Forming Equivalent Equations Chapter 2 Linear Equation and Inequalities Michael Giessing giessing@math.utah.edu An equation can be rewriten into an equivalent form by: 1. Simplify either side 2. Using the Golden Rule of Algebra 3. Interchanging sides University of Utah Linear Equation and Inequalities – p.1/32 The Golden Rule (of Algebra) Do unto one side as thou hath done unto the other. Linear Equation and Inequalities – p.2/32 Examples Simplify the left hand side. x(4 − x) = 2x → 4x − x2 = 2x Added 4 to both sides. x(4 − x) = 2x → x(4 − x) + 4 = 2x + 4 Interchangesd sides. x(4 − x) = 2x → 2x = x(4 − x) Linear Equation and Inequalities – p.3/32 The Golden Rule and Addition You can think of an equation as a scale. If you do something to one side you must do the same thing to the other side to keep things in balance. You must do all operations to entire expressions! You can add anything to both sides. This includes negative numbers and variables y+π y + π + (−3) 2x + 3 2x + x = = = = y2 → y 2 + (−3) x→ −x + x Linear Equation and Inequalities – p.5/32 Linear Equation and Inequalities – p.4/32 The Golden Rule and Multiplication You can mulitiply both side by any number. You must multiply the entire side by the number not just some of the terms on that side. 2l + 6 = 4l → 30(2l + 6) = 30 × 4l You can multiply by any multiplicative inverse. Which is the same thing as saying you can divide by any number. 4 1 4 1 = 5n → × = × 5n n 4 n 4 There is a number that has no multiplicative inverse. What is it? Linear Equation and Inequalities – p.6/32 The Golden Rule and Multiplication by a variable Never divide by Zero Zero does not have a multiplicative inverse. To see this try to find Zero’s inverse.(Remember that an inverse is the number x such that ax = 1. For most numbers x = 1/a) 0 ∗ x = 1; x =? Unsual things happen if you divide by zero. You can alway multiply both sides by a variable. x2 + y 2 = 4 → x(x2 + y 2 ) = 4x You can divide by x as long as x 6= 0. It is easy to make this mistake. Be cautious whenever you divide by a variable. Linear Equation and Inequalities – p.7/32 The Golden Rule and Function We will use the Golden Rule with more than just addition and multiplication. We will use it with many other operations called functions. Always remember to apply the operation to the entire left hand side(LHS) and right hand side( RHS). Linear Equation and Inequalities – p.8/32 Word Problems 1. What do you need to know to solve the problem? Write this down in english. 2. Assing numbers to the the known parts. Assign a letter to the unknown parts 3. Translate this into an algebraic equation or inequality. 4. Solve. 5. Make sure that your solution answers the original question Linear Equation and Inequalities – p.9/32 Example Linear Equation and Inequalities – p.10/32 Example Continued You purchased my new bike for $1500.00. The bike shop was having 30% off sale the day you purchased the bike. What was the original price of the bike? 1. The price you paid for the bike is 30% off the original price. Restated, the price you paid for the bike is the orignal price minus 30% of the original price. 2. 1500.00 =x-.3x 1500.00=.7x Simplifying the RHS 1500.00/.7=.7x/.7 Golden Rule of Algebra 3. 2142.86=x Simplifying both sides x=2142.86 Interchanging the sides Linear Equation and Inequalities – p.11/32 The answer $2142.86 answers the question and seems reasonable. Lets check it to make sure everything worked. We need to see if 30% off of $2142.86 is $1500.00. 2142.86 − .3 × 2142.86 = 1500.00 Linear Equation and Inequalities – p.12/32 Geometry Formulas Useful Formulas Business Shape Square Rectangle Circle Triangle Selling price = Cost + Mark up Mark up = Mark up rate × Cost Selling price = List price - Discount Rates in Mixtures Area A = s2 A = lw A = πr 2 A = bh/2 Perimeter P = 4s P = 2l + 2w P = 2πr P =a+b+c first rate × Amount + second rate × Amount = Final Rate × Amount I expect you to learn all of these. Rate problems Distance = rate × time I expect you to know all of these. Linear Equation and Inequalities – p.13/32 3-D Geometry Formulas Linear Equation and Inequalities – p.14/32 Problem Solving with Formulas Shape Volume Cube V = s3 Rectangular Prism V = lwh Surface Area SA = 6s2 SA = 2lw + 2wh +2hl 2 Cylinder V = πr h SA = 2πrh + 2πr 2 Sphere A = 43 πr3 SA = 4πr 2 1. Determine which formulas fit your problem. This formulas should contain the information given in the problem and the quantities you are looking for. 2. Plug the information from the problem into these formulas. 3. Solve these equations for the unkown values. You will need one equation per variable. I expect you to learn all of these. Linear Equation and Inequalities – p.15/32 Example 5 on page 83 Linear Equation and Inequalities – p.16/32 Step 2: Plug Students are traveling in two cars to a football game 150 miles away. The first car leaves on time and travels at an average speed of 48 mph. The second car starts .5 hours later and travels at an average speed of 58 mph. At these speeds, how long will it take the second car to catch up to the first car? Step 1: What formula? We will use d = rt. We want to find a time when the two cars have gone the same distance. We will start time when the first car leaves. Linear Equation and Inequalities – p.17/32 For the first car d1 = 48t. If we put some time t into that we get the first cars position. We need to write the same thing for the second car. Remember that car two leaves .5 hours late so the distance it has traveled is d2 = 58(t − .5). We want to find the time when the cars are in the same place so d1 = d2 . That give us three equations and three unknowns. Linear Equation and Inequalities – p.18/32 Step 3: Solve We start from d1 = d2 . We know that d1 = 48t and d2 = 58(t − .5). We substitute this in and get 48t = 58(t − .5). Now we solve this. 48t = 58(t − .5) 48t = 58t − .5 distributive prop 48t − 58t = 58t − .5 − 58t Golden Rule −10t = −.5 Simplify LHS t = −.5/ − 10 Golden Rule t = .05 Simplify RHS Inequalities Linear Equation and Inequalities – p.19/32 Properties of Inequalities Linear Equation and Inequalities – p.20/32 Multiplication By a Negative Inequalities obey the golden rule but, the direction of the inequality changes whenever you multiply or divide by a negative number. Example: 2 < 3 → 3 × 2 < 3 × 3 → Golden Rule 6 < 9 Observe that 2 < 3 but, −2 > −3. We can get the second inequality form the first by multiplying the first by −1 on both sides. Example: 2 < 3 → −3 × 2 > −3 × 3 → Golden Rule −6 > −9 Linear Equation and Inequalities – p.21/32 Solving an Equations Inequalities and Division Example Positive 6 6/2 3 Negative 6 6/(−2) −3 Linear Equation and Inequalities – p.22/32 Absolute Value Lets solve |x| = 2. If x ≥ 0, then we have x = 2. If x < 0, then −x = 2 → x = −1 < 10 < 10/2 < 5 → → Golden Rule < 10 → > 10/(−2) → Golden Rule > −5 Linear Equation and Inequalities – p.23/32 Now lets solve all problems like this one. This is called abstraction. Let say we were given some positive real number a. (Why does it need to be positive?) Then we could find x so that |x| = a. This works just like before. If x ≥ 0, then x = a. IF x < 0, then −x = a → x = −a. Linear Equation and Inequalities – p.24/32 Solving a Concrete Example Solving the Abstract Problem |x| 1.2 |x| a −1.2 1.2 −a Linear Equation and Inequalities – p.25/32 Nonstandard Form (Ex. 3 pg. 108) Solve |2x-1|+3=8. First we try to simplify as much as possible. Using the golden rule we subtract 3 from both sides. |2x − 1| + 3 − 3 = 8 − 3 → |2x − 1| = 5 If 2x − 1 ≥ 0 then 2x − 1 = 5. We will add 1 to both sides. Then we divide both sides by 2. (Use the Golden Rule twice) a Linear Equation and Inequalities – p.26/32 Continued If 2x − 1 < 0, then −(2x − 1) = 5. We will apply the Golden Rule three times. −(2x−1) = 5 → 2x−1 = −5 → 2x = −4 → x = −2 So, x = 3 and x = −2 are solutions to the equation. 2x − 1 + 1 = 5 + 1 → 2x/2 = 6/2 → x = 3 Linear Equation and Inequalities – p.27/32 Linear Equation and Inequalities – p.28/32 Absolute Inequalities Picture, Picture blah |2x−1| 5 −2 1/2 3 Linear Equation and Inequalities – p.29/32 Linear Equation and Inequalities – p.30/32 |x| 1.2 |x| 1.2 −1.2 1.2 Linear Equation and Inequalities – p.31/32 −1.2 1.2 Linear Equation and Inequalities – p.32/32