides evenly: 40 5 ÷ 81 = 5 so 34 · 5 = 40 5. Put x and y3 in absolute values because they’re raised to even powers that match the index of the radical. You can’t drop those absolute value bars because x1 and y3 have odd powers and could be negative. ngous Book of Algebra Problems 280 The Humo Chapter Thirteen — Radical Expressions and Equations Rational Exponents Fractional powers are radicals in disguise 13.13 Write xa/b as a radical expression. Assume that a and b are natural numbers. Rational exponents (that is, fractional exponents) indicate radical expressions. The denominator of the rational power (b in this problem) represents the index of the radical, and the numerator of the power represents either the exponential power of the radicand or the power to which the entire expression is raised: and . 13.14 Write 43/2 as a radical expression and simplify it. According to Problem 13.13, b = 2. You can use either formula to rewrite xa/b. You’ll probably use more often because xa tends to be a large number that’s hard to simplify. . In this problem, x = 4, a = 3, and 13.15 Write 2501/3 as a radical expression and simplify it. According to Problem 13.13, b = 3. . In this problem, x = 250, a = 1, and Notice that 125 is a perfect cube and a factor of 250. 13.16 Write as an exponential expression with base 7. The radicand 49 is equal to 72, so . According to Problem 13.13, . In this problem, x = 7, a = 2, and b = 5. The Humongous Book of Algebra Problems 281 Chapter Thirteen — Radical Expressions and Equations 13.17 According to Problem 13.1, . Verify the statement using rational exponents. Begin by writing Do this the same way Proble m 13.16 turned into an exponent ial expression with base 7. as an exponential expression with base 4. Rewrite the exponential expression using a rational exponent. 13.18 Write (12x 3y 2)1/2 as a radical expression and simplify. The entire quantity 12x 3y 2 is raised to the rational exponent radical expression with radicand 12x 3y 2 and index 2. , so construct a Simplify the radical expression by identifying perfect square factors. 13.19 Write (9x)1/2 y7/4 as a radical expression and simplify it. Note that 9 and x are both raised to the power. (9x)1/2 y 7/4 = 91/2 x 1/2 y 7/4 Rewrite the expression as a product of radical expressions. ngous Book of Algebra Problems 282 The Humo Chapter Thirteen — Radical Expressions and Equations To simplify , divide the power of y by the index of the radical: 7 ÷ 4 = 1 with remainder 3, so . Substitute into the expression. Group together the radicals of the expression. 13.20 Identify the value of x that makes the statement 16x/4 = 8 true. Rewrite the left side of the equation as a radical expression. In this case, is preferred to . The solution x answers the question, “Two raised to what power produces a value of 8?” The answer is x = 3. Radical Operations Add, subtract, multiply, and divide roots Solving the “nonpreferred” way to write the radical equation involves logarithms, which aren’t covered until Chapter 18. The preferred version is easier because x isn’t inside the radical. 13.21 What condition must be met by radical expressions in order to calculate a sum or difference? Radical expressions can be added or subtracted only when they contain “like radicals,” which means that the radicands and indices are equal. This requirement is similar to the “like terms” requirement placed on polynomial addition and subtraction, wherein terms are required to have equivalent variable expressions before they can be combined. 13.22 Simplify the expression: “Indices” is just the plural form of “index.” . Combine the coefficients of the like radicals to compute the sum: 2 – 5 + 14 = 11. You’re allowed to add these because they all contain the same radical expression: . The Humongous Book of Algebra Problems 283 Chapter Thirteen — Radical Expressions and Equations 13.23 Simplify the expression: . The radicals cannot be added because the radicands are not equal. However, can be simplified. The expression now consists of like radicals. Calculate the sum. You can subtract the coefficients because the radicals match: 2 – 3 = –1. Instead of writing a coefficient of –1, just use the negative sign by itself. 13.24 Simplify the expression: . The cube roots contain factors that are perfect cubes; simplify both. 13.25 Simplify the expression: . Apply the distributive property. Problem 13.5 said you could split a product inside one radical into separate radicals. This problem tells you it works backward as well; two radicals that are multiplied can be merged into one big radical as long as they have the same index. The product of two roots with the same index is equal to the root of the product: . Simplify the expression. ngous Book of Algebra Problems 284 The Humo Chapter Thirteen — Radical Expressions and Equations 13.26 Simplify the expression: . Express the squared quantity as a product. Calculate the product using the technique described in Problem 11.19. Combine like radicals. Recall that If you don’t understand why you need absolute values, look at Problem 13.10. . 13.27 Simplify the expression: Apply the distributive property. Combine like radicals and simplify. 13.28 Simplify the expression and express it using radical notation. According to a property of exponents, (xa)b = xab. When something raised to a power (like x1/2 or y3/4) is raised to a power (like 2), multiply the powers together. Calculate the product. The Humongous Book of Algebra Problems 285 Chapter Thirteen — Radical Expressions and Equations Write y 7/2 as a radical expression and simplify. 13.29 Calculate the quotient In other words, instead of having separate radica ls in the numerator and denominator, put the entire fraction un der one radical. . Express the quotient as a fraction. The quotient of two roots with the same index is equal to the root of the quotient. Eliminate the negative exponent by moving x –2 into the denominator. Simplify the expression. ngous Book of Algebra Problems 286 The Humo Chapter Thirteen — Radical Expressions and Equations 13.30 Rationalize the expression: . A rationalized expression does not contain a root in the denominator of the fraction. Rewrite the square root of this quotient as a quotient of square roots. To eliminate by . The square root in the denominator, multiply the numerator and denominator Anything divided by itself equals 1 (including divided by itself), and multiplying by 1 doesn’t change the fraction’s value. is eliminated by replacing it with the perfect square 25. 13.31 Simplify the expression and rationalize the answer. Write the quotient of cube roots as the cube root of a quotient and simplify the fraction. This fraction is made up of the coefficients in the numerator a nd denominator. The re no explicit numer is ator up top, so there’s an implie coefficient of 1. d Eliminate the negative exponent by moving y –2 into the denominator. To rationalize the expression, the cube root in the denominator must be eliminated. To accomplish this, multiply the numerator and denominator of the fraction by . The denominator starts as 3y2 . Multiplying this by two more 3s, and one more y gives you 33y3. The goal is to raise everything inside the radical to a power that matches the index of the radical, in this case 3. The Humongous Book of Algebra Problems 287 Chapter Thirteen — Radical Expressions and Equations Solving Radical Equations Use exponents to cancel out radicals 13.32 Solve the equation for x. To solve for x, you must eliminate the square root from the equation. To eliminate a root with index 2, raise both sides of the equation to the second power. Squaring a square root, cubing a cube root, or more generally raising an nth root to the n power will eliminate the radical symbol, leaving only the radicand. Solve for x by subtracting 2 from both sides of the equation. Don’t square both sides of th e equation to get rid of the square ro ot until you’ve got th e radical all by itse lf on one side of th e equal sign. 13.33 Solve the equation for x. Isolate the radical expression by adding 1 to both sides of the equation. Eliminate the square root by squaring both sides of the equation. 13.34 Solve the equation Isolate the radical expression ngous Book of Algebra Problems 288 The Humo for x and verify your answer. left of the equal sign. Chapter Thirteen — Radical Expressions and Equations Eliminate the radical expression and solve for x. To verify the answer, substitute x = –4 into the original equation. The result should be a true statement. 13.35 Solve the equation The index of the radical is 3, so raise both sid es to the third pow er. Don’t forget to cu be the number on th e right side of the equation. for x. Isolate the radical expression left of the equal sign. Eliminate the square root by squaring both sides of the equation. 13.36 Solve the equation for x: Add . to both sides of the equation to separate the radical expressions. Eliminate the square roots by squaring both sides of the equation. If you squared both sides without moving one of the radicals, you’d get which is not nearly as easy to deal with. The Humongous Book of Algebra Problems , 289 Chapter Thirteen — Radical Expressions and Equations Complex Numbers Numbers that contain i, which equals 13.37 Simplify the expression: . Write –49 as the product of –1 and a perfect square. You can’t take the square root of a negative, because nothing times itself equals a negative number. That means negative square roots (and fourth roots, sixth roots,…all even roots) are imaginary numbers. Radical expressions with an even index that contain negative radicands represent imaginary numbers; they are defined using the value . 13.38 Simplify the expression: . Identify the largest factor of –128 that is a perfect square to simplify the radical expression. Substitute How can 2 i = –1? Usually, it’s impossible to square something and get a negative value. However, i is an imaginary number, so it’s not governed by the same rules real numbers are. into the expression and simplify. 13.39 Simplify the expression: i 13. If , then ngous Book of Algebra Problems 290 The Humo . Express i 13 as a product of i 2 factors. There are six –1’s here. Tha t’ three pairs of (–1) s (–1): Chapter Thirteen — Radical Expressions and Equations Note: Problems 13.40–13.42 refer to complex numbers a = 2 – 3i and b = –4 + 5i. 13.40 Calculate a + b, the sum of the complex numbers, and a – b, the difference of the complex numbers. Adding and subtracting complex numbers is very similar to adding and subtracting polynomials—combine like terms. To calculate a – b, multiply b by –1. Note: Problems 13.40–13.42 refer to complex numbers a = 2 – 3i and b = –4 + 5i. 13.41 Calculate a · b. Multiply complex numbers the same way you would multiply two binomials. According to Problem 13.39, . Substitute that value into the expression. If you’re not sure how to multiply binomials, look at Problem 11.19. The Humongous Book of Algebra Problems 291 Chapter Thirteen — Radical Expressions and Equations The conjugate of –4 + 5i is –4 – 5i. All you do is change the middle sign to its opposite and leave everything else alone. Here’s the benefit: a complex number times its conjugate always produces a real number with no i’s in it. Note: Problems 13.40–13.42 refer to complex numbers a = 2 – 3i and b = –4 + 5i. 13.42 Calculate a ÷ b. Express the quotient as a fraction. Multiply the numerator and denominator of the fraction by the conjugate of the denominator. Calculate the products in the numerator and denominator. Substitute i 2 = –1 into the expression and simplify. Each term in the numerator becomes its own fraction. Both of the new fractions have the same denominator as the big fraction (41). Complex numbers are usually expressed in the form c + di, so rewrite the expression as a sum of two fractions with denominator 9. ngous Book of Algebra Problems 292 The Humo Chapter Thirteen — Radical Expressions and Equations 13.43 Calculate the product: (10 + i)(6 – i). Apply the technique described in Problem 13.41. 13.44 Calculate the quotient: . Multiply the numerator and denominator by the conjugate of 1 – 2i. Substitute i 2 = –1 into the expression. The Humongous Book of Algebra Problems 293 Chapter 14 Quadratic Equations and Inequalities 2 g x n i n t a i n o c s n o i t a u q e Solve The techniques used to solve linear equations presented in Chapter 4 are significantly different than those used to solve quadratic equations. Whereas the primary means by which linear equations are solved still apply (that is, it is still permissible to add or subtract the same quantity from both sides of an equation), quadratic equations require you to factor, complete the square, or apply the quadratic formula in order to isolate x and solve the equation. This chapter also discusses techniques used to solve and graph quadratic inequalities. Chapter 4 explains how to solve line ar equations—equations like x + 4 = 12 , where the highest power of the variable is 1. Chapter 7 explains how to solve and graph line ar inequalities, things like x + 4 > 12 . This chapter ups the ante and expla ins how to solve quadratic equatio ns and inequalities, which contain a va riable raised to the second power. Quadratic equations have UP TO two different real number solutions , and something called the discrimin ant helps you figure out exactly how many there are without calculating them. You can use one of three methods to solve a quadratic equa tion: factoring, completing the squar e, and the quadratic formula. Factori ng is probably the easiest, but it works only when you can actually fa ctor the polynomial. The quadratic formula and completing the square always work but they are a little more complicated. Finally, this chapter explains how to solve inequalities containing x2. You need to use something called “critic al numbers,” and they’re explained as well. You’ll need them again in Chap ter 21 to solve rational inequalities. Chapter Fourteen — Quadratic Equations and Inequalities Solving Quadratics by Factoring You can’t multiply two numbers and get zero unless one (or both) of those numbers is 0. That’s a property that is unique to 0. For example, if xy = 4, then there’s no guarantee that either x = 4 or y = 4. You could set x = 2 and y = 2, or maybe x = 8 and y = 0.5. Use techniques from Chapter 12 to solve equations 14.1 Solve the equation: xy = 0. According to the zero product property, a product can only equal 0 if at least one of the factors (in this case either x or y) equals 0. Therefore, xy = 0 if and only if x = 0 or y = 0. 14.2 Solve the equation: x(x – 3) = 0. The product of x and (x – 3) is equal to 0. According to the zero product property (explained in Problem 14.1), one (or both) of the factors must equal 0. x = 0 or x – 3 = 0 Solve x – 3 = 0 by adding 3 to both sides of the equation. x = 0 or x = 3 There are two possible solutions to | the equation. Use the word “or” to separate them because plugging either x = 0 OR x = 3 into the equation produces a true statement. Mathematically, saying “x = 0 AND x = 3” means x has to equal both of those things at the same time, and that doesn’t make sense. The solution to the equation x(x – 3) = 0 is x = 0 or x = 3. 14.3 Solve the equation: (x + 2)(2x – 9) = 0. According to the zero product property, the product left of the equal sign only equals 0 if either (x + 2) or (2x – 9) equals 0. Set both factors equal to 0 and solve the equations. The solution to the equation (x + 2)(2x – 9) = 0 is x = –2 or . Note: Problems 14.4–14.5 demonstrate two different ways to solve the equation x2 = 16. 14.4 Solve the equation using square roots. Problems 13.32–13.36 demonstrate that squaring both sides of an equation that contains a square root eliminates the root. Conversely, taking the square root of both sides of an equation containing a perfect square eliminates the perfect square. Back in Chapter 13 both sides AFTE , you only squared R you isolated th e square root on one side of only square root the equal sign. Similarly, both sides when the perfect square is isolate d on one side of the equal sign. ngous Book of Algebra Problems 296 The Humo Chapter Fourteen — Quadratic Equations and Inequalities Notice the “±” sign. Whenever you take the root of both sides of an equation and the index of that root is even, you must introduce the ± sign on one side of the equation, usually the side not containing the variable. The solution to the equation is x = –4 or x = 4. You have to include ± or you’ll exclude va lid answers. 4 is not the only squared num ber that equals 16. –4 works, too. Note: Problems 14.4–14.5 demonstrate two different ways to solve the equation x2 = 16. 14.5 Solve the equation by factoring. To apply the zero product property (the principle used to solve the equations in Problems 14.1–14.3), one side of the equation must equal 0. Change the right side of the equation to 0 by subtracting 16 from both sides. See Problems 12.26–12.30 to practice factoring the difference of perfect squares. Factor the left side of the equation, a difference of perfect squares. (x + 4)(x – 4) = 0 Apply the zero product property by setting each factor equal to 0 and solving the resulting equations. This solution echoes the solution generated by Problem 14.4: x = –4 or x = 4. 14.6 Solve the equation: x 2 – 11x + 28 = 0. The right side of the equation equals 0, so factor the quadratic trinomial. Because the leading coefficient of x 2 – 11x + 28 is 1, use the techniques described in Problems 12.35–12.41. The first step to solving a quadratic by factoring is to make sure one side of the equation (usually the right side) equals 0. That way, after you factor, you can use the zero product property to set the factors equal to 0. Apply the zero product property and solve the equations that result. The solution to the equation is x = 4 or x = 7. The coefficient of x2 is 1, so to factor, you’ll need to find two numbers that add up to –11 and multiply to give you +28. The numbers are –4 and –7. The Humongous Book of Algebra Problems 297 Chapter Fourteen — Quadratic Equations and Inequalities 14.7 Solve the equation: x 2 = –7x + 78. The coefficient of x2 is 1, so you need two numbers with a sum of 7 and a product of –78. The product’s negative, so the signs are different. The sum is positive, so the positive number is bigger. Try 8 – 1, 9 – 2, 10 – 3, … until you reach 13 – 6 = 7, because 13(–6) = –78. To solve a quadratic equation by factoring, only 0 should appear right of the equal sign. Add 7x to, and subtract 78 from, both sides of the equation. Factor the quadratic trinomial. (x + 13)(x – 6) = 0 Apply the zero product property. The solution to the equation is x = –13 or x = 6. 14.8 Solve the equation: 10x 2 + 13x – 3 = 0. Factor the quadratic trinomial by decomposition, as illustrated by Problems 12.42–12.44. Use decomposition because the coefficient of x 2 is n’t You’re looking for 1. tw numbers that ha o ve a sum of +13 and a product of –30. T he numbers are 15 and –2 . Apply the zero product property to solve the equation. The solution to the equation is or . 14.9 Solve the equation: 36x 2 + 12x + 1 = 0. Factor the quadratic trinomial by decomposition. ngous Book of Algebra Problems 298 The Humo Chapter Fourteen — Quadratic Equations and Inequalities Apply the zero product property. The solution to the equation is the double root . 14.10 Solve the equation: 12y 3 + 324 = 0. The terms 12y 3 and 324 share a greatest common factor: 12. Factor it out of the sum. There’s only one equation (instead of two separated by the word “or”) because both factors are equal. There’s no need to solve the same equation twice. 12(y 3 + 27) = 0 It’s called a “double root” because you get this answer TW ICE in the same problem, as the factor (6x + 1) is repeated. Apply the zero product property. Discard the left equation, as 12 ≠ 0. The solution to the equation is y = –3. 14.11 Solve the equation: 6x 3 – 3x 2 – 4x = –2. Add 2 to both sides of the equation and factor by grouping. Apply the zero product property to solve the equation. It’s okay to take the root of a negative number w the index of the hen root is odd: (–3)(–3)(–3) = Negative roots w –27. ith an even index, howev er are imaginary nu , mbers. To review factoring by grouping, look at Problems 12.20–12.25. The Humongous Book of Algebra Problems 299 Chapter Fourteen — Quadratic Equations and Inequalities The equation is allowed to have up to three real number solutions because the highest power of x in the original equation was 3. Rationalize by multiplying the numerator and denominator by The solution to the equation is , , or . . Completing the Square Make a trinomial into a perfect square 14.12 Given the quadratic expression x 2 – 8x + n, identify the value of n that makes the trinomial a perfect square and verify your answer. x2 – 8x + 16 is a perfect square because it’s equal to some quantity multiplied times itself: (x – 4)(x – 4). It’s the same logic that makes 25 a perfect square: 5 ˙ 5 = 25. Divide the coefficient of x by 2. –8 ÷ 2 = –4 Square the quotient. (–4)2 = 16 Substituting n = 16 into the expression creates the perfect square x 2 – 8x + 16. To verify that the quadratic is a perfect square, factor it. x 2 – 8x + 16 = (x – 4)(x – 4) = (x – 4)2 14.13 Given the expression x 2 – 3x + b, identify the value of b that makes the trinomial a perfect square. This trick only works if you have an x2-term with a coefficient of 1 and an x-term. Apply the technique outlined in Problem 14.12: divide the coefficient of x by 2 (that is, multiply it by ) and square the result. 14.14 Solve the equation by completing the square: x 2 – 6x = 0. Convert the quadratic binomial x 2 – 6x into a perfect square quadratic trinomial by dividing the coefficient of x by 2 (–6 ÷ 2 = –3) and squaring the result ([–3]2 = 9). x 2 – 6x + 9 = 0 + 9 Notice that you cannot add 9 only to the left side of the equation. Any real number added to one side of the equation must be added to the other side as well. x 2 – 6x + 9 = 9 ngous Book of Algebra Problems 300 The Humo Chapter Fourteen — Quadratic Equations and Inequalities Factor the left side of the equation. (x – 3)2 = 9 Solve for x by taking the square root of both sides of the equation. Remember that this operation requires you to include “±” immediately right of the equal sign. Solve both equations that result. The solution to the equation is x = 0 or x = 6. 14.15 Solve by completing the square: x + 2x – 7 = 0. 2 Move the constant to the right side of the equation by adding 7 to both sides. x 2 + 2x = 7 Here’s a tip: When you factor the trinomial, you’ll always get (x + something) 2 . That “something” is half of the original x-coefficient, the number you squared to get 9 a few steps ago. The equation x – 3 = ±3 represents two different equati ons: x – 3 = +3 or x – 3 = –3. Divide the coefficient of x by 2 and square the result: (2 ÷ 2)2 = 12 = 1. Add 1 to both sides of the equation. Factor the perfect square. (x + 1)2 = 8 Solve for x by taking the square root of both sides of the equation. This leaves an x2-term and an x-term on the left side of the equation, so you can come up with a constant to make a perfect square like in Problems 14.12–14.14. Solve for x by subtracting 1 from both sides of the equation. The solution to the equation is or . The Humongous Book of Algebra Problems 301 Chapter Fourteen — Quadratic Equations and Inequalities 14.16 Solve by completing the square: x 2 – 5x + 1 = 0. Move the constant to the right side of the equation by subtracting 1 from both sides. x 2 – 5x = –1 Square half of the x-coefficient: sides of the equation. ; add the result to both Factor the left side of the equation. Don’t forget the factoring trick: this number is always half of the original xcoefficient. Solve the equation for x. You can add these fractions because they have the same denominator. The solution to the equation is ngous Book of Algebra Problems 302 The Humo or . Chapter Fourteen — Quadratic Equations and Inequalities 14.17 According to Problem 14.7, the solution to the equation x 2 = –7x + 78 is x = –13 or x = 6. Verify the solution by completing the square. Add 7x to both sides of the equation. x 2 + 7x = 78 Divide the x-coefficient by 2 and square the result: Add . to both sides of the equation. Factor the left side of the equation and take the square root of both sides. 361 is not an obvious perfect square, so don’t kick your se if you panicked w lf hen you had to simplif y this radical. Isolate x to solve the equation. Completing the square verifies that the solution to the equation is x = –13 or x = 6. The Humongous Book of Algebra Problems 303 Chapter Fourteen — Quadratic Equations and Inequalities You can divide all the terms of an equation by any nonzero real number (dividing by zero is an algebraic no-no), and it won’t change the solution to the equation. 14.18 Solve by completing the square: 2x 2 – 20x + 4 = 0. To apply the technique described in Problems 14.12–14.17, the coefficient of x 2 must be 1. Divide every term in the equation by 2 so that the coefficient of x 2 is 1. Complete the square. Half of the x-coefficient is –10 ÷ 2 = –5. Square that: (–5) 2 = +25. Add 25 to both sides of the equation. The solution to the equation is or . 14.19 Solve by completing the square: 3x 2 + 2x + 9 = 0. Divide each term of the equation by 3 so that the coefficient of x 2 equals 1. Complete the square to solve the equation. Take half of th e x-coefficient: . Square that: . ngous Book of Algebra Problems 304 The Humo Chapter Fourteen — Quadratic Equations and Inequalities Recall that . The solutions to the equation are complex numbers: or . Quadratic Formula Use an equation’s coefficients to calculate the solution 14.20 According to the quadratic formula, what are the solutions to the equation ax 2 + bx + c = 0, assuming a ≠ 0? To solve the equation ax 2 + bx + c = 0 using the quadratic formula, substitute the coefficients a, b, and c into the formula below. This is the quadratic formula. Make sure to memorize it. If the equation is equal to 0, then a is the coefficient of x2, b is the coefficient of x, and c is the constant. Make sure that one side of the equation equals 0 before you use the formula! 14.21 Solve using the quadratic formula: x2 – 5x – 14 = 0. Substitute a = 1, b = –5, and c = –14 into the quadratic formula. If x 2 or x doesn’t have coefficient writte a n in front of it, the coefficient is 1. The solution to the equation is expressions. or . Simplify the rational The solution to the equation x 2 – 5x – 14 = 0 is x = –2 or x = 7. The Humongous Book of Algebra Problems 305 Chapter Fourteen — Quadratic Equations and Inequalities 14.22 Solve using the quadratic formula: 2x 2 – 7x + 3 = 0. Substitute a = 2, b = –7, and c = 3 into the quadratic formula. The solution to the equation is or x = 3. 14.23 Solve using the quadratic formula: –5x 2 + x + 8 = 0 Substitute a = –5, b = 1, and c = 8 into the quadratic formula. You can’t simplify this radical. The only factors of 161 (besides the obvious factors 1 and 161) are 7 and 23, and neither of those are perfect squares. ngous Book of Algebra Problems 306 The Humo Chapter Fourteen — Quadratic Equations and Inequalities The solution is or . 14.24 Solve using the quadratic formula: x2 + 2 = –8x. To apply the quadratic formula, one side of the equation must equal 0. Add 8x to both sides of the equation. x 2 + 8x + 2 = 0 Substitute a = 1, b = 8, and c = 2 into the quadratic formula. Reduce the fraction to lowest terms. The solution is or . To get these answers, fa ctor –1 out of the num er a to rs and denominators and cancel out –1’s: The Humongous Book of Algebra Problems 307 Chapter Fourteen — Quadratic Equations and Inequalities If you have trouble factoring, use the quadratic formula to solve quadratic equations. As this problem demonstrates, you’ll get the same answer either way. 14.25 According to Problem 14.8, the solution to the equation 10x 2 + 13x – 3 = 0 is or . Verify the solution using the quadratic formula. Substitute a = 10, b = 13, and c = –3 into the quadratic formula. The quadratic formula verifies the solution: or . 14.26 According to Problem 14.19, the solution to the equation 3x 2 + 2x + 9 = 0 is or . Verify the solution using the quadratic formula. Substitute a = 3, b = 2, and c = 9 into the quadratic formula. ngous Book of Algebra Problems 308 The Humo Chapter Fourteen — Quadratic Equations and Inequalities Reduce the fraction to lowest terms. The quadratic formula verifies the solution generated by completing the square: 14.27 Generate the quadratic formula by completing the square to solve the equation ax 2 + bx + c = 0, assuming a > 0. To complete the square, the coefficient of x 2 must be 1. Divide each term of the equation by a. Subtract the constant from both sides of the equation. Even though c and a look like variables, they’re constants. The coefficients of a quadratic equation (a, b, and c) don’t suddenly change in the middle of a problem—you don’t know what those coefficients are, so you use a, b, and c to represent them. You can make the x2-term of any quadratic equa tion positive, so this is a pretty reasonabl e assumption. If th e x2-coefficient is negative, just divide all of the terms by –1. 0 divided by any nonzero number equals 0. The Humongous Book of Algebra Problems 309 Chapter Fourteen — Quadratic Equations and Inequalities This value (half of the xcoefficient) goes inside the factored form of the perfect square. Multiply the x-coefficient by Add and square the result: . to both sides of the equation. Factor the perfect square and solve for x. The problem states that a > 0, so the absolute value bars may be omitted: 310 The Humongous Book of Algebra Problems . Chapter Fourteen — Quadratic Equations and Inequalities 14.28 Solve the equation for x: . Square both sides of the equation in an attempt to eliminate the radical expressions. does NOT equal . Use the FOIL method to multiply Isolate the radical expression right of the equal sign by subtracting x and 11 from both sides of the equation. Square both sides of the equation to eliminate the radical expression. Subtract 36x and 72 from both sides of the equation so that the right side of the equation is equal to 0. Divide each term of the equation by 9. The Humongous Book of Algebra Problems 311 Chapter Fourteen — Quadratic Equations and Inequalities Solve the equation using the quadratic formula. Getting to the last steps from the steps above them is quite advanced, so don’t worry about simplifying something like . Instead, type the expressions into a calculator. The decimal you get for the left side won’t equal the decimal you get for the right side, and that’s enough to show that the equation is false. “Roots” is another words for “solutions,” so the book’s asking what you can predict about the real number solutions of an equation. Don’t get these roots confused with radicals; square ROOTS and the ROOTS of an equation mean totally different things. 312 Reduce the fraction to lowest terms. Squaring both sides of an equation might introduce false answers, so verify the solutions by substituting them into the original equation. Neither nor satisfies the equation, so the equation has no real number solution. Applying the Discriminant What b2– 4ac tells you about an equation 14.29 The discriminant of the equation ax 2 + bx + c = 0 is defined as the value b 2 – 4ac. What conclusions can be drawn about the real roots of a quadratic equation based on the sign of its discriminant? According to the fundamental theorem of algebra, all quadratic equations have two roots, because all quadratic equations have degree two. The discriminant is calculated to predict how many (if any) of those roots will be real numbers. The Humongous Book of Algebra Problems Chapter Fourteen — Quadratic Equations and Inequalities A positive discriminant indicates that the equation has two unique real roots. In other words, there are two different, real number solutions to the quadratic. The solutions may be rational (which means that the equation can be solved by factoring) or irrational (which means that the equation must be solved by completing the square or the quadratic equation), but both solutions are real numbers. A zero discriminant is an indication of a double root. The quadratic still has two solutions, but the solutions are equal. For instance, the quadratic equation (x + 5)(x + 5) = 0 has two roots: x = –5 and x = –5. Because a single real root is repeated, x = –5 is described as a double root. And they’ll contain the imaginary number . If a quadratic equation has a negative discriminant, there are no real number solutions. That does not mean the equation has no solutions, but instead that the solutions are complex numbers. 14.30 Given the equation 3x 2 + 7x + 1 = 0, use the discriminant to predict how many of the roots (if any) are real numbers, then calculate the roots to verify the prediction made by the discriminant. The equation has form ax 2 + bx + c = 0, so substitute a = 3, b = 7, and c = 1 into the discriminant formula. According to Problem 14.29, the equation 3x 2 + 7x + 1 = 0 has two positive real roots because the discriminant is positive (37 > 0). Calculate the roots by applying the quadratic formula. The solution to the equation is or In case you didn’t recognize it, the discrimina nt b2 – 4ac is the pa rt of the quadrati c formula that’s in sid the radical. Whe e n yo plug everything in u to quadratic formul the a, there’s no need to calculate b2 – 4a c again—just plug in the discriminant , 37. . The Humongous Book of Algebra Problems 313 Chapter Fourteen — Quadratic Equations and Inequalities Note: Problems 14.31–14.32 refer to the equation x2 + 2x + 6 = 0. 14.31 Use the discriminant to classify the roots of the quadratic and then calculate the roots to verify the answer. RULE OF THUMB: Complex roots always come in pairs called “complex conjugates.” The only difference between them is the sign in the middle. For example, if an equation has complex root –2 + 7i, then it will also have complex root –2 – 7i. Substitute a = 1, b = 2, and c = 6 into the discriminant formula. Because the discriminant is negative, the equation has no real roots. Instead, the equation has two complex roots. Apply the quadratic formula to calculate them. Reduce the fraction to lowest terms. The solution to the equation is In other words, explain WHY you end up getting 0, 1, or 2 real roots based on the sign of the discriminant you calculate in Problem 14.31. Hint: It’s not enough to say, “Because that’s how discriminants work.” 314 or . Note: Problems 14.31–14.32 refer to the equation x2 + 2x + 6 = 0. 14.32 Describe the causal relationship between the discriminant and the nature of the equation’s roots. The discriminant of a quadratic equation, b 2 – 4ac, is the radicand in the quadratic formula. Because the equation x 2 + 2x + 6 = 0 has a negative discriminant, when the quadratic formula is applied, the radical expression therein contains a negative value (as demonstrated in Problem 14.31). Square roots with a negative radicand have imaginary values, and therefore produce complex roots. The Humongous Book of Algebra Problems Chapter Fourteen — Quadratic Equations and Inequalities Note: Problems 14.33–14.34 refer to the equation 9x2 – 12x = –4. 14.33 Use the discriminant to classify the roots of the quadratic and then calculate the roots to verify the answer. Rewrite the equation in form ax 2 + bx + c = 0 by adding 4 to both sides. 9x 2 – 12x + 4 = 0 Substitute a = 9, b = –12, and c = 4 into the discriminant formula. Because the discriminant equals 0, the equation 9x 2 – 12x = –4 has a double root. Apply the quadratic formula to calculate it. The solution to the equation is . Note: Problems 14.33–14.34 refer to the equation 9x2 – 12x = –4. 14.34 Describe the causal relationship between the discriminant and the nature of the equation’s roots. When the discriminant of a quadratic equation is 0, the radical expression in the quadratic formula is equal to 0, as demonstrated in Problem 14.33. Therefore, is a double root, the only solution to the equation. When the square root equals 0, you lose the “plus or minus ” part of the form ula that actually c reates two different a nswers; one that’s –b plus th square root of th e e discriminant and on that’s –b minus th e e square root of th e discriminant. Tha t’s why you end up w ith only one answer instead of two. The Humongous Book of Algebra Problems 315 Chapter Fourteen — Quadratic Equations and Inequalities The right side of the inequality needs to equal 0 before you do anything else. Algebraic expressions run the risk of being undefined either when they’re fractions (because the denominator could be 0) or when they contain radicals with an even index (because you could have a negative inside). Neither of those things occur in a quadratic expression, so don’t worry about quadratics being undefined. Focus on when they equal 0. For more information on cl osed (solid) and open (h ol points on a graph low) —and how they’re rela ted inequality signs— to look at Problem 7.13. One-Variable Quadratic Inequalities Inequalities that contain x2 Note: Problems 14.35–14.37 refer to the inequality x2 – x – 12 ≤ 0. 14.35 Identify the critical numbers of the quadratic expression. The critical numbers of an expression are the values that make the expression equal zero or cause it to be undefined. No real number, when substituted into a quadratic expression, can cause a quadratic to be undefined, so calculate the x-values for which x 2 – x – 12 = 0. Solve the equation by factoring, using the method outlined in Problems 14.1–14.10. The critical numbers of x 2 – x – 12 are x = –3 and x = 4. Note: Problems 14.35–14.37 refer to the inequality x2 – x – 12 ≤ 0. 14.36 Solve the inequality. According to Problem 14.35, the critical numbers of the quadratic are x = –3 and x = 4. When graphed, those two values split the number line into three intervals, as illustrated by Figure 14-1. Figure 14-1: T he critical numbers x = –3 and x = 4 divide the number line into three intervals. Note that the critical numbers are plotted as closed points due to the inequality sign ≤ in the original problem. To decide whether to include the intervals pictured in Figure 14-1 as part of the solution to the inequality, choose one “test value” from each, such as x = –4, x = 0, and x = 5. Substitute each of the test values into the inequality. If a test value satisfies the inequality (that is, results in a true inequality statement), then the entire interval from which that test value was taken represents valid solutions to the inequality. x = –4 represents the left int erval of Figure 14.1 (because –4 ≤ –3), x = 0 represents the middle inter val (because 0 is between –3 and 4), and x = 5 represents the right interval (because 5 ≥ 4). 316 The Humongous Book of Algebra Problems Chapter Fourteen — Quadratic Equations and Inequalities Of the three test values, only x = 0 satisfies the inequality. Therefore, the solution is –3 ≤ x ≤ 4, the interval that contains x = 0. Note: Problems 14.35–14.37 refer to the inequality x2 – x – 12 ≤ 0. 14.37 Graph the solution to the inequality. According to Problem 14.36, the solution to the inequality is –3 ≤ x ≤ 4. Graph the solution by darkening that interval on the number line and using solid endpoints, as illustrated by Figure 14-2. Use solid dots on the number line when the inequality contains ≤ or ≥. Figure 14-2: The graph of the inequality x2 – x – 12 ≤ 0. You could divide by +3, or you could choose not to divide by anything at all. The final answer isn’t affected. Note: Problems 14.38–14.39 refer to the inequality 12x – 12x2 ≥ 3. 14.38 Solve the inequality. Rewrite the inequality in form ax 2 + bx + c ≥ 0. –12x 2 + 12x – 3 ≥ 0 Divide each term of the inequality by –3. Calculate critical numbers by setting the quadratic expression equal to 0 and solving for x. Remember to reverse the inequality symbol when you multiply or divide by a negative. Plot the critical number on a number line using a solid point, as illustrated by Figure 14-3. The Humongous Book of Algebra Problems 317 Chapter Fourteen — Quadratic Equations and Inequalities is a critical number, so when yo into 4x2 – 4x + 1, u plug it you get 0. Figure 14-3: The quadratic equation 12x – 12x2 ≥ 3 has only one critical point, The critical point splits the number line in Figure 14-3 into two intervals: and . Choose test points from both intervals (such as x = 0 and x = 1) to identify the solution to the inequality. Neither interval contains solutions to the inequality. The only valid solution is . Note: Problems 14.38–14.39 refer to the inequality 12x – 12x2 ≥ 3. 14.39 Graph the solution to the inequality. According to Problem 14.38, the solution to the inequality is the single value . To graph the solution, plot a single solid point at , as illustrated by Figure 14-4. Figure 14-4: The solution to the inequality 12x – 12x2 ≥ 3 is a single point on the number line. 318 The Humongous Book of Algebra Problems . Chapter Fourteen — Quadratic Equations and Inequalities Note: Problems 14.40–14.41 refer to the inequality 2x2 + 3x – 1 > 0. 14.40 Solve the inequality. Identify critical numbers using the quadratic formula, as the quadratic cannot be factored. Plot the critical numbers, and Type the roots into a calculator to get approxima te decimal values so you can figure out whe re to plot the points. , on a number line, as illustrated by Figure 14-5, and choose test values (such as x = –2, x = 0, and x = 2) from the three resulting intervals. Figure 14-5: Plot the critical numbers of the quadratic using open points, as the statement contains the > inequality symbol. The solution to the equation is that contain x = –2 and x = 2. or , the intervals The Humongous Book of Algebra Problems 319 Chapter Fourteen — Quadratic Equations and Inequalities Note: Problems 14.40–14.41 refer to the inequality 2x2 + 3x – 1 > 0. 14.41 Graph the solution to the inequality. To graph the solution, graph the inequalities and on the same number line, as illustrated by Figure 14-6. Figure 14-6: The graph of 2x2 + 3x – 1 > 0. Note: Problems 14.42–14.43 refer to the inequality 4x3 + 5x2 – 6x < 0. 14.42 Solve the inequality. Calculate the critical numbers by factoring the expression. This isn’t a quadratic equation, but you’ll get a quadratic expression once you factor out the greatest common factor, x. Factor by decomposition. Set each factor equal to zero and solve for x to get critical numbers x = 0, x = –2, and . As illustrated by Figure 14-7, those critical numbers separate the number line into four intervals: x < –2, –2 < x < 0, , and . Figure 14-7: Three different critical numbers will separate a number line into four intervals. ngous Book of Algebra Problems 320 The Humo Chapter Fourteen — Quadratic Equations and Inequalities Substitute test values from the intervals (such as x = –3, x = –1, , and x = 1) into the inequality to identify the solution. The solution to the inequality is x < –2 or It’s easier to plug the test values into the factored version of the inequality: x(x + 2)(4x – 3) < 0 instead of 4x3 + 5x2 – 6x < 0. . Note: Problems 14.42–14.43 refer to the inequality 4x3 + 5x2 – 6x < 0. 14.43 Graph the solution to the inequality. According to Problem 14.42, the solution to the inequality consists of two intervals: x < –2 and . Plot them both on the same number line, as illustrated by Figure 14-8. Figure 14-8: The solution to the inequality 4x3 + 5x2 – 6x < 0. The Humongous Book of Algebra Problems 321 Chapter 15 functions utput per input Named expressions that give one o Though a fundamental concept of mathematics, the function is little more than an expression that meets a specific criterion—each independent variable input results in exactly one dependent variable output. Cosmetically, the function differs from a general expression because it is assigned a name, often f(x). This chapter investigates the distinction that makes a mathematical relation a function, operations performed on functions, the composition of functions, inverse functions, and piecewise-defined functions. Many books describe a function as a machine: you plug in an x-value, and you get out an f(x)-value. That description works, but it might be easier to think of a function as an expression you can refer to by name . For instance, let’s say f(x) = x2 + 1. After you define it, every time you refer to f(x), everyone knows you’re talking about x2 + 1. There’s other benefits as well. Writin g f(2) means “plug x = 2 into the function f(x).” So if f(x) = x2 + 1, the n f(2) = 22 + 1 = 5. Functions are more than just name d expressions, however. A function ha s to meet one requirement—every x-v alue you plug into it has to produce exactly one output. In other words, if you plug x = 1 into a function, you ’ll never get two different answers. Chapter Fifteen — Functions Relations and Functions The left number in each ordered pair is the input: 1, 2, 5, 8; p is not defined for any other values, so p(3), for example, is undefined. “Abscissa” is the fancy word for x-value of a point, and “ordinate” means the y-value. The abscissa of (1,9) is 1 and the ordinate is 9. None of the x-values in p are repeated—each on appears once. Tha ly t means each x-va lue is paired with on ly one y-value, so it ’s a function. What makes a function a function? Note: P roblems 15.1–15.2 refer to relation p defined below. p : {(1,9), (2,6), (5,1), (8,6)} 15.1 Evaluate p(1) and p(8). A relation defines a relationship between two sets of numbers. Relation p is defined as a list of coordinate pair (a,b), such that each input value a is paired with an output value b. To evaluate p(1), locate the ordered pair with abscissa 1 and report the ordinate. Relation p contains the ordered pair (1,9), so p(1) = 9. Note that p(1) ≠ 5. Although the point (5,1) belongs to the relation, 1 is the ordinate of that coordinate pair, not the abscissa. To evaluate p(8), identify the ordered pair with abscissa 8: (8,6). The ordinate of that pair is 6, so p(8) = 6. Note: Problems 15.1–15.2 refer to relation p defined below. p : {(1,9), (2,6), (5,1), (8,6)} 15.2 Is p a function? Why or why not? A relation is a function if and only if every abscissa value is paired with exactly one ordinate value. Relation p states that p(1) = 9, p(2) = 5, p(5) = 1, and p(8) = 6. Each input value (1, 2, 5, and 8) is paired with only one output value, so p is a function. Note: Problems 15.3–15.4 refer to relation q defined below. q : {(–4, –1), (–2,3), (0,6), (–2,9), (–6, –13)} 15.3 Evaluate q(0) and q(–1). To evaluate q(0), identify the coordinate pair with an abscissa of 0: (0,6). Relation q defines q(0) as the ordinate of that pair: q(0) = 6. Q does contain the point (–4, –1), but –1 is the output (ordinate) of that pair, not the input. Because none of the ordered pair constituting relation q have an abscissa of –1, q(–1) is undefined. Note: P roblems 15.3–15.4 refer to relation q defined below. q : {(–4, –1), (–2,3), (0,6), (–2,9), (–6, –13)} 15.4 Is q a function? Why or why not? Every input of a function can only produce one output. In this case, the input –2 produces two outputs, 3 and 9. To be a function, each of a relation’s input values must be paired with only one corresponding output value. Notice that two coordinate pairs in relation q contain the same abscissa: (–2,3) and (–2,9). Therefore, q(–2) = 3 and q(–2) = 9. Relation q pairs abscissa value –2 with two different ordinate values, so q is not a function. ngous Book of Algebra Problems 324 The Humo Chapter Fifteen — Functions Note: Problems 15.5–15.6 refer to the relation . 15.5 Evaluate r(–3) and r(4). To evaluate r(–3) and r(4), substitute x = –3 and x = 4 into the relation. Note: Problems 15.5–15.6 refer to the relation . 15.6 Assuming x is a real number, is r(x) a function? Why or why not? Relation r(x) is a function, as every real number has exactly one absolute value. The absolute value of a real number greater than or equal to zero equals the number itself, and the absolute value of a real number less than zero is defined as the opposite of that number. Note: Problems 15.7–15.8 refer to the relation s(x) = 5 ± x. 15.7 Evaluate s(0) and s(–1). To evaluate s(0), substitute x = 0 into s(x) = 5 ± x. Evaluate s(–1). The expression contains a ± symbol, so consider both possibilities as you evaluate s(1). s(0) = 5 ± 0 had a ± sign in it, too, but adding 0 to 5 and subtracting 0 from 5 both produce the same result: 5. The Humongous Book of Algebra Problems 325 Chapter Fifteen — Functions Note: Problems 15.7–15.8 refer to the relation s(x) = 5 ± x. 15.8 Is s(x) a function? Why or why not? According to Problem 15.7, s(–1) = 4 or s(–1) = 6. For a relation to be a function, each value substituted into the relation must produce only one result. In this case, substituting x = –1 into s(x) produces two results (4 and 6), so s(x) is not a function. Operations on Functions +, –, ˙, and ÷ functions This is NOT the distributive property—you’re not distributing x through (f + g) to get f(x) + g(x). It’s just notation: (f + g)(x) = f(x) + g(x), (f – g)(x) = f(x) – g(x), (fg)(x) = f(x) ˙ g(x), and . Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and h(x) = 2x2 – 13x + 20. 15.9 Express (f + g)(x) in terms of x. The function (f + g)(x) represents the sum f(x) + g(x). Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and h(x) = 2x2 – 13x + 20. 15.10 Evaluate (f + 4g – h)( –2). The function (f + 4g – h)(x) represents the sum f(x) + 4g(x) – h(x). Evaluate f(x), g(x), and h(x) for x = –2. Use the values of f(–2), g(–2), and h(–2) to calculate (f + 4g – h)( –2). ngous Book of Algebra Problems 326 The Humo Chapter Fifteen — Functions Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and h(x) = 2x2 – 13x + 20. 15.11 Evaluate (fg)(6). The function (fg)(x) is defined as the product f(x) · g(x). Therefore, (fg)(6) = f(6) · g(6). Evaluate f(x) and g(x) for x = 6. Use the values of f(6) and g(6) to calculate (fg)(6). Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and h(x) = 2x2 – 13x + 20. 15.12 Express (fg)(x) in terms of x and evaluate that expression for x = 6 to verify the solution to Problem 15.11. The function (fg)(x) is defined as the product f(x) · g(x). Calculate the product. Evaluate the expression for x = 6. To calculate (fg)(6), you can either plug 6 into f(x) and g(x) and then multiply those results (like Problem 15.11) or multiply f(x) and g(x) first and THEN plug in x = 6 (like Problem 15.12). The Humongous Book of Algebra Problems 327 Chapter Fifteen — Functions Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and h(x) = 2x2 – 13x + 20. 15.13 Express in terms of x. The function Chapters 20 and 21 go into a lot more detail about rational expressions (fractions that contain polynomials), including how to simplify them. The original denominator of the fraction wa s x – 4, so when x = 4 the denominator is 4 – 4 = 0. Dividing by 0 is no t allowed. Sure yo u ca simplify the fract n ion to get 2x – 5, but is STILL undefine d at x = 4. is defined as the quotient of h(x) and f(x). Reduce the fraction to lowest terms by factoring the numerator. Note: Problems 15.9–15.14 refer to the following functions: f(x) = x – 4, g(x) = 3x + 2, and h(x) = 2x2 – 13x + 20. 15.14 Evaluate . Substitute x = –1 into g(x) and h(x). Calculate the quotient of h(–1) and g(–1). ngous Book of Algebra Problems 328 The Humo Chapter Fifteen — Functions Note: Problems 15.15–15.17 refer to the following functions: . and 15.15 Evaluate (p + r)(4). Calculate the sum p(4) + r(4). You can’t add these yet because they’re not like radicals— they don’t have the same numbers under the radical symbol. Simplify the radical expressions. Therefore, . Note: Problems 15.15–15.17 refer to the following functions: . and 15.16 Express (pr)(x) in terms of x and evaluate (pr)(–2). Multiply functions p(x) and r(x). If you’re multiplyin g two radicals wit h the same index, you ca multiply the cont n ents of the radicals insi de big radical sign w one it matching index. h a Evaluate (pr)(–2) by substituting x = –2 into (pr)(x). The Humongous Book of Algebra Problems 329 Chapter Fifteen — Functions Note: Problems 15.15–15.17 refer to the following functions: . 15.17 Evaluate and . Evaluate r(1) and p(1) and then calculate the quotient. Simplify the expression. Composition of Functions Plug one function into another Note: Problems 15.18–15.19 refer to the functions f(x) = x2 – 1 and g(x) = 3x. 15.18 Evaluate f(5). Substitute x = 5 into f(x). ngous Book of Algebra Problems 330 The Humo Chapter Fifteen — Functions Note: Problems 15.18–15.19 refer to the functions f(x) = x2 – 1 and g(x) = 3x. 15.19 Express f(g(x)) in terms of x. Problem 15.18 requires you to substitute x = 5 into f(x). Here, you substitute g(x) into f(x). Recall that g(x) = 3x. Note: Problems 15.20–15.21 refer to the functions 15.20 Express is equivalent to j(h(x)). Replace the x in 7x 2 – 3 with Substitute into the function. Note: Problems 15.20–15.21 refer to the functions Note that evaluating j(–2). . in terms of x. The expression h(x). 15.21 Evaluate and and “Composing” functions means plugging functions into each other. In Problem 15.19, you plug g(x) into f(x). That can be written f(g(x)), which is read “f of g of x”, or , which is real “f circle g of x.” . Keep the letters in the same order—j comes first in and j(h(x)). Composition of functions is not necessarily commutative, so there’s no guarantee that j(h(x)) and h(j(x)) are equal. Squaring a square root cancels it out, leaving only the radicand x + 7 behind. . . To evaluate the composite function, begin by The Humongous Book of Algebra Problems 331 Chapter Fifteen — Functions Substitute j(–2) = 25 into h(j(–2)). h(j(–2)) = h(25) Evaluate h(25). Simplify the radical expression. Therefore, . Note: Problems 15.22–15.24 refer to the following functions: f(x) = x2 – 4x + 3, h(x) = 2x – 1. 15.22 Express in terms of x and evaluate Note that g(x) into f(x). , . ; to write the function in terms of x, substitute Evaluate f(g(6)). h(x) is plugged into g(x), which is plugged into f(x). Sta with the innermos rt t function (the fu nction inside the most parentheses) and work your way ou t. Note: Problems 15.22–15.24 refer to the following functions: f(x) = x2 – 4x + 3, h(x) = 2x – 1. 15.23 Evaluate f(g(h(5))). Begin by evaluating h(5). h(5) = 2(5) – 1 = 10 – 1 = 9 Substitute h(5) = 9 into the expression: f(g(h(5))) = f(g(9)). Evaluate g(9). ngous Book of Algebra Problems 332 The Humo , Chapter Fifteen — Functions Substitute g(9) = 3 into the expression: f(g(9)) = f(3). Evaluate f(3). Therefore, f(g(h(5))) = 0. Note: Problems 15.22–15.24 refer to the following functions: f(x) = x2 – 4x + 3, h(x) = 2x – 1. , 15.24 Express f(h(g(x))) in terms of x. Assume x > 0. Begin by substituting g(x) into h(x). Substitute into f(h(g(x))). You don’t need to put this x in absolute values because the problem says to assume x is positive (x > 0). Otherwise, . 15.25 If r(s(x)) = (x 3 – 1)2 and r(x) = x 2, then express s(r(x)) in terms of x. The function r(s(x)) is defined the expression x 3 – 1 squared, and r(x) = x 2, so s(x) = x 3 – 1. In the function r(s(x)), s(x) is plugged into r(x). You know that r(x) = x2, so anything you plug into r(x) gets squared. To figure out s(x), ask yourself this question: “What expression, when squared, is equal to (x3 – 1) 2?”. The answer is s(x) = x3 – 1. The Humongous Book of Algebra Problems 333 Chapter Fifteen — Functions 15.26 If and , evaluate g(g(7)). Substitute g(x) into f(x) to generate f(g(x)). This is true because of the transitive property, which says that two things equal to the same thing are equal to each other. If and , then Solve the equation for g(x) by squaring both sides of the equation to eliminate the radicals. . and mean the same thing: . Rather than expressing g(x) as a fraction containing a sum, express it as a sum of fractions. To evaluate g(g(7)), begin by substituting x = 7 into g(x). Therefore, ngous Book of Algebra Problems 334 The Humo . Chapter Fifteen — Functions Inverse Functions Functions that cancel each other out 15.27 What is the defining characteristic of inverse functions? If functions f(x) and g(x) are inverse functions, then composing one function with the other—and vice versa—results in x: f(g(x)) = g(f(x)) = x. 15.28 Describe the relationship between the points on the graph of a function and the points on the graph of its inverse. If the graph of function f(x) contains the point (a,b), then the graph of its inverse function contains the point (b,a). Reversing the coordinates of one graph produces the coordinates for the graph of its inverse. 15.29 Does an inverse function exist for the function f(x) graphed in Figure 15-1? Why or why not? In other words, inverse functions cancel each other out. Plugging g(x) into f( x) f(x) into g(x) make or s f( and g(x) go away, x) leaving only x behind. “f–1(x)” means “the inverse of f(x).” It does NOT mean “f(x) raised to the –1 power.” That would be written [f(x)]–1. Figure 15-1: No information about f(x) is provided other than this graph. Function f(x) does not have an inverse function because it fails the horizontal line test, which states that a function is one-to-one if and only if any horizontal line drawn on its graph will, at most, intersect the graph only once. Notice that any horizontal line drawn on Figure 15-1 between y = 1 and y = 3 will intersect the graph three times. The Humongous Book of Algebra Problems 335 Chapter Fifteen — Functions Every input value of a function corresponds to exactly one output value, but the output values of one-to-one functions correspond to exactly one input value as well. If a horizontal line intersects a graph multiple times, the x-values of those intersection points represent input values that are paired with an identical output (the shared y-value represented by the horizontal line). Consider the horizontal line y = h on the graph of f(x) in Figure 15-2. It intersects the graph at three points, with x-values a, b, and c. Therefore, f(a) = h, f(b) = h, and f(c) = h. RU LE OF TH UMB: If ANY horizontal line intersects a graph in more than one place, the graph fails the horizontal line test. That means the graph is not one-to-one, and only one-to-one functions have inverses. Figure 15-2: The horizontal line y = h intersects f(x) when x = a, x = b, and x = c. As stated in Problem 15.28, the coordinate pairs of functions and their inverses are reversed; therefore, , , and . However, each input value of a function can correspond to only one output value, so the relation is not a function. ngous Book of Algebra Problems 336 The Humo Chapter Fifteen — Functions 15.30 Does an inverse function exist for the function g(x) graphed in Figure 15-3? Why or why not? Figure 15-3: No information about g(x) is provided other than this graph. Function g(x) in Figure 15-3 passes the horizontal line test, as it is monotonically increasing. According to the logic presented in Problem 15.29, g(x) is a one-to-one function, and therefore the inverse function g–1(x) exists. “Monotonic” means “doesn’t change direction.” As you trace the function’s graph from left to right, it always travels upward. At no time does the graph turn and start to travel downwards again. The Humongous Book of Algebra Problems 337 Chapter Fifteen — Functions 15.31 Given the graph of function h(x) in Figure 15-4, draw the graph of the inverse function, h –1(x). These are the points that h(x) passes through, after you reverse the coordinates. For example, h(x) passes through (5,–2) so h–1(x) passes through (–2,5). Figure 15-4: The graph of h(x) passes through points (–4,3), (0,2), (3,0), (5,–2). According to Problem 15.28, the graphs of a function and its inverse contain reversed coordinate pair—if (a,b) belongs to the graph of h(x), then the graph of h –1(x) contains the point (b,a). In this case, the graph of h –1(x) passes through points (–2,5), (0,3), (2,0), and (3,–4). As a result the graphs are symmetric about the line y = x, as illustrated by Figure 15-5. Figure 15-5: The graphs of h–1(x) and h(x) are reflections of each other across the line y = x. ngous Book of Algebra Problems 338 The Humo Chapter Fifteen — Functions 15.32 Verify that and are inverse functions. As stated in Problem 15.27, if f(x) and g(x) are inverse functions, then f(g(x)) = g(f(x)) = x. Substitute g(x) into f(x) (and vice versa) to verify that both result in x. Because f(g(x)) = g(f(x)) = x, f(x) and g(x) are inverse functions. Note: Problems 15.33–15.34 refer to the function j(x) = 3(x + 4) – 1. 15.33 Identify the inverse function, j–1(x). Begin by rewriting j(x) as y. y = 3(x + 4) – 1 As Problem 15.28 states, the points on the graph of a function and the points on the graph of its inverse contain reversed coordinate pairs. To reflect this property of inverses, reverse x and y in the equation. x = 3(y + 4) – 1 Solve the equation for y. If a fraction contains a sum or difference in th e numerator, you sh ou rewrite it as the ld sum or difference of two fractions (that have the same denom ina as the big fract tor ion did), just like in Pr oblem 15.26. Replace y with j –1(x) to identify the inverse function. The Humongous Book of Algebra Problems 339 Chapter Fifteen — Functions Note: Problems 15.33–15.34 refer to the function j(x) = 3(x + 4) – 1. 15.34 Verify that j(x) and the function j –1(x) generated in Problem 15.33 are inverses. Verify that j(j –1(x)) = j –1(j(x)) = x. The domain is the collection of numbers you’re allowed to plug into a function. You can plug any real number x into the function f(x) = (x + 2) 2, so the domain of f(x) is all real numbers. Look at Problems 16.9–16.20 to practice identifying the domain of a function. Because j(j –1(x)) = j –1(j(x)) = x, j(x) and j –1(x) are inverse functions. Note: Problems 15.35–15.36 refer to the function f(x) = (x + 2)2. 15.35 Why does no inverse function exist for f(x)? Explain how to alter the domain of the function such that an inverse does exist. The graph of f(x), illustrated in Figure 15-6, fails the horizontal line test, so no inverse function exists. Figure 15-6: Any horizontal line above the x-axis intersects the graph of f(x) twice. ngous Book of Algebra Problems 340 The Humo Chapter Fifteen — Functions However, if you limit the domain of the function, an inverse does exist. For instance, if you consider only the portion of the function for which x ≥ –2, the corresponding graph (illustrated in Figure 15-7) does pass the horizontal line test, and therefore an inverse exists. The function also has an inverse for x ≤ –2 . Figure 15-7: The portion of the graph of f(x) for which x ≥ –2 increases monotonically and thus has an inverse. Note: Problems 15.35–15.36 refer to the function f(x) = (x + 2)2. 15.36 Identify the inverse of the modified function identified in Problem 15.35. According to Problem 15.35, the function f(x) = (x + 2)2 has an inverse when x ≥ –2. Rewrite f(x) as y. y = (x + 2)2 Follow the procedure outlined in Problem 15.33 to calculate the inverse: Reverse x and y, solve the resulting equation for y, and then replace y with f –1(x). Therefore, the inverse function is either or . Notice that the graph of f(x) passes through point (0,4) in Figure 15-7. Therefore, the graph of f –1(x) must pass through the point (4,0). Substitute x = 4 into both inverse function candidates and determine which results in f(4) = 0. Because when When you take the root of both sides of an equation and the index of the root is even, multiply the side of the equation that DOESN’T contain the variable you’re isolating by “±1.” Reverse x and y in a point on the graph of f(x) to get a point on the graph of f-1(x). , that is the inverse function. The Humongous Book of Algebra Problems 341 Chapter Fifteen — Functions 15.37 Identify the inverse function of Rewrite g(x) as y to get solve for y. Normally, x = 0 is the best number to test the function and find its inverse. That would make g(0) = –4 in this problem. Unfortunately, BOTH possible inverse functions equal 0 when you plug –4 into them, so that’s why you have to go with something else (like x = 1). . . Reverse x and y in the equation and Either or is the inverse function of g(x). To determine which is correct, evaluate g(x) for an x-value (such as x = 1) to produce a coordinate pair. Because , it must follow that . Substitute into both potential inverse functions to determine which produces the correct value. Because function. ngous Book of Algebra Problems 342 The Humo when , that is the inverse Chapter Fifteen — Functions Piecewise-Defined Functions Function rules that change based on the x-input 15.38 Given the piecewise-defined function f(x) below, evaluate f(–2), f(0), and f(6). A piecewise-defined function consists of two or more individual functions (in this case g(x), h(x), and j(x)), each of which represents values of the function for specific values of x. Here, f(x) = g(x) when x < 0. For instance, when x = –2, f(–2) = g(–2). When x = 0, f(x) = h(x); therefore, f(0) = h(0). Finally, f(x) = j(x) when x > 0, so f(6) = j(6). f(x) changes based on INPUT values. The little statements right of the functions (x ‹ 0, x = 0, and x › 0) te ll you which functi on to plug x into. Note: Problems 15.39–15.41 refer to the piecewise-defined function k(x) below. 15.39 Evaluate k(–3). Note the three domain intervals defined for k(x): x ≤ –3, –3 < x < 1, and x ≥ 1. The only domain interval that contains x = –3 is x ≤ –3. Therefore, k(x) = 3x + 4 when x = –3. –3 is less than O R equal to –3; 3 is not less than itself, bu t equal to itself, a it is nd only one of those two cases has to be true. Note: Problems 15.39–15.41 refer to the piecewise-defined function k(x) below. 15.40 Evaluate k(1). When x ≥ 1, k(x) = 5. Because 1 ≥ 1, k(1) = 5. No matter what x-value you plug in that’s greater than or equal to 1, the output will be 5. The Humongous Book of Algebra Problems 343 Chapter Fifteen — Functions Note: Problems 15.39–15.41 refer to the piecewise-defined function k(x) here. 15.41 Graph k(x). Use the techniques outlined in Chapter 5, “Graphing Linear Equations in Two Variables,” to construct the graphs of y = 3x + 4, , and y = 5. Figure 15-8 illustrates the graphs as dotted lines. Figure 15-8: The graphs of the individual functions that comprise k(x). They are drawn as dotted lines because only specific pieces of each graph (to be identified in later steps) contribute to the graph of k(x). Remember, ≤ and ≥ mean solid dots on the graph , and < and > mea n hollow dots. Darken the portions of each graph that represent portions of k(x). In other words, darken the interval of y = 3x + 4 for which x ≤ –3, the portion of the graph left of the vertical line x = –3. The inequality symbol in the statement x ≤ 3 indicates that a closed point should be used at the endpoint x = –3, as illustrated in Figure 15-9. Similarly, darken the portion of y = 5 for which x ≥ 1 and indicate a closed endpoint. Finally, darken the portion of between x = –3 and x = 1. The corresponding inequality statement (–3 < x < 1) indicates that the interval has open endpoints. The graphs are illustrated in Figure 15-9. ngous Book of Algebra Problems 344 The Humo Chapter Fifteen — Functions Figure 15-9: The darkened interval of each graph represents a portion of the graph of k(x). To graph k(x), draw the darkened portions of each linear graph in Figure 15-9 on a single coordinate plane, as illustrated by Figure 15-10. Figure 15-10: The graph of k(x). The Humongous Book of Algebra Problems 345 Chapter Fifteen — Functions Note: Problems 15.42–15.43 refer to the relation f(x) defined below. 15.42 Evaluate f(–1). When x ≤ 2, f(x) = 3x 2 – 6x + 5. Note: Problems 15.42–15.43 refer to the relation f(x) defined below. 15.43 For what value of a is f(x) is a function? Substituting any real number x into f(x) produces a single output except when x = 2. The domain restrictions in this function are x ≤ 2 and x ≥ 2; the value x = 2 satisfies both conditions, so x = 2 can either be substituted into 3x 2 – 6x + 5 or . Therefore, both expressions must be equal when x = 2 to ensure that f(x) is a function. If f(x) is a function, each input can only have one output. That mea ns plugging x = 2 into f(x) has to produce O NE real number outp ut. Unfortunately, tw o different rules a pply to x = 2, so both of them have to eq ual the same thing when you plug in x = 2. Substitute x = 2 into the equation and solve for a. ngous Book of Algebra Problems 346 The Humo Chapter 16 Graphing functions Drawing graphs that aren’t lines Graphing linear equations—and therefore graphing linear functions—is a straightforward task. There are numerous techniques by which the graphs can be constructed (described in Chapter 5), but the behavior of a linear graph is predictable. Other functions, however, present a challenge. Cubic, absolute value, and square root functions, to name a few, have distinctly different graphs. Rather than design algorithms by which each classification of function should be graphed, it is more useful to use one of two techniques to plot function graphs: completing a table of values (that is, the brute force technique) to plot an extremely accurate graph or applying function transformations to create a relatively accurate sketch of the graph quickly. Graphing functions is tricky. It would n’t be if all functions were lines, because let’s face it, linear graphs are the easiest to create. All you nee d is one point and the slope of the line an d you’re in business! However, most fu nctions aren’t linear, so you need a few mo re graphing strategies. There are basically two ways to dra w a function graph. One is to plug in a lot of x-values into a function f(x) to see what numbers you get as a result. Plug in enough x’s and you can connect the dots to make a graph. The downsid e? A lot of calculations and a big time invest ment. There is a faster way to graph that involves memorizing a few basic func tion graphs and then using transformat ions to stretch, squish, move, and flip them to make new graphs. For example, af ter you memorize the simple graph of y = x3, you can graph things like f(x) = x3 + 3 6, g(x) = (x – 7) , and h(x) = –2x3 in the blink of an eye. Chapter Sixteen — Graphing Functions Graphing with a Table of Values Plug in a bunch of things for x Note: Problems 16.1–16.2 refer to the function f(x) = (x – 3)2. 16.1 Complete the table of values. In other words, plug 0, 1, 2, 3, 4, 5, and 6 into f(x) one at a time and simplify. Evaluate f(x) for each x-value in the left column. Note: Problems 16.1–16.2 refer to the function f(x) = (x – 3)2. 16.2 Graph f(x) using the table of values completed in Problem 16.1. The numbers in the left column of the table of values are the x-values of each point, and the numbers in the right column are the matching y’s. According to the table of values in Problem 16.1, f(0) = 9, f(1) = 4, f(2) = 1, f(3) = 0, f(4) = 1, f(5) = 4, and f(6) = 9. Therefore, the graph of f(x), presented in Figure 16-1, passes through the following points: (0,9), (1,4), (2,1), (3,0), (4,1), (5,4), and (6,9). ngous Book of Algebra Problems 348 The Humo Chapter Sixteen — Graphing Functions Figure 16-1: The graph of f(x) = (x – 3)2. Note: Problems 16.3–16.4 refer to the function . 16.3 Complete the table of values. Evaluate g(x) for x = –2, x = –1, x = 0, x = 1, and x = 2. The Humongous Book of Algebra Problems 349 Chapter Sixteen — Graphing Functions Note: Problems 16.3–16.4 refer to the function . 16.4 Graph g(x) using the table of values completed in Problem 16.3. According to the table of values in Problem 16.3, the graph of g(x), presented in Figure 16-2, passes through the following points: (–2,0), (–1,1), (0,2), (1,1), and (2,0). Figure 16-2: The graph of Note: Problems 16.5–16.6 refer to the function 16.5 Complete the table of values. ngous Book of Algebra Problems 350 The Humo . . Chapter Sixteen — Graphing Functions Evaluate f(x) for the x-values in the left column of the table. Note: Problems 16.5–16.6 refer to the function . 16.6 Graph f(x) using the table of values completed in Problem 16.5. According to the table of values in Problem 16.5, the graph of f(x), presented in Figure 16-3, passes through the following points: (–3,–4), , (1,4), and , (–1,0), . Figure 16-3: The graph of . The Humongous Book of Algebra Problems 351 Chapter Sixteen — Graphing Functions Note: Problems 16.7–16.8 refer to the function . 16.7 Complete the table of values. Evaluate g(x) for the x-values in the left column of the table. ngous Book of Algebra Problems 352 The Humo Chapter Sixteen — Graphing Functions Note: Problems 16.7–16.8 refer to the function . 16.8 Graph g(x) using the table of values completed in Problem 16.7. According to the table of values in Problem 16.7, the graph of g(x) passes through the following points: , (–2,–1), (–1,–2), (1,2), (2,1), and Additional points may be used to determine the shape of the graph, illustrated in Figure 16-4, if necessary. . There’s a better way to graph functions w x in the denomina ith Check out Proble tor. ms 16.30, 16.33, and 16.34. Unless you already knew that functi on with a denomina s tor of x looked like Fi gure 16-4, the six poin ts you got from the tabl e of values probabl y aren’t enough to draw the graph. When that happens, pick mor e values of x (like –4, –5, –6, 4, 5, and 6) a nd plot more points. Figure 16-4: The graph of . The Humongous Book of Algebra Problems 353 Chapter Sixteen — Graphing Functions Domain and Range of a Function This time the book doesn’t give you x-values to plug into the function. You don’t need them—just make sure you use enough points to visualize the shape of the graph. What can you plug in? What comes out? Note: Problems 16.9–16.11 refer to the function . 16.9 Graph f(x) using a table of values. Construct a table of values. As illustrated by Figure 16-5, the graph of f(x) passes through the coordinate pairs generated by the table of values: (–4,0), (–3,1), , , (0,2), and . Figure 16-5: The graph of ngous Book of Algebra Problems 354 The Humo . Chapter Sixteen — Graphing Functions Note: Problems 16.9–16.11 refer to the function . 16.10 Identify the domain of f(x). The domain of a function is the set of real numbers for which f(x) is defined. A square root, and in fact any root with an even index, is defined only when the radicand is nonnegative. Therefore, f(x) is defined only when x + 4 ≥ 0. Solve the inequality. Another method by which to identify the domain of a function is via its graph. Consider the graph of f(x) in Figure 16-5. Any vertical line drawn on the coordinate plane that intersects the graph represents a member of the domain. The vertical line x = –4 is the leftmost vertical line that intersects f(x), so x = –4 belongs to the domain of the function. All vertical lines to the right of x = –4 intersect f(x) as well, so the domain of f(x) is x ≥ –4. Note: Problems 16.9–16.11 refer to the function In other words, the domain is the collection of numbers you’re allowed to plug into the function. The radicand is the expression inside the radical, in this case x + 4. . 16.11 Identify the range of f(x). Problem 16.10 states that vertical lines intersecting the graph of a function represent members of that function’s domain. Similarly, horizontal lines that intersect a graph represent members of the range. Consider the graph of f(x) in Figure 16-5. The horizontal line y = 0 and all of the horizontal lines above it intersect f(x), so the range of the function is f(x) ≥ 0. Instead of writing y ≥ 0 for the range, use f(x) ≥ 0 instead. Technically, the function doesn’t contain y, so the answer shouldn’t contain y either. A function’s range is the collection of all its possible outputs. For instance, if r(5) = –12 for some function r(x), then 5 is a member of the domain and –12 is a member of the range. It might not look like the horizontal lines y = 4, y = 5, and y = 6 intersect the graph, but they do. A square root graph increases steadily (if slowly) forever and ever as the graph travels right. The Humongous Book of Algebra Problems 355 Chapter Sixteen — Graphing Functions Note: Problems 16.12–16.14 refer to the function . 16.12 Graph g(x) using a table of values. Construct a table of values. As illustrated by Figure 16-6, the graph of g(x) passes through the coordinate pairs generated by the table of values: (0,5), (1,4), (2,3), (3,2), (4,3), (5,4), and (6,5). Figure 16-6: The graph of Note: Problems 16.12–16.14 refer to the function . . 16.13 Identify the domain of g(x). Consider the graph of g(x) in Figure 16-6. Any vertical line drawn on the coordinate plane will intersect the graph. Therefore, all real numbers comprise the domain of g(x). ngous Book of Algebra Problems 356 The Humo Chapter Sixteen — Graphing Functions Note: Problems 16.12–16.14 refer to the function . At the point (3,2) 16.14 Identify the range of g(x). Consider the graph of g(x) in Figure 16-6. The horizontal line y = 2 intersects the graph, as do all the horizontal lines above y = 2. Therefore, the range is g(x) ≥ 2. 16.15 Identify the domain of . Rational functions (that is, functions defined as fractions) are defined for all real numbers, except values that make their denominators equal zero. Set the denominator of j(x) equal to 0 and solve the equation. Function j(x) is undefined when x = 8. Therefore, the domain of j(x) is all real numbers except 8. 16.16 Identify the domain of . Problems 16.10 and 16.13 use the graph of a function to find its domain, and Problems 16.15– 16. 20 show you how to identify the domain without a graph. Range is a different story— graphing is often the best way to identify the range, especially when the functions are complicated. A square root is undefined when its radicand is negative. Therefore, the expressions x – 9 and 4x + 3 both must be greater than or equal to 0. If you plug x = 8 into j(x), you get . Function h(x) is only defined when x ≥ 9 and x ≥ 9. 16.17 Identify the domain of , so the domain of h(x) is You’re not allowed to divide by 0, so x = 8 is excluded from the domain. . The domain of a rational function consists of all real numbers except the values that cause the denominator to equal zero. Set the denominator equal to zero and solve the equation to identify the values that must be explicitly excluded from the domain. x2 – x – 4 = 0 Solve the equation using the quadratic formula. If a number is greater than or equal to 9, it’s automatically greater than or equal to , so there’s no need to include the fraction in the answer. The Humongous Book of Algebra Problems 357 Chapter Sixteen — Graphing Functions The domain of p(x) is all real numbers except When you reverse the sides of an inequality, you have to reverse the inequality sign as well, changing ≥ to ≤. 16.18 Identify the domain of and . . Function k(x) is undefined when the radicand in the numerator is negative, so identify the values of x for which 2 – x is nonnegative. The function is also undefined when the denominator is equal to zero. Identify those values to explicitly exclude them from the domain. , and according to the numerator, only numbers less than or equal to 2 belong in the domain. is less than 2, so you have to exclude it, but is excluded automatically. The denominator of k(x) equals zero when the function is x ≤ 2, 16.19 Identify the domain of or . The domain of . . Function q(x) is undefined when the denominator is equal to zero or when the radicand is negative. When you’re finding the domain, the two big things you look for are zero in the denominator and negatives inside radicals (when the radicals have an even index—negatives inside a cube root, for example, are fine). To identify values of x for which the radicand is positive, solve the inequality 3x 2 + 8x – 3 > 0 using the technique described in Problems 14.35–14.43. Begin by solving the equation 3x 2 + 8x – 3 = 0 to identify the critical numbers of the radicand. ngous Book of Algebra Problems 358 The Humo Chapter Sixteen — Graphing Functions The expression 3x 2 + 8x – 3 is positive when x < –3 or , so those intervals comprise the domain of q(x). Note that the answer “x ≤ –3 or x incorrect, as that domain includes the critical numbers –3 and the denominator to equal zero. Factor by decomposition— find two numbers that have a sum of 8 and a product of –9 (in other words 9 and –1) and go from there. Look at Problems 12.42–12.44 for more information about factoring by decomposition. ” is , which cause 16.20 Identify the domain and range of r(x) given its graph in Figure 16-7. See Problems 13.35–13.43 to review quadratic inequalities Figure 16-7: The graph of a function identified only as r(x). Vertical lines drawn on the coordinate plane that intersect the graph indicate members of the function’s domain. Any vertical line drawn on Figure 16-7 intersects r(x) except for the line x = 1, as the point (1,2) is specifically excluded from the graph. Therefore, the domain of r(x) is all real numbers except x = 1. Horizontal lines drawn on the coordinate plane that intersect the graph indicate members of the function’s range. The highest point reached by the The Humongous Book of Algebra Problems 359 Chapter Sixteen — Graphing Functions graph of r(x) is (–4,5), and any horizontal line below (and including) y = 5 intersects the graph. This is even true for y = 2. Although the graph excludes point (1,2), the horizontal line y = 2 passes through the graph of r(x) at two other points. Therefore, the range is r(x) ≤ 5. Symmetry Pieces of a graph are reflections of each other 16.21 Complete the graph of f(x) in Figure 16-8 assuming it is y-symmetric. If you graphed f(x) on a sheet of paper and folded the paper along the y-axis, the left and right sides of f(x) would overlap. Figure 16-8: A portion of the graph of f(x). A y-symmetric function is a reflection of itself across the y-axis, so if the graph passes through point (x,y), it passes through the corresponding point (–x,y) as well. For instance, the graph of f(x) in Figure 16-8 passes through point (4,–3), so it must pass through (–4,–3) as well. Figure 16-9 presents the completed graph of f(x). ngous Book of Algebra Problems 360 The Humo Chapter Sixteen — Graphing Functions Figure 16-9: The graph of f(x) is y-symmetric, so the segments of the graph left and right of the y-axis are reflections of one another. 16.22 Complete the graph of g(x) in Figure 16-10 assuming that it is symmetric about the origin. Figure 16-10: A portion of the graph of g(x). The Humongous Book of Algebra Problems 361 Chapter Sixteen — Graphing Functions Start at the origin and trace your finger left along the graph. Whatever the function does as you travel left, it does the opposite as you travel right of the origin. In this case, as you travel left from the origin, the function goes up to (–3,5) and then down to (–6,0). The right side does the opposite— it goes DO WN to (3,–5) and then UP to (6, 0). Origin-symmetric functions have this property: If the graph passes through point (x,y), then it passes through the corresponding point (–x,–y) as well. For instance, the graph of g(x) passes through point (–3,5), so it must pass through the point (3,–5) as well. The completed graph is presented in Figure 16-11. Figure 16-11: The graph of g(x) is origin-symmetric. ngous Book of Algebra Problems 362 The Humo Chapter Sixteen — Graphing Functions 16.23 Complete the graph of h(x) in Figure 16-12 assuming that it is symmetric about the x-axis. Figure 16-12: A portion of the graph of h(x). If a relation is symmetric about the x-axis, then the segments of its graph above and below the x-axis are reflections of one other. Therefore, an x-symmetric function passing through point (x,y) will pass through the corresponding point (x,–y) as well. For instance, h(x) passes through point (1,2), so it also must pass through point (1,–2). The completed graph is presented in Figure 16-13. h(x) isn’t a function becaus e it fails the vertic a line test (as do m l ost x-symmetric gra phs). The vertical line te states that vert st ical lines drawn on gr aph functions can inte s of rs those graphs only ect on at most. Vertica ce l lines right of x = –5 w ill intersect h(x) tw ice, h(x) is not a func so tion. The Humongous Book of Algebra Problems 363 Chapter Sixteen — Graphing Functions Here’s how to figure out if f(x) is odd: replace all of the x’s with –x. You should end up with the exact opposite of f(x). In this case you get –4x3 + x, which is the opposite of 4x3 – x because the pairs of corresponding terms (–4x3 and 4x3, +x and –x) are opposites. If (x,y) and (–x,y) are on the graph of g(x), then g(x) = y and g(–x) = y. If you substitute –x into g(x), you should end up with the same expression (y). Figure 16-13: The x-symmetric graph of h(x). 16.24 Verify that f(x) = 4x 3 – x is an odd function. A function is odd if and only if its graph is origin-symmetric. According to Problem 16.22, an origin-symmetric graph that passes through point (x,y) also passes through point (–x,–y). In other words, substituting –x into f(x) should result in –f(x). Because f(–x) = –f(x), function f(x) is odd. 16.25 Verify that is an even function. A function is even if and only if its graph is y-symmetric. According to Problem 16.21, a y-symmetric graph that passes through point (x,y) must pass through the corresponding point (–x,y) as well. In other words, substituting x and –x into g(x) produces the same result, so g(–x) = g(x). ngous Book of Algebra Problems 364 The Humo Chapter Sixteen — Graphing Functions Because g(x) = , g(x) is an even function. Fundamental Function Graphs Even functions usually have even powers in them like this, and odd functions usually have odd powers. However, the powers alone are not enough evidence to classify the function as even or odd. The graphs you need to understand most 16.26 Graph the function f(x) = x 2 using a table of values. Identify the domain and range of the function, as well as the type of symmetry (if any) the graph exhibits. Figure 16-14 illustrates the graph of f(x) = x 2 and the table of values used to generate the graph. You should memorize the graphs in Problems 16.26–16.30. If you have to use a table of values to draw these five graphs every time, the rest of the chapter will take a whole lot longer. Figure 16-14: The graph of a quadratic function is a parabola. The domain of f(x) is all real numbers; the range is f(x) ≥ 0, as squaring a real number always produces a nonnegative result. The graph is symmetric about the y-axis, as squaring a real number x and its opposite –x both produce the same result: f(–x) = f(x). The Humongous Book of Algebra Problems 365 Chapter Sixteen — Graphing Functions 16.27 Graph the function g(x) = x 3 using a table of values. Identify the domain and range of the function, as well as the type of symmetry (if any) the graph exhibits. Figure 16-15 illustrates the graph of f(x) = x 3 and the table of values used to generate the graph. Figure 16-15: The graph of the cubic function, g(x) = x3. The domain and range of g(x) both consist of all real numbers—you may cube both positive and negative real numbers, and doing so produces both positive and negative results, respectively. The graph is origin-symmetric, as cubing a number x and its opposite –x produce opposite results: g(–x) = –g(x). 16.28 Graph the function using a table of values. Identify the domain and range of the function, as well as the type of symmetry (if any) the graph exhibits. Figure 16-16 illustrates the graph of generate the graph. ngous Book of Algebra Problems 366 The Humo and the table of values used to Chapter Sixteen — Graphing Functions Figure 16-16: T he graph of the absolute value function has a cusp—a sharp point or corner—at x = 0. The domain of h(x) is all real numbers, and the range is h(x) ≥ 0—every real number has an absolute value, and it is a nonnegative number. The graph of h(x) is symmetric about the y-axis, as the absolute value of a real number x and its opposite –x are equal: h(–x) = h(x). 16.29 Graph the function using a table of values. Identify the domain and range of the function, as well as the type of symmetry (if any) the graph exhibits. Figure 16-17 illustrates the graph of generate the graph. and the table of values used to The Humongous Book of Algebra Problems 367 Chapter Sixteen — Graphing Functions Figure 16-17: The graph of the square root function is located within the first quadrant of the coordinate plane. The domain of j(x) is x ≥ 0, and the range is j(x) ≥ 0—you can only take the square root of a nonnegative number, and the result is a nonnegative number. The graph exhibits no x-, y-, or origin-symmetry. 16.30 Graph the function using a table of values. Identify the domain and range of the function, as well as the type of symmetry (if any) the graph exhibits. Figure 16-18 illustrates the graph of generate the graph. and the table of values used to Figure 16-18: The graph of k(x) intersects neither the x-axis nor the y-axis. ngous Book of Algebra Problems 368 The Humo Chapter Sixteen — Graphing Functions The domain and range of k(x) both consist of all real numbers except for 0. The graph is symmetric about the origin, as the reciprocal of a number x and its opposite –x are opposites: k(–x) = –k(x). Graphing Functions Using Transformations Move, stretch, squish, and flip graphs 16.31 Transform the graph of f(x) = x 2 to graph the function g(x) = x 2 – 5. The only difference between functions f(x) and g(x) is a constant, –5. Adding or subtracting a constant c from a function affects its graph—all points on the graph are shifted c units up or down, respectfully, on the coordinate plane. In this problem, the graph of g(x) is exactly 5 units below the graph of f(x), as illustrated by Figure 16-19. Figure 16-19: The graphs of f(x) = x2 and g(x) = x2 – 5 are equivalent, except that g(x) is five units below f(x). 16.32 Transform the graph of f(x) = x 2 to graph the function h(x) = (x – 3)2. Whereas only x is squared in function f(x), the expression x – 3 is squared in h(x). When a constant c is added to, or subtracted from, the input x of a function, the corresponding graph is shifted –c units to the right or left, as illustrated by Figure 16-20. You can’t have 0 in the denominator, so the domain must exclude 0. There’s no number you can divide into 1 to get a quotient of 0, so you have to exclude 0 from the range as well. If you subtract 3 from the input of a function, it moves the graph 3 units to the RIGHT. Adding 3 to the input would move it 3 units to the LEFT. That’s the opposi te of what you might think— negative numbers move things right and positi numbers move th ve ings left. The Humongous Book of Algebra Problems 369 Chapter Sixteen — Graphing Functions Figure 16-20: Move the graph of f(x) = x2 three units to the right to create the graph of h(x) = (x – 3)2. 16.33 Transform the graph of to graph the function . The function g(x) is generated by adding 1 to f(x). According to Problem 16.31, adding the constant 1 to a function moves its graph up one unit, as illustrated by Figure 16-21. Figure 16-21: Move the graph of ngous Book of Algebra Problems 370 The Humo . up one unit to generate the graph of Chapter Sixteen — Graphing Functions 16.34 Transform the graph of to graph the function . The reciprocal function f(x) divides 1 by the number substituted into the function, x; in h(x), a constant (1) is added to the input (x). Problem 16.32 states that adding a constant c to the input of a function shifts the graph –c units horizontally, so the graph of h(x) is equivalent to the graph of f(x) shifted one unit to the left, as illustrated by Figure 16-22. Figure 16-22: The graphs of 16.35 Graph the function and Let’s say you know the graph of a function f(x). That means you automatically know the graph of f(x) + 2 (move the graph of f(x) up 2 units), f(x) – 3 (move it down 3 units), f(x + 6) (move it left 6 units), and f(x – 4) (move it right 4 units). . . Transform the graph of , presented in Figure 16-17. Multiplying a function by a constant—in this case multiplying by 3—stretches its graph vertically. Each point on the graph of j(x) is three times as far from the x-axis as the corresponding point on . The ˆ symbol used in is meant to disting ui it from the given sh function j(x). The symbol itself mea ns nothing. It’s desc ribing a different (but ve similar) function ry usin a different (but g very similar) name. Multiply the y-values of the po ints by 3. The graph of contain points (0,0), (1,1) a nd (4,2), so the gr s the j(x) contains the po aph of ints (0,0 3), (1,1 ˙ 3) and (4,2 3). ˙ , ˙ The Humongous Book of Algebra Problems 371 Chapter Sixteen — Graphing Functions Figure 16-23: Each point on the graph of is three times as far from the x-axis as the corresponding point on the graph of . The graph of passes through the points (–1,–1), (1,1), and (2,8). Multiply the y-values of each point by to get points on f(x): , , 16.36 Graph the function . According to Problem 16.35, multiplying a function by a constant affects how far each point on the graph is from the x-axis. In this case, each point on the graph of is one-half the distance from the x-axis as the corresponding point on . Furthermore, multiplying a function by a negative number reflects the graph across the x-axis, as illustrated by Figure 16-24. and (2,–4). Figure 16-24: The graph of ngous Book of Algebra Problems 372 The Humo . Chapter Sixteen — Graphing Functions 16.37 Graph the function . Consider the function . Multiplying the input x of a function by –1 to get reflects the graph about the y-axis. For instance, contains points (1,1) and (4,2) so contains points (–1,1) and (–4,2), as illustrated by Figure 16-25. Figure 16-25: The graph of . 16.38 Graph the function h(x) = (x – 4)2 + 2. To transform into h(x), you subtract 4 from the input x and add 2 to the result. These operations shift the graph of four units right and two units up, as illustrated by Figure 16-26. Figure 16-26: The graphs of h(x) = (x – 4)2 + 2 and . The Humongous Book of Algebra Problems 373 Chapter Sixteen — Graphing Functions 16.39 Graph the function Multiplying x by a negative flips the graph across the y-axis. Multiplying the function itself by a negative flips the graph across the x-axis. . To transform the basic square root function into f(x), multiply the function by –1, add 1 to the input, and subtract 4 from the result. Each of those transformations is accompanied by a change in the graph: multiplying by –1 reflects graph across the x-axis, adding 1 shifts the graph one unit to the left, and subtracting 4 shifts the graph 4 units down. Figure 16-27 illustrates the graph of fˆ ( x ) and the transformed graph of f(x). Figure 16-27: The graphs of and . Absolute Value Functions These graphs might have sharp points 16.40 Graph the function . To transform the basic absolute value function into g(x), multiply it by 2 (which stretches the graph vertically, moving points twice as far from the x-axis), add 3 to the input (which shifts the graph 3 units left), and subtract 5 from the result (which shifts the graph down 5 units). The graph of g(x) is illustrated in Figure 16-28. 374 The Humongous Book of Algebra Problems Chapter Sixteen — Graphing Functions Figure 16-28: The graphs of 16.41 Graph the function and . . To identify the x-value of the vertex (the sharp point) on a linear absolute value graph, set the absolute value expression equal to zero and solve. Substitute into the function to calculate the y-value of the vertex. The vertex of the graph is an x-value greater than . Substitute an x-value less than and into h(x) to identify a point left of the vertex and a point right of the vertex (such as x = –5 and x = 0). Graph transformations are not always the easiest way to graph an absolut e value function. T he worked fine in Prob y 16.40, but when th lem e coefficient of x isn ’t 1 (like in Problem 16 .4 the quickest way 1), to graph is to find th e vertex and then plot one point on either side of it. The Humongous Book of Algebra Problems 375 Chapter Sixteen — Graphing Functions According to these calculations, the graph of h(x) also passes through points (–5,0) and (0,–4). Draw two rays, each starting at the vertex and extending through the points identified above to complete the graph, as illustrated by Figure 16-29. In other words, j(x) = . Figure 16-29: The graph of 16.42 Graph the function When you graph a function that’s completely inside absolute values, ignore the absolute values at first and draw the graph. Then erase any part of the graph that’s below the x-axis and redraw its mirror image above the x-axis. . . The absolute value expression in j(x) can be graphed using transformations: . Because is contained within absolute values, its output must be nonnegative. Consider the graphs of j(x) and in Figure 16-30; the graph of j(x) is generated by reflecting the portion of below the x-axis (that is, when ) across the axis so that the entirety of the graph of j(x) is nonnegative). ngous Book of Algebra Problems 376 The Humo Chapter Sixteen — Graphing Functions Figure 16-30: No portion of the graph of j(x) lies below the x-axis. Note: Problems 16.43–16.44 refer to the graph of f(x) in Figure 16-31. Figure 16-31: The graph of a function f(x). 16.43 Graph . As Problem 16.42 explains, the graph of the absolute value of a function is generated by graphing the original function—in this case f(x)—and then reflecting any negative segments of the graph across the x-axis. The graph of f(x) is illustrated in Figure 16-32. The Humongous Book of Algebra Problems 377 Chapter Sixteen — Graphing Functions Figure 16-32: R eflect any portion of the graph of f(x) that is below the x-axis across the axis to plot the graph of . Note: Problems 16.43–16.44 refer to the graph of f(x) in Figure 16-31. 16.44 Graph To graph , first graph f(x). Then erase any part of the graph left of the y-axis. Replace the missing left side with a mirror image of the right side. Think of it like this: is a function that gives you the same thing when you plug in a number and its opposite: , so f(–2) = f(2). That means the graph has to be ysymmetric. . As demonstrated by Problems 16.42 and 16.43, taking the absolute value of a function replaces the segments of its graph below the x-axis with reflections of those segments across the x-axis. Similarly, taking the absolute value of the input (x) of a function replaces the segment of its graph that is left of the yaxis with a reflection of the remaining graph across the y-axis, as illustrated by Figure 16-33. Figure 16-33: R eflect the portion of f(x) for which x ≥ 0 across the y-axis to generate the graph of ngous Book of Algebra Problems 378 The Humo . Chapter 17 Calculating Roots of Functions Roots = solutions = x-intercepts Chapter 7 was dedicated primarily to the solution of linear equations (equations with degree one) and Chapter 14 almost entirely focused on the solution of quadratic equations (which have degree two), so it is moderately surprising that equations (and functions) of degree three and higher do not each require separate solution techniques. Instead, solutions can be determined analytically by means of a common set of theorems, including the Fundamental Theorem of Algebra, the remainder theorem, the leading coefficient test, Descartes’ rule of signs, and the rational root test. There might be a quadratic formu la, but there’s no cubic formula that solves cubic equations using only the ir coefficients—the same goes for equations of degree four, degree fiv e, and so on. One thing you should kno w: Instead of calculating the solutions of equations, most of the time you calculate the roots of functions. Essentially, it means the same thing—you just use different words. Transforming an equation into a fu nction is simple—just solve the equa tion for 0. For example, to solve the equation 2 x = 9, sub tra ct 9 fro m bot h sides (to get x2 – 9 = 0), and instead of writing 0, give the function a name (so you en d up with f(x) = x2 – 9). The solutions of the equation x2 = 9 and the roots of the function f(x) = x2 – 9 are equal: x = –3 or x = 3. To calculate the roots of functions, use a combination of different tes ts to learn about the function and its gra ph. Do its ends point in the same dir ection, or do they go different directions, like one going up and one going down ? How many positive roots are there? How many are negative? How in the world do you factor a polynomial like x3 + 13x2 + 31x – 45 ? Chapter Seventeen — Calculating Roots of Functions Identifying Rational Roots Factoring polynomials given a head start 17.1 Given a polynomial of degree n that has real number coefficients, what is guaranteed by the Fundamental Theorem of Algebra? The Fundamental Theorem of Algebra states that the polynomial will have exactly n complex roots. It is an existence theorem, merely guaranteeing that those roots exist but providing no means by which to calculate them. Neither can the theorem be used to further classify the roots, such as to determine how many of the complex roots are real numbers or how many are rational. 17.2 According to the remainder theorem, if the quotient (ax 3 + bx 2 + cx + d) ÷ A “polynomial written in terms of x” is a polynom ia containing x’s and l no other variables. The number left over when you divide a polynomial by (x – m) is equal to the number you get out when you plug x = m into the polynomial. If you need synthetic division practice, check out Problems 11.38–11.45. (x – m) has remainder r, what conclusion can be drawn? Assume that a, b, c, d, and m are real numbers. The remainder theorem states that substituting x = m into a polynomial written in terms of x produces a value exactly equal to the remainder when that polynomial is divided by x – m. In this example, dividing ax 3 + bx 2 + cx + d by x – m produces remainder r, so the remainder theorem states that substituting x = m into the expression results in the same value. a(m)3 + b(m)2 + c(m) + d = r Note: Problems 17.3–17.4 refer to the polynomial x3 – 3x2 + 7x – 4. 17.3 Evaluate the expression for x = 5 using the remainder theorem. Verify your answer. According to the remainder theorem, the remainder of (x 3 – 3x 2 + 7x – 4) ÷ (x – 5) is equal to the expression x 3 – 3x 2 + 7x – 4 evaluated for x = 5. Calculate the quotient using synthetic division. Therefore, x 3 – 3x 2 + 7x – 4 = 81 when x = 5. Verify this by actually substituting x = 5 into the expression and simplifying. ngous Book of Algebra Problems 380 The Humo Chapter Seventeen — Calculating Roots of Functions Note: Problems 17.3–17.4 refer to the polynomial x3 – 3x2 + 7x – 4. 17.4 Evaluate the expression for x = –2 using the remainder theorem. Calculate the remainder of (x 3 – 3x 2 + 7x – 4) ÷ (x + 2) using synthetic division. Therefore, (–2)3 – 3(–2)2 + 7(–2) – 4 = –38. Instead of plugging x = –2 into the expression, plug it into a synthetic division box—the signs match. Use the opposite of –2 to write the divisor: x + 2. Note: Problems 17.5–17.9 refer to the function f(x) = x3 + 13x2 + 31x – 45. 17.5 Evaluate f(1) using the remainder theorem. Apply synthetic division. The remainder is 0, so f(1) = 0. Note: Problems 17.5–17.9 refer to the function f(x) = x3 + 13x2 + 31x – 45. 17.6 Explain why x = 1 is a root of f(x). If f(c) = 0, then c is a root of function f(x). According to Problem 17.5, f(1) = 0, so x = 1 is a root of the function. Roots are also called “zeroes” because plugging them into the function makes the function equal 0. Note: Problems 17.5–17.9 refer to the function f(x) = x3 + 13x2 + 31x – 45. 17.7 Explain why x – 1 is factor of x3 + 13x 2 + 31x – 45. According to Problem 17.5, f(x) ÷ (x – 1) has remainder 0. Because f(x) is evenly divisible by x – 1, by definition, x – 1 is a factor of f(x). C is a factor of D if C divides evenly into D. In other words, D ÷ C has no remainder. The Humongous Book of Algebra Problems 381 Chapter Seventeen — Calculating Roots of Functions Note: Problems 17.5–17.9 refer to the function f(x) = x3 + 13x2 + 31x – 45. To factor the function, take th e root you’re given (in this case 1), divid e out synthetically it , make a quadratic poly no out of the numbe mial rs get (in this case you x2 + 14x + 45), a nd then try to factor th at quadratic. 17.8 Factor f(x). According to Problem 17.5, (x 3 + 13x 2 + 31x – 45) ÷ (x – 1) = x 2 + 14x + 45. Therefore, x 3 + 13x 2 + 31x – 45 = (x – 1)(x 2 + 14x + 45). Factor the quadratic expression. (x – 1)(x 2 + 14x + 45) = (x – 1)(x + 9)(x + 5) Therefore, the factored form of x 3 + 13x 2 + 31x – 45 is (x – 1)(x + 9)(x + 5). Note: Problems 17.5–17.9 refer to the function f(x) = x3 + 13x2 + 31x – 45. 17.9 Identify all three roots of f(x). According to Problem 17.8, f(x) = (x – 1)(x + 9)(x + 5). Substitute f(x) = 0 into the equation, use the zero product property to set each factor equal to 0, and solve the individual equations. Problems 17.28–17.34 explain how to factor a cubic (or a function with a higher degree) when you’re not given a root (like –3) to start with. The number in the box matches the root (–3), and the number in the corresponding factor always has the opposite sign (x + 3). The roots of f(x) are x = –9, x = –5, and x = 1. Note: Problems 17.10–17.11 refer to the function g(x) = 12x3 + 25x2 – 38x – 15. 17.10 Factor g(x) given that –3 is one of its roots. If –3 is a root of g(x), then (12x 3 + 25x 2 – 38x – 15) ÷ (x + 3) has remainder 0. Apply synthetic division. Therefore, (12x 3 + 25x 2 – 38x – 15) ÷ (x + 3) = 12x 2 – 11x – 5. Factor the quadratic by decomposition. Therefore, g(x) = (x + 3)(4x – 5)(3x + 1). ngous Book of Algebra Problems 382 The Humo Chapter Seventeen — Calculating Roots of Functions Note: Problems 17.10–17.11 refer to the function g(x) = 12x3 + 25x2 – 38x – 15. 17.11 Identify all three roots of g(x). According to Problem 17.10, g(x) = (x + 3)(4x – 5)(3x + 1). Let g(x) = 0, set each factor equal to zero, and solve the resulting equations. The roots of g(x) are x = –3, , and . The roots of g(x) are the same as the solutions to the equation 12x3 + 25x2 – 38x – 15 = 0, which is the equation h(x) = 0. Note: Problems 17.12–17.13 refer to the function h(x) = 7x3 – 30x2 + 9x – 4. 17.12 Factor h(x) given that 4 is one of its roots. Calculate (7x 3 – 30x 2 + 9x – 4) ÷ (x – 4) using synthetic division. The quotient is 7x 2 – 2x + 1, so h(x) = (x – 4)(7x 2 – 2x + 1). The quadratic expression is prime. Prime = “unfactor able, so you can’t fact ” or 7x 2 – 2x + 1. Note: Problems 17.12–17.13 refer to the function h(x) = 7x3 – 30x2 + 9x – 4. 17.13 Identify all three roots of h(x). According to Problem 17.12, h(x) = (x – 4)(7x 2 – 2x + 1). Let h(x) = 0 and set both factors equal to zero. x–4=0 7x 2 – 2x + 1 = 0 The solution to the left equation is x = 4. Solving the right equation requires the quadratic formula. The Humongous Book of Algebra Problems 383 Chapter Seventeen — Calculating Roots of Functions Reduce the fraction to lowest terms. Write each complex root as the sum or difference of two fractions with the same denominator (instead of writing them as bigger fractions that contain a sum or difference in the numerator). The roots of h(x) are x = 4, , and . Leading Coefficient Test The ends of a function describe the ends of its graph 17.14 Explain how the leading coefficient test classifies the end behavior of a End behavior is what the function graph does at its far left and right sides. Polynomial functions will either increase or decrease without bound (shoot upward or shoot downward) at the edges of the graph. Look for the term containing x raised to the highest power. That power is the degree, and the coefficient of that term is the leading coefficient. function. The leading coefficient test classifies the end behavior of a function based on its degree and the sign of its leading coefficient (that is, the coefficient of the term containing the variable raised to the highest power). If the degree of the function is even, the left and right ends of the function behave the same way. Furthermore, a positive leading coefficient indicates that both ends of its graph “go up” (that is, increase without bound), whereas a negative leading coefficient indicates that both ends of the graph “go down” (that is, decrease without bound). The end behavior of a function with an odd degree differs—the ends behave oppositely. Specifically, the graph of a function with a positive leading coefficient has a left end that goes down and a right end that goes up. The converse is true for functions with a negative leading coefficient. Here’s the leadin g (LC stands for le coefficient test in a nutshell ading coefficien t): * Even degree a nd + LC: Both en ds go up. * Even degree a nd * Odd degree a – LC: Both ends go down. nd + LC: Left sid e go * Odd degree a nd – LC: Left sid es down; right side goes up. e goes up; right si de goes down. ngous Book of Algebra Problems 384 The Humo Chapter Seventeen — Calculating Roots of Functions 17.15 Describe the end behavior of f(x) = x 3 – 2 using the leading coefficient test. Verify your answer by plotting the graph of f(x). Function f(x) consists of two terms: x 3 and –2. The term containing x raised to the highest degree is x 3, so the degree of f(x) is 3, and the leading coefficient is 1. According to the leading coefficient test, the graph of a function with an odd degree and a positive leading coefficient has a left end that goes down and a right end that goes up. Plot the graph of f(x) to verify this conclusion, as illustrated by Figure 17-1. There’s no coefficient written in front of x3, so its coefficient is 1. Figure 17-1: T he graph of f(x) = x3 – 2 is the graph of y = x3 shifted down two units. The negative side of the graph decreases without bound, and the positive side of the graph increases without bound. Put simply, the left end “goes down” and the right end “goes up.” 17.16 Describe the end behavior of g(x) = x 2 + 2x + 1 using the leading coefficient test. Verify your answer by plotting the graph of g(x). The highest power of x in g(x) appears in the term x 2. Thus the degree of g(x) is 2, and the leading coefficient is 1. According to the leading coefficient test, the graph of a function with an even degree and a positive leading coefficient goes up at both ends. This conclusion is verified by the graph of g(x) in Figure 17-2. You can graph g(x) using a table of values, but if you factor you get g(x) = (x + 1)2 . You can graph that using a transformation—it’s the graph of x2 shifted left one unit. The Humongous Book of Algebra Problems 385 Chapter Seventeen — Calculating Roots of Functions Figure 17-2: T he graph of g(x) = x2 + 2x + 1 has right and left ends that “go up” (that is, increase without bound). 17.17 Describe the end behavior of h(x) = –3x 4 + 5x 2 – 2 using the leading coefficient test. Function h(x) has degree 4 and leading coefficient –3. According to the leading coefficient test, the graph of a function with an even degree and a negative leading coefficient goes down at both its right and left ends. The terms of j(x) aren’t written in order of their exponents, from highest to 3 lowest. Even though –x is written in the middle of the function instead of in the front, it still has the highest exponent. It’s okay to rewrite the function as j(x) = –x3 + 9x2 – 6 if you want. 17.18 Describe the end behavior of j(x) = 9x 2 – x 3 – 6 using the leading coefficient test. Function j(x) has degree 3 and its leading coefficient is –1. According to the leading coefficient test, the graph of a function with an odd degree and a negative leading coefficient has a left end that goes up and a right end that goes down. 17.19 Describe the end behavior of k(x) = x(2x 4 + 11x 2 – 1) using the leading coefficient test. Write k(x) as a polynomial, rather than a product of polynomials, by distributing x. k(x) = 2x 5 + 11x 3 – x The function has degree 5 and leading coefficient 2. According to the leading coefficient test, the graph of a function with an odd degree and a positive leading coefficient has a left end that goes down and a right end that goes up. ngous Book of Algebra Problems 386 The Humo Chapter Seventeen — Calculating Roots of Functions 17.20 Consider the function p(x) = a(x + b)2 + c, defined such that a, b, and c are nonzero real numbers. Describe the end behavior of p(x) if the vertex of its graph is (2,–3) and abc < 0. To plot the graph of p(x), transform the graph of y = x 2, which has vertex (0,0). The graph of p(x) has points a units farther away from the x-axis and is shifted –b units horizontally and c units vertically. Stretching the graph of y = x 2 by a multiple of a does not affect its vertex: (0,0 · a) = (0,0). However, the vertex is affected by the vertical and horizontal shifts; the vertex of p(x) is (0 – b, 0 + c). The problem states that p(x) has vertex (2,–3). Therefore, (0 – b, 0 + c) = (2,–3). Set the corresponding x- and y-values equal to calculate b and c. Substitute b = –2 and c = –3 into p(x) and expand the expression. The degree of p(x) is 2, and the leading coefficient is a. The problem states that abc < 0. Substitute b and c into that expression. Quadratics like p(x) have U-shaped graphs called parabola s. If the ends of th e parabola go up, th e vertex is the low est point of the U, th e bottom of the va lle If the ends of th y. e parabola go dow n, the vertex is the high est point of the upside-down U. To review how graph transformations work, check out Problems 16.31–16. 40. If b were equal to 7, for example, you would shift the graph of y = x2 seven units to the left. The leading coefficient of p(x) is negative. According to the leading coefficient test, the graph of a function with an even degree and a negative leading coefficient has left and right ends that go down. The leading coefficient of p(x) is a, and you just proved that it’s negative (a ‹ 0). The Humongous Book of Algebra Problems 387 Chapter Seventeen — Calculating Roots of Functions Go from left to right: 3x 4 and 5x3 are both positive, so no sign change; 5x 3 a nd –6x2 have differe nt signs, so 1 sign cha ng so far; –6x2 and x e have different signs, so 2 total sign changes ; and x and –9 ha ve different signs, so 3 total sign changes. Descartes’ Rule of Signs Sign changes help enumerate real roots 21.21 Describe how to apply Descartes’ rule of signs to determine the number of positive and negative roots of a function. To predict the number of positive roots of a function f(x), count the number of times its consecutive terms change sign and then subtract multiples of 2. For instance, the terms of the function g(x) = 3x 4 + 5x 3 – 6x 2 + x – 9 change sign three times, so it either has three positive roots or 1 positive root. To predict how many negative roots a function has, substitute –x into the function and count the number of sign changes in f(–x). That number (or an even integer fewer than that number) represents the total number of negative roots of f(x). Note: Problems 17.22–17.23 refer to the function f(x) = x2 – 7x + 13. This is the tricky part of Descartes’ rule of signs. The number of sign changes is not always the number of roots—it could have 2 fewer than the total, or 2 fewer than that, or 2 fewer than that, and so on. For example, if some function j(x) has 6 sign changes, it could have 6, 4, 2, or 0 roots. You don’t have to worry about saying “or two fewer than that, or two fewer than that, and so on.” in this case, because you can’t have fewer than no roots! 17.22 Apply Descartes’ rule of signs to predict the total number of positive roots of f(x). Compare the signs of consecutive terms, working from left to right. Note that x 2 and –7x have different signs, as do –7x and 13, for a total of two sign changes. According to Descartes’ rule of signs, f(x) may have two positive roots. However, f(x) may also have two fewer than that: 2 – 2 = 0. Therefore, f(x) has either two or zero positive roots. Note: Problems 17.22–17.23 refer to the function f(x) = x2 – 7x + 13. 17.23 Apply Descartes’ rule of signs to predict the total number of negative roots of f(x). Substitute –x into the function and simplify. Compare the signs of consecutive terms, working from left to right. The first pair of terms (x 2 and 7x) have the same sign, as do the final pair of terms (7x and 13). Because the terms of f(–x) do not change sign, f(x) has no negative roots. ngous Book of Algebra Problems 388 The Humo Chapter Seventeen — Calculating Roots of Functions Note: Problems 17.24–17.25 refer to the function g(x) = x3 + 9x2 – 6x – 5. 17.24 Apply Descartes’ rule of signs to predict the total number of positive roots of g(x). The consecutive terms of g(x) change sign only once as you work from left to right: x 3 and 9x 2 have the same sign, as do the terms –6x and –5. The only sign change occurs between the terms 9x 2 and –6x. According to Descartes’ rule of signs, g(x) has exactly one positive root. Note: Problems 17.24–17.25 refer to the function g(x) = x3 + 9x2 – 6x – 5. 17.25 Apply Descartes’ rule of signs to predict the total number of negative roots of Again, you don’t have to worry about saying “or two fewer than that.” You only have to throw in the “2 fewer than that” possibilities when there are two or more total sign changes. g(x). Identify the function g(–x) and simplify it. The terms of g(–x) change sign twice (once between –x 3 and 9x 2 and once between 6x and –5). According to Descartes’ rule of signs, g(x) has either two or zero negative roots. Note: Problems 17.26–17.27 refer to the function h(x) = –9x3 + 7x2 + 2x5 + 4 – 4x4 – 10x. 17.26 Apply Descartes’ rule of signs to predict the total number of positive roots of h(x). Before applying Descartes’ rule of signs, rewrite the polynomial so that the powers of x appear in order, from greatest to least. h(x) = 2x 5 – 4x 4 – 9x 3 + 7x 2 – 10x + 4 The function contains four sign changes, so h(x) has either four, two, or zero roots. The order of the terms affects how many sign changes there will be, so make sure the powers of x are listed in order, from the largest to the smallest. Note: Problems 17.26–17.27 refer to the function h(x) = –9x3 + 7x2 + 2x5 + 4 – 4x4 – 10x. 17.27 Apply Descartes’ rule of signs to predict the total number of negative roots of h(x). As stated in Problem 17.26, the terms must be rewritten so that the powers of x are in descending order. h(x) = 2x 5 – 4x 4 – 9x 3 + 7x 2 – 10x + 4 The Humongous Book of Algebra Problems 389 Chapter Seventeen — Calculating Roots of Functions Substitute –x into h(x). The rational root test gives yo u a list of possible roots. One or more num be from the list coul rs d roots of the func be tion. Maybe none of th em are, but if the fu nc has ANY roots th tion at can be expresse d as a fraction, they’ll be in this list. Factors are numbers that divide in evenly; 6 is a factor of 12 because 12 ÷ 6 = 2 with no remainder. The signs of the numbers don’t matter. Write each factor of the constant (1 and 2) over the only factor of the leading coefficient (1). The signs of consecutive terms change only once, between –4x 4 and 9x 3, so h(x) has exactly one negative root. Rational Root Test Find possible roots given nothing but a function Note: Problems 17.28–17.30 refer to the function f(x) = x3 – x2 – 4x – 2. 17.28 Apply the rational root test to identify possible rational roots of f(x). To apply the rational root test, first list all the factors of the leading coefficient and factors of the constant. In this example, f(x) has a leading coefficient of 1 (whose only factor is 1) and a constant of –2 (whose factors are 1 and 2). All rational numbers , where c is a factor of the constant and l is a factor of the leading coefficient, are possible roots of f(x). According to the rational root test, –2, –1, 1, and 2 are possible roots of f(x). Note: Problems 17.28–17.30 refer to the function f(x) = x3 – x2 – 4x – 2. 17.29 Demonstrate that x = –2 is not a root of f(x). Problem 17.28 identifies –2 as a possible root of the function. If –2 is a root of f(x), then f(–2) = 0. Evaluate f(–2). Because f(–2) ≠ 0, x = –2 is not a root of f(x). ngous Book of Algebra Problems 390 The Humo Chapter Seventeen — Calculating Roots of Functions Note: Problems 17.28–17.30 refer to the function f(x) = x3 – x2 – 4x – 2. 17.30 Demonstrate that x = –1 is a root of f(x) and use the root to factor the function. Problem 17.28 states that –1 is a possible root of f(x). Evaluate f(–1) to verify that it is. Because –1 is a root of f(x), x + 1 is a factor of f(x). Calculate the quotient f(x) ÷ (x + 1). Synthetically dividing f(x) by –1 is the equivalent of lo ng dividing f(x) by (x + 1). Therefore, f(x) = (x + 1)(x 2 – 2x – 2). The quadratic factor x 2 – 2x – 2 is prime, so f(x) cannot be factored any further. Note: Problems 17.31–17.32 refer to the function g(x) = 6x3 – 35x2 + 47x – 12. 17.31 Apply the rational root test to identify possible rational roots of g(x). Function g(x) has constant –12 (with factors 1, 2, 3, 4, 6, and 12) and leading coefficient 6 (with factors 1, 2, 3, and 6). All possible combinations of constant factors divided by leading coefficient factors (and their opposites) represent possible rational roots of g(x). Start with the fi factor of the co rst ns divide it by all th tant (1) and e leading coefficien factors of the t (1, 2, 3, and 6): Don’t forget the ± one. Then move sign in front of each on to the second fa of the constant and do the sam ctor e thing: Keep going until yo u ru of the constant. n out of factors The Humongous Book of Algebra Problems 391 Chapter Seventeen — Calculating Roots of Functions Reduce the fractions. Note that numerous roots are repeated. Eliminate the repeated roots from the list. Reorder the list of possible roots from least to greatest. Note: Problems 17.31–17.32 refer to the function g(x) = 6x3 – 35x2 + 47x – 12. 17.32 Identify one root of g(x) and use it to factor the function. Working from right to left is a good idea because you can avoid dealing with fractions for as long as possible. If you set each of the factors equal to 0 and solve, you get the three rational roots of the function: Only three of the 24 possible rational roots identified by Problem 17.31 are roots of the function—only those three values make f(x) equal zero when substituted into the function. Substitute the possible roots into g(x) until you identify one of the roots. If you work from right to left on the list, g(12) ≠ 0, g(–12) ≠ 0, g(6) ≠ 0, and g(–6) ≠ 0. However, g(4) = 0, so 4 is a root of the function. Because 4 is a root of g(x), g(x) must be divisible by x – 4. Perform synthetic division. Therefore, g(x) = (x – 4)(6x 2 – 11x + 3). Factor the quadratic expression by decomposition. Express g(x) in factored form. ngous Book of Algebra Problems 392 The Humo Chapter Seventeen — Calculating Roots of Functions Note: Problems 17.33–17.34 refer to the function h(x) = 3x3 – 28x2 + 4x + 45. 17.33 Apply the rational root test to identify possible rational roots of h(x). The function has constant 45, the factors of which are 1, 3, 5, 9, 15, and 45; the leading coefficient of h(x) is 3, which has factors 1 and 3. List all possible fractions such that a factor of 45 is the numerator and a factor of 3 is the denominator. Reduce the fractions and eliminate repeated values. List the numbers in order, from least to greatest, according to the absolute value of each. The result is the list of possible rational roots. Note: Problems 17.33–17.34 refer to the function h(x) = 3x3 – 28x2 + 4x + 45. 17.34 Factor h(x). To factor the function, you need to identify one of its roots. Of the possible roots identified by Problem 17.33, only 9 is a root of h(x), as h(9) = 0. Apply synthetic division to divide h(x) by x – 9. Therefore, h(x) = (x – 9)(3x 2 – x – 5). The quadratic 3x 2 – x – 5 is prime, so h(x) cannot be factored any further. In other words, ignore the ± sign when you rearrange the possible roots. There are 16, not 8, possible roots because the positive and negative versions of each number are included. That doesn’t mean h(x) only has the single root 9. If you use the quadratic formula to solve 3x2 – x – 5 = 0, you get the other two roots. Both of them will contain a square root and will be irrational, which means two things: (1) the rational root test can’t predict them; and (2) you can’t factor the quadratic. The Humongous Book of Algebra Problems 393 Chapter Seventeen — Calculating Roots of Functions Synthesizing Root Identification Strategies In other words, every term has the opposite sign of th e terms beside it. The maximum numbe r of sign changes you ca have is one fewer n than the numbe r of terms in the polynomial. Factoring big polynomials from the ground up Note: Problems 17.35–17.37 refer to the function f(x) = 2x3 – 12x2 + 15x – 25. 17.35 Apply Descartes’ rule of signs to predict the total number of positive and negative roots of f(x). Adjacent terms of f(x) are opposites, so there are three sign changes in f(x). According to Descartes’ rule of signs, f(x) has either three roots or one root. To determine the possible number of negative roots, substitute –x into f(x). Every term in f(–x) is negative, so there are no sign changes. Therefore, f(x) has no negative roots. Note: Problems 17.35–17.37 refer to the function f(x) = 2x3 – 12x2 + 15x – 25. 17.36 Apply the rational root test to list the possible roots of f(x). Modify the list in light of the information provided by Problem 17.35. If you’re wondering why there are no ± signs in this list, it’s not a typo—keep reading. The constant of h(x) is –25, which has factors 1, 5, and 25; the leading coefficient is 2, which has factors 1 and 2. Apply the rational root test. There are only six possible rational roots. Note that the list does not consider negative roots, because Problem 17.35 concluded that h(x) had no negative roots. Of the numbers in the list of possible rational roots, only 5 is a root of f(x), as f(5) = 0. ngous Book of Algebra Problems 394 The Humo Chapter Seventeen — Calculating Roots of Functions Note: Problems 17.35–17.37 refer to the function f(x) = 2x3 – 12x2 + 15x – 25. 17.37 According to the Fundamental Theorem of Algebra, a polynomial of degree n has exactly n complex roots. The degree of f(x) is 3, indicating that it should have exactly three complex roots. Problem 17.36 identified only one root; identify the two remaining roots. Complex roots include real numbers and imaginary numbers. According to Problem 17.36, 5 is a root of f(x). Therefore, f(x) is evenly divisible by (x – 5). Calculate the quotient using synthetic division. Thus f(x) = (x – 5)(2x 2 – 2x + 5). Identify the remaining roots by solving the equation 2x 2 – 2x + 5 = 0 via the quadratic formula. The roots of f(x) are 5, 1 – 3i, and 1 + 3i. Note: Problems 17.38–17.39 refer to the function g(x) = x4 – 2x3 – 7x2 + 8x + 12. 17.38 Apply Descartes’ rule of signs to predict the total number of positive and negative roots of g(x). Consecutive terms of g(x) change sign twice, so g(x) has either 2 or 0 positive roots. The terms of g(–x) = x 4 + 2x 3 – 7x 2 – 8x + 12 change signs twice as well, so g(x) has either 2 or 0 negative roots. g(–x) = (–x)4 – 2(–x) 3 – 7(–x)2 + 8(–x) + 12 = x4 – 2(–x3) – 7x2 – 8x + 12. The Humongous Book of Algebra Problems 395 Chapter Seventeen — Calculating Roots of Functions Instead of picking random numbers from this list, you m want to use a gr ay aphing calculator to sket ch the graph of g(x) . Pay attention to its xintercepts, beca use the roots of a functi on also the x-interc are epts of its graph. The gr of g(x) APPEARS aph to pass through x = –2 , x = –1, x = 2, and x = 3. So 3 is a root of g(x) because 3 4 – 2(3) 3 – 7(3) 2 + 8(3) + 12 = 0. When you just synthetically divided, you got x3 – 3x2 – 4x – 12 , and 3 has to be a root of that polynomial as well. This time, synthetically divide the cubic by the root instead of the original function g(x). There’s no x term because the quotient has an x-coefficient of 0. 396 Note: Problems 17.38–17.39 refer to the function g(x) = x4 – 2x3 – 7x2 + 8x + 12. 17.39 Identify all four roots of g(x). According to the rational root test, the possible roots of g(x) are ±1, ±2, ±3, ±4, ±6, and ±12. Notice that –1 is a root of g(x). Therefore, g(x) = (x + 1)(x 3 – 3x 2 – 4x + 12). Factor the cubic by identifying another root of g(x), such as 3. Thus g(x) = (x + 1)(x 3 – 3x 2 – 4x + 12) = (x + 1)(x – 3)(x 2 – 4). Factor the remaining quadratic, a difference of perfect squares: x 2 – 4 = (x + 2)(x – 2). Therefore, g(x) = (x + 1)(x – 3)(x + 2)(x – 2). Set all four factors equal to zero and solve to identify the roots of g(x): –2, –1, 2, and 3. Note: Problems 17.40–17.43 refer to the function h(x) = c(x – 4)2 + 1. Assume c is a rational number and c < 0. 17.40 Use the leading coefficient test to explain why h(x) has exactly two real roots. As explained in Problem 17.20, the graph of h(x) is the graph of y = x 2 after three transformations are performed. Each point on the graph of h(x) is c times further away from the x-axis than the graph of y = x 2. Furthermore, the graph of h(x) is shifted four units to the right and one unit up on the coordinate plane. Whereas the graph of y = x 2 has vertex (0,0), the vertex of h(x) is (4,1). The problem states that c < 0, so the graph of y = x 2 must also be reflected across the x-axis. According to the leading coefficient test, both ends of the graph of h(x) go down, because h(x) has an even degree (2) and a negative leading coefficient (c). The graph of h(x) is a downward-pointing parabola whose maximum value (1) occurs at its vertex (4,1). The domain of the function is all real numbers, because no value of x causes h(x) to be undefined. Its ends point down, so the graph must intersect the x-axis twice. Those two x-intercepts are the roots of h(x). He in a nutshell. The re’s the answer gr aph of h(x) is a sort of U-sh it’s an upside-dow aped). The ends of the parabola parabola (which is n U-shaped gr point down, w no higher than y = 1. The function aph. The vertex is (4,1), so the gr hich means aph reaches only goes DO W N fr cuts through tha om there, so the t U twice (as will x-axis all horizontal lines below y = 1). The Humongous Book of Algebra Problems Chapter Seventeen — Calculating Roots of Functions Note: Problems 17.40–17.43 refer to the function h(x) = c(x – 4)2 + 1. Assume c is a rational number and c < 0. 17.41 For what values of c might h(x) have two or zero positive roots, according to Descartes’ rule of signs? Expand h(x) by squaring (x – 4) and distributing c. –8 is negative and so is c. Multiplying two negatives gives you a positive. The function will have either two or zero positive roots when there are exactly two sign changes in h(x). Consider the coefficients of h(x) one at a time. The first term has a negative coefficient (c < 0,) and the coefficient of –8cx is positive, so there is a sign change between the first two terms. The constant term is represented by 16c + 1. For two sign changes to occur, 16c + 1 must be a negative number, as the preceding term –8cx is positive. Solve the equation 16c + 1 < 0 for c. When , the consecutive terms of h(x) exhibit two sign changes, which indicates that h(x) has either two or zero positive roots according to Descartes’ rule of signs. Note: Problems 17.40–17.43 refer to the function h(x) = c(x – 4)2 + 1. Assume c is a rational number and c < 0. 16c and 1 look like different terms but they should be considered a single, constant term, because neither contains x. It’s sort of like the polynomial x2 + 6x + 7 + 1, which is the sum of four things, but there’s actually only three terms after you add the constants 7 and 1. 17.42 For what values of c does h(x) have exactly one negative root, according to Descartes’ rule of signs? Substitute –x into h(x) and expand the function. The constant 16c + 1 equals zero when . Because c < 0, the first term of h(–x) is negative, as is the second term. If h(x) has one negative root, then h(–x) should exhibit one sign change—the constant term 16c + 1 must be positive. According to Problem 17.41, the constant term is negative when . Therefore, the constant is positive when . The Humongous Book of Algebra Problems 397 Chapter Seventeen — Calculating Roots of Functions Note: P roblems 17.40–17.43 refer to the function h(x) = c(x – 4)2 + 1. Assume c is a rational number and c < 0. 17.43 Apply the rational root test to identify all possible rational roots of h(x). As explained in Problems 17.40–17.41, the constant of h(x) is 16c + 1, and the leading coefficient is c. According to the rational root test, all rational roots of h(x) have the form , such that m is a factor of 16c + 1 and n is a factor of c. A more specific answer cannot be given, because the value of c is unknown. ngous Book of Algebra Problems 398 The Humo Chapter 18 Logarithmic Functions Contains enough logs to build yourself a cabin In preceding chapters, the vast majority of the expressions that contained exponential powers consisted of a variable raised to a real number exponent, such as x 5 or y 2. Solving equations containing such powers required a radical with an index equal to the exponent to isolate x. Functions containing variables in the exponent, such as 5x and 2y, are very different than polynomials. Chapter 19 explores equations containing such expressions, but to solve those equations, you must first understand the logarithmic function, which allows you to isolate the variable exponent. To solve an equation like x2 = 9, take the square root of both sides to get x = ±3. Radicals allow you to iso late x by eliminating the attached exponent. This chapter deals with exponents that contain variables. If you want to solve 2x = 9, you need a way to isolat e x and get rid of the 2. You can’t just divide by 2—it’s not a coefficie nt, it’s a base—so you need to use logarithms. Chapter Eighteen — Logarithmic Functions Evaluating Logarithmic Expressions Given loga b = c, find a, b, or c This expression is read “log base a of b equals c.” 18.1 Express the logarithmic equation loga b = c as an exponential equation. The logarithmic equation loga b = c is equivalent to the exponential equation a c = b. 18.2 Identify the value of n that completes the equation: log2 n = 3. According to Problem 18.1, log2 n = 3 is equivalent to the exponential equation 23 = n. 18.3 Identify the value of n that completes the equation: log6 n = –2. Rewrite the logarithmic equation as an exponential equation. A negative power means “the reciprocal of.” The reciprocal of 62 is 6 –2 = n Eliminate the negative exponent and simplify. . 18.4 Identify the value of n that completes the equation: Rewrite the logarithmic equation as an exponential equation. If you need to review how fractional exponents work, flip back to Problems 13.13 –13.17. 271/3 = n Rewrite 271/3 as a radical expression and simplify. The cube root of 27 is 3 because 3 cube d equals 27. ngous Book of Algebra Problems 400 The Humo . Chapter Eighteen — Logarithmic Functions 18.5 Identify the value of n that completes the equation: log4 16 = n. Rewrite the logarithmic equation as an exponential equation. 4n = 16 Write both sides of the equation as powers of 4. 18.6 Identify the value of n that completes the equation: log25 5 = n. Rewrite the logarithmic equation as an exponential equation. 25n = 5 If two equal bases are raised to exponents and the results are equal, the exponents must be equal as well. In other words, you can drop the bases and set the powers equal. Write both sides of the equation as powers of 5 and solve for n. 18.7 Identify the value of n that completes the equation: log10 10 = n. 5 2 is raised to the n power. When something to a power is raised to a power, multiply the exponents: (5 2)n = 5 2 ˙ n = 5 2n. Rewrite the logarithmic equation as an exponential equation. 10n = 10 Write both sides of the equation as powers of 10 and solve for n. The expression logc c always equals 1; c can be any pos-itive number except 1. 18.8 Identify the value of n that completes the equation: logn 1,000 = 3. Rewrite the logarithmic equation as an exponential equation. n 3 = 1,000 Solve for n by taking the cube root of both sides of the equation. The Humongous Book of Algebra Problems 401 Chapter Eighteen — Logarithmic Functions 18.9 Identify the value of n that completes the equation: logn 1 = 0. Rewrite the logarithmic equation as an exponential equation. n0 = 1 A logarithmic base n may be any positive number except 1; all positive numbers raised to the zero power equal 1. 18.10 Explain why a logarithmic function cannot have a base of 1. Consider the function f(x) = log1 x. Evaluate f(1). f(1) = log1 1 Rewrite the logarithmic equation as an exponential equation. 1f(1) = 1 Numerous values substituted into f(1) satisfy this equation. Setting f(1) = 2, f(1) = 5, and f(1) = –3 all produce true statements. In fact, f(1) could equal any real number and the equation 1f(1) = 1 would be true. However, a function might only produce one output value for each input. In this case, the input x = 1 does not have a single corresponding value of f(1), so f(x) is not a function. Graphs of Logarithmic Functions All log functions have the same basic shape Note: Problems 18.11–18.14 refer to the function f(x) = log2 x. 18.11 Complete this table of values. Rewrite log2 x = f(x) as an exponential function: 2f(x) = x. To complete the table of values, answer this question for each value of x: “2 raised to what power is equal to x?” The answer is the corresponding value of f(x). For instance, to evaluate equal to , answer the question, “2 raised to what power is ?” The correct answer is –2, so ngous Book of Algebra Problems 402 The Humo . Chapter Eighteen — Logarithmic Functions Note: Problems 18.11–18.14 refer to the function f(x) = log 2 x. 18.12 Use the table of values generated by Problem 18.11 to graph f(x). According to Problem 18.11, the graph of f(x) passes through the following points: , , (1,0), (2,1), (4,2), and (8,3). The graph of f(x) is presented in Figure 18-1. If you raise 2 to the y-coordinate of each point, you get the x-coordinate. For example, the point (8,3) is on the graph because 23 = 8. Figure 18-1: The graph of f(x) = log2 x. Note: Problems 18.11–18.14 refer to the function f(x) = log2 x. 18.13 Identify the domain, range, and intercept(s) of f(x). Consider the graph of f(x) in Figure 18-1. Any vertical line drawn on that coordinate plane that intersects f(x) is a member of the domain of f(x). Notice that the vertical line y = 0 does not intersect f(x), and neither does any vertical line left of the y-axis. Therefore, the domain of f(x) is x > 0. The graph gets infinitely close to the vertical line y = 0 though. It gets as close as you can possibly get without actually touching it. Lines like that are called asymptotes— they’re covered in Chapter 20. The Humongous Book of Algebra Problems 403 Chapter Eighteen — Logarithmic Functions Any log function f(x) = loga x will intersect the x-axis at (1,0) because a0 = 1 no matter what a is. Reflecting across the x-axis and shifting the graph horizontally are types of transformations. They’re explained in Problems 16.31– 16.40. Any horizontal line drawn on the coordinate plane that intersects the graph represents a member of the range. All horizontal lines drawn on f(x) will intersect the graph because f(x) increases without bound (its right end goes “up”), even if it does so slowly. Therefore, the range of f(x) is all real numbers. The graph of f(x) does not intersect the y-axis (as previously stated), but it does intersect the x-axis; the x-intercept of f(x) is 1. Note: Problems 18.11–18.14 refer to the function f(x) = log2 x. 18.14 Graph the function . To transform f(x) = log2 x into , multiply f(x) by –1 (which reflects the graph across the x-axis) and add 3 to the input (which shifts the graph 3 units to the left). The graph of is presented in Figure 18-2. Figure 18-2: The graph of is the graph of f(x) = log2 x reflected across the x-axis and shifted three units to the left. Note that has vertical asymptote x = –3. ngous Book of Algebra Problems 404 The Humo Chapter Eighteen — Logarithmic Functions Note: Problems 18.15–18.16 refer to the function g(x) = log3 x. 18.15 Complete the table of values below and use it to graph g(x). Express g(x) = log3 x as an exponential function: 3g(x) = x. Complete the table of values by reporting the power to which 3 must be raised in order to generate the values in the left column. For instance, g(9) = 2 because 32 = 9. As illustrated in Figure 18-3, the graph of g(x) passes through the following points: , , (1,0), (3,1), and (9,2). Figure 18-3: The graph of g(x) = log3 x. The Humongous Book of Algebra Problems 405 Chapter Eighteen — Logarithmic Functions Note: Problems 18.15–18.16 refer to the function g(x) = log3 x. 18.16 Graph . To transform g(x) = log3 x into , insert –x into the function (which reflects the graph about the y-axis) and add 1 to the function (which shifts its graph up one unit). The graph of is presented in Figure 18-4. Figure 18-4: T he graph of is the graph of g(x) = log3 x reflected about the y-axis and shifted up one unit. Common and Natural Logarithms What the bases equal when no bases are written Note: Problems 18.17–18.20 refer to the function f(x) = log x. 18.17 Express the logarithmic equation as an exponential equation. The function f(x) = log x answers this question: “Raising 10 to what power results in the value x?” The function log x with no explicitly defined base is classified as a common logarithm and has an implied base of 10. Therefore, f(x) = log10 x. Express the logarithmic equation as an exponential equation. ngous Book of Algebra Problems 406 The Humo Chapter Eighteen — Logarithmic Functions Note: Problems 18.17–18.20 refer to the function f(x) = log x. 18.18 Evaluate f(12) using a calculator, and round the answer to the thousandths place. Different calculators provide different default degrees of decimal accuracy, but all scientific and graphic calculators should report at least three or four places behind the decimal. In other words, round the answer to three decimal places. The actual dec 1. 0791812460 4762 imal is more like 4827 . that means 101.07918124604762487227… 72 … = 12 . log 12 ≈ 1.079 Note: Problems 18.17–18.20 refer to the function f(x) = log x. 18.19 Solve the equation for x: log x = –1. The equation contains the common logarithm log x (with no explicitly stated base). Therefore, the implied base of the logarithm is 10. log10 x = –1 Express the logarithmic equation as an exponential equation and solve it for x. Note: Problems 18.17–18.20 refer to the function f(x) = log x. 18.20 Identify consecutive integers a and b such that a < log 781 < b. The expression log 781 is a common logarithm, with implied base 10: log 781 = log10 781. The expression log10 781 has the same value as the exponent to which 10 must be raised to produce a result of 781. Consider the following powers of 10: 102 = 100 and 103 = 1,000. Therefore, log 100 = 2 and log 1,000 = 3. The domain of f(x) is all positive real numbers, and the graph of f(x) increases monotonically over its entire domain; hence log 100 < log 781 < log 1,000 and 2 < log 781 < 3. Therefore, a = 2 and b = 3. If you trace the graph of f(x) from left to right, you notice that the graph always goes up. That means the bigger the x-value, the bigger the value of log x. Therefore, log 781 is bigger than log 100 (which equals 2) but smaller than log 1,000 (which equals 3). The Humongous Book of Algebra Problems 407 Chapter Eighteen — Logarithmic Functions Natural logs are written “ln,” and are the only logarithmic function not writ te using the “log” nota n tion like (“log x” or “log 3 9”). e is one of those “predefined” math decimals that stretches on forever and doesn’t repeat, like π. You don’t have to memorize its value (e ≈ 2.71828182846…) because exact answers are written in terms of e and approximate decimal answers are handled by a calculator (which has the value of e programmed into it). Note: Problems 18.21–18.23 refer to the function g(x) = ln x. 18.21 Express the logarithmic function as an exponential function. The common logarithm function, as described in Problems 18.17–18.20, is written without an explicit base: log x. The natural logarithm function, written ln x, also has an implied base but is easily distinguishable from the common log because of its naming convention. The implied base of a natural logarithm is e, Euler’s number. Therefore, ln x = loge x. Rewrite the equation as an exponential equation. Note: Problems 18.21–18.23 refer to the function g(x) = ln x. 18.22 Evaluate using a calculator and round the answer to the thousandths place. Common and natural logarithms are usually represented by different buttons on scientific and graphing calculators. Note: Problems 18.21–18.23 refer to the function g(x) = ln x. 18.23 Evaluate g(e) without using a calculator. Substitute e into g(x). The actual value is more like -1.098612288668109691... . g(e) = ln e Rewrite the natural logarithm, explicitly stating base e. g(e) = loge e Rewrite the logarithmic equation as an exponential equation and write both sides of the equation as powers of e. ngous Book of Algebra Problems 408 The Humo Chapter Eighteen — Logarithmic Functions 18.24 Simplify the expression: ln e 4 – ln e –7. Rewrite the expression as a – b, such that a = ln e 4 and b = ln e –7. Explicitly state the bases of the natural logarithms and express the logarithmic equations as exponential equations. Evaluate the expression a – b given the above values. Change of Base Formula Calculate log values that have weird bases 18.25 Explain how to use the change of base formula to calculate logb a. Assume that b is a positive real number not equal to 1. According to the logarithmic change of base formula, . The quotient of common logarithms is equivalent to the quotient of natural logarithms. Note: Problems 18.26–18.27 refer to the real number c = log7 5. 18.26 Calculate c using the common logarithm version of the change of base formula and round the answer to the thousandths place. According to the change of base formula, . Therefore, Notice that ln e4 = 4 and ln e–7 = –7. If you plug e to some power into the natural log, the log and the e cancel out leaving behind only the power. Sound familiar? x Because ln x and e are inverse functions, plugging one into the other causes the functions to cancel out. More on that in Problems 19.19– 19.35. To calculate any logarithmic value, divide the lo of the argument g (a) by the log of the base (b). You can also divide the natural log of a the natural log of by b— you’ll get the sa me answer. . Evaluate both logs using a calculator and then calculate the quotient. Don’t round anything until you get the final answer. Every time you round a decimal during a problem, it makes the final answer less accurate. The Humongous Book of Algebra Problems 409 Chapter Eighteen — Logarithmic Functions Note: Problems 18.26–18.27 refer to the real number c = log7 5. 18.27 Calculate c using the natural logarithm version of the change of base formula If you can use any base, why stick with common and natural logs? Most calculators have a LOG button and an LN button. and round the answer to the thousandths place. Demonstrate that the solution matches the solution to Problem 18.26. The change of base formula may be applied using common logarithms (as Problem 18.26 demonstrates) or natural logarithms: . In fact, you can use any logarithmic base in the formula, as long as you use the same base in the numerator and denominator. The quotient of natural logs generates the same value of c as the quotient of common logs calculated in Problem 18.26. Note: Problems 18.28–18.29 refer to the function f(x) = log3 x. You don’t need to use the chang e of base formula here. You can us e the technique fr om Problems 18.5–18. 7. 410 18.28 Evaluate f(81) using the change of base formula. Substitute x = 81 into the function. f(81) = log3 81 Apply the change of base formula. As demonstrated in Problems 18.26–18.27, the common logarithm and natural logarithm versions of the formula produce the same value. The common logarithm quotient is presented here. The Humongous Book of Algebra Problems Chapter Eighteen — Logarithmic Functions Note: Problems 18.28–18.29 refer to the function f(x) = log3 x. 18.29 Evaluate f(14) using the change of base formula and round the answer to the thousandths place. The natural logarithm version of the change of base formula is presented here. 18.30 Evaluate log6 2 using the change of base formula and round the answer to the thousandths place. The common and natural logarithm versions of the change of base formula produce the same result; the common logarithm solution is presented here. 18.31 Solve the equation 3x = 8 using the change of base formula and round the answer to the thousandths place. Express the exponential equation as a logarithmic equation. log3 8 = x Calculate x using the change of base formula. To transform 3x = 8 into a log, make the base of the exponent (3) the base of the log, take the log of the right side of the equation (8), and set it equal to the power (x). The Humongous Book of Algebra Problems 411 Chapter Eighteen — Logarithmic Functions 18.32 Solve the equation 9x = 13 using the change of base formula and round the answer to the thousandths place. Express the exponential equation as a logarithmic equation and solve. 18.33 Solve the equation 7 2x + 1 = 6 using the change of base formula and round the answer to the thousandths place. Isolate 72x on the left side of the equation. Rewrite the exponential equation as a logarithmic equation. log7 5 = 2x Calculate log7 5 using the change of base formula. Or divide both sides by 2 Multiply both sides of the equation by to solve for x. Logarithmic Properties Expanding, contracting, and simplifying log expressions Which are presented one at a time in Problems 18.34, 18.36, and 18.38 412 18.34 Expand the expression: ln (2x). Logarithmic expressions can be rewritten according to three fundamental properties. Specifically, a single logarithm can be expressed as multiple logarithms and vice versa. The first property states that the logarithm of a product can be expressed as the sum of the logarithms of its factors: loga (xy) = loga x + loga y. The Humongous Book of Algebra Problems Chapter Eighteen — Logarithmic Functions In this example, the logarithm of the product 2x may be expressed as the sum of the logarithms of its factors, 2 and x. ln (2x) = ln 2 + ln x 18.35 Demonstrate the logarithmic property presented in Problem 18.34 by verifying that ln 10 = ln 2 + ln 5. Because 2(5) = 10, the natural logarithm of the product (10) should equal the sum of the natural logarithms of the factors (2 and 5). Verify using a calculator. 18.36 Expand the expression: . A property of logarithms states that the logarithm of a quotient is equal to the difference of the logarithms of the dividend and divisor: . 18.37 Demonstrate the logarithmic property presented in Problem 18.36 by There are no properties for the log of a sum. For example, ln (2 + x) ≠ (ln 2)(ln x). You also can’t “distribute” the letters “ln.” For example, ln (x + 2) ≠ ln x + ln 2. This is not only true for natural logs but it also works for any base, as long as all the bases in the expression match. In other words, log 10 = log 2 + log 5 and log12 10 = log12 2 + log2 5. verifying that log 4 = log 12 – log 3. Because , the logarithm of the quotient (4) is equal to the difference of the logarithms of the dividend (12) and the divisor (3). In other words, the log of a fraction equals the log of the numerator minus the log of the denominator. Note: Problems 18.38–18.39 refer to the expression log y3. 18.38 Expand the logarithmic expression. A property of logarithms states that the logarithm of a quantity raised to an exponent n is equal to the product of n and the logarithm of the base: loga xn = n(loga x). log y 3 = 3(log y) The Humongous Book of Algebra Problems 413 Chapter Eighteen — Logarithmic Functions Note: Problems 18.38–18.39 refer to the expression log y3. 18.39 Prove that the expansion generated by Problem 18.38 is valid by applying one of the two other logarithmic properties. According to Problem 18.34, loga (xy) = loga x + loga y; the logarithm of a product is equal to the sum of the logarithms of its factors. The property is equally valid for a product of three factors. loga (xyz) = loga x + loga y + loga z The “argument” of a log is whate ve you’re taking the r lo of. In other word g s, the argument of log 2 5 is 5. Rewrite the expression log y 3 by expanding the argument y 3 of the logarithm. log y 3 = log (y · y · y) Apply the logarithmic property just described to express log (y · y · y) as a sum. = log y + log y + log y Add like terms. The three “log y” expression s are the most alik e of any like terms— they’re exactly the sam e. equation log y + The log log y = 3(log y) is y + true for the same rea son “dog + dog + dog = 3 dogs” is true. Set log2 2 = c and rewrite the equation using exponents. = 3 log y Therefore, log y 3 = 3 log y. 18.40 Expand and simplify the logarithmic expression: log2 (2y4). Apply the property described in Problem 18.34 to express the logarithm of a product as the sum of the logarithms of its factors. log2 (2y4) = log2 2 + log2 y4 Apply the property presented in Problem 18.38 to express the logarithm of an argument that is raised to a power as the product of that power and the logarithm of the argument: log2 y 4 = 4 log2 y. = log2 2 + 4 log2 y. Simplify the expression, noting that log2 2 = 1. = 1 + 4 log2 y Therefore, log2 (2y4) = 1 + 4 log2 y. 18.41 Expand the logarithmic expression: . Express the quotient as a difference of logarithms. Distribute the – sign to ln 5 and ln y6. 414 Express each of the products (4x 2 and 5y 6) as a sum of logarithms. The Humongous Book of Algebra Problems = (ln 4 + ln x 2) – (ln 5 + ln y 6) Chapter Eighteen — Logarithmic Functions The logarithmic arguments x 2 and y 6 contain exponents. Rewrite those expressions as a product of the power and the logarithm of the base: ln x 2 = 2 ln x and ln y 6 = 6 ln y. = ln 4 + 2 ln x – ln 5 – 6 ln y Therefore, . 18.42 Rewrite the logarithmic expression as a single logarithm: log x + 9 log y. According to Problem 18.38, n(loga x) = loga xn . Therefore, 9 log y = log y 9. log x + 9 log y = log x + log y 9 Do the exact opposite of what you did in Problems 18.40 and 18.41. “Contract” the expression into one log instead of expanding it into many logs. Express the sum of the logarithms as the logarithm of a product. = log (xy 9) Therefore, log x + 9 log y = log (xy 9). 18.43 Rewrite the logarithmic expression as a single logarithm: 3 log x – 2 log y – log z. Transform coefficients 3 and 2 into exponents of the respective logarithmic arguments. Instead of pulling the exponent OUT of the log as a coefficient, stick the coefficient INTO the expression as an exponent. 3 log x – 2 log y – log z = log x 3 – log y 2 – log z Express the difference (log x 3 – log y 2) as a logarithmic quotient, according to the property . Again, express the difference as the logarithm of a quotient. Therefore, . This is the log property from Problem 18.34 —it’s just reversed. You’re changing a sum into a product instead of a product into a sum. From Problem 18.36 To divide a fraction like by z, just write z in th e denominator. The Humongous Book of Algebra Problems 415 Chapter Eighteen — Logarithmic Functions 18.44 Rewrite the logarithmic expression as a single logarithm: . Transform coefficients arguments. Raising something to the one-half power is the same as taking the square root of it. See Problems 13.13– 13.20 for more info. and 4 into exponents of the respective logarithmic Express the difference as the logarithm of a quotient: . Express the sum as the logarithm of a product. Therefore, 416 The Humongous Book of Algebra Problems . Chapter 19 exponential Functions Functions with a variable in the exponent This chapter explores exponential functions, inverses of logarithmic functions. Specifically, f(x) = ax and g(x) = loga x are inverse functions (assuming a > 0 and a ≠ 1). Therefore, you can use logarithms to solve exponential equations (and vice versa). The chapter concludes with an investigation of exponential functions used to model growth and decay. In Chapter 18, you learn that log 36 = 2 because you have to raise the 6 base of the log (6) to the second pow er to get 36. It’s easier to understan d how logs work when you think of the m in terms of exponents, and that’s no accident. Logarithmic and expon ential functions are actually invers es when their bases match. For exam ple, the inverse of f(x) = log x is . 6 This chapter begins a lot like Chap ter 18, by exploring the graphs of exponential functions. Then it moves into composing the inverse functions— taking the log of an exponential fu nction and vice versa—and applies that to solving exponential and logar ithmic equations. It ends with exponential growth and decay, which uses the formula f(t) = Nekt. Chapter Nineteen — Exponential Functions Graphing Exponential Functions Graphs that start close to y = 0 and climb fast Note: Problems 19.1–19.5 refer to the function j(x) = 2 x. 19.1 Complete the table of values below. Evaluate j(x) for each of the x-values in the left column. Note: Problems 19.1–19.5 refer to the function j(x) = 2 x. 19.2 Graph j(x) using the table of values generated in Problem 19.1. According to Problem 19.1, the graph of j(x) passes through the following points: , , (0,1), (1,2), (2,4), and (3,8). The graph of j(x) is presented in Figure 19-1. 418 The Humongous Book of Algebra Problems Chapter Nineteen — Exponential Functions Figure 19-1: The graph of j(x) = 2 x. Note: Problems 19.1–19.5 refer to the function j(x) = 2 x. 19.3 Identify the domain, range, and intercept(s) of j(x). Consider the graph of j(x) in Figure 19-1. Every vertical line drawn on the coordinate plane that intersects the graph represents a member of the domain of j(x). Any vertical line drawn on the graph will intersect j(x), so the domain of j(x) is all real numbers. Every horizontal line that intersects the graph of j(x) represents a member of the range of the function. The line y = 0, the x-axis, does not intersect the graph, and neither does any horizontal line below it. Only horizontal lines above y = 0 intersect the graph, so the range of j(x) is j(x) > 0. The graph of j(x) intersects the y-axis at point (0,1), because j(0) = 20 = 1. Therefore, the y-intercept of j(x) is 1. Note: Problems 19.1–19.5 refer to the function j(x) = 2 x. 19.4 Demonstrate that f(x) = log2 x (from Problems 18.11–18.12) and j(x) are The x-axis is an asymptote of the graph, so j(x gets infinitely clos ) e to the x-axis but ne ver touches it. That means j(x) doesn’t have an x-intercept. inverses based on their graphs. According to Problem 18.11, the graph of f(x) = log2 x passes through points , , (1,0), (2,1), (4,2), and (8,3). According to Problem 19.1, the graph of j(x) = 2x passes through points , , (0,1), (1,2), (2,4), and (3,8). Reversing the coordinate pairs of one graph produces the coordinate pairs of the other, a characteristic of inverse functions. Furthermore, consider the graphs of f(x) and j(x), plotted on the same coordinate plane in Figure 19-2. The Humongous Book of Algebra Problems 419 Chapter Nineteen — Exponential Functions The table of values for f(x) and j(x) prove that six coordinate pairs are reversed copies of each other. When functions are reflections of each other (and the dotted line y = x in Figure 19-2 is the mirror), all of the points on the graphs are reversed copies of each other. Figure 19-2: T he graphs of f(x) = log2 x and j(x) = 2x are reflections of each other across the dotted line y = x. The graph of j(x) is the graph of f(x) reflected across the line y = x (and vice versa), indicating that reversing all the points on one graph will generate the points on the other. Note: Problems 19.1–19.5 refer to the function j(x) = 2 x. 19.5 Graph . To transform the function j(x) = 2x into , multiply the input by –1 to get –x (which reflects the graph of j(x) across the y-axis) and subtract 3 (which shifts the graph down three units). The graph of is presented in Figure 19-3. ngous Book of Algebra Problems 420 The Humo Chapter Nineteen — Exponential Functions Figure 19-3: The graph of is the graph of j(x) = 2x reflected across the y-axis and shifted down three units. Note: Problems 19.6–19.9 refer to the function g(x) = ex. 19.6 Use a calculator to complete the table of values below, rounding all values of g(x) to the thousandths place. Evaluate the natural exponential function ex for each value of x in the left column. Euler’s number (e) is the base for the natural exponential function ex, just like it’s the base for the natural logarithmic function loge x = ln x. The Humongous Book of Algebra Problems 421 Chapter Nineteen — Exponential Functions Note: Problems 19.6–19.9 refer to the function g(x) = ex. 19.7 Graph g(x). According to Problem 19.6, the graph of g(x) passes through the following points: (–2, 0.135), (–1, 0.368), (0,1), (1, 2.718), (2, 7.389), and (3, 20.086). The graph of g(x) is presented in Figure 19-4. A positive num to a negative ex ber raised ponent makes a fraction that ge ts sm aller as x gets more negative: Figure 19-4: The graph of the natural exponential function, g(x) = ex. Note: Problems 19.6–19.9 refer to the function g(x) = ex. 19.8 Identify similarities between the graphs of j(x) = 2x (in Figure 19-1) and g(x) = ex (in Figure 19-4). Both functions have the same domain (all real numbers), and both graphs are entirely contained in the first and second quadrants—neither passes below nor intersects the x-axis. This behavior is not unexpected, as a positive number raised to any real number exponent produces a positive number. Any positive number raised to the zero power equals 1. Both graphs have values close to zero when x is negative and both increase steeply when x is positive. Both graphs have a y-intercept of 1, as do all exponential functions of the form f(x) = ax (if a > 0 and a ≠ 1). ngous Book of Algebra Problems 422 The Humo Chapter Nineteen — Exponential Functions Note: Problems 19.6–19.9 refer to the function g(x) = ex. 19.9 Graph . To transform the function g(x) = ex into , multiply g(x) by –1 (which reflects the graph across the x-axis) and subtract 4 from the input (which shifts the graph four units to the right). The graph of ĝ(x) is presented in Figure 19-5. Figure 19-5: The graph of is the graph of g(x) = ex reflected across the x-axis and shifted four units to the right. Composing Exponential and Logarithmic Functions They cancel each other out Note: Problems 19.10–19.11 prove that h(x) = log3 x and k(x) = 3x are inverse functions. 19.10 Demonstrate that h(k(x)) = x. Substitute k(x) = 3x into h(x) = log3 x. If f(g(x)) = g(f(x)) = x, then f(x) and g(x) are inverse functions. Problem 19.10 proves that f(g(x)) = x, and Problem 19.11 proves that g(f(x)) = x. The Humongous Book of Algebra Problems 423 Chapter Nineteen — Exponential Functions Express the logarithmic equation as an exponential equation. 3h(k(x)) = 3x If equivalent exponential expressions have equal bases, then the exponents must be equal as well. h(k(x)) = x Note: Problems 19.10–19.11 prove that h(x) = log3 x and k(x) = 3x are inverse functions. 19.11 Demonstrate that k(h(x)) = x. Substitute h(x) = log3 x into k(x) = 3x . Express the exponential equation as a logarithmic equation. If two equivalent logarithms have the same base, then the arguments of those logarithms must also be equal. k(h(x)) = x If you plug an exponential function into a log with the same base, everything disappears except for the exponent. 19.12 Simplify the expression: log6 65x . Substituting 65x (an exponential function with base 6) into its inverse (a logarithmic function with base 6) produces an expression equal to the exponent: logb bf(x) = f(x). log6 65x = 5x 19.13 Simplify the expression: If the power of an exponentia l function is a log with the same ba se everything disapp , ears except for whate ver’s inside the logari thm, in this case x + 1. ˙ . Substituting a logarithmic function with base 2 into an exponential function with base 2 produces an expression equal to the argument of the logarithm: . Note: Problems 19.14–19.15 refer to the expression ln (9ex). 19.14 Apply a logarithmic property to expand the logarithmic expression. According to Problem 18.34, the logarithm of a product is equal to the sum of the logarithms of its factors: loga (xy) = loga x + loga y. ln (9ex ) = ln 9 + ln ex ngous Book of Algebra Problems 424 The Humo Chapter Nineteen — Exponential Functions Note: Problems 19.14–19.15 refer to the expression ln (9ex). 19.15 Simplify the expanded logarithm expression generated by Problem 19.14. The natural exponential function ex and the natural logarithm function ln x have the same base, e. Therefore, substituting ex into ln x results in an expression equal to the exponent: ln ex = x. f(x) = ln 9 + ln ex = ln 9 + x Therefore, ln (9ex ) = x + ln 9. 19.16 Apply a logarithmic property to simplify the expression: . According to Problem 18.36, the logarithm of a quotient is equal to the Common logs (like log 10x – 4) have an implied base of 10. In other words, x–4 log 10x – 4 = log10 10 . You have an exponential function with base 10 plugged into a log of base 10. They cancel each other out, leaving behind the exponent x – 4. . difference of the logarithms of the dividend and divisor: Simplify the expression log 10x – 4. = x – 4 – log 3 19.17 Apply a logarithmic property to simplify the expression: . Express the quotient as the difference of the logarithm of the numerator and the logarithm of the denominator. Wonder where this number came from? Set log 4 2 = c and create the exponential equation 4c = 2. Write both sides of the equation as powers of 2 and solve for c: Substituting 4x into its inverse function, a logarithm with base 4, eliminates both functions: log4 4x = x. 19.18 Apply a logarithmic property to simplify the expression: . According to Problem 18.38, the logarithm of a quantity raised to an exponent n is equal to the product of n and the logarithm of the base: n · loga x = loga xn . Rewrite the natural logarithm by transforming its coefficient (4) into the exponent of the argument : . Express the square root as rational exponent and simplify: . To raise y1/2 to the fourth power multiply the power , s: . The Humongous Book of Algebra Problems 425 Chapter Nineteen — Exponential Functions The natural exponential function and the natural logarithmic function are inverses, so composing them eliminates them from the equation. = y2 Therefore, . Exponential and Logarithmic Equations Cancel logs with exponentials and vice versa 19.19 Solve the equation for x: 25x = 125. Express both sides of the equation as a power of five: 25 = 52 and 125 = 53. Two equivalent exponential expressions with equal bases must also have exponents that are equal. 2x = 3 Solve for x. 19.20 Solve the equation for x: 8x = 128. Express both sides of the equation as a power of two: 8 = 23 and 128 = 27. ngous Book of Algebra Problems 426 The Humo Chapter Nineteen — Exponential Functions 19.21 Solve the equation for x: 3x = 19. To eliminate the exponential function 3 —thereby isolating x and solving the equation—apply its inverse function (log3 x) to both sides of the equation. x 19.22 Solve the equation for x: 2 – 5ex = –17. Isolate ex on the left side of the equation. Take the natural logarithm of both sides of the equation to eliminate the natural exponential function and solve for x. 19.23 Solve the equation for x: e 2x – 6ex = 0. Express e 2x as (ex )2. (ex)2 – 6ex = 0 Factor ex out of the expression left of the equal sign. ex (ex – 6) = 0 Apply the zero product property, setting both factors equal to zero. ex = 0 or ex – 6 = 0 Solve the equations for x. Note that the natural logarithm function is not defined for x = 0, so x = ln 0 is not a valid solution. The only solution to the equation is x = ln 6. You can’t do this problem the way you did Problems 19.19 and 19.20 because you can’t write 3 and 19 using a common base. That means you have to remove x from the exponent (using log 3 ) and solve for it. Even though it looks ugly, the exact solution is log3 19. Don’t use a calculator and/or the change of base formula to give an approximate decimal answer unless a problem specifically tells you to. If you multiply the powers of (ex) 2, you get e2x Why write e2x as x.2 (e When you do, it’s ) ? a easier to factor xlittle e of the expression out in next step—you fa the ctor one ex out of (ex) 2 = (ex)(ex), lea ving one behind. Look at the graphs of the log functions in Problems 18.12 and 18.15. They don’t intersect the y-axis (x = 0) or any vertical line left of the y-axis. The Humongous Book of Algebra Problems 427 Chapter Nineteen — Exponential Functions 19.24 Solve the equation for x: e 2x – 9ex = –14. Like Problem 19.23, this equation can be solved by factoring. Therefore, like the preceding problem, one side of the equation must equal zero in order to apply the zero product property. Add 14 to both sides of the equation. “Exponentiate with base e” means turn something into an exponent with base e. When you exponentiate to solve a log equation, isolate the log containing x on one side of the equal sign and then make both sides exponents that have the same base as the log. The exact answer is e4. Don’t use a calculator to approximate the value of an answer containing e unless you’re specifically instructed to do so. e 2x – 9ex + 14 = 0 Express e 2x as (ex )2 and solve the equation by factoring. The solution to the equation is x = ln 2 or x = ln 7. 19.25 Solve the equation for x: ln x = 4. In this equation, x is the argument of a natural logarithm—a logarithm with base e. To eliminate the logarithmic expression, exponentiate both sides of the equation with base e. e ln x = e 4 Because ex and ln x are inverses, composing the functions eliminates both of them from the equation. x = e4 19.26 Solve the equation for x: log3 x = 2. To eliminate the logarithmic expression, and thereby isolate x on the left side of the equation, exponentiate both sides of the equation with the base of the logarithm, 3. 19.27 Solve the equation for x: ln (3x + 7) = 13. To eliminate the natural logarithm, exponentiate both sides of the equation with base e. ngous Book of Algebra Problems 428 The Humo Chapter Nineteen — Exponential Functions Solve for x. 19.28 Solve the equation for x: log5 (2x) + 1 = 0. Because logarithms have Isolate the logarithmic expression left of the equal sign. restricted domains (you can’t take the log log5 (2x) = –1 of a negative number Eliminate the logarithmic expression with base 5 by exponentiating both sides or zero), it’s a good idea of the equation with base 5. to plug your final answers into the original equation to check them: Note: Problems 19.29–19.30 refer to the equation log x + log (x – 2) = log 15. 19.29 Solve the equation for x. A property of logarithms states that the sum of two logarithms is equal to the logarithm of the product of the arguments: log x + log (x – 2) = log [x(x – 2)]. The same base as the common logs in the equation. Eliminate the logarithmic expressions by exponentiating both sides of the equation with base 10. Solve the equation by factoring. Before you can solve by factoring, you need to set the equatio n equal to zero. That’s why you subtract 15 from both sides of the equation. You can’t take the log of a negative number. Substituting x = –3 into the original equation produces an invalid statement: log (–3) + log (–5) = log 15. Therefore, x = –3 is not a solution to the equation. The only valid solution is x = 5. The Humongous Book of Algebra Problems 429 Chapter Nineteen — Exponential Functions Note: Problems 19.29–19.30 refer to the equation log x + log (x – 2) = log 15. 19.30 Verify the solution(s) generated by Problem 19.29. Substitute x = 5 into the equation. According to a logarithmic property, the sum of two logarithms is equal to the logarithm of the product of the arguments: log 5 + log 3 = log (5 · 3). Because substituting x = 5 into the equation generates a true statement (log 15 = log 15), x = 5 is a solution to the equation. 19.31 Solve the equation for x: ln (x + 4) + ln (x – 9) = ln 30. Use the log property from Problems 19. 29 and 19.30. Express the sum of the logarithms left of the equal sign as a single logarithm. Eliminate the natural logarithms by exponentiating both sides of the equation with base e. Solve the equation by factoring. ln (–2) + ln (–15) = ln 30 The only valid solution to the equation is x = 11, as substituting x = –6 into the equation produces in an invalid statement. ngous Book of Algebra Problems 430 The Humo Chapter Nineteen — Exponential Functions 19.32 Solve the equation for x: log2 (x + 3) – log2 9 = 4. Express the difference of the logarithms left of the equal sign as the logarithm of the quotient of the arguments. Eliminate the logarithm from the equation by exponentiating both sides of the equation with base 2. Note: Problems 19.33–19.34 refer to the equation (ln x)2 – 2 ln x3 = –5. 19.33 Solve the equation for x. Apply a logarithmic property to express –2 ln x as –6 ln x. 3 (ln x)2 – 6 ln x = –5 Solve the equation by factoring. The solution to the equation is x = e or x = e 5. If you’re taking the log of something to a power, you can pull that power out in front of the log as a coefficient. This log already has a coefficient of –2 , so when you pull the exponent 3 out front, multiply it by the coefficient that’s already there: –2 ln x3 = –2 ˙ 3 ln x ˙ = –6 ln x. Factoring this works just like factoring a regular quadratic polynomial: w2 – 6w + 5 = (w – 5)(w – 1). The only difference is that w = ln x. The Humongous Book of Algebra Problems 431 Chapter Nineteen — Exponential Functions Note: Problems 19.33–19.34 refer to the equation (ln x)2 – 2 ln x3 = –5. ln x and e cancel each other out, leaving behind only the exponent of e. When something has no exponent written, the implied exponent is 1, so (ln e) 2 = (ln e1) 2 = (1) 2 . x 19.34 Verify the solution(s) generated in Problem 19.33. Substitute x = e and x = e 5 into the equation, one at a time, and verify that both of the resulting statements are true. 19.35 Solve the equation for x: log (x 2 + 1) = 1. Eliminate the logarithm property to rewrite the left side of the equation as a single logarithm. Eliminate the logarithm by exponentiating both sides of the equation with base 10. Solve for x. Both x = –3 and x = 3 are valid solutions. ngous Book of Algebra Problems 432 The Humo Chapter Nineteen — Exponential Functions Exponential Growth and Decay Use f(t) = Nekt to measure things like population Note: Problems 19.36–19.39 refer to the laboratory experiment described here. 19.36 A group of scientists conduct an experiment to determine the effectiveness of an agent designed to accelerate the growth of a specific bacterium. They introduce 40 bacterial colonies into a growth medium, and exactly two hours later the number of colonies has grown to 200. In fact, the number of colonies grows exponentially for the first 5 hours of the experiment. Design a function f(t) that models the number of bacterial colonies t hours after the experiment has begun (assuming 0 ≤ t ≤ 5) that includes constant of proportionality k. A population exhibiting exponential growth, or exponential decay, is modeled using the function f(t) = Nekt . The variables in the formula are defined as follows: N = original population (in this case N = 40 colonies), t = number of hours elapsed, and k = the constant of proportionality. f(t) = 40ekt Note: Problems 19.36–19.39 refer to the laboratory experiment described in Problem 19.36. 19.37 Calculate k. Exactly t = 2 hours after the start of the experiment, the number of bacterial colonies was f(2) = 200. Substitute t = 2 and f(2) = 200 into the population model f(t) = 40ekt from Problem 19.36. Exponential growth is observed only for the first 5 hours, so t can’t be higher than 5; t can’t be below 0 because the experiment begins at time t = 0. Don’t try to guess what k is or plug a numbe r from the origina l problem in for k. You almost always ha ve calculate k base to d the given inform on ation, which is exactly what you do in Problem 19.37. Isolate the exponential expression on the right side of the equal sign. Solve for k by taking the natural logarithm of both sides of the equation. The Humongous Book of Algebra Problems 433 Chapter Nineteen — Exponential Functions Note: Problems 19.36–19.39 refer to a laboratory experiment described in Problem 19.36. 19.38 Calculate the number of colonies that were present 5 hours after the To calculate the population 5 hours after the experiment start s, substitute t = 5 into the population m odel. experiment began, rounding the answer to the nearest whole number. According to Problem 19.37, . Substitute k into the population model defined in Problem 19.36. Evaluate f(5). Use a calculator to calculate the exponent of e: . After 5 hours, the 40 original colonies have grown to 2,236 colonies. Note: Problems 19.36–19.39 refer to a laboratory experiment described in Problem 19.36. 19.39 Approximately how many minutes does it take the bacterial colonies to grow in number from 40 to 120? Round the answer to the nearest whole minute. According to Problem 19.38, the population is modeled by the function . Calculate t when f(t) = 120. Eliminate the natural exponential function by taking the natural logarithm of each side of the equation. Multiply both sides of the equation by the reciprocal of this number to solve for t. ngous Book of Algebra Problems 434 The Humo Chapter Nineteen — Exponential Functions Convert t ≈ 1.365212389 hours into minutes. 60(1.365212389) ≈ 81.91274334 There were a total of 120 bacterial colonies approximately 82 minutes after the experiment began. Note: Problems 19.40–19.43 refer to the radioactive decay of carbon-14. There are 60 minutes in an hour, so multiply t by 60 to calculate how many minutes are in 1.365212389 hours. 19.40 Upon the death of an organism, the amount of carbon-14 present in that organism decays exponentially, with a half-life of approximately 5,730 years. Construct a function g(t) that models the amount of carbon-14 present in an organism t years after its demise, including the constant of proportionality accurate to eight decimal places. Apply the exponential growth and decay formula g(t) = Nekt , such that N = the original amount of carbon-14 in the living organism. According to the problem, carbon-14 has a half-life of 5,730 years. Therefore, when t = 5,730, . Substitute t and g(t) into the function and solve for k. Multiply both sides of the equation by to isolate the exponential expression. Half-life is the amount of time it takes something to decay to half of its original amount. Carbon-14 has a half-life of 5,730 years, so a 10 0 kg sample of carbon14 will decay to 10 0 ÷ 2 = 50 kg in 5,730 years. Take the natural logarithm of both sides of the equation to isolate k. The original versio n of g(t): g(t) = Nekt. Substitute k into g(t). g(t) = Ne –0.00012097t The Humongous Book of Algebra Problems 435 Chapter Nineteen — Exponential Functions Note: P roblems 19.40–19.43 refer to the radioactive decay of carbon-14, as modeled by the function g(t) generated in Problem 19.40. 19.41 Approximately how long does it take N grams of carbon-14 to decay by twothirds? Report the answer rounded to the nearest whole year. After some time t has elapsed, . Substitute this value into g(t) = Ne –0.00012097t (the half-life model from Problem 19.40) and solve for t. If the mass decreased by two-thirds, that means one-third of the original mass N is left. It takes approximately 9,082 years for a mass of carbon-14 to decay by two-thirds. Note: P roblems 19.40–19.43 refer to the radioactive decay of carbon-14, as modeled by the function g(t) generated in Problem 19.40. 19.42 Approximately how long does it take N grams of carbon-14 to decay by threefourths? The half-life of carbon-14 is approximately 5,730 years. Therefore, in 5,730 years, a mass N of carbon-14 will decay to a mass of . In an additional 5,730 years, the mass will again decrease by half: . Thus, one-fourth of the mass is left after 2(5,730) = 11,460 years. ngous Book of Algebra Problems 436 The Humo Chapter Nineteen — Exponential Functions Note: Problems 19.40–19.43 refer to the radioactive decay of carbon-14, as modeled by the function g(t) generated in Problem 19.40. 19.43 If an organism contains 15 grams of carbon-14 at the time of its death, how much carbon-14 will the organism contain 100 years later? Round the answer to the thousandths place. Substitute N = 15 and t = 100 into the half-life model g(t) defined in Problem 19.40 and evaluate g(100). The organism will contain approximately 14.820 grams of carbon-14 one hundred years after its death. The Humongous Book of Algebra Problems 437 Chapter 20 Rational Expressions Fractions with lots of variables in them The foundational methods by which a rational number is reduced to lowest terms are used to reduce rational expressions as well. By extension, the methods by which rational expressions are combined (that is, added, subtracted, multiplied, and divided) are based on the same principles that govern the combination of rational numbers. The similarities between rational numbers and rational expressions, however, do not preclude the need to thoroughly investigate the latter. Rational numbers (fractions) and rat ional expressions (fractions with x in them) are a lot alike. For exam ple, you simplify a rational expressio n by factoring the numerator and de nominator and then canceling out matching factors across the fract ion bar—the same way you simplify fractions containing only numbers. Ev en though the processes are the sa me, there are tiny differences you lea rn about in this chapter. Most of this chapter is familiar. Wa nt to add or subtract rational expressions? You need a common de nominator. Want to multiply them? No problem; go ahead. Want to divide rational expressions? Keep the first fraction the same, change div ision to multiplication, and flip the second fraction. Basically, you’re ex tending the things you learned about fractions in Chapter 2, to include fra ctions with variables inside. The chapter ends with graphing rat ional expressions, which is a little bit tricky, especially the bit about obliqu e asymptotes, so make sure you pay particular attention to that. Chapter Twenty — Rational Expressions Simplifying Rational Expressions Reducing fractions by factoring Note: Problems 20.1–20.2 refer to the expression . 20.1 Simplify the rational expression and identify the restriction on the answer. The quotient of exponential expressions with the same base is equal to the base raised to the difference of the exponents: The numerator contains x2 and th denominator cont e x (which equals x1 ains ), so subtract the powers of x. This is the restriction the problem mentioned. The fraction is equal to 5x only when x ≠ 0. Why? See Problem 20.2. Avoid dividing by 0 at all costs. 0 ÷ 0 is called an indeterminate value, and any other number divided by 0 is said to be undefined. In either case, 0 in the denominator is bad. Restrictions are the x-values that make the denominator zero . Therefore, . when x ≠ 0. Note: Problems 20.1–20.2 refer to the expression . 20.2 Explain why the restriction stated in Problem 20.1 is necessary. The expression is not defined for x = 0; substituting x = 0 into the expression results in , which is indeterminate. However, 5x is defined when x = 0. The expressions and 5x are truly not equivalent if their values differ when x = 0, so stating they are equal must include the disclaimer that the equality is not guaranteed at x = 0. 20.3 Simplify the expression and indicate restrictions, if any, on the result. Write the rational expression as a product of three fractions, the quotient of the coefficients, the quotient of the x-terms, and the quotient of the y-terms. Express the coefficients as prime factorizations to reduce the corresponding fraction. See Problems 2.30 –2 .32 for mor information about e prime factorizations. ngous Book of Algebra Problems 440 The Humo Chapter Twenty — Rational Expressions Apply the exponential property referenced in Problem 20.1. Place variable expressions with positive exponents in the numerator and expressions with negative exponents in the denominator. Therefore, 20.4 Simplify the expression: when x ≠ 0. and indicate restrictions, if any, on the result. Factor the numerator, a difference of perfect squares. The numerator and denominator contain the common factor (x + 7). To reduce the fraction, eliminate the shared factor. Therefore, 20.5 Simplify the expression After you place the y–4 in the denominator where it belongs, change the negative exponent to a positive. You don’t need to include y ≠ 0 because both of the expressions would be undefined in that case. You need to include restrictions only on variables that would make one of the fraction denominators equal 0 but not the other. The difference of perfect squares formula (from Problem 12.26) is (a2 – b2) = (a + b)(a – b). when x ≠ –7. and indicate restrictions, if any, on the result. Factor the numerator and denominator; each contains a greatest common factor. Whereas 5 is a prime number, 10 is not. Express 10 as a prime factorization: 10 = 2 · 5. The fraction is undefined when x = –7 because substituting –7 into the denominator gives you (–7) + 7 = 0, and you can’t divide by 0. The Humongous Book of Algebra Problems 441 Chapter Twenty — Rational Expressions Factor the quadratic expression in the denominator: x 2 + 6x – 27 = (x + 9)(x – 3). Either of these fractions is an acceptable final answer. You eliminate the factor (x – 3) from the denom inator. Set it equal to 0 and solve to get the restriction: x = 3. The numerator and denominator contain common factors 5 and (x – 3). Eliminate the common factors to reduce the rational expression. when x ≠ 3. Therefore, 20.6 Simplify the expression and indicate restrictions, if any, on the result. The terms of the numerator have the greatest common factor y. Factoring it out of the expression results in a difference of perfect squares: y 3 – 4y = y(y 2 – 4). The formula is a3 – b3 = 2 (a – b)(a2 + ab + b ). Check out Problems 12.31 and 12.33 for more information about factoring differences of perfect cubes. Factor the difference of perfect squares in the numerator and the difference of perfect cubes in the denominator. Reduce the fraction to lowest terms. Therefore, ngous Book of Algebra Problems 442 The Humo when y ≠ 2. Chapter Twenty — Rational Expressions 20.7 Simplify the expression and indicate restrictions, if any, on the result. Factor the numerator and denominator by decomposition. Look at Problem 12.42 for an explanation of factoring by decomposition. You don’t have to use this technique to factor if you’d rather “play around” and figure out the factors through trial and error. The numerator and denominator share common factor (3x + 2). Simplify the expression by eliminating the common factor. Therefore, 20.8 Simplify the expression when When you eliminate factors (like 3x + 2) from the denominator, account for them using restrictions. In this case, set 3x + 2 = 0 and solve for x. . and indicate restrictions, if any, on the result. Factoring the denominator is significantly easier if you first factor the numerator: 2x 2 + 11x + 15 = (2x + 5)(x + 3). The fraction can only be reduced if the numerator and denominator share one or more common factors. To determine whether (x + 3) is a factor of 2x 3 + 9x 2 + 4x – 15, apply synthetic division. If (x + 3) is a factor, then x = –3 is a root of the polynomial. Pu t –3 in the synthe tic division box. Thus (2x 3 + 9x 2 + 4x – 15) ÷ (x + 3) = 2x 2 + 3x – 5. The Humongous Book of Algebra Problems 443 Chapter Twenty — Rational Expressions Factor the remaining quadratic: 2x 2 + 3x – 5 = (2x + 5)(x – 1). Eliminate common factors (x + 3) and (2x + 5) to simplify the rational expression. when x ≠ –3 and Therefore, . Adding and Subtracting Rational Expressions Use common denominators The first denominator contains four factors (2, 3, x, and y); the second denominator contains one factor (y). That’s four unique factors. (The y is repeated.) Each should be raised to its highest power. Note: Problems 20.9–20.10 refer to the expression . 20.9 Identify the least common denominator of and Consider the denominators: y 2 and 6xy = 2 · 3 · xy. The least common denominator must include every factor of either denominator raised to the highest power of that factor. (See Problem 2.30 for a more thorough explanation of the algorithm used to calculate the least common denominator.) Note: Problems 20.9–20.10 refer to the expression Dividing each fraction’s denominator into the LCD tells you what to multiply each fraction by to get the LCD. . . 20.10 Simplify the expression. Divide the least common denominator, as calculated in Problem 20.9, by the denominators of each fraction. Multiply the numerator and denominator of by y to express the fraction using the least common denominator. Similarly, multiply the numerator and denominator of ngous Book of Algebra Problems 444 The Humo by 6x. Chapter Twenty — Rational Expressions The fractions now have a common denominator. Simplify the expression by adding the numerators and dividing by the common denominator. Note: Problems 20.11–20.12 refer to the expression . 20.11 Identify the least common denominator. Express the coefficients of 12x 3 and 28x 2 as a product of prime numbers: 12x 3 = 22 · 3 · x 3 and 28x 2 = 22 · 7 · x 2. Together, the denominators consist of four unique factors: 2, 3, 7, and x. The least common denominator is the product of those factors, each raised to its highest power. Note: Problems 20.11–20.12 refer to the expression Look at Problem 2.32 if you’re not sure how to factor 12 and 28. . 20.12 Simplify the expression. Divide the least common denominator by the denominators of the fractions in the expression. Multiply the numerator and denominator of each fraction in the expression by the above values. You don’t have to show this step if you can do it in your head. Multiply the top and bottom of the left fractio n by 7; multiply the to pa bottom of the ri nd ght fraction by 3x. The Humongous Book of Algebra Problems 445 Chapter Twenty — Rational Expressions The fractions share the same denominator, so combine the numerators. Note: Problems 20.13–20.14 refer to the expression . 20.13 Identify the least common denominator. Factor the denominators. Don’t multiply the three binomials together yet—leaving them like this makes Problem 20.14 easier. The denominators of the expression consist of three unique factors: (x + 5), (x – 5), and (x + 10). The least common denominator is the product of all three: (x + 5)(x – 5)(x + 10). Note: Problems 20.13–20.14 refer to the expression . 20.14 Simplify the expression. Divide the least common denominator, as identified in Problem 20.13, by each of the denominators in the expression. To express the fractions using the least common denominator, multiply the numerator and denominator of denominator of ngous Book of Algebra Problems 446 The Humo by x + 10; multiply the numerator and by x – 5. Chapter Twenty — Rational Expressions Combine the numerators to simplify the expression. . Therefore, 20.15 Simplify the expression: This minus sign gets distributed to the numerator of the right fraction. You don’t have to multiply (x + 5)(x – 5) (x + 10) to get this polynomial—you can leave it in factored form. . Express each fraction using the least common denominator (x – 4)(x + 8). Simplify the expression by combining the numerators. Therefore, . The Humongous Book of Algebra Problems 447 Chapter Twenty — Rational Expressions 20.16 Simplify the expression: . Factor the denominators. There’s no need to divide the LCD by each denominator. You know the LCD is x(x – 4)(x – 6). The left fraction has a denominator of x(x – 4)—all it needs is (x – 6). The right fraction already has (x – 4) and (x – 6) in the denominator— it needs an x. The denominators of the expression consist of three unique factors, the product of which is the least common denominator: x(x – 4)(x – 6). Rewrite the fractions using the least common denominator. Simplify the expression by combining the numerators. Reduce the fraction to lowest terms. Therefore, ngous Book of Algebra Problems 448 The Humo . Chapter Twenty — Rational Expressions 20.17 Simplify the expression: . Factor the denominators. The denominators consist of two unique factors: (x + 6) and (x – 6). The least common denominator is (x + 6)2 (x – 6)2. Write each fraction using the least common denominator. Both factors are raised to the first power and the second power somewhere in the denominators of the expression. Choose the higher power for the LCD. Expand the numerator of each fraction. Combine the numerators to simplify the expression. Therefore (x + 6) 2 (x – 6) 2 = (x2 + 12x + 36) (x2 – 12x + 36) = x 4 – 72x2 + 1, 29 6 The Humongous Book of Algebra Problems 449 Chapter Twenty — Rational Expressions Note: Problems 20.18–20.19 refer to the function f(x) defined below. 20.18 Identify the least common denominator d(x) of the function f(x) and express f(x) using the least common denominator. Factor the denominators in each of the fractions. The denominators consist of three unique factors, and the least common denominator is the product of all three: (x + 9)(x – 8)(x + 1). Express f(x) using the least common denominator. Combine the numerators to simplify f(x). Group the terms of the numerator according to the power to which x is raised in each. Factor x 2 and x out of the corresponding quantities. ngous Book of Algebra Problems 450 The Humo Chapter Twenty — Rational Expressions Note: Problems 20.18–20.19 refer to the function f(x) defined below. 20.19 If , given function d(x) defined in Problem 20.18, calculate the values of a, b, and c. According to Problem 20.18, . The problem states that . Set the rational functions equal. Two equivalent fractions with equivalent denominators must have equivalent numerators. 4x 3 + 7x 2 + 37x – 103 = ax 3 + (a + b)x 2 + (1 + 9b – c)x + (–31 + 8c) For the cubic polynomials to be equal, each of the corresponding terms must be equal. Set the x 3 -terms equal and solve for a. Set the x 2-terms equal, substitute a = 4 into the equation, and solve for b. |Because f(x) equals both of th e fractions (the on e that starts with 3 ax in the numerator a nd the one that starts w it 4x 3), those fract h ions are equal to each ot her. You can drop the denominators and set the numerators equal. In other words, the x3 terms are equal 2 (4x3 = ax3), the x terms are equal (7x2 = [a + b]x2), the x–terms are equal, and the constants are equal. Set the x-terms equal, substitute b = 3 into the equation, and solve for c. The Humongous Book of Algebra Problems 451 Chapter Twenty — Rational Expressions Therefore, a = 4, b = 3, and c = –9. Multiplying and Dividing Rational Expressions Common denominators not necessary 20.20 Simplify the expression: . To multiply two rational expressions, divide the product of the numerators by the product of the denominators. Reduce the fraction to lowest terms by eliminating the common factor x from the numerator and denominator. 20.21 Simplify the expression: You know it’s not possible to reduce the fina l answer because the numerator a nd denominator did n’t have any factors in co mmon in the previous st ep. . The product of rational expressions is equal to the product of the numerators divided by the product of the denominators. ngous Book of Algebra Problems 452 The Humo Chapter Twenty — Rational Expressions 20.22 Simplify the expression: . Multiply the rational expressions. Factor the numerator and denominator and simplify the expression. Therefore, . 20.23 Simplify the expression: You might be wondering what happened to the restrictions (like x ≠ 2) from the beginning of the chapter. There’s no need to include them unless the problem tells you to. Keep in mind, however, that this equation is not true when x = –2, x = 0, or x = 2—when factors you eliminated equal 0. . Express the quotient as a product. Multiply the rational expressions and simplify the result. This is introduced in Problems 2.38–2.40. To calculate , take the reciprocal of the fraction you’re dividing by and change the division symbol to multiplication: . 20.24 Simplify the expression: . Express the quotient as a product. The Humongous Book of Algebra Problems 453 Chapter Twenty — Rational Expressions Factor the quadratic expression: x 2 – 2x – 3 = (x – 3)(x + 1). Either of these answers is fine—whether you expand (x + 1) 2 into x2 + 2x + 1 or leave it factored doesn’t matter. Reduce the fraction to lowest terms. 20.25 Simplify the expression: . Express the product as a quotient and factor the expressions. Eliminate factors shared by the numerator and denominator. Therefore, 20.26 Simplify the expression: . . Express the product as a quotient and factor the expressions. 8x 3 + 27 is a sum of perfect cubes a nd 4x2 – 9 is a diffe renc of perfect square e s. ngous Book of Algebra Problems 454 The Humo Chapter Twenty — Rational Expressions Eliminate factors shared by the numerator and denominator. Therefore, 20.27 Simplify the expression: . . Express the quotient as a product. Factor the quadratic expressions as well as the difference of perfect squares: x 4 – 256 = (x 2 + 16)(x 2 – 16). Note that the factor x 2 – 16 is a difference of perfect squares as well. Therefore, x 4 – 256 = (x 2 + 16)(x + 4)(x – 4). Notice that (x – 4) is a factor of x 3 – 4x 2 + 16x – 64. Therefore, x 3 – 4x 2 + 16x – 64 = (x – 4)(x 2 + 16). Write the preceding rational expression using the factored form of the cubic. Therefore, The goal of the problem is to simplify the fraction, so one of the factors in the numerator is probably going to match one of the factors of x3 – 4x2 + 16x – 64. In other words, the first two factors you should try to divide in synthetically are (x + 4) and (x – 4). . The Humongous Book of Algebra Problems 455 Chapter Twenty — Rational Expressions Divide the first two fractions (by rewriting them as a multiplication problem and then simplifying) and multiply the answer by the fraction on the right. 20.28 Simplify the expression: . According to the order of operations, multiplication and division should be performed in the same step, from left to right. Express the quotient as a product. Factor the quadratic expressions and simplify. 20.29 Simplify the expression: . According to the order of operations, multiplication must be completed before addition. Express the product as a single fraction reduced to lowest terms. 3x and 15 are both divisible by 3, so factor it out: 3x + 15 = 3(x + 5) . The least common denominator of the expression is x 4(x – 5). Rewrite the expression using the least common denominator and simplify. Combine the numerators of the fractions. ngous Book of Algebra Problems 456 The Humo Chapter Twenty — Rational Expressions Simplifying Complex Fractions Reduce fractions that contains fractions 20.30 Simplify the complex fraction: A fraction is really a division problem— the numerator divided by the denominator. . Express the complex fraction as a rational quotient. Just like in Problems 20.23– 20.29. Write the rational quotient as a product and simplify. 20.31 Simplify the complex fraction: . Express the complex fraction as a quotient and simplify the expression. The Humongous Book of Algebra Problems 457 Chapter Twenty — Rational Expressions 20.32 Simplify the complex fraction: . Express the complex fraction as a quotient and simplify the expression. Reduce the fraction to lowest terms. 20.33 Simplify the complex fraction: . Express the complex fraction as a quotient. xy – y has a greatest common factor of y, and x2 – 9 is a difference of perfect squares. ngous Book of Algebra Problems 458 The Humo Chapter Twenty — Rational Expressions Reduce the fraction to lowest terms. 20.34 Simplify the complex fraction: . Express the complex fraction as a quotient. Factor the quadratic expressions and simplify the product. You can’t cancel out the x’s here to get , because x and –7 aren’t multiplied in the denominator. Graphing Rational Functions Rational functions have asymptotes 20.35 Given a function , describe how to identify the vertical and horizontal asymptotes of its graph. If x = c is a vertical asymptote of the graph, then d(c) = 0 but n(c) ≠ 0. To determine the horizontal asymptotes of f(x), if any exist, consider the degrees of the numerator and denominator. Let a be the degree of n(x) and b represent the degree of d(x). Set the denominator equal to zero and solve. The solutions represent vertical asymptotes as long as they don’t also make the numerator equal 0 when you plug them into n(x). The Humongous Book of Algebra Problems 459 Chapter Twenty — Rational Expressions ◆ If a > b, then f(x) has no horizontal asymptotes. ◆ If a < b, then the y-axis is the horizontal asymptote of f(x). If the highest powers of x in the numerator and denominator are equal, then the coefficient attached to the highest power in the numerator divided by the coefficient attached to the highest power of the denominator is the horizontal asymptote. ◆ If a = b, then the horizontal asymptote of f(x) is . Note: Problems 20.36–20.38 refer to the function 20.36 Identify the vertical asymptote to the graph of g(x). Set the denominator of g(x) equal to 0 and solve for x. The line x = –2 is a vertical asymptote to the graph of g(x) because g(–2) is undefined. Note: Problems 20.36–20.38 refer to the function A function is undefined when its denominator equals zero and its numerator doesn’t. In this case, g(–2) = . . . 20.37 Identify the horizontal asymptote to the graph of g(x). The degree of the numerator of g(x) is 0, and the degree of the denominator is 1. According to Problem 20.35, when the degree of the numerator is less than the degree of the denominator, the y-axis, g(x) = 0, is the horizontal asymptote to the graph. Note: Problems 20.36–20.38 refer to the function . 20.38 Graph g(x). The highest power of x in the denominator is x1, so its degree is 1. There are no variables in the numerator, so it has degree 0. It’s more correct to say the equa tion of the y-axis is g( x) instead y = Â�0, be = 0 ca technically ther use e aren’t any y’s in the eq ua Write g(x) where tion. you’d normally write y. To transform the function function, which shifts the graph of is presented in Figure 20-1. See Problem 16.34 for more information. ngous Book of Algebra Problems 460 The Humo into g(x), you add two to the input of the two units to the left. The graph of g(x) Chapter Twenty — Rational Expressions Figure 20-1: The graph of has vertical asymptote x = –2 and horizontal asymptote g(x) = 0. Note: Problems 20.39–20.41 refer to the function . 20.39 Identify the vertical asymptote(s) to the graph of h(x). Factor the numerator and denominator of h(x). Set the factors of the denominator equal to zero and solve for x. Because is undefined, If you plug x = –6 into h(x), it is a vertical asymptote to the graph of h(x). makes the numer and denominator ator zero. For x = –6 to equal vertical asymptot be a e, to make the den it has ominator equal zero but no t the numerator. The Humongous Book of Algebra Problems 461 Chapter Twenty — Rational Expressions Note: Problems 20.39–20.41 refer to the function . 20.40 Identify the horizontal asymptote to the graph of h(x), if one exists. The numerator and denominator both have degree two. According to Problem 20.35, the horizontal asymptote is the quotient of the leading coefficients of the numerator and denominator. Therefore, the graph of h(x) has horizontal asymptote Plug the x’s into the factored version of h(x) from Problem 20.39. It makes evaluating the functions a little easier. . Note: Problems 20.39–20.41 refer to the function . 20.41 Use a table of values to plot the graph of h(x). Construct a table of values that includes x-values near the vertical asymptote, because the most dramatic changes in the graph of a rational function occur near its vertical asymptotes. The graph of h(x) is presented in Figure 20-2. ngous Book of Algebra Problems 462 The Humo Chapter Twenty — Rational Expressions Figure 20-2: The graph of and horizontal asymptote has vertical asymptote . Note: Problems 20.42–20.44 refer to the function . 20.42 Identify all horizontal and vertical asymptotes to the graph of f(x). Because f(–1) is undefined, x = –1 is a vertical asymptote to the graph of f(x). The graph of f(x) has no horizontal asymptotes because the degree of its numerator is greater than the degree of its denominator. (See Problem 20.35.) Note: Problems 20.42–20.44 refer to the function . Only rational functions with numerators that Divide the denominator x + 1 into the numerator x 2 – 2x + 5 using synthetic or are exactly one long division. degree higher than their denominators may have an oblique asymptote, but not all of them do. Oblique asymptotes are sometimes Thus, . The equation of the oblique called “slant asymptotes.” asymptote is the quotient, omitting the remainder: y = x – 3. 20.43 Identify the equation of the oblique asymptote to the graph of f(x). The Humongous Book of Algebra Problems 463 Chapter Twenty — Rational Expressions Note: Problems 20.42–20.44 refer to the function . 20.44 Use a table of values to plot the graph of f(x). According to Problem 20.42. Ensure that the table of values includes x-values near x = –1, the vertical asymptote of the graph. The graph of f(x) is presented in Figure 20-3. Figure 20-3: The graph of oblique asymptote y = x – 3. ngous Book of Algebra Problems 464 The Humo has vertical asymptote x = –1 and Chapter 21 Rational Equations and Inequalities m Chapter 20 Solve equations using the skills fro The preceding chapter explored the arithmetic manipulation of rational expressions; in this chapter, those skills are harnessed to solve rational equations. In addition, this chapter explores direct and indirect variation, as well as solving and graphing rational inequalities. Make sure you’ve worked through Ch apter 20 before you start this chapter, so that you know how to ad d, subtract, multiply, and divide fractions containing polynomials. You need those skills to solve equations featuring those sorts of fractions. This chapter also discusses variatio n in the form of word problems. You describe a direct or indirect varia tion relationship using a rational equation and then solve it. Finally, this chapter explains how to solve inequalities that contain ration al expressions. Basically, you use a slig htly modified version of the techniqu e described in Problems 14.35–14.43 for solving quadratic inequalities. Chapter Twenty-One — Rational Equations and Inequalities Proportions and Cross Multiplication When two fractions are equal, “X” marks the solution An equation that states two fractions are equal is called a proportion. 21.1 Apply cross multiplication to the proportion and generate an equivalent statement. Cross multiplying a proportion eliminates the fractions from the equation. Multiply the numerator of one fraction by the denominator of the other and set the results equal. 21.2 Solve the proportion for x: . Cross multiply (as described in Problem 21.1) to eliminate fractions from the equation. Distribute 3 to the terms inside the parentheses: 3(x – 3) 3(x) + 3(–3) = 3x – = 9. 21.3 Solve the proportion for x: . Cross multiply to eliminate fractions from the equation. 21.4 Solve the proportion for y: . Cross multiply to eliminate fractions from the equation. ngous Book of Algebra Problems 466 The Humo Chapter Twenty-One — Rational Equations and Inequalities 21.5 Solve the proportion for x: . Cross multiply to eliminate fractions from the equation. 21.6 Solve the proportion for x: . Cross multiply to eliminate fractions from the equation. Subtract x2 from both sides of the equation. 21.7 Solve the equation for x: Add . to both sides of the equation to create a proportion. Cross multiply to eliminate fractions from the proportion. Unlike Problem 21.6, the x2terms won’t go away, which means you end up with a quadratic equation. Move all the terms to the left side of the equation so that the right side equals zero. The Humongous Book of Algebra Problems 467 Chapter Twenty-One — Rational Equations and Inequalities Apply the quadratic formula. Reduce the fraction to lowest terms. 21.8 Solve the proportion for x: Cross multiply to eliminate fractions from the equation. Distribute x and then distribute 1 to each term of x2 – x – 2: . ngous Book of Algebra Problems 468 The Humo Chapter Twenty-One — Rational Equations and Inequalities Note: Problems 21.9–21.10 refer to the equation . 21.9 Combine the expressions on the left side of the equation. The least common denominator of and is x – 1. Write the fractions using the least common denominator and simplify the expression. Note: Problems 21.9–21.10 refer to the equation . 21.10 Solve the equation for x. Substitute Replace the left side of the equation with the sum calculated in Problem 21.9. into the equation. Cross multiply to eliminate fractions from the equation. Solve the quadratic by factoring. The solution to the equation is x = –2 or x = –1. The Humongous Book of Algebra Problems 469 Chapter Twenty-One — Rational Equations and Inequalities Solving Rational Equations Ditch the fractions or cross multiply to solve Note: Problems 21.11–21.12 present different methods to solve the equation This is the method used in Problems 21. 2– 21.10. This is the strategy you’ll use to solve rational equations that aren’t simple proportions. When cross multiplication won’t do the trick, eliminate the fractions using the least common denominator and go from there. 21.11 Apply cross multiplication to solve the proportion. Apply the cross multiplication algorithm defined in Problem 21.1. Note: Problems 21.11–21.12 present different methods to solve the equation NOT . . 21.12 Eliminate fractions by multiplying the entire equation by the least common denominator and solve. Multiply each term of the equation by 7(5x – 1), the least common denominator. Distribute 4 on the right side of the equation and solve for x. You’re not trying to get com m denominators, yo on u’re multiplying by th e le common denomin ast ato words, you’re mul r. In other tiplying each term by . , The solution to the equation verifies the solution in Problem 21.11. ngous Book of Algebra Problems 470 The Humo Chapter Twenty-One — Rational Equations and Inequalities 21.13 Verify the solution to the equation from Problem 21.2 by multiplying the entire equation by the least common denominator and solving for x. Multiply both fractions in the proportion by 22 = 4, the least common denominator. The solutions to Problems 21.2 and 21.13 are equal , so both methods of eliminating the fractions in a rational equation (multiplying by the least common denominator and applying cross multiplication) produce the same solution. 21.14 Solve the equation for x: . Multiply the entire equation by 3x, the least common denominator. Subtract x2 from both sides of the equation. That leaves 15 on the left side and x on the right. Solve the equation for x. 15 = x 21.15 Solve the equation for x: . Multiply the entire equation by x – 3, the least common denominator, to eliminate the fraction. Solve the quadratic equation by factoring. x – 3 is the least common denominator because it is the only denominator (excluding the implied denominators of 1 in the terms 4 and 2x). The Humongous Book of Algebra Problems 471 Chapter Twenty-One — Rational Equations and Inequalities Add 12 to, |and subtract 5x from, both sides of the equation so that only zero appears left of the equal sign. Then, flip-flop the sides of the equation so zero appears on the right side instead. The solution to the equation is 21.16 Verify the solution to the equation or x = 4. from Problems 21.9–21.10 using the least common denominator to eliminate fractions before solving. Multiply the entire equation by 2(x – 1), the least common denominator. Solve the equation by factoring. This solution matches the solution in Problem 21.10: x = –2 or x = –1. 21.17 Verify the solution(s) to the equation from Problem 21.7 using the least common denominator to eliminate fractions before solving. Multiply the entire equation by (x + 4)(x – 1), the least common denominator. Distribute –1 through the second quantity and simplify. ngous Book of Algebra Problems 472 The Humo Chapter Twenty-One — Rational Equations and Inequalities Apply the quadratic formula. Here’s how to simplify this expr ession: Note that the solution verifies the solution to Problem 21.7. 21.18 Solve the equation for x: . Factor the quadratic denominator: x 2 – 12x + 20 = (x – 10)(x – 2). Multiply the entire equation by (x – 10)(x – 2), the least common denominator, to eliminate fractions. Solve for x. The Humongous Book of Algebra Problems 473 Chapter Twenty-One — Rational Equations and Inequalities 21.19 Solve the equation for x: . Factor the quadratic denominator: x 2 – 1 = (x + 1)(x – 1). Multiply the entire equation by (x + 1)(x – 1), the least common denominator, to eliminate fractions. It looks like x = 1 is a solution , but if you plug it into the original equa tion to check it, two of the three denominato rs become 0, and yo u’re not allowed to divide by zero. Solve for x. There are no valid solutions to the equation. 21.20 Solve the equation for x: . Multiply the entire equation by (x + 2)(x – 5), the least common denominator, to eliminate the fractions. Set the equation equal to 0 by adding x2 to, and subtracting 3x and 10 from, both sides of the equation. 474 Apply the quadratic formula. The Humongous Book of Algebra Problems Chapter Twenty-One — Rational Equations and Inequalities Direct and Indirect Variation Turn a word problem into a rational equation 21.21 Assume that the value of x varies directly with the value of y according to the constant of proportionality k. Identify two equations that describe the relationship between x and y. If x and y vary proportionally, then y = kx and Varying proportionally means the same thing as varying directly. . Note: Problems 21.22–21.23 refer to the direct variation relationship described below. 21.22 A professional sports team notes that the ambient crowd noise at home games is directly proportional to the attendance at those games. During one game, the cheering of a maximum capacity crowd of 75,000 fans averages 90 decibels. Identify the constant of proportionality k. Let a represent attendance and n represent the noise (in decibels) generated by that population. If n varies directly with a, then (according to Problem 21.21) n = ka. Substitute n = 90 and a = 75,000 into the equation and solve for k. Let’s say k = 2. According to the equation y = kx, y is always twice as big as x. Because y is always two times as large as x, y ÷ x always equals 2. Note: Problems 21.22–21.23 refer to the direct variation relationship described in Problem 21.22. 21.23 Approximately how loud is a crowd of 62,000 fans? Substitute a = 62,000 and k = 0.0012 into the variation equation n = ka to calculate n. This value of k (and the equation n = ka that you plug it into) come from Problem 21.22. The Humongous Book of Algebra Problems 475 Chapter Twenty-One — Rational Equations and Inequalities 21.24 Assume x and y vary proportionally. If x = 16 when y = –3, calculate the value of y when x = 6. If x and y vary proportionally, then x = ky. Substitute x = 16 and y = –3 into the equation and solve for k. To cancel out the coefficient of y, multiply both sides of the equation by its reciprocal. To determine the value of y when x = 6, substitute x = 6 and proportionality equation. into the 21.25 Assume that the growth of a vine is directly proportional to the time it is exposed to light. If the vine is exposed to 72 hours of light and grows 1.5 inches, how long will the vine grow when exposed to 200 hours of light? Round the answer to the hundredths place. Let l represent the length the vine grows when exposed to h hours of light. The values vary proportionally, so l = kh, where k is a constant of proportionality. Substitute l = 1.5 and h = 72 into the equation to calculate k. Rounding 4.166666… to the hundredths place gives you 4.17. To determine how long the vine grows after 200 hours of light, substitute h = 200 and into the equation l = kh. The vine grows approximately 4.17 inches when exposed to 200 hours of light. ngous Book of Algebra Problems 476 The Humo Chapter Twenty-One — Rational Equations and Inequalities 21.26 Assume that the value of x varies inversely with the value of y according to the constant of variation k. Identify two equations that describe the relationship between x and y. If x and y vary inversely, then Inverse variation is also called indirect variation. and xy = k. Note: Problems 21.27–21.28 refer to the inverse variation relationship described below. 21.27 A computer security company determines that the chance a computer is infected with a virus is inversely proportional to the number of times the virus protection software is updated during the year. If a user updates the software 250 times in one year, there is a 1.25% chance that the system will become infected with a virus. Calculate the corresponding constant of variation k. Let u represent the number of updates applied in one year and v represent the corresponding chance that the computer will be infected during that year. Because v varies inversely with u, uv = k (according to Problem 21.26). Substitute u = 250 and v = 0.0125 into the equation to calculate k. Note: Problems 21.27–21.28 refer to the inverse variation relationship described in Problem 21.27. 21.28 What are the chances a computer will be infected with a virus during a year in which the antivirus software is updated once a month? Round the answer to the nearest whole percentage. If x and y vary inversely, then the product of x and y is constant. For example, let’s say x = 2 and y = 5, so xy = 10. If y gets bigger (y = 10) then x has to get smaller (x = 1) to keep the product constant (1 ˙ 10 = 10). You can also refer to the constant of proportionality in direct variation problems as the constant of variation. Substitute u = 12 and k = 3.125 into the equation uv = k and solve for v. There is a 26% chance the computer will be infected with a virus during that year. The value of v should be a perc en tage converted into a decimal, so dro p percentage sign the and move the decim al places to the left two : v = 1. 25% = 0. 0125 . Convert the decimal into a percentage by moving the decimal point two places to the right: 0.260416 = 26.0416%. The Humongous Book of Algebra Problems 477 Chapter Twenty-One — Rational Equations and Inequalities 21.29 Assume x and y vary inversely. If x = –9 when y = 4, calculate the value of x when y = –18. If x and y vary inversely, then xy = k, where k is the constant of variation. Substitute x = –9 and y = 4 into the equation to calculate k. To determine the value of x when y = –18, substitute y = –18 and k = –36 into the equation xy = k and solve for x. 21.30 A police commissioner determines that the number of robberies committed in a downtown tourist district is inversely proportional to the number of police officers assigned to patrol the area on foot. In June, 80 officers were assigned patrols and 35 robberies were reported. If budget cuts to the city’s budget dictate that only 65 officers will be assigned in July, approximately how many robberies will occur during that month? Round the answer to the nearest whole number. The words inversely proportional indicate inverse variation. Let p represent the number of officers on patrol and r represent the corresponding number of robberies during a given month. Because p and r are inversely proportional, pr = k, where k is the constant of variation. Substitute p = 80 and r = 35 into the equation to calculate k. To predict the number of robberies in July, substitute k = 2,800 and p = 65 into the equation pr = k and solve for r. Approximately 43 robberies will occur during July. ngous Book of Algebra Problems 478 The Humo Chapter Twenty-One — Rational Equations and Inequalities Solving Rational Inequalities Critical numbers, test points, and shading Note: Problems 21.31–21.33 refer to the inequality . 21.31 Identify the critical numbers of the rational expression. The critical numbers of an expression are the values for which the expression equals zero or is undefined. A rational expression equals zero when its numerator equals zero and is undefined when its denominator equals zero. Set the numerator and denominator of the fraction equal to zero and solve the resulting equations. The critical numbers of the rational expression are x = –2 and x = 3. Note: Problems 21.31–21.33 refer to the inequality . 21.32 Identify the intervals of the real number line that represent solutions to the inequality. The critical numbers x = –2 and x = 3 (calculated in Problem 21.31) split the real number line into three intervals, as illustrated by Figure 21-1. Figure 21-1: The critical numbers x = –2 and x = 3 split the real number line into three intervals: x < –2, –2 < x < 3, and x > 3. Use open points to mark the critical numbers, because the inequality sign “>” does not allow for the possibility of equality. Choose one test value from within each interval (such as x = –5 from x < –2, x = 0 from –2 < x < 3, and x = 5 from x > 3) and substitute those values into the inequality. The solution to the inequality is x < –2 or x > 3. The symbols < and > always mean “do not incl ude the boundaries,” whether you use hollow dots on a number line or dotted lines on a coordinate plane. The test points from these intervals produced true statements, so that’s why they make up the solution. Use the word “or” between the solution intervals, because there aren’t any numbers that are fewer than –2 and greater than 3. The Humongous Book of Algebra Problems 479 Chapter Twenty-One — Rational Equations and Inequalities Note: Problems 21.31–21.33 refer to the inequality . 21.33 Graph the inequality. Darken the segments of the number line identified by Problem 21.32 as solutions to the inequality: x < –2 and x > 3. The graph is presented in Figure 21-2. Figure 21-2: The graph of . Note: Problems 21.34–21.36 refer to the inequality . 21.34 Identify the critical numbers of the expression. Factor the numerator. The critical numbers of a rational expression are the x-values that cause either the numerator or denominator to equal zero. Set each of the factors equal to zero and solve for x. Normally ≤ and ≥ mean solid dots, but x = 5 makes the denominator zero, which is not allowed. Anything that makes the denominator zero becomes an open dot on the number lin e no matter what the inequality symbol is. Note: Problems 21.34–21.36 refer to the inequality . 21.35 Solve the inequality. According to Problem 21.34, the critical numbers of the rational expression are x = –4, , and x = 5. Plot the values on a number line, noting that x = 5 is represented by an open point, whereas the other two critical numbers are plotted as closed points. As illustrated by Figure 21-3, the critical numbers divide the number line into four intervals. ngous Book of Algebra Problems 480 The Humo Chapter Twenty-One — Rational Equations and Inequalities Figure 21-3: The critical numbers separate the number line into four intervals: x ≤ –4, , , and x > 5. Substitute a test value from each interval (such as x = –6, x = 0, x = 3, and x = 8) into the inequality to identify the solutions. The solution to the inequality is The inequality symbols attached to x = 5 are ‹ and ›, instead of ≤ and ≤, because x = 5 is an open dot on the graph. or x > 5. Note: Problems 21.34–21.36 refer to the inequality . 21.36 Graph the inequality. Darken the intervals of the number line identified by Problem 21.35 as solutions to the inequality: Figure 21-4: The graph of and x > 5. The graph is presented in Figure 21-4. . The Humongous Book of Algebra Problems 481 Chapter Twenty-One — Rational Equations and Inequalities When you solve a rational inequality, you want the right side of the statement to be zero. It’s not like an equation, where you add to both sides and cross multiply. Note: Problems 21.37–21.40 refer to the inequality . 21.37 Write the left side of the inequality as a single rational expression. Express the left side of the inequality using the least common denominator 2(x – 1). If you don’t know how to do this, look at Problems 20.9–20.19. Note: Problems 21.37–21.40 refer to the inequality . 21.38 Identify the critical numbers of the rational expression generated in Problem 21.37. Except the factor 2 from the denominator, because 2 by itself can’t equal 0. Factor the numerator. Set each of the factors equal to zero and solve the resulting equations. As illustrated by Figure 21-5, the critical numbers x = –2, x = 1, and x = 3 split the number line into four intervals: x ≤ –2, –2 ≤ x < 1, 1 < x ≤ 3, and x ≥ 3. Figure 21-5: The critical numbers x = –2, x = 1, and x = 3 split the number line into four intervals. Plot x = 1 as an open point because it causes the denominator to equal zero, making the rational expression undefined. ngous Book of Algebra Problems 482 The Humo Chapter Twenty-One — Rational Equations and Inequalities Note: Problems 21.37–21.40 refer to the inequality . 21.39 Solve the inequality. Substitute a test value from each of the intervals identified by Problem 21.38 (such as x = –3, x = 0, x = 2, and x = 4) into the inequality to identify the solution. The test values are plugged into the factored version of the fraction from Problem 21.38, but you could also use the original inequality. The solution to the inequality is x ≤ –2 or 1 < x ≤ 3. Note: Problems 21.37–21.40 refer to the inequality . 21.40 Graph the inequality. Darken the segments of the number line identified by Problem 21.39 as solutions to the inequality: x ≤ –2 and 1 < x ≤ 3. The graph is presented in Figure 21-6. Figure 21-6: The graph of the inequality Note: Problems 21.41–21.44 refer to the inequality . . 21.41 Combine the rational expressions into a single rational expression. Subtract from both sides of the inequality. Combine the fractions using the least common denominator x(x + 3). So that the right side of the inequality equals zero. The Humongous Book of Algebra Problems 483 Chapter Twenty-One — Rational Equations and Inequalities Multiply both sides of the inequality by –1. That changes –2x2 into 2x2, whi ch is slightly easier to factor. Don’t forg et reverse the ineq to uality sign because you’ re multiplying both sides of the inequality by a negative number. Note: Problems 21.41–21.44 refer to the inequality . 21.42 Identify the critical numbers of the rational expression generated by Problem 21.41. Factor the quadratic in the numerator. Set each of the factors in the numerator and denominator equal to zero and solve the resulting equations to identify the critical numbers. The critical numbers of the rational expression are x = –3, x = 3. ngous Book of Algebra Problems 484 The Humo , x = 0, and Chapter Twenty-One — Rational Equations and Inequalities Note: Problems 21.41–21.44 refer to the inequality . 21.43 Solve the inequality. The critical numbers identified in Problem 21.42 (x = –3, , x = 0, and x = 3), split the number line into five intervals, as illustrated by Figure 21-7. All the dots shou ld be open, because th inequality sign is e <. Figure 21-7: The critical numbers split the number line into five distinct intervals: x < –3, ,- 3 < x < 0 , 0 < x < 3, and x > 3. 2 Choose one test point from each interval (such as x = –4, x = –2, x = –1, x = 1, and x = 4) to identify the intervals that represent solutions to the inequality. The solution to the inequality is x < –3 or or x > 3. The Humongous Book of Algebra Problems 485 Chapter Twenty-One — Rational Equations and Inequalities Note: Problems 21.41–21.44 refer to the inequality . 21.44 Graph the inequality. Darken the intervals identified by Problem 21.43 as the solutions to the inequality: x < –3 or Figure 21-8: The graph of ngous Book of Algebra Problems 486 The Humo or x > 3. The graph is presented in Figure 21-8. . Chapter 22 Conic sections Parabolas, Circles, Ellipses, and Hyperbolas The conic sections are a collection of equations that contain x and y raised to the second power. Among them, only parabolas contain a single variable that is squared, whereas circles, ellipses, and hyperbolas contain two squared variables. In this chapter, the standard form and defining geometric characteristics of each conic section are explored. This chapter starts out with a discu ssion of parabolas, which are the graphs of quadratic equations that contain x and y. Your goal will be to write parabolas in standard form an d identity some of their parts, such as the vertex, the focus, and the dir ectrix. Warning! To put conic sections in sta ndard form, you need to be able to complete the square. If you don’t kno w how, review Problems 14.12–14.19. Chapter Twenty-Two — Conic Sections Parabolas Vertex, axis of symmetry, focus, and directrix Note: Problems 22.1–22.3 refer to the equation y = x2 – 2x + 5. You’ll need to complete the square for the expression x2 – 2x. Move all of the terms that don’t contain x to the left side of the equation. To complete the square, divid the x-coefficient e two (–2 ÷ 2 = –1) by a square the result nd ([–1] 2 = 1). To keep everything balanc ed have to add 1 to , you the left and right sid es of the equation. 22.1 Write the equation of the parabola in standard form. The standard form of a parabola containing an x 2-term is y = a(x – h)2 + k, where (h,k) is the vertex and a is a real number. Subtract 5 from both sides of the equation in order to isolate the x-terms right of the equal sign. y – 5 = x 2 – 2x To complete the square on the right side of the equation, add 1 to both sides of the equation. Factor the quadratic expression. y – 4 = (x – 1)2 Solve for y. y = (x – 1)2 + 4 Note: Problems 22.1–22.3 refer to the equation y = x2 – 2x + 5. 22.2 Identify the vertex and axis of symmetry. The standard form of a parabola containing an x 2-term is y = a(x – h)2 + k. According to Problem 22.1, the standard form of the parabola is y = (x – 1)2 + 4. Therefore, a = 1, h = 1, and k = 4. This is the number in front of the parentheses. When there is no number, a = 1. The vertex of the parabola is (h,k) = (1,4). The axis of symmetry, the vertical line that passes through the vertex, has equation x = h; the axis of symmetry for this parabola is x = 1. This is the OPPOSITE of the number in parentheses. The opposite of –1 is 1. This is the number added to the squared quantity. ngous Book of Algebra Problems 488 The Humo Chapter Twenty-Two — Conic Sections Note: Problems 22.1–22.3 refer to the equation y = x2 – 2x + 5. 22.3 Graph the parabola. To transform the equation y = x 2 into the standard form equation y = (x – 1)2 + 4, subtract 1 from the input x (which shifts the graph of y = x 2 one unit to the right) and add 4 (which shifts the graph up four units). The graph is presented in Figure 22-1. Figure 22-1: The graph of y = x2 – 2x + 5 has vertex (1,4) and axis of symmetry x = 1. Note: Problems 22.4–22.6 refer to the equation x = y2 – 6y + 7. 22.4 Write the equation of the parabola in standard form. The standard form of a parabola containing a y 2-term is x = a(y – k)2 + h, where (h,k) is the vertex and a is a real number. Subtract 7 from both sides of the equation to isolate the y-terms right of the equal sign. If you’re not sure how to graph a parabola using transformations, look at Problems 16.3 1 and 16.32. Divide the y-coefficient by two (–6 ÷ 2 = –3) and square the result ([–3] 2 = 9). Add that number to both sides of the equation to keep everything balanced. x – 7 = y 2 – 6y Complete the square on the right side of the equation by adding 9 to both sides. Factor the quadratic expression. x + 2 = (y – 3)2 Solve for x. x = (y – 3)2 – 2 The Humongous Book of Algebra Problems 489 Chapter Twenty-Two — Conic Sections When the parabola has a y2-term instead of an x2-term, k is the opposite of the number inside the parentheses and h is the number added to or subtracted from the squared quantity. Note: Problems 22.4–22.6 refer to the equation x = y2 – 6y + 7. 22.5 Identify the vertex and axis of symmetry. The standard form of a parabola containing a y 2-term is x = a(y – k)2 + h. According to Problem 22.4, the standard form of the parabola is x = (y – 3)2 – 2. Therefore, a = 1, h = –2, and k = 3. The vertex of the parabola is (h,k) = (–2,3) and the axis of symmetry (y = k) is y = 3. Note: Problems 22.4–22.6 refer to the equation x = y2 – 6y + 7. 22.6 Graph the parabola. When the parabola contain 2 s the axis of symm y , etry is a horizontal lin e (n a vertical line lik ot e in Problem 22 .2) th at passes through th e vertex. Plug the numbers from the left column into y and calculate x. Graph the parabola using a table of values that includes y-values near the axis of symmetry y = 3. According to the table of values, the graph passes through the following points: (–2,3), (–1,2), (–1,4), (2,1), (2,5), and (7,0). The graph is presented in Figure 22-2. The numbers in the left column are the y-values, not the x-values. That means the top line of the table generates the point (2,5), not (5,2). Figure 22-2: The graph of the parabola x = y2 – 6y + 7 has vertex (–2,3) and axis of symmetry y = 3. ngous Book of Algebra Problems 490 The Humo Chapter Twenty-Two — Conic Sections Note: Problems 22.7–22.9 refer to the equation 3x2 – 12x + y + 21 = 0. 22.7 Write the equation of the parabola in standard form. By subtracting 3x from and adding 12x to both sides. 2 Isolate the terms containing x on the right side of the equation. y + 21 = –3x 2 + 12x In order to complete the square, the coefficient of the x 2-term must be 1. Factor –3, the coefficient of the x 2-term, out of the terms right of the equal sign. y + 21 = –3(x 2 – 4x) Complete the square by adding 4 to the parenthetical quantity; add –3(4) = –12 to the left side of the equation to maintain equality. Factor the quadratic expression. y + 9 = –3(x – 2)2 Solve for y. y = –3(x – 2)2 – 9 Note: Problems 22.7–22.9 refer to the equation 3x2 – 12x + y + 21 = 0. If you factor a negative number (–3) out of a posi tive number (12), the result is negative (– 4). You add 4 inside parentheses, but those parentheses are multiplied by –3, so you’re actually adding –3(4) = –12 to the right side of the equation. Add –12 to the left side as well. 22.8 Identify the focus of the parabola. The standard form of a parabola containing an x 2-term is y = a(x – h)2 + k. According to Problem 22.7, the standard form of the parabola is y = –3(x – 2)2 – 9. Therefore, a = –3. Let focus. represent the distance between the vertex (h,k) = (2,–9) and the Every point on a parabola is exactly the same distance from its focus as it is from its directrix (which is identified in Problem 22 .9). When a > 0, the focus is c units above the vertex: (h,k + c). When a < 0, the focus is c units below the vertex: (h,k – c). Calculate the coordinates of the focus. The focus is always inside the “cup” of the parabola. The focus is . The Humongous Book of Algebra Problems 491 Chapter Twenty-Two — Conic Sections Note: Problems 22.7–22.9 refer to the equation 3x2 – 12x + y + 21 = 0. 22.9 Identify the directrix of the parabola. The focus of a parabola is a point, and the directrix is a line. The directrix of a parabola that contains an x 2-term is a horizontal line. When a > 0, the directrix is units below the vertex: y = k – c. When a < 0, the directrix is c units above the vertex: y = k + c. According to Problem 22.8, . Because a < 0 (a = –3, according to Problem 22.8), the directrix is c units above the vertex. The distance between the vertex and the focus is equal to the distance between the vertex and the directrix. Both of those distances are labeled c. The directrix is . 22.10 Identify the focus and directrix of the parabola: 2y 2 – x – 5y + 9 = 0. Isolate the terms containing y on the right side of the equation. –x + 9 = –2y 2 + 5y Factor the coefficient of y 2 out of the terms right of the equal sign. Complete the square. Square half of th e y-coefficient: . Add that insid e the parenthese However, the cont s. ents of the pare ntheses are multi plied by –2 , so you’re actually a dding to the right side. Add it to th e left side of th e equation as wel l. ngous Book of Algebra Problems 492 The Humo Chapter Twenty-Two — Conic Sections Multiply the entire equation by –1 and solve for x. The equation is in standard form x = a(y – k)2 + h; therefore, a = 2, , and . Calculate c. Because a > 0, the focus is right of the vertex: (h + c,k). This parabola contains y2 , so the rules about the focus and directrix in Problems 22.8 and 22.9 have to be adjusted. Change “above” to “right of” and “below” to “left of.” Instead of adding c to the y-value of the focus, add it to the x-value. The directrix is the vertical line c units left of the vertex: x = h – c. The focus of the parabola is and the directrix is 22.11 Write the equation of the parabola with vertex . and directrix x = 0 in standard form. If the vertex is , then and k = 0. The directrix is a vertical line The ends of this parabola poin left. Its directrix t vertical line, so th is a you it opens either at tells left. The axis of right or symmetry is the horizontal line passing through the vert ex: y = 0. one half of a unit right of the vertex, so the focus is the point one half of a unit left of the vertex along the axis of symmetry: (–1,0). Note that c represents the distance between the vertex and either the directrix or the focus—the distances are equivalent—so Substitute into the formula . to calculate a. The Humongous Book of Algebra Problems 493 Chapter Twenty-Two — Conic Sections Apply cross multiplication to solve the proportion. Substitute equation. , , and k = 0 into the appropriate standard form Use this one instead of y = a(x – h) 2 + k because the directrix is vertical and the axis of symmetry is horizontal. Circles Center, radius, and diameter Note: Problems 22.12–22.13 refer to the circle with center (2,–1) and radius 4. 22.12 Graph the circle. The points are (2,3), (2,–5), (–2 ,–1), and (6,–1). Plot the center point (2,–1) on the coordinate plane. Next, plot the points that are four units above, below, left, and right of the center. Draw the circle that passes through those four points, as illustrated by Figure 22-3. Figure 22-3: Each of the points on the graph of the circle is exactly four units away from the center point (2,–1). ngous Book of Algebra Problems 494 The Humo Chapter Twenty-Two — Conic Sections Note: Problems 22.12–22.13 refer to the circle with center (2,–1) and radius 4. 22.13 Write the equation of the circle in standard form. The standard form for the equation of a circle is (x – h)2 + (y – k)2 = r2, where (h,k) is the center of the circle and r is the radius. Substitute h = 2, k = –1, and r = 4 into the equation. Note: Problems 22.14–22.16 refer to the equation x2 + y2 + 8x – 6y = 0. 22.14 Write the equation of the circle in standard form. Use parentheses to group the x-terms and the y-terms. (x 2 + 8x) + (y 2 – 6y) = 0 Complete the square twice, once for each parenthetical quantity. Note: Problems 22.14–22.16 refer to the equation x2 + y2 + 8x – 6y = 0. 22.15 Identify the center and radius of the circle. The standard form of a circle is (x – h)2 + (y – k)2 = r2. According to Problem 22.14, the standard form of this circle is (x + 4)2 + (y – 3)2 = 25. Therefore, h = –4, k = 3, and r2 = 25. Solve the equation to calculate r. The solution r = –5 is discarded, as the radius of a circle must be a positive number. The center of the circle is (h,k) = (–4,3) and the radius is r = 5. Square half of the x-coefficient: (8 ÷ 2) 2 = 4 2 = 16 . Add that numbe r inside the left se t of parentheses a nd make sure to ad d it to the right sid e of the equation, too. Then, square ha lf of the y-coefficien t: (– 6 ÷ 2) 2 = (–3) 2 = 9. Add that to the second set of parentheses and the right side of the equation. h is the opposite of the number in the x parentheses and k is the opposite of the number in the y parentheses. The Humongous Book of Algebra Problems 495 Chapter Twenty-Two — Conic Sections Note: Problems 22.14–22.16 refer to the equation x2 + y2 + 8x – 6y = 0. 22.16 Graph the circle. According to Problem 22.15, the center of the circle is (–4,3) and the radius is 5. Plot the center and then plot the points five units above, below, left, and right of the center. Draw a circle that intersects the four points surrounding the center, as illustrated in Figure 22-4. Figure 22-4: The graph of the circle x2 + y2 + 8x – 6y = 0 has center (–4,3) and radius 5. 22.17 Write the equation of the circle 3x 2 + 3y 2 + 12x + 15y – 1 = 0 in standard form and identify the center and radius. The x2- and y -terms have the same coefficient in the equations of circles (in this case 3). 2 Writing the equation in standard form requires you to complete the square. Divide each of the terms in the equation by 3 so that the coefficients of x 2 and y 2 are 1. Move the constant to the right side of the equation by adding ngous Book of Algebra Problems 496 The Humo to both sides. Chapter Twenty-Two — Conic Sections Place the x- and y-terms in separate groups. Complete the square twice, once for each group of terms on the left side of the equation. The equation is in standard form (x – h)2 + (y – k)2 = r2 so the center of the circle is . Calculate the radius. To rationalize the denominator, multiply the numerator and denominator by . Note: Problems 22.18–22.19 refer to the circle with center (–6,4) that passes through point (–2,3). 22.18 Use the distance formula to calculate the radius of the circle. The distance between the center of a circle and any point on that circle is equal to the radius. Substitute x1 = –6, y1 = 4, x 2 = –2, and y 2 = 3 into the distance formula to calculate the radius. x1 and y1 refer to the x- and y-values of point #1, which could be (–6,4) OR (–2,3). In other words, substituting x1 = –2, y1 = 3, x2 = –6, and y2 = 4 into the distance formula gives you the same final answer. The Humongous Book of Algebra Problems 497 Chapter Twenty-Two — Conic Sections Note: Problems 22.18–22.19 refer to the circle with center (–6,4) that passes through point (–2,3). 22.19 Write the equation of the circle in standard form. The center of the circle is (–6,4), so h = –6 and k = 4. According to Problem 22.18, the radius of the circle is . Substitute h, k, and r into the standard form equation. Note: Problems 22.20–22.21 refer to the circle that has a diameter with endpoints (3,–1) and (–3,7). The x-value of the midpoint is the average of the endpoints’ x-values, so add them up and divide by 2. Do the same thing with the y’s to complete the midpoint. 22.20 Use the midpoint formula to identify the center of the circle. The midpoint of a diameter is the center of the corresponding circle. Apply the midpoint formula to the endpoints of the diameter to identify the center (h,k) of the circle. Note: Problems 22.20–22.21 refer to the circle that has a diameter with endpoints (3,–1) and (–3,7). 22.21 Calculate the radius of the circle and write the equation in standard form. Apply the distance formula, defined in Problem 22.18, to calculate the radius. Find the distance betwee n the center (0,3) and either of the en dpo For example, you ints. co set x1 = 0 and y uld 1 = 3 (the center) and se x2 = 3 and y = –1 t (the 2 first of the endpo ints). ngous Book of Algebra Problems 498 The Humo Chapter Twenty-Two — Conic Sections Substitute h = 0, k = 3, and r = 5 into the standard form equation. Ellipses Major and minor axes, center, foci, and eccentricity 22.22 Identify both standard forms of an ellipse and describe the values of the constants. The standard form of an ellipse with a horizontal major axis is ; the standard form of an ellipse with a vertical major axis is . The center of the ellipse is (h,k), a represents the distance between the center and a vertex along the major axis, and b represents the distance between the center and an endpoint of the minor axis. The axes of an ellipse are the perpendicular segments that pass through the center and extend to the right, left, top, and bottom “edges” of the ellipse. The longer of the two is called the ma jor axis, and the other one’s the minor axis. Note: Problems 22.23–22.24 refer to the ellipse with center (–1,0), vertical major axis of length 10, and horizontal minor axis of length 6. 22.23 Graph the ellipse. If the vertical major axis is ten units long, then it extends five units above and five units below the center. Similarly, a horizontal minor axis of length six extends three units left and three units right of the center, as illustrated by Figure 22-5. Five units up = (–1,5); five units down = (–1,–5); three units left = (–4,0); and three units right = (2, 0). Figure 22-5: This ellipse has the center (–1,0), a major axis ten units long, and a minor axis six units long. The Humongous Book of Algebra Problems 499 Chapter Twenty-Two — Conic Sections Note: Problems 22.23–22.24 refer to the ellipse with center (–1,0), vertical major axis of length 10, and horizontal minor axis of length 6. 22.24 Write the equation of the ellipse in standard form. a 2 appears beneath the yexpression when th ma jor axis is vert e ic and appears bene al ath the x-expression w hen the ma jor axis is horizontal. According to Problem 22.22, the standard form of an ellipse with a vertical major axis is . The center of the ellipse is (–1,0), so h = –1 and k = 0. The values of a and b are half of the lengths of the major and minor axes, respectively. Therefore, a = 10 ÷ 2 = 5 and b = 6 ÷ 2 = 3. Substitute h, k, a, and b into the standard form equation. Note: Problems 22.25–22.27 refer to the ellipse with equation x2 + 9y2 – 6x – 36y + 36 = 0. 22.25 Write the equation in standard form. Group the x-terms together in a set of parentheses, group the y-terms in a second set of parentheses, and move the constant to the right side of the equation. (x 2 – 6x) + (9y 2 – 36y) = 0 – 36 But it’s not, so factor the coefficient (9) out of both y-terms. In order to complete the square for the expression 9y 2 – 36y, the coefficient of the squared term must be 1. (x 2 – 6x) + 9(y 2 – 4y) = –36 Complete the square for each of the parenthetical quantities. To complete the square, you have to add 9 to the x-parentheses and 4 to the yparentheses. However, the y-parentheses are multiplied by 9, so you’re actually adding 9(4), so make sure to add 9(4) = 36 to the right side of the equation too. The equation of an ellipse in standard form contains only the constant 1 on the right side of the equation. Divide the entire equation by 9, the constant currently right of the equal sign. ngous Book of Algebra Problems 500 The Humo Chapter Twenty-Two — Conic Sections Note: Problems 22.25–22.27 refer to the ellipse with equation x2 + 9y2 – 6x – 36y + 36 = 0. 22.26 Graph the ellipse. According to Problem 22.25, the standard form of the ellipse is . The center of the ellipse is (3,2). The constant beneath the x-expression (9) is greater than the constant beneath the y-expression (1), so the major axis of the ellipse is horizontal. Let a 2 be the greater of the two denominators and b 2 be the lesser of the denominators: a 2 = 9 and b 2 = 1, so a = 3 and b = 1. To graph the ellipse, plot the points three units right and left of the center (because a = 3 and the major axis is horizontal) and the points one unit above and below the center (because b = 1 and the minor axis is vertical). The ellipse is graphed in Figure 22-6. The xcoordinate of the center is the opposite of the number in the xexpression (x – 3) 2 . The y-coordinate of the center is the opposite of the constant in the y-expression (y – 2)2 . Figure 22-6: The graph of the ellipse x2 + 9y2 – 6x – 36y + 36 = 0 has center (3,2), a horizontal major axis of length 6, and a vertical minor axis of length 2. The Humongous Book of Algebra Problems 501 Chapter Twenty-Two — Conic Sections Note: Problems 22.25–22.27 refer to the ellipse with equation x2 + 9y2 – 6x – 36y + 36 = 0. 22.27 Identify the vertices of the ellipse. The vertices of an ellipse are the endpoints of the major axis. Consider the graph of the ellipse in Figure 22-6. The major axis is the horizontal segment with endpoints (0,2) and (6,2), so those points are the vertices of the ellipse. Note: Problems 22.28–22.30 refer to the ellipse with equation 9x2 + 4y2 + 90x – 48y + 333 = 0. 22.28 Write the equation in standard form. Group the x-terms as one quantity, the y-terms as another quantity, and move the constant to the right side of the equation. (9x 2 + 90x) + (4y 2 – 48y) = 0 – 333 Factor the coefficients of the squared terms out of each quantity. 9(x 2 + 10x) + 4(y 2 – 12y) = –333 Complete the square for each quantity. Divide the entire equation by 36 to set the right side of the equation equal to 1. Note: Problems 22.28–22.30 refer to the ellipse with equation 9x2 + 4y2 + 90x – 48y + 333 = 0. 22.29 Identify the center, vertices, and foci of the ellipse. If the ma jor axis is horizontal, the vertices are (h – a,k) and (h + a,k). The denominator beneath the y-expression is greater than the denominator beneath the x-expression, so the major axis of the ellipse is vertical, a 2 = 9, and b 2 = 4. Thus a = 3 and b = 2. The center of the ellipse is comprised of the opposites of the constants within the squared expressions: (h,k) = (–5,6). Because the major axis is vertical, the vertices are the points a = 3 units above and below the center. ngous Book of Algebra Problems 502 The Humo Chapter Twenty-Two — Conic Sections The foci are the two points c units away from the center along the major axis such that . Calculate c. If you go up and down from the center to reach the vertices, you also go up and down (but not quite as far) to reach the foci. Calculate the coordinates of the foci. Note: Problems 22.28–22.30 refer to the ellipse with equation 9x2 + 4y2 + 90x – 48y + 333 = 0. If the major axis is horizontal, the foci are (h – c,k) and (h + c,k). 22.30 Calculate the eccentricity of the ellipse. The eccentricity of an ellipse is defined as , where c is the distance between the center and a focus and a is the distance between the center and a vertex. According to Problem 22.29, a = 3 and . Note: Problems 22.31–22.32 refer to the ellipse with center (4,–1) that passes through the origin if the major axis is horizontal and ab = 8. 22.31 Write the equation of the ellipse in terms of a. The standard form of an ellipse with a horizontal major axis is . Solve the equation ab = 8 for b. Substitute h = 4, k = –1, and Simplify the complex fraction by multiplying the top and bottom by the reciprocal of the denominator: into the equation and simplify. The Humongous Book of Algebra Problems 503 Chapter Twenty-Two — Conic Sections Note: Problems 22.31–22.32 refer to the ellipse with center (4,–1) that passes through the origin if the major axis is horizontal and ab = 8. 22.32 Write the equation of the ellipse in standard form. The ellipse passes through the origin, the point (x,y) = (0,0). Substitute these values into the equation so that the only unknown that’s left is a. Substitute x = 0 and y = 0 into the equation from Problem 22.31 written in terms of a. Multiply the entire equation by the least common denominator, 64a 2, to eliminate fractions. Use the zero product property. The only way (a2 – 32)(a2 – 32) could equal zero is if the factors equal zero. Don’t worry about the ± sign, because a, b, and c are positive values. Solve the equation by factoring. Substitute ngous Book of Algebra Problems 504 The Humo into the equation to calculate b. Chapter Twenty-Two — Conic Sections Rationalize the denominator. Substitute h = 4, k = –1, , and for an ellipse with a horizontal major axis. into the standard form equation The Humongous Book of Algebra Problems 505 Chapter Twenty-Two — Conic Sections Hyperbolas Transverse and conjugate axes, foci, vertices, and asymptotes Note: Problems 22.33–22.36 refer to the hyperbola graphed in Figure 22-7 below. Figure 22-7: The graph of a hyperbola with vertices (–4,3) and (4,3). 22.33 Identify the center of the hyperbola. The center, (h,k), of a hyperbola is the midpoint of the segment connecting the vertices. Substitute x1 = –4, y1 = 3, x 2 = 4, and y 2 = 3 into the midpoint formula. The center of the hyperbola is (0,3). If you know the endpoints of a VERTICAL segment, you can find its length by subtracting the yvalues and taking the absolute value. Note: Problems 22.33–22.36 refer to Figure 22-7, the hyperbola graphed in Problem 22.33. 22.34 Calculate the lengths of the transverse and conjugate axes. The transverse axis of a hyperbola is the segment whose endpoints are the vertices. According to Figure 22-7, the vertices of the hyperbola are (–4,3) and (4,3). The length, lt, of the horizontal transverse axis is equal to the absolute value of the difference of the x-values. ngous Book of Algebra Problems 506 The Humo Chapter Twenty-Two — Conic Sections Therefore, the transverse axis is 8 units long. To calculate the length of the conjugate axis—a segment perpendicular to the transverse axis at the center of the hyperbola—draw vertical segments that extend from the vertices to the asymptotes of the graph. Plot the intersection points of those two vertical segments and the asymptotes. Connect the four intersection points to form a rectangle, as illustrated by Figure 22-8. The transverse axis is horizontal, so draw vertical lin es up and down from both vertices th at stop at the dotte d asymptotes. Tho se points are the co rn of the rectangle ers in Figure 22-8. Figure 22-8: The transverse axis bisects the rectangle horizontally, with endpoints at the vertices. The conjugate axis bisects the hyperbola vertically, with endpoints (0,2) and (0,4). The length, lc , of the vertical conjugate axis is the absolute value of the difference of the endpoints’ y-values. The center is the midpoint of Note: Problems 22.33–22.36 refer to Figure 22-7, the hyperbola graphed in Problem 22.33. both the transverse and conjugate axes. 22.35 Write the equation of the hyperbola in standard form. Dividing the length of the transverse axis in half The standard form of a hyperbola with a horizontal transverse axis is gives you a, and dividing the length of the , such that the center is (h,k), the distance between the conjugate axis in half center and a vertex is a, and the distance between the center and an endpoint of gives you b. the conjugate axis is b. According to Problem 22.33, the transverse axis is 8 units long, so a = 8 ÷ 2 = 4. The conjugate axis is 2 units long, so b = 2 ÷ 2 = 1. According to Problem 22.33, the center of the hyperbola is (0,3), so h = 0 and k = 3. Substitute a = 4, b = 1, h = 0, and k = 3 into the standard form equation. The Humongous Book of Algebra Problems 507 Chapter Twenty-Two — Conic Sections Note: Problems 22.33–22.36 refer to Figure 22-7, the hyperbola graphed in Problem 22.33. 22.36 Write the equations of the asymptotes in slope-intercept form. Consider the rectangle in Figure 22-8; its corners are also points on the linear asymptotes. For instance, the asymptote passing through the upper-left and lower-right corners of the rectangle passes through points (–4,4) and (4,2). Apply the slope formula to calculate the slope of that asymptote. This b is the y-intercept value (b = 3), NOT the b used to represent the distance between the center of the hyperbola and an endpoint of the conjugate axis. The slope of the line is . The asymptote intersects the y-axis at the center of the hyperbola, (0,3); therefore, its y-intercept is 3. Apply the slopeintercept form of a line. The remaining asymptote has slope equation Asymptotes don’t always have the same yintercept. These happen to because they intersect on the y-axis. and y-intercept 3. Thus, it has . Note: Problems 22.37–22.40 refer to the hyperbola with equation x2 – 4y2 + 2x + 32y – 27 = 0. 22.37 Write the equation in standard form. Group the x-terms in a set of parentheses, group the y-terms in a second set of parentheses, and move the constant to the right side of the equation. (x 2 + 2x) + (–4y 2 + 32y) = 27 The coefficients of both squared terms must equal 1, so factor the coefficient of y 2 out of the corresponding quantity. (x 2 + 2x) – 4(y 2 – 8y) = 27 Complete the square for each parenthetical quantity. Add 1 to the x-expression and 16 to the yexpression. However, the y-expression is multiplied by –4, so you’re actually adding –4(16) = –64 to both sides of the equation. The right side of the equation of a hyperbola in standard form must equal 1, so divide both sides of the equation by –36. ngous Book of Algebra Problems 508 The Humo Chapter Twenty-Two — Conic Sections The positive rational expression must precede the negative rational expression in the standard form of a hyperbola. Put the positive fraction in front of the negative fraction. Note: Problems 22.37–22.40 refer to the hyperbola with equation x2 – 4y2 + 2x + 32y – 27 = 0. 22.38 Identify the center and vertices and sketch the graph of the hyperbola. According to Problem 22.37, the standard form of the equation is . Because the fraction containing the y-expression is positive, the transverse axis of this hyperbola is vertical. The standard form of a hyperbola with a vertical transverse axis is . Therefore, the center of the hyperbola is (h,k) = (–1,4), a = 3, and b = 6. Note that a is not necessarily greater than b, unlike in the equations of ellipses. Plot the points three units above and below the center and the points six units left and right of the center. Draw a rectangle whose sides pass through those four points, as illustrated by Figure 22-9. a2 is below the y-expression, so you go a units up and down ( because = 3 y’s vertical distance). Similarly 2 measure , b is below the x-expression, so you go b = 6 units left and right (because x’s measure horizontal distance). The hyperbola opens up and down. When the x-expression is positive (like in Problems 22.33–22.36) the transverse axis is horizontal and the hyperbola opens left and right. a is the square root of the positive fraction’s denominator and b of root re is the squa tion’s the negative frac denominator . It doesn’t matter which is larger. The Humongous Book of Algebra Problems 509 Chapter Twenty-Two — Conic Sections If you need a more exact graph, solve the equation for x and make a table of values that includes yvalues greater than 7 and less than 1 (the y-values of the vertices). Figure 22-9: The segment that bisects this rectangle vertically is the transverse axis of the hyperbola. The segment that bisects the rectangle horizontally is the conjugate axis. Draw the asymptotes of the hyperbola, the pair of lines that extends through opposite corners of the rectangle in Figure 22-9. The points a = 3 units above and below the center, along the transverse axis, are the vertices of the hyperbola: (–1,1) and (–1,7). The graph passes through each vertex, bends away from the center point, and approaches (but does not intersect) the asymptotes, as illustrated in Figure 22-10. Figure 22-10: The graph of x2 – 4y2 + 2x + 32y – 27 = 0 has a vertical transverse axis and vertices (–1,1) and (–1,7). Note that the rectangle used to construct the graph is not part of the graph. 510 The Humongous Book of Algebra Problems Chapter Twenty-Two — Conic Sections Note: Problems 22.37–22.40 refer to the hyperbola with equation x2 – 4y2 + 2x + 32y – 27 = 0. 22.39 Write the equations of the asymptotes in slope-intercept form. One of the asymptotes in Figure 22-10 passes through points (–7,7) and (5,1). Substitute x 1 = –7, y1 = 7, x 2 = 5, and y 2 = 1 into the slope formula. Substitute and x- and y-values from one of the points through which the asymptote passes (such as x = 5 and y = 1) into the slope-intercept equation to calculate the y-intercept. Substitute and into the slope-intercept formula to generate the equation of the asymptote. The remaining asymptote has the opposite slope, . Calculate its y-intercept by substituting its slope and the coordinates of one of the points through which it passes (such as x = 5 and y = 7) to calculate its y-intercept. The Humongous Book of Algebra Problems 511 Chapter Twenty-Two — Conic Sections Substitute and into the slope-intercept formula to generate the equation of the asymptote. The vertices are above and below the center of this hyperbola, so the foci will be above and below the center as well. Unlike ellipses, the foci of hyperbolas are farther away from the center than the vertices. Like the foci of an ellipse with a vertical major axis The equations of the asymptotes are and . Note: Problems 22.37–22.40 refer to the hyperbola with equation x2 – 4y2 + 2x + 32y – 27 = 0. 22.40 Identify the foci of the hyperbola. The foci of a hyperbola are the points units away from the center in both directions along the transverse axis. According to Problem 22.38, a = 3 and b = 6. Calculate c. The foci of a hyperbola with a vertical transverse axis are (h,k – c) and (h,k + c). Note: Problems 22.41–22.43 refer to the hyperbola with equation x2 – 25y2 – 4x – 21 = 0. 22.41 Write the equation in standard form. Group the x-terms and move the constant to the right side of the equation. (x 2 – 4x) – 25y 2 = 0 + 21 Complete the square for the x-expression. The right side of the standard form equation must be 1, so divide the entire equation by 25. 512 The Humongous Book of Algebra Problems Chapter Twenty-Two — Conic Sections Note: Problems 22.41–22.43 refer to the hyperbola with equation x2 – 25y2 – 4x – 21 = 0. 22.42 Identify the center and vertices and sketch the graph of the hyperbola. According to Problem 22.41, the standard form of the hyperbola is . The positive rational expression left of the equal sign is written in terms of x, so the transverse axis of the hyperbola is horizontal. The standard form of a hyperbola with a horizontal transverse axis is . Therefore, the center of the hyperbola is (h,k) = (2,0), These are the op posites of the cons ta in the squared qu nts a There’s no consta ntities. nt y2 expression, so th in the e coordinate of the ycenter is 0. a = 5, and b = 1. The vertices of the ellipse are a units away from the center in both directions along the transverse axis. The transverse axis is horizontal so the vertices are a units left and right of the center: (h – a,k) and (h + a,k). Plot the vertices and the endpoints of the conjugate axis, (2,1) and (2,–1). Draw a rectangle that passes through the endpoints of the axes, as illustrated by Figure 22-11. Figure 22-11: The rectangle passes through the endpoints of the transverse axis, (–3,0) and (7,0), and the endpoints of the conjugate axis, (2,1) and (2,–1). a is the square root of the positive fraction’s denominator and b is the square root of the negative fraction’s denominator. The con jugate axis is perpendicular to the transverse axis, so it’s vertical. The endpoints are b = 1 unit above and below the center: (h,k + b) and (h,k – b). The asymptotes of the hyperbola extend through opposite corners of the rectangle in Figure 22-11. The graph passes through the vertices, bends away from the center point, and approaches (but does not intersect) the asymptotes, as illustrated by Figure 22-12. The Humongous Book of Algebra Problems 513 Chapter Twenty-Two — Conic Sections Figure 22-12: The graph of x2 – 25y2 – 4x – 21 = 0. Note: Problems 22.41–22.43 refer to the hyperbola with equation x2 – 25y2 – 4x – 21 = 0. 22.43 Identify the foci of the hyperbola. away from the center in both directions along the The foci are transverse axis. According to Problem 22.42, a = 5 and b = 1. Calculate c. The transverse axis is horizontal, so the foci are (h – c,k) and (h + c,k). 514 The Humongous Book of Algebra Problems Chapter 23 Word Problems l of consecutive If two trains leave the station ful integers, how much interest is earned? Word (or story) problems encourage a systematic approach to the identification of unknown values. Though there are many categories of word problems, the most prolific are included in this chapter. The methods for each differ slightly, but all word problems are approached in the same manner: identify and define the unknown quantities, discern a relationship between the unknown and the known quantities in the problem, define the relationship using one or more equations, and solve the equations. A lot of people hate word problems, because compared to other algebra problems, they are chock-full of wo rds. Why is that a big deal? Just wh en you get used to equations, they’re gone. Actually, word problems still boi l down to solving equations, with one big difference: You have to come up with the equations on your own—they’re not given to you. However, after you figure out how to set up those equa tions, the answers aren’t far behind. Chapter Twenty-Three — Word Problems Determining Unknown Values Integer and age problems The problem says that the su of the numbers is m so add the numbe 64, and 3x) together rs (x and set the sum equa l to 64. Consecutive integers are right next to each other on the number line, like 4 and 5 or –12 and –11. The next consecutive integer after x is x + 1. For example, if x = 9, then the next consecutive integer is 9 + 1 = 10. The counting numbers are 1, 2, 3, 4, 5, ..., basically the positive integers. Negative integers are excluded, as is zero. 23.1 What two integers have a sum of 64, if one of them is three times the other? Let x be the smaller of the two integers. The other integer is three times as large, so it is equal to 3x. x + 3x = 64 Solve the equation for x. Recall that x represents the smaller of the two integers. The second integer is 3x = 3(16) = 48. The two integers described by the problem are 16 and 48. 23.2 What two consecutive integers have a sum of 85? Let x be one of the integers and x + 1 be the other. Create an equation stating that the sum of the integers is 85. x + (x + 1) = 85 Solve the equation for x. The integers are x = 42 and x + 1 = 42 + 1 = 43. 23.3 What two consecutive odd counting numbers have a product of 1,443? Let x be one of the unidentified numbers and let x + 2 be the other. Create an equation stating that the product of the numbers is 1,443. x(x + 2) = 1,443 Distribute x. To go from one odd number to the next, you have to skip over the even number between them, so add 2 to x instead of 1. 516 x 2 + 2x = 1,443 Subtract 1,443 from both sides of the quadratic equation. The Humongous Book of Algebra Problems x 2 + 2x – 1,443 = 0 Chapter Twenty-Three — Word Problems Apply the quadratic formula to solve for x. Evaluate both values of x. Discard the solution x = –39 because it is not a counting number. The odd counting numbers are x = 37 and x + 2 = 37 + 2 = 39. The problem states that the numbers you’re lo oking for are counting numbers, which means they’ve go t to be positive. 23.4 The product of x = 8 and an unknown integer y is 29 less than the sum of x and y. Calculate y. The product of x and y is xy, and the sum of x and y is x + y. Construct an equation stating that the product is 29 less than the sum. xy = x + y – 29 Substitute x = 8 into the equation and solve for y. The Humongous Book of Algebra Problems 517 Chapter Twenty-Three — Word Problems 23.5 What two positive integers have a sum of 25 and a product of 154? You’ve got two different equations, and unlike Problems 23.1–23.4, both of the equations contain two variables. To solve the problem, you need to find an x and a y that make both of the equations true. Let x be one of the positive integers and y be the other. The sum of the numbers is 25, so x + y = 25. The product of the numbers is 154, so xy = 154. Thus, the solution to the problem is the solution to the system of equations. To solve the system by substitution, solve the first equation for x. Substitute x = 25 – y into the second equation of the system. Distribute y. 25y – y 2 = 154 Set one side of the equation equal to zero. Multiply 25y – y2 = 154 by –1 and add 154 to both sides to get this equation. y 2 – 25y + 154 = 0 Apply the quadratic formula to solve the equation. Evaluate both values of x. The positive integers are 11 and 14. 518 The Humongous Book of Algebra Problems Chapter Twenty-Three — Word Problems 23.6 Six years ago, Nick was three times as old as Erin. Two years later, Nick was twice as old as Erin. How old is Erin now? Let x be Nick’s current age and let y be Erin’s current age. Six years ago, Nick’s age was x – 6 and Erin’s age was y – 6. At that time, Nick was three times as old as Erin. Two years later (four years ago), Nick’s age was x – 4 and Erin’s age was y – 4. At that time, Nick was twice as old as Erin. This reads “x – 6” (Nick’s age six years ago) “= ” (is) “ 3(y – 6)” (three tim Erin’s age six yea es rs ago). Distribute th e3 and then put th e linear equation in stand a form (x and y le rd ft of the equal sign, co ns on the right side, tant and a positive x-term). The solution to this problem is the solution to the system of equations describing Nick and Erin’s ages four and six years ago. Solve the system by elimination. Erin is currently eight years old. Multiply the equation x – 2y = –4 by –1 and then add the equations together to get –y = –8. To solve for y, multiply both sides by –1. The Humongous Book of Algebra Problems 519 Chapter Twenty-Three — Word Problems 23.7 Sara is six years older than Ted. Thirteen years ago, Sara was four times as old Use as few variables as possible. Instead of saying that Sara’s age is y, you can say her age is x + 6 (six years older than Ted). You couldn’t do that in Problem 23.6 because Nick and Erin’s CURRENT ages were never compared. as Ted. How old is Ted now? Let x be Ted’s current age. Sara is six years older than Ted, so Sara’s age is x + 6. Thirteen years ago, Ted’s age was x – 13 and Sara’s age was (x + 6) – 13 = x – 7. At that time, Sara was four times as old as Ted. Ted is currently 15 years old. 23.8 In ten years, Mel’s age will be five less than twice Alice’s age. If their current This says “x – 7” (Sara’s age 13 years ago) “=” (is) “4(x – 13)” (four times Ted’s age 13 years ago). That 7 is there because you’re subtracting 13 from Sara’s age (just like you did with Ted’s) and her age started out as x + 6. ages total 65, how much older is Mel? Let x be Mel’s current age and y be Alice’s current age. In ten years, Mel’s age will be x + 10 and Alice’s will be y + 10. At that time, Mel’s age will be five less than twice Alice’s age. Construct an equation stating that the sum of Mel’s and Alice’s current ages is 65. x + y = 65 This reads “x + 10 ” (Mel’s age in ten years) “=” (is) “2(y + 10) – (five less than tw 5” o times Alice’s age in ten years). Distribu te the 2 and put th e linear equation in standard form. The solution to this problem is the solution to the system of equations. ngous Book of Algebra Problems 520 The Humo Chapter Twenty-Three — Word Problems Solve the system using elimination. Multiply the second equation by –1, add the equations, and solve for y. Alice is currently 20 years old. To determine Mel’s age, substitute y = 20 into either equation of the system and solve for x. Mel is currently 45 years old, 25 years older than Alice. Make sure you answer the question posed by the word problem. In this case, you’re supposed to figure out how much older Mel is than Alice, so subtract their ages: x – y = 45 – 20 = 25. Calculating Interest Simple, compound, and continuously compounding 23.9 How much simple interest accrues on a loan of $1,200 at an annual rate of 4.5% over a three-year period? The formula for simple interest is i = prt, where i is interest, p is the principal, r is the annual interest rate expressed as a decimal, and t is the length of time (in years) the principal accrues interest. Substitute p = 1,200, r = 0.045, and t = 3 into the formula to calculate i. The total accrued interest is $162. 23.10 If you wish to borrow $3,000 from a friend who will charge you simple interest at an annual rate of 7%, and you cannot exceed $3,300 in total debt. What is the maximum length of time you have to repay the debt and the accrued interest? Round the answer to the nearest whole day. The principal is the dollar amount that earns interest. In this case, the principal is $1,200. To change 4.5% into a decimal, move the decimal point two places to the left: 0.045. You intend to borrow $3,000 but your total debt cannot exceed $3,300. Therefore, the maximum total interest you can afford to pay is $3,300 – $3,000 = $300. Substitute i = 300, p = 3,000, and r = 0.07 into the simple interest formula to calculate the number of years t it would take to accrue $300 in interest. The Humongous Book of Algebra Problems 521 Chapter Twenty-Three — Word Problems Because r is an annual interest rate, t is measured in years. There are 365 days in a (non leap) year, so multiply t by 365. To convert t into days, multiply by 365. 365(1.42857142857) ≈ 521.428571429 ≈ 521 The maximum length of time you can afford to borrow the money is 521 days. 23.11 Describe the difference between simple and compound interest. Including the interest you’ve already earned on your initial investment. Each time it compounds, all the interest you’ve earned is added to your initial investment, and you start earning interest on that. The balance is the original deposit (principal) plus all the interest it earned. In other words, the balance is the total amount of money in your account. Simple interest accrues only on the principal, whereas compound interest transfers interest accrued into principal each time it is compounded. Consider a savings account that pays only simple interest. No matter what length of time the money remains in the account, interest is earned only on the principal deposited initially. However, a compound interest account would earn interest on all the money in the account. 23.12 If $500 is deposited into a savings account with a 3.75% annual interest rate compounded monthly, what is the balance of the account ten years later? Round the answer to the hundredths place. The formula for compound interest is , where b is the balance, p is the principal, r is the annual interest rate expressed as a decimal, n is the number of times the interest is compounded in one year, and t is the length of time (in years) interest is earned. Substitute p = 500, r = 0.0375, n = 12, and t = 10 into the formula to calculate b. The balance of the account is $727.07. Interest is compounded monthly, a total of n = 12 times per year. ngous Book of Algebra Problems 522 The Humo Chapter Twenty-Three — Word Problems 23.13 How long will it take an initial investment of $4,000 to grow to $5,000 at an annual interest rate of 8% compounded quarterly? Express the answer in years, rounded to the thousandths place. Substitute b = 5,000, p = 4,000, r = 0.08, and n = 4 into the compound interest formula. The interest is compounded quarterly, n = 4 times per year. Solve for t. Take the natural logarithm of both sides of the equation. According to a property of logarithms, the power of a logarithmic argument can be removed from the argument and multiplied by the logarithm itself. This is the property ln xa = a ln x from Problem 19.1 8. It will take approximately 2.817 years for the initial investment of $4,000 to grow by $1,000. The Humongous Book of Algebra Problems 523 Chapter Twenty-Three — Word Problems The more times in a year that interest is compounded, the more interest you earn. You can’t compound interest more often than “continuously.” You need a calculator to evaluate e 0.11. 23.14 If $900 is deposited into a savings account with an annual interest rate of 5.5%, how much interest is earned over a two-year period if interest is compounded continuously? Round the answer to the hundredths place. The formula for continuously compounding interest is b = pert , where b is the balance, p is the principal, e is Euler’s number (described in Problem 18.21), r is the annual interest rate expressed as a decimal, and t is the length of time (in years) interest accrues. Substitute p = 900, r = 0.055, and t = 2 into the formula to calculate b. The total balance of the account after two years is $1,004.65. To determine the total amount of interest earned, subtract the balance from the principal. i = $1,004.65 – $900 = $104.65 The problem doesn’t ask for the total balance. It asks how much money was earned on the $900 investment, so subtract 900 from the balance to calculate the interest. Recognize the formula b = pert? It’s the exponential growth formula first defined in Problem 19.36. Continuously compounding interest makes the investment grow exponentially. A total of $104.65 in interest is earned. 23.15 You wish to deposit $3,500 in a savings account with an annual interest rate of 6.25%. How much more interest will you earn over a 20-year period if interest is compounded continuously rather than annually? Round each balance and the final answer to the hundredths place. Substitute p = 3,500, r = 0.0625, and t = 20 into the continuously compounding interest formula and calculate b. The continuously compounding interest account earns $12,216.20 – $3,500 = $8,716.20 interest. Substitute p = 3,500, r = 0.0625, and t = 20 into the compound interest formula to calculate the balance of the account if interest is compounded annually (n = 1 time per year). ngous Book of Algebra Problems 524 The Humo Chapter Twenty-Three — Word Problems The total interest accrued when compounded annually is $11,766.49 – $3,500 = $8,266.49. Subtract this amount from the total interest earned when compounded continuously (calculated earlier in the problem). $8,716.20 – $8,266.49 = $449.71 The principal earns $449.71 more interest in a continuously compounding account than it does in an account that compounds interest annually. Geometric Formulas Area, volume, perimeter, and so on 23.16 Calculate the radius of a circle that has a circumference of 10 inches. The formula for the circumference C of a circle with radius r is C = 2Ï•r. Substitute C = 10 into the formula and solve for r. The radius of the circle is The perimeter is the sum of the sides of a geomet figure. A square ha ric s four equal sides, so to calculate its perim et multiply the length er, of one side by 4. inches. 23.17 Calculate the area of a square that has a perimeter of 50 cm. The formula for the perimeter of a square with side length s is P = 4s. Substitute P = 50 into the formula and solve for s. The Humongous Book of Algebra Problems 525 Chapter Twenty-Three — Word Problems The area of any rectangle (including a square, which is technically also a rectangle) is equal to its length times its width. The length and width of a square are equal, so the area is the length of one side squared. Each side of the square is inches long. Substitute into the formula for the area of a square: A = s2. The area of the square is in2. 23.18 Calculate the area of a circle that has a diameter 8 cm long. The radius of a circle is equal to half of its diameter: 8 ÷ 2 = 4. Substitute r = 4 into the formula for the area of a circle: A = Ï•r2. A = Ï•(4)2 = 16Ï• The area of the circle is 16Ï• cm2. You can write the answer in decimal form: (12 .5) 2 = 156.25 in2. If the length of the square’s side is measured in inches, then the area is measured in square inches (in2). 23.19 If the base of a triangle is one unit longer than its height, and the area of the triangle is 10 in2, what is the length of the base? Let h represent the height of a triangle and b represent the base. The area of the corresponding triangle is . Note that the height of the triangle is one unit shorter than the base: h = b – 1. Substitute A = 10 and h = b – 1 into the formula. Make sure to include units in your final answers. The length is measured in centimeters, so the area is measured in square centimeters (cm2). You could also se tb=h+1 and plug that in to the formula. You’ll end up calc ul height. Plug that ating the height into b = h + 1 (in othe rw calculate the le ords, add 1) to ngth of the base . ngous Book of Algebra Problems 526 The Humo Multiply both sides of the equation by 2 to eliminate fractions. Chapter Twenty-Three — Word Problems Set the left side of the quadratic equation equal to zero and solve by factoring. The sides of a triangle, like the sides of any geometric figure, must have positive lengths, so discard the solution b = –4. The base of the triangle is 5 in long. 23.20 If the perimeter of a rectangle is 28 cm and the length of the rectangle is one less than twice the width, what is the area of the rectangle? The perimeter of a rectangle with length l and width w is P = 2l + 2w; the area is A = lw. The length of this rectangle is one less that twice the width, so l = 2w – 1. Substitute l and P into the perimeter formula and solve for w. The opposite sides of a rectangle are eq ual, so there are two si L cm long and tw des o sides W cm long. Calcul ate the perimeter by adding up the le ngths of all the sides: P=L+L+W+W = 2L + 2W . Substitute w = 5 into the length expression. Substitute w = 5 and l = 9 into the area formula A = lw = (9)(5) = 45 The area of the rectangle is 45 cm2. The Humongous Book of Algebra Problems 527 Chapter Twenty-Three — Word Problems 23.21 Calculate the radius (in inches) of a right circular cylinder that is one foot tall The answer’s supposed to be in terms of inches, but the height is reported in feet. Multiply h by 12 to convert 1 foot into 12 inches. and has a volume of 96Ï• in3. The volume of a right circular cylinder with radius r and height h is V = Ï•r2h. Substitute V = 96Ï• and h = 12 into the formula and solve for r. The radius, like any geometric length, must be a positive number, so discard the solution . The radius of the cylinder is inches. 23.22 Calculate the dimensions of a right rectangular prism with volume 60 in3 if it has a height of 5 in and its length is three more than four times its width. “Right rectangular prism” is a fancy way of saying “box.” Dividing every term of an equation by the same nonzero number doesn’t affect the solution, and the smaller coefficients you get as a result make solving the equation a little easier. The volume of a right rectangular prism with length l, width w, and height h is V = lwh. Note that the length of this prism is three more than four times the width, so l = 4w + 3. Substitute V = 60, h = 5, and l = 4w + 3 into the formula to solve for w. Divide the entire equation by 5, the greatest common factor. Apply the quadratic formula to solve the equation. ngous Book of Algebra Problems 528 The Humo Chapter Twenty-Three — Word Problems Discard the negative solution. The width of the prism is Substitute w into the length formula defined above. in. Use common denominators to simplify the expression. The dimensions of the prism are h = 5 in, in, and in. Speed and Distance Distance equals rate times time 23.23 Calculate the average speed (in km/h) of a turtle that travels 2.5 km in four hours. The formula for distance traveled at an average rate r for a length of time t is d = rt. Substitute d = 2.5 and t = 4 into the formula and calculate r. The turtle traveled at a rate of r = 0.625 km/h. The Humongous Book of Algebra Problems 529 Chapter Twenty-Three — Word Problems The units need to match up. The speed is written in terms of miles per HO UR so time needs to be measured in hours. Either that or convert 6 mph into miles per minute and keep t = 75. You’ll get the same value for d. 23.24 How far will a jogger moving at an average speed of 6 mph travel in 75 minutes? The speed is reported in miles per hour, so express time in terms of hours, rather than minutes: hours. Substitute r = 6 and into the distance formula to calculate d. The jogger travels miles. 23.25 If it takes Bus A 3.25 hours to complete a fixed route at an average speed of 35 mph, how long will it take Bus B (in hours) to complete the same route at an average speed of 30 mph? Round the answer to the hundredths place. Let d A be the distance Bus A travels (in miles) at rate rA mph for time tA hours. Similarly, let d B be the distance Bus B travels at rate r B mph for time t B hours. The distance formula for Bus A is d A = rAtA , and the distance formula for Bus B is d B = rBt B. The buses travel the same distance, so dA = d B. Substitute dA = rAt A and d B = rBt B into that equation. Substitute rA = 35, tA = 3.25, and r B = 30 into the equation and solve for t B. Multiply 0.79 (the decimal pa rt of the answer) by 60 to approximate m inutes: (0.79)(60) ≈ 47.4. Bus took just over 3 ho B ur and 47 minutes to s complete the rout e. Bus B will complete the route in approximately 3.79 hours. ngous Book of Algebra Problems 530 The Humo Chapter Twenty-Three — Word Problems 23.26 If Don travels 320 km at a speed 5 km/h faster than Mike, who travels 300 km during the same period of time, what was the average speed of both drivers? Let d D represent the distance traveled (in kilometers) by Don at a rate of rD km/h for t D hours. Let d M represent the distance traveled by Mike at a rate of rM km/h for t M hours. The distance formula for Don’s trip is d D = r Dt D, an the distance formula for Mike’s trip is d M = rMt M. Solve each of the distance formulas for t. Both drivers travel for the same period of time, so t D = t M. Substitute the above rational expressions into the equation. In Problem 23.25, the distances trave led were equal, and yo substituted into u the equation d = d . In A this problem, theB tim traveled is equa e l. You’re solving for t so that you can plug into the equation tD = tM . Don’s speed is 5 km/h faster than Mike’s speed, so r D = rM + 5. Substitute d D = 320, d M = 300, and r D = rM + 5 into the equation. Cross multiply and solve for rM. Mike drove at an average speed of 75 km/h, and Don drove at an average speed of 80 km/h. Don’s speed is rD = rM + 5 = 75 + 5 = 80 km/h. The Humongous Book of Algebra Problems 531 Chapter Twenty-Three — Word Problems 23.27 Jack and Jill are exercising on a 400-meter-long track. Jack walks at an average speed of 2 m/sec. After Jack has completed one full lap, Jill starts from the same position Jack did, jogging along the same path at an average speed of 5 m/sec. How long does Jill have to run to catch up to and pass Jack? Let d JA represent the distance (in meters) Jack travels at an average rate of rJA m/sec for t JA seconds. Let d JI represent the distance Jill travels at an average rate of rJI m/sec for t JI seconds. The formula for Jack’s distance walked is d JA = rJAtJA, and the formula for Jill’s distance jogged is d JI = rJIt JI. After Jill has caught up to Jack, she will have traveled the same distance he has. At that moment, d JA = d JI. Substitute the distance formulas into the equation. d = rt = (2)(20 0) = 40 0 Jill begins jogging after Jack has completed a lap, a distance of 400 m. It will take Jack 400 ÷ 2 = 200 seconds to travel that distance at 2 m/sec. Therefore, Jack’s travel time is 200 seconds longer than Jill’s: t JA = t JI + 200. Substitute rJA = 2, rJI = 5, and t JA = t JI + 200 into the equation. Which is minutes? Jill will catch up to Jack in seconds. 23.28 It takes Phil 10 minutes to swim from the shore to a buoy, at an average speed of 3.5 ft/sec. How long will it take Phil to swim back to shore along the same route if his speed is slowed to 2 ft/sec by the currents in the water? Express the answer in minutes and seconds. Let d1 = r 1t1 represent the distance from the shore to the buoy and d 2 = r2t 2 represent the distance from the buoy to the shore. According to the problem, Phil swims the same route each way, so the distances traveled each way are equal. Because d1 = r1t1, you can substitute r1t1 for d1 in the equation d1 = d2 . Same thing for d2 = r2t2 . ngous Book of Algebra Problems 532 The Humo Chapter Twenty-Three — Word Problems Time is expressed in minutes but speed is expressed in ft/sec. Convert time into seconds, thereby making the units compatible: 10 minutes = 10(60) = 600 seconds. Substitute r1 = 3.5, t1 = 600, and r2 = 2 into the equation and solve for t 2. It takes Phil 1,050 seconds to swim from the buoy to the shore. Divide 1,050 by 60 to calculate the number of minutes Phil swam. 1,050 ÷ 60 = 17.5 minutes Multiply by 60 to convert the decimal part of the answer back to seconds: (60)(0.5) = 30. Phil’s swim from the buoy to the shore lasted 17 minutes and 30 seconds. 23.29 Two cars are placed at opposite ends of a straight, 0.75-mile-long test track to determine the effectiveness of driver’s side air bags in a head-on collision. The first car begins driving toward the second at an average speed of 30 mph, and 10 seconds later, the second car begins driving toward the first car at an average speed of 40 mph. How long does the first car drive before colliding head-on with the second car? Report the answer in hours. Let d1 = r 1t1 represent the distance traveled by the first car and d 2 = r2t 2 represent the distance traveled by the second car. The cars begin at opposite ends of the 0.75-mile-long track and will collide when the sum of the distances traveled by both cars is 0.75 miles. The second car starts driving 10 seconds after the first car, so at the time of collision, the first car will have traveled for 10 seconds longer. Notice that the speeds of the cars are expressed as miles per hour, so convert 10 seconds into hours: . Therefore, t1 = t2 + 10 seconds, and Substitute r 1 = 30, r2 = 40, and into the equation. hours. To convert from seconds to minutes, you divide by 60. To convert from minutes to hours, divide by 60 again. If you want to convert straight from seconds to hours, divide by (60)(60) = 3,600. The Humongous Book of Algebra Problems 533 Chapter Twenty-Three — Word Problems Express 0.75 as a fraction: 0.75 is read “seventy-five HUNDREDths.” To convert the decimal into a fraction, divide 75 by one HUNDRED. . Multiply the entire equation by 12, the least common denominator, to eliminate fractions. Solve for t 2. Multiply by 3,600 to convert to seconds: . The first car will travel hours before the collision. Mixture and Combination Measuring ingredients in a mixture 23.30 A 12-ounce cup of fruit punch contains 70% fruit juice and 30% water. If combined with a 16-ounce bottle of fruit punch that contains 10% fruit juice and 90% water, what portion of the mixture is juice? Report the answer as a percentage rounded to the hundredths place. To solve a mixture problem, multiply the amount of each ingredient by the value that describes it (such as its concentration) and add the results. Set the sum equal to the total amount of the mixture multiplied by its descriptive value. ngous Book of Algebra Problems 534 The Humo Chapter Twenty-Three — Word Problems In this problem, the ingredients are 12- and 16-ounce containers of punch, which contain 70% and 10% real juice, respectively. Convert the percentages into decimals (70% = 0.7 and 10% = 0.1) and substitute the values into the equation blueprint above. Let x represent the unknown value, the portion of the final mixture that is real juice. The mixture is approximately 35.71% fruit juice. 23.31 If a 5-pound bag of mixed nuts that is 45% peanuts by weight is combined with a 12-pound bag of mixed nuts that is 40% peanuts by weight, what percentage of the mixture is peanuts (by weight)? Report the answer as a percentage rounded to the hundredths place. To convert from a decimal to a percent, move the decimal point two places to the right. Don’t round to the hundredths place until after you move the decimal. Apply the blueprint for mixture and combination problems. Multiply the first ingredient, a 5-pound bag, by 0.45 (its descriptive value 45% expressed as a decimal). Multiply the second ingredient, a 12-pound bag, by its descriptive value, 0.4. The mixture weighs 5 + 12 = 17 pounds; let x represent the unknown descriptive value, the percentage of the mixture that is peanuts by weight. The mixture is approximately 41.47% peanuts by weight. 23.32 How much extra-virgin olive oil, containing 0.8% acidity, must be added to 4 gallons of olive oil with an acidity of 2.5% so that the mixture has an acidity of 2%? Report the answer in gallons, rounded to the hundredths place. Apply the blueprint for mixture and combination problems. The Humongous Book of Algebra Problems 535 Chapter Twenty-Three — Word Problems Even though the percentage 0.8% contains a decimal, you still need to move that decimal point two places to the left to convert the percentage into a decimal value: 0.8% = 0.0 08. Let x represent the unknown amount (in gallons) of the first ingredient, the extra-virgin olive oil; multiply x by its acidity (0.008). Add the amount of the second ingredient (4 gallons) multiplied by its acidity (0.025). The total mixture is x + 4 gallons and has an acidity of 0.02. Solve for x. In order to give the mixture an acidity of 2%, 1.67 gallons of extra-virgin olive oil are required. 23.33 Two machines at a manufacturing plant are used to fulfill an order for widgets. Machine A produces 350 widgets, and machine B produces 800. Upon receiving the order, the customer notes that 138 of the widgets have manufacturing defects. If engineers determine that approximately 1.5% of the widgets made by machine B are defective, what percentage of the widgets made by machine A are defective? 138 out of the 1,150 total widge had defects, whi ts ch is 0.12 = 12% of the order. Let w A = 350 represent the number of widgets made by machine A, w B = 800 represent the number of widgets made by machine B, rA represent the percentage of widgets made by machine A that contain defects, rB = .015 represent the percentage of widgets made by machine B that contain defects, wo = 800 + 350 = 1,150 represent the total number of widgets ordered, and ro = 138 ÷ 1,150 = 0.12 represent the percentage of the widgets ordered that were defective. Apply the blueprint for mixture and combination problems. Solve for rA. Approximately 36% of the widgets made by machine A were defective. ngous Book of Algebra Problems 536 The Humo Chapter Twenty-Three — Word Problems 23.34 A collection of 49 nickels and pennies is worth 85 cents. How many of the coins are nickels? Let n represent the number of nickels. The collection contains 49 total coins, so there are p = 49 – n pennies. Multiply the total number of each coin by its worth (expressed as a decimal) and set the sum equal to 0.85 = 85 cents. If there are n nickels and a total of 49 coins overall, then subtract the number of nickels from 49 to find out how many pennies there are. Solve for n. There are nine nickels among the 49 coins. 23.35 A jar of dimes and quarters is worth $32.45. If the number of quarters is 13 less than three times the number of dimes, how many quarters are in the jar? Let d represent the number of dimes in the jar and q = 3d – 13 represent the number of quarters. Multiply the total number of each coin by its value and set the sum equal to $32.45. 3d – 13 is thirteen less than three times the number of dimes. Solve for d. The collection of coins contains 42 dimes. Calculate the number of quarters. The collection of coins contains 113 quarters. The Humongous Book of Algebra Problems 537 Chapter Twenty-Three — Word Problems There are q quarters and 2q dimes. The combined number of quarters and dimes is q + 2q = 3q. The total number of coins is 44, so subtract the combined number of quarters and dimes from 44 to calculate the number of pennies. 23.36 A collection of 44 coins containing pennies, dimes, and quarters is worth $5.90. There are twice as many dimes as quarters. How many pennies are in the collection? Let q represent the number of quarters, d = 2q represent the number of dimes, and p = 44 – 3q represent the number of pennies. Multiply the total number of each coin by its value and set the sum equal to 5.90. Solve for q. This is the expression for pennies defined in the beginning of the problem: p = 44 – 3q. There are q = 13 quarters, 2(13) = 26 dimes, and 44 – 3(13) = 44 – 39 = 5 pennies. Work How much time does it save to work together? 23.37 Two computer programmers are instructed to write code for an application. Working by herself, Donna could complete the task in five days, but when working with Chris, she is able to complete the program in three days. How long would it take Chris to complete the task by himself? To set up a work problem, create one fraction for each of the individuals involved. The numerator of each fraction is the length of time the individual worked, and the denominator is the length of time it would take that individual to complete the task working alone. Add the fractions and set the sum equal to one. The sum of the fractions always equals 1. Chris and Donna worked together for the same length of time, three days. Donna can complete the task alone in five days. Let x represent the length of time it would take Chris to complete the task alone. ngous Book of Algebra Problems 538 The Humo Chapter Twenty-Three — Word Problems Multiply the entire equation by 5x, the least common denominator, to eliminate fractions. Solve for x. It would take Chris days to complete the task by himself. Note: Problems 23.38–23.39 refer to a pair of carpenters named Lisa and Karen. Lisa can build a bookshelf in 30 minutes, and Karen can build the same bookshelf in 45 minutes. 23.38 How long will it take Lisa and Karen to build the bookshelf together? Let x represent the length of time Lisa and Karen work together to complete the bookshelf. Create a rational expression for each carpenter, dividing x (the length of time worked by both) by the length of time it would take each to complete the bookshelf alone. Add the fractions and set the sum equal to one. Multiply the entire equation by 90, the least common denominator, to eliminate fractions. “1” represents “1 complete booksh elf.” The fractions on the left side of th equation represen e t the portions of th at complete job tha t ea person is responsi ch ble for. Working together, Lisa and Karen complete the bookshelf in 18 minutes. The Humongous Book of Algebra Problems 539 Chapter Twenty-Three — Word Problems Note: Problems 23.38–23.39 refer to a pair of carpenters named Lisa and Karen. Lisa can build a bookshelf in 30 minutes. 23.39 Karen is injured and can no longer complete the bookshelf alone in 45 minutes. It now takes Lisa and Karen 28 minutes 48 seconds to complete the bookshelf working together. How long would it take Karen to complete the bookshelf working alone? Divide 48 seconds by 60 to convert into minutes: . It takes the pair of carpenters 28.8 minutes to build the bookshelf together, and Lisa can complete the bookshelf in 30 minutes by herself. Let x represent the length of time it would take Karen to build the bookshelf. Multiply the entire equation by 30x, the least common denominator, to eliminate fractions. Solve for x. It would take Karen 720 minutes, or hours, to complete the task alone. Note: Problems 23.40–23.41 refer to Rob and Matt, who go into business together mowing lawns during the summer. Rob works twice as fast as Matt. 23.40 If it takes Rob and Matt 90 minutes to mow one lawn together, how long would it take each of them to mow it separately? Let x be the length of time it takes Rob to mow the lawn alone, and let 2x be the time Matt needs to complete the same task. They spend 90 minutes working together. ngous Book of Algebra Problems 540 The Humo Chapter Twenty-Three — Word Problems Multiply the entire equation by 2x, the least common denominator, to eliminate fractions. Rob could mow the lawn alone in 135 minutes, and it would take Matt 2(135) = 270 minutes. Note: Problems 23.40–23.41 refer to Rob and Matt, who go into business together mowing lawns during the summer. Rob works twice as fast as Matt. 23.41 Rob and Matt are mowing the lawn described in Problem 23.40, a job they can complete in 90 minutes working together. Twenty minutes after they begin, Rob has to leave and does not return. How long will it take Matt to complete the remainder of the job alone? Rob and Matt begin work together, but Rob works for only 20 minutes. If x represents the amount of time it will take Matt to complete the job after Rob leaves, Matt works a total of x + 20 minutes. According to Problem 23.40, Rob can mow the lawn by himself in 135 minutes and it takes Matt 270 minutes. He works 20 minutes with Rob and then x minutes after Rob leaves. Multiply the entire equation by 270, the least common denominator, to eliminate fractions. It will take Matt 210 minutes, or hours, to finish mowing the lawn. The Humongous Book of Algebra Problems 541 Chapter Twenty-Three — Word Problems Note: Problems 23.42–23.43 refer to an aquarium that contains a plecostomus, an algaeeating fish. Under normal circumstances, this plecostomus can completely clean the algae from the sides of the aquarium in seven days. 23.42 Due to poor maintenance, algae are blooming in the fish tank at an accelerated rate. In fact, if the plecostomus were absent, the algae would cover the sides of the aquarium in four days. How long will it take algae to cover the sides of the tank despite the efforts of the fish? Instead of adding the fractions, you subtract th e fish fraction from the algae fract ion. The fish is activ ely working against the algae, so all of it s work is slowing down the algae growth. Let x represent the time it will take algae to cover the sides of the aquarium. Both the fish and the algae “work” for the entire time period. Multiply the entire fraction by 28, the least common denominator, to eliminate fractions. In days, the sides of the tank will be covered with algae. Note: Problems 23.42–23.43 refer to an aquarium that contains a plecostomus, an algaeeating fish. Under normal circumstances, this plecostomus can completely clean the algae from the sides of the aquarium in seven days. 23.43 The owner of the aquarium described in Problem 22.42 cleans the tank The algae could take over the tank in 4 days. Neither fish can work that fast—one takes 5 days to clean the aquarium and one takes 7—but working together, they can eliminate the algae. and thereby restores the ecosystem. However, two months later, the algae begin to bloom again. To combat the algae problem this time, the aquarium owner adds a second plecostomus to the tank. Working alone under normal circumstances, the new fish could clean the sides of the tank in five days. How long will it take both fish, working together, to clean the sides of the aquarium as they combat an algae bloom that (if left unchecked) would once again cover the sides of the tank in four days? Let x represent the time it takes both fish working together to clean the sides of the tank. The pair of fish and the algae “work” for that entire period of time. Note that the algae is working against the progress of the fish, so its fraction should be subtracted from, rather than added to, the fractions representing the work of the fish. ngous Book of Algebra Problems 542 The Humo Chapter Twenty-Three — Word Problems Multiply the entire equation by 140, the least common denominator, to eliminate fractions. Working together, it takes the fish days to clean the sides of the aquarium despite the algae bloom. The Humongous Book of Algebra Problems 543 Appendix A s e i t r e p o r P c i Algebra Associative Property of Addition: (a + b) + c = a + (b + c) Associative Property of Multiplication: (ab)c = a(bc) Commutative Property of Addition: a + b = b + a Commutative Property of Multiplication: ab = ba Additive Identity: a + 0 = 0 + a = a Multiplicative Identity: a(1) = 1(a) = a Additive Inverse: a + (–a) = –a + a = 0 Multiplicative Inverse: Distributive Property: a(b + c) = ab +Â€ac Symmetric Property: If a = b, then b = a. Transitive Property: If a = b and b = c, then a = c. Appendix B Important graphs and graph transformations y = x2 y = x3 y = |x| Appendix B 548 The Humongous Book of Algebra Problems Appendix B How constants transform a function graph The Humongous Book of Algebra Problems 549 Appendix C Key Algebra Formulas Linear Equations Slope-intercept form: y = mx + b Point-slope form: y – y1 = m(x – x1) Standard form: Ax + By = C Matrix Formulas 2 × 2 determinant: 3 × 3 determinant: Cramer’s Rule: Solution to the system is and FOIL Method: (a + b)(c + d) = ac + ad + bc + bd Factoring Formulas Difference of perfect squares: a2 – b2 = (a + b)(a – b) Sum of perfect cubes: a3Â€+Â€b3 = (a + b)(a2 – ab + b2) Difference of perfect cubes: a3 – b3 = (a – b)(a2 + ab + b2) Appendix C Quadratic formula: Variation Direct: Inverse: xy = k Standard Forms of Conic Sections Parabola with vertical axis of symmetry: y = a(x – h) 2 + k Parabola with horizontal axis of symmetry: x = a(y – k) 2 + h Circle: (x – h) 2 + (y – k) 2 = R2 Ellipse with horizontal major axis: Ellipse with vertical major axis: Eccentricity of an ellipse: Hyperbola with horizontal transverse axis: Hyperbola with vertical transverse axis: Interest Simple interest: i = prt Compound interest: Continuously compounding interest: b = pert The Humongous Book of Algebra Problems 552 Appendix C Geometric Formulas Area of a circle: A = πr2 Circumference of a circle: C = 2πr Area of a rectangle: A = LW Perimeter of a rectangle: P = 2L + 2W Area of a triangle: Volume of a right rectangular prism: V = LWH 2 Volume of a right rectangular cylinder: V = πr h Distance = rate × time Mixture and Combination: (ingredient 1)(its value) + (in gredient 2)(its value) = (total mixtu re)(its value) Work: The Humongous Book of Algebra Problems 553 Index Alphabetical List of concepts with Problem numbers This comprehensive index organizes the concepts and skills discussed within the book alphabetically. Each entry is accompanied by one or more problem numbers, in which the topics are most prominently featured. ok. For example, 8.2 is the bo e th in s, ge pa not , ms ble pro to er All of these numbers ref second problem in Chapter 8. A–B abscissa: 15.1–15.4 absolute value functions: 15.4–15.5, 16.3–16.4, 16.12–16.14, 16.28, 16.40–16.44 linear equations: 4.29–4.26, 5.40–5.44, 7.35 (of) signed numbers: 1.17–1.21, 1.27–1.30 additive identity: 1.36, 1.40, 10.3 age word problems: 23.6–23.8 area: 23.17–23.20 associative property: 1.32, 1.35, 1.43 asymptotes horizontal: 20.35, 20.37, 20.40, 20.42 (of) hyperbolas: 22.36, 22.38–22.39, 22.42 oblique: 20.43 vertical: 20.35–20.36, 20.39, 20.42 augmented matrices: 10.1–10.2, 10.19, 10.23, 10.25–10.26, 10.32, 10.34, 10.36, 10.38, 10.40 axioms of algebra: see properties of algebra axis of symmetry: 22.2, 22.6 base (of a triangle): 23.19 C carbon dating: 19.40–19.43 center (of a) circle: 22.12, 22.15, 22.17, 22.20 (of an) ellipse: 22.22–22.24, 22.26, 22.29 (of a) hyperbola: 22.33, 22.38, 22.40, 22.42– 22.43 change of base formula: 18.25–18.33 circle: 22.12–22.21, 23.16, 23.18 circumference: 23.16 cofactor: 9.32–9.33 cofactor expansion: 9.34–9.37 coin word problems: 23.34–23.36 combination word problems: 23.30–23.33 common logarithms: 18.17–18.20 commutative property: 1.34–1.35 completing the square: 14.12–14.19, 14.27, 22.1, 22.4, 22.7, 22.10, 22.14–22.15, 22.17, 22.25, 22.28, 22.37, 22.41 complex fractions: 2.41–2.43, 20.30–20.34, 22.31 complex numbers: 1.7–1.10, 13.37–13.44, 14.19, 14.26, 14.31 composite numbers: 1.4 Index — Alphabetical List of Concepts with Problem Numbers composition of functions: 15.18–15.27, 15.32, 15.34, 19.10–19.11 compound interest: 23.11–23.13, 23.15 conic sections circle: 22.12–22.21 ellipse: 22.22–22.32 hyperbola: 22.33–22.43 parabola: 22.1–22.11 conjugate (of a complex number): 13.42, 13.44 conjugate axis: 22.34–22.35, 22.38, 22.42 consecutive integer word problems: 23.1–23.5 constant: 11.6, 11.8 constant of proportionality: 21.21–21.25 constant of variation: 21.21–21.30 continuously compounding interest: 23.14–23.15 coordinate plane: 5.1, 5.4–5.6 counting numbers: see natural numbers Cramer’s Rule: 9.38, 9.40–9.44 critical numbers: 14.35, 14.38, 14.40, 14.42, 21.31, 21.34, 21.38, 21.42 cross multiplication: 21.1–21.8, 21.10–21.11, 23.26 cylinder: 23.21 D diameter: 22.21 decimals converting to/from fractions: 2.3, 2.5, 2.18–2.21, 4.26–4.28 converting to/from percentages: 2.1–2.2 Descartes’ rule of signs: 17.21–17.27, 17.35, 17.38, 17.41–17.42 determinant (of a) 2Â€×Â€2 matrix: 9.26–9.28 (of a) 3Â€×Â€3 matrix: 9.29–9.30, 9.34–9.36 (of a) 4Â€×Â€4 matrix: 9.37 diameter: 23.18 difference of perfect cubes: 12.31, 12.33–12.34, 20.6 difference of perfect squares: 12.26–12.30, 12.34, 14.5, 17.39, 20.4, 20.6, 20.13, 20.17, 20.22, 20.26–20.27, 20.33 direct variation: 21.21–21.25 directrix: 22.9–22.11 discriminant: 3.43, 14.29–14.34 ngous Book of Algebra Problems 556 The Humo distance formula: 22.18, 22.21 distance word problems: 23.23–23.29 distributive property: 1.38–1.39, 3.23–3.29 domain: 16.10, 16.13, 16.15–16.20, 16.26–16.30, 18.13, 19.3 double root: 14.9, 14.33–14.34, 14.38 E eccentricity: 22.30 elements (of a matrix): 9.2–9.3, 9.5–9.6 elimination technique: 8.19–8.28, 23.6, 23.8 ellipse: 22.22–22.32 end behavior: see leading coefficient test equations absolute value: 5.40–5.44 exponential: 13.20, 19.19–19.24, 19.37, 19.39– 19.41, 23.13 linear: 5.7–5.9, 5.11, 5.13–5.14, 5.21, 5.23, 5.25, 5.39, 6.21–6.22, 6.24, 6.26, 6.32, 6.38, 6.41, 6.43, 8.1–8.7 logarithmic: 18.2–18.10, 18.19, 18.31–18.33, 19.25–19.35 matrix: 10.37, 10.39–10.41 quadratic: 14.1–14.42 radical: 13.32–13.36, 14.28 rational: 4.13–4.14, 4.19, 4.23, 21.2–21.20, 23.37–23.43 systems of: 8.1–8.28, 9.39–9.44, 10.19–10.24, 10.26, 10.33, 23.5–23.6, 23.8 Euler’s number: 18.21 even functions: 16.25 even numbers: 1.3 expressions evaluating: 3.40–3.44 writing: 3.1–3.7 exponential growth and decay: 19.36–19.43, 23.14–23.15 exponentiation: 19.25–19.29, 19.31–19.33, 19.35 exponents basic: 3.8–3.18, 3.35 equations involving: 13.20, 19.19–19.24, 19.37, 19.39–19.41, 23.13 expressions involving: 19.13, 19.18 Index — Alphabetical List of Concepts with Problem Numbers F factoring (by) decomposition: 12.42–12.44, 14.8–14.9, 14.42, 17.10, 17.32 difference of perfect cubes: 12.31, 12.33–12.34, 20.6 difference of perfect squares: 12.26–12.30, 12.34, 14.5, 17.39, 20.4, 20.6, 20.13, 20.17, 20.22, 20.26–20.27, 20.33 (by) grouping: 12.20–12.25, 14.11 integers: 2.30, 2.32, 12.1–12.8, 12.10, 12.15 polynomials: 12.14, 12.16–12.44, 17.7–17.8, 17.10, 17.12, 17.30, 17.32, 17.34, 17.39 sum of perfect cubes: 12.31–12.32, 12.34, 20.26 trinomials 12.35–12.44, 14.6–14.9, 14.35, 14.38 focus (of an) ellipse: 22.29 (of a) hyperbola: 22.40, 22.43 (of a) parabola: 22.8, 22.10 FOIL method: 11.19–11.25 fractions adding and subtracting: 2.25, 2.27–2.29, 2.31– 2.32 complex: 2.41–2.43 expressed as decimals: 2.3, 2.5, 2.18–2.21 expressed as percentages: 2.4, 2.6 improper: 2.8–2.10 multiplying and dividing: 2.33–2.43 reducing to lowest terms: 2.11, 2.13, 2.15–2.18 functions absolute value functions: 16.3–16.4, 16.12– 16.14, 16.28, 16.40–16.44 composition: 15.18–15.27, 15.32, 15.34, 19.10– 19.11 domain of: 16.10, 16.13, 16.15–16.20, 16.26– 16.30 evaluating: 15.1, 15.3, 15.5, 15.7, 15.10–15.12, 15.14–15.18, 15.21–15.23, 15.26, 15.40, 15.42, 17.3–17.6 even: 16.25 identifying: 15.1–15.8, 15.43 inverse: 15.27–15.37, 19.4, 19.10–19.11 logarithmic: 18.11–18.12, 18.14–18.16 odd: 16.24 operations on: 15.9–15.17 piecewise-defined: 15.38–15.43 range of: 16.11, 16.14, 16.20, 16.26–16.30 rational functions: 16.7–16.8, 16.15, 16.17, 16.30, 16.33–16.34 square root functions: 16.9–16.11, 16.16, 16.29, 16.35, 16.37, 16.39 symmetry: 16.21–16.30 Fundamental Theorem of Algebra: 17.1, 17.37 Fundamental Theorem of Arithmetic: 12.1 G Gauss-Jordan elimination: see reduced-row echelon form Gaussian elimination: see row echelon form geometric word problems: 23.16–23.22 graphing absolute value functions: 16.3–16.4, 16.12– 16.14, 16.28, 16.40–16.44 circles: 22.12, 22.16 ellipses: 22.23, 22.26 exponential functions: 19.1–19.2, 19.4–19.9 hyperbolas: 22.38, 22.42 linear absolute value equations: 5.40–5.44 linear equations: 5.7–5.9, 5.11, 5.13–5.14, 5.21, 5.23, 5.25, 5.39, 6.21–6.22, 6.24, 6.26, 6.32, 6.38, 6.41, 6.43, 8.1–8.7 linear inequalities: 7.5–7.43, 8.29–8.36, 8.40 logarithms: 18.11–18.12, 18.14–18.16 parabolas: 16.1–16.2, 16.26, 16.31, 16.38, 17.16, 17.40, 22.3, 22.6 quadratic functions: 16.1–16.2, 16.26, 16.31, 16.38 quadratic inequalities: 14.37, 14.39, 14.41, 14.43 rational functions: 16.7–16.8, 16.15, 16.17, 16.30, 16.33–16.34, 20.38, 20.41, 20.44 rational inequalities: 21.33, 21.36, 21.40, 21.44 square roots: 16.9–16.11, 16.16, 16.29, 16.35, 16.37, 16.39 (using a) table of values: 16.1–16.9, 16.12, 16.26–16.30, 18.11–18.12, 18.15, 20.41, 20.44 (using) transformations: 16.31–16.40, 17.15– 17.16, 17.20, 17.40, 18.14, 18.16, 19.5, 19.9, 20.38 The Humongous Book of Algebra Problems 557 Index — Alphabetical List of Concepts with Problem Numbers greatest common factor: 2.12, 2.14, 2.16–2.17, 12.9, 12.11–12.16, 12.29–12.30, 12.33 grouping symbols: 1.22–1.31 H half-life: 19.40–19.43 horizontal asymptotes: 20.35, 20.37, 20.40, 20.42 horizontal line test: 15.29–15.30, 15.35 hyperbola: 22.33–22.43 I–J–K i: see complex numbers identity matrix: 10.4–10.7, 10.34–10.38, 10.40– 10.41 identity properties: 1.36 improper fractions: 2.8–2.10, 2.19 index (of a radical): 13.3–13.4 indirect variation: see inverse variation inequalities compound: 7.21–7.30, 7.36 containing absolute values: 7.27–7.34, 7.36–7.39 graphing in the coordinate plane: 7.40–7.43 graphing on a number line: 7.13–7.17, 7.22, 7.24, 7.26, 7.28–7.30, 7.32–7.34 linear, in one variable: 7.5–7.12, 7.17–7.20, 7.23, 7.25 linear, in two variables: 7.40–7.43 rational: 21.31–21.44 inequality symbols: 7.1–7.4 integer word problems: 23.1–23.5 integers: 1.2, 1.8–1.9 intercepts: 5.18–5.25, 5.37, 6.9, 6.17, 6.44, 18.13, 19.3 interest compound: 23.11–23.13, 23.15 continuously compounding: 23.14–23.15 simple: 23.9–23.11 inverse functions: 15.27–15.37, 19.4, 19.10–19.11 inverse matrix: 10.34–10.41 inverse properties: 1.40–1.41 inverse variation: 21.26–21.30 irrational numbers: 1.6, 1.8 ngous Book of Algebra Problems 558 The Humo L leading coefficient: 11.6, 11.8, 20.36 leading coefficient test: 17.14–17.20, 17.40 least common denominator: 2.22–2.24, 2.26, 2.28–2.30, 3.6, 20.9–20.18, 21.9, 21.12–21.20, 21.31, 21.41 like radicals: 13.21–13.24, 15.15 like terms: 3.6, 11.9 linear equations graphing: 5.7–5.9, 5.11, 5.13–5.14, 5.21, 5.23, 5.25, 5.39, 6.21–6.22, 6.24, 6.26, 6.32, 6.38, 6.41, 6.43 in one variable: 4.1–4.28, 4.37–4.43 point-slope form of: 6.1–6.10, 6.20, 6.37, 6.43 slope of: 5.26–5.39, 6.28, 6.30–6.31, 6.37, 6.40, 6.43–6.44 slope-intercept form of: 6.11–6.20, 6.21–6.26, 6.32, 6.38, 6.41, 6.32, 22.36, 22.39 standard form of: 5.37–5.38, 6.29–6.36, 6.39, 6.42, 6.44 linear programming: 8.36–8.43 logarithms change of base formula: 18.25–18.33 common logarithms: 18.17–18.20 equations: 18.1–18.10, 18.19, 18.31–18.33, 19.25–19.35 expressions: 19.12, 19.14–19.18 graphing: 18.11–18.12, 18.14–18.16 natural logarithms: 18.21–18.24 properties: 18.34–18.44, 19.14, 19.16–19.18, 19.28, 19.31–19.32, 19.35 long division: 11.29–11.36 M–N major axis: 22.22–22.24, 22.26, 22.29, 22.31– 22.32 matrices adding and subtracting: 9.7–9.15, 10.3 augmented: 10.1–10.2, 10.19, 10.23, 10.25– 10.26, 10.32, 10.34, 10.36, 10.38, 10.40 cofactors: 9.32–9.33 determinants: 9.26–9.30, 9.34–9.37 elements: 9.2–9.3, 9.5–9.6 Index — Alphabetical List of Concepts with Problem Numbers equations: 10.37, 10.39–10.41 identity: 10.4–10.7, 10.34–10.38, 10.40–10.41 inverse: 10.34–10.41 minors: 9.31–9.33 multiplying: 9.16–9.25, 10.4, 10.7, 10.35, 10.37, 10.41 order: 9.1, 9.4 reduced-row echelon form: 10.18–10.34, 10.36, 10.38, 10.40 row echelon form: 10.18–10.21, 10.23, 10.25– 10.29, 10.32 row operations: 10.8–10.17 scalar multiplication: 9.10–9.15 midpoint formula: 22.20, 22.33 minor (of a matrix): 9.31–9.33 minor axis: 22.23–22.24, 22.26 mixed numbers: 2.7–2.10 mixture word problems: 23.30–23.33 multiplicative identity: 1.36, 1.41, 10.4, 10.6 natural logarithms: 18.21–18.24 natural numbers: 1.1–1.2, 1.8–1.9 number line: 5.1–5.3 O oblique asymptotes: 20.43 odd functions: 16.24 odd numbers: 1.3, 1.9, 23.3 optimizing expressions: 8.36–8.43 order (of a matrix): 9.1, 9.4 order of operations: 1.22–1.30, 1.39, 3.30–3.39 ordinate: 15.1–15.4 P parabolas: 16.1–16.2, 16.26, 16.31, 16.38, 17.16, 17.20, 17.40, 22.1–22.11 parallel lines: 5.33–5.34, 6.37, 8.7 percentages: 2.2, 2.4, 2.6 perimeter: 23.17, 23.20 perpendicular lines: 5.35–5.36, 6.40 pi (p): 1.6 piecewise-defined functions: 15.38–15.43 point-slope form of a line: 6.1–6.10, 6.20, 6.37, 6.43 polynomials adding and subtracting: 11.9–11.18 classifying: 11.1–11.8 leading coefficient of: 11.6, 11.8 long division of: 11.29–11.36 multiplying: 11.19–11.28, 11.31 synthetic division of: 11.37–11.45 population modeling: 19.36–19.39 prime factorization: 2.30, 2.32, 12.1–12.8, 12.10, 12.15, 20.3, 20.5, 20.11 prime numbers: 1.4, 1.9, 12.1 principal: 23.9–23.15 prism: 23.22 properties of algebra: 1.31Â�–1.44 proportional variation: see direct variation proportions: 21.1–21.11 Q quadratic equations: (solving by) completing the square: 14.14–14.19, 14.27 (solving by) factoring: 14.1–14.3, 14.5–14.9, 14.11, 14.35, 14.38, 14.42 (solving using the) quadratic formula: 14.20– 14.28, 14.30–14.34, 14.40, 23.5, 23.22 (solving using) radicals: 14.4, 14.10 quadratic formula: 14.20–14.28, 14.30–14.34, 14.40, 17.13, 17.37, 21.7, 21.17, 21.20, 23.5, 23.22 quadratic inequalities graphing: 14.37, 14.39, 14.41, 14.43 solving: 14.35–14.36, 14.38, 14.40, 14.42 R radical equations: 13.32–13.36, 14.28 radical expressions adding/subtracting: 13.21–13.24, 15.15 classifying: 13.1, 13.3 dividing: 13.29–13.31 The Humongous Book of Algebra Problems 559 Index — Alphabetical List of Concepts with Problem Numbers multiplying: 13.25–13.28 simplifying: 13.2, 13.4–13.12, 13.14–13.20, 13.22–13.31, 13.37–13.38 radicand: 13.1, 13.3 radius: 22.12, 22.15, 22.17–22.18, 22.21, 23.16, 23.18, 23.21 range: 16.11, 16.14, 16.20, 16.26–16.30, 18.13, 19.3 rational equations: 4.13–4.14, 4.19, 4.23, 21.2– 21.20, 23.37–23.43 rational exponents: 13.13–13.20, 18.4 rational expressions adding and subtracting: 20.9–20.19, 21.9, 21.37, 21.41 dividing: 20.23–20.28 graphing: 20.38, 20.41, 20.44 multiplying: 20.20–20.22, 20.28–20.29 (containing) signed numbers: 1.26–1.28, 1.30 simplifying: 15.13, 20.1–20.34 rational functions domain of: 16.15, 16.17, 16.30 graphing: 16.7–16.8, 16.33–16.34 rational inequalities graphing: 21.33, 21.36, 21.40, 21.44 solving: 21.31–21.32, 21.34–21.35, 21.37–21.39, 21.41–21.43 rational numbers: 1.5–1.6, 1.8–1.10 rational root test: 17.28, 17.31, 17.33, 17.36, 17.43 real numbers: 1.7–1.10 rectangle: 23.17, 23.20 reduced-row echelon form: 10.18–10.34, 10.36, 10.38, 10.40 reducing fractions: 2.11, 2.13, 2.15–2.17, 2.35 relations: 15.1–15.8 remainder theorem: 17.2–17.5 roots (of functions): 17.6, 17.9, 17.11, 17.13, 17.29, 17.30, 17.32, 17.34, 17.37, 17.39 row echelon form: 10.18–10.21, 10.23, 10.25– 10.29, 10.32 row operations: 10.8–10.17 ngous Book of Algebra Problems 560 The Humo S scalar multiplication: 9.10–9.15 scientific notation: 3.19–3.22 set notation: 7.35–7.39 signed numbers adding and subtracting: 1.11–1.13, 1.17–1.21, 1.31 multiplying and dividing: 1.14–1.16, 1.33 simple interest: 23.9–23.11 slant asymptote: see oblique asymptotes slope: 5.26–5.39, 6.28, 6.30–6.31, 6.37, 6.40, 6.43–6.44, 22.36, 22.39 slope-intercept form of a line: 6.11–6.20, 6.21– 6.26, 6.32, 6.38, 6.41–6.42, 7.40–7.43, 22.36, 22.39 solving equations: see equations speed word problems: 23.23–23.29 square: 23.17 square root functions: 16.9–16.11, 16.16, 16.29, 16.35, 16.37, 16.39 standard form (of a) circle: 22.13–22.14, 22.17, 22.19, 22.21 (of an) ellipse: 22.22, 22.24–22.25, 22.28, 22.31–22.32 (of a) hyperbola: 22.35, 22.37–22.38, 22.41– 22.42 (of a) linear equation: 5.37–5.38, 6.29–6.36, 6.39, 6.42, 6.44 (of a) parabola: 22.1–22.2, 22.4–22.5, 22.7, 22.10–22.11 story problems: see word problems substitution technique: 8.8–8.18, 23.5 sum of perfect cubes: 12.31–12.32, 12.34, 20.26 symmetric property: 1.37, 1.44 symmetry: 16.21–16.30 synthetic division: 11.37–11.45, 17.3–17.5, 17.10, 17.12, 17.30, 17.32, 17.34, 17.37, 17.39, 20.8, 20.27, 20.43 systems of equations graphing: 8.1–8.7 solving via Cramer’s Rule: 9.38–9.44 Index — Alphabetical List of Concepts with Problem Numbers solving via elimination: 8.18–8.28, 9.39, 23.6, 23.8 solving via matrices: 10.19–10.24, 10.26–10.33 solving via substitution: 8.8–8.18, 23.5 systems of inequalities: 8.29–8.36, 8.40 T table of values: 5.10, 5.12–5.17, 5.40, 16.1–16.9, 16.12, 16.26–16.30, 18.11–18.12, 18.15, 19.1, 19.6, 20.41, 20.44, 22.6 transformations: 16.31–16.40, 17.15–17.16, 17.20, 17.40, 18.14, 18.16, 19.5, 19.9, 20.38 transitive property: 1.42, 1.44 transverse axis: 22.34–22.35, 22.38, 22.40, 22.42– 22.43 triangle: 23.19 W–X–Y–Z whole numbers: 1.1, 1.8–1.9 word problems age: 23.6–23.8 coins: 23.34–23.36 distance: 23.23–23.29 geometric: 23.16–23.22 integer: 23.1–23.5 interest: 23.9–23.15 mixture and combination: 23.30–23.33 work: 23.37–23.43 work word problems: 23.37–23.43 zero matrix: 10.3 zero product property: 14.1, 22.32 U–V variation direct (proportional): 21.21–21.25 inverse (indirect): 21.26–21.30 vertex (of an) absolute value graph: 5.41–5.44, 16.41 (of an) ellipse: 22.27, 22.29 (of a) hyperbola: 22.34, 22.38, 22.42 (of a) parabola: 17.20, 17.40, 22.2, 22.5, 22.8, 22.11 vertical asymptotes: 20.35–20.36, 20.39, 20.42 vertical line test: 16.23 volume: 23.21–23.22 The Humongous Book of Algebra Problems 561 Also by W. Michael Kelley… ISBN: 978-1-59257-512-1 The only way to learn calculus is to do calculus problems. Packed with 1000 problems—limits, continuity, derivatives, integrals, tangent lines, velocity, acceleration, area, volume, infinite series, epsilon-delta proofs, formal Riemann sums, and more! A member of Penguin Group (USA) Inc. penguin.com

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