Solving Linear Systems Solving Linear Systems 3-6 3-6 in Three Variables in Three Variables Warm Up Lesson Presentation Lesson Quiz Holt Algebra Holt Algebra 22 3-6 Solving Linear Systems in Three Variables Warm Up Solve each system of equations algebraically. 1. x = 4y + 10 4x + 2y = 4 (2, –2) 2. 6x – 5y = 9 2x – y =1 (–1,–3) Classify each system and determine the number of solutions. 3x – y = 8 x = 3y – 1 4. 3. 6x – 2y = 2 6x – 12y = –4 inconsistent; none Holt Algebra 2 consistent, independent; one 3-6 Solving Linear Systems in Three Variables Objectives Represent solutions to systems of equations in three dimensions graphically. Solve systems of equations in three dimensions algebraically. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Systems of three equations with three variables are often called 3-by-3 systems. In general, to find a single solution to any system of equations, you need as many equations as you have variables. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Recall from Lesson 3-5 that the graph of a linear equation in three variables is a plane. When you graph a system of three linear equations in three dimensions, the result is three planes that may or may not intersect. The solution to the system is the set of points where all three planes intersect. These systems may have one, infinitely many, or no solution. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Identifying the exact solution from a graph of a 3-by-3 system can be very difficult. However, you can use the methods of elimination and substitution to reduce a 3-by-3 system to a 2-by-2 system and then use the methods that you learned in Lesson 3-2. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Example 1: Solving a Linear System in Three Variables Use elimination to solve the system of equations. 5x – 2y – 3z = –7 1 2x – 3y + z = –16 2 3x + 4y – 2z = 7 3 Step 1 Eliminate one variable. In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1 and z is easy to eliminate from the other equations. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Example 1 Continued 1 2 5x – 2y – 3z = –7 5x – 2y – 3z = –7 3(2x –3y + z = –16) Multiply equation -2 by 3, and add 6x – 9y + 3z = –48 to equation 1 . 11x – 11y Use equations in x and y. 3 and 2 = –55 4 to create a second equation 1 3 2 3x + 4y – 2z = 7 3x + 4y – 2z = 7 Multiply equation 2(2x –3y + z = –16) 4x – 6y + 2z = –32 -2 by 2, and add 7x – 2y Holt Algebra 2 = –25 to equation 5 3 . 3-6 Solving Linear Systems in Three Variables Example 1 Continued You now have a 2-by-2 system. Holt Algebra 2 11x – 11y = –55 4 7x – 2y = –25 5 3-6 Solving Linear Systems in Three Variables Example 1 Continued Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate y by using methods from Lesson 3-2. 4 5 equation –2(11x – 11y = –55) –22x + 22y = 110 1 Multiply -4 by –2, and 11(7x – 2y = –25) 77x – 22y = –275 equation -5 by 11 55x 1 = –165 x = –3 Holt Algebra 2 and add. Solve for x. 3-6 Solving Linear Systems in Three Variables Example 1 Continued Step 3 Use one of the equations in your 2-by-2 system to solve for y. 4 11x – 11y = –55 1 11(–3) – 11y = –55 Substitute –3 for x. 1 y=2 Holt Algebra 2 Solve for y. 3-6 Solving Linear Systems in Three Variables Example 1 Continued Step 4 Substitute for x and y in one of the original equations to solve for z. 2 2x – 3y + z = –16 2(–3) – 3(2) + z = –16 z = –4 The solution is (–3, 2, –4). Holt Algebra 2 Substitute 1 –3 for x and 2 for y. 1Solve for y. 3-6 Solving Linear Systems in Three Variables Check It Out! Example 1 Use elimination to solve the system of equations. –x + y + 2z = 7 1 2x + 3y + z = 1 2 –3x – 4y + z = 4 3 Step 1 Eliminate one variable. In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 1 Continued 1 2 –x + y + 2z = 7 Multiply equation –x + y + 2z = 7 –2(2x + 3y + z = 1) –4x – 6y – 2z = –2 -2 by –2, and add 1 –5x – 5y Use equations in x and y. 1 and 3 =5 to equation . 4 to create a second equation 1 1 3 –x + y + 2z = 7 –2(–3x – 4y + z = 4) –x + y + 2z = 7 Multiply equation 6x + 8y – 2z = –8 -3 by –2, and add to equation 5x + 9y Holt Algebra 2 = –1 5 1 . 3-6 Solving Linear Systems in Three Variables Check It Out! Example 1 Continued You now have a 2-by-2 system. –5x – 5y = 5 5x + 9y = –1 Holt Algebra 2 4 5 3-6 Solving Linear Systems in Three Variables Check It Out! Example 1 Continued Step 2 Eliminate another variable. Then solve for the remaining variable. You can eliminate x by using methods from Lesson 3-2. 4 5 –5x – 5y = 5 5x + 9y = –1 4y = 4 y=1 Holt Algebra 2 Add equation Solve for y. 5 1 to equation 4 . 3-6 Solving Linear Systems in Three Variables Check It Out! Example 1 Step 3 Use one of the equations in your 2-by-2 system to solve for x. 4 –5x – 5y = 5 Substitute 1 for y. –5x – 5(1) = 5 1 –5x – 5 = 5 –5x = 10 x = –2 Holt Algebra 2 Solve for x. 1 3-6 Solving Linear Systems in Three Variables Check It Out! Example 1 Step 4 Substitute for x and y in one of the original equations to solve for z. 2 2x +3y + z = 1 2(–2) +3(1) + z = 1 –4 + 3 + z = 1 z=2 Substitute –2 for x and 1 for y. Solve for 1z. 1 The solution is (–2, 1, 2). Holt Algebra 2 3-6 Solving Linear Systems in Three Variables You can also use substitution to solve a 3-by-3 system. Again, the first step is to reduce the 3-by-3 system to a 2-by-2 system. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Example 2: Business Application The table shows the number of each type of ticket sold and the total sales amount for each night of the school play. Find the price of each type of ticket. Orchestra Mezzanine Balcony Total Sales Fri 200 30 40 $1470 Sat 250 60 50 $1950 Sun 150 30 0 $1050 Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Example 2 Continued Step 1 Let x represent the price of an orchestra seat, y represent the price of a mezzanine seat, and z represent the present of a balcony seat. Write a system of equations to represent the data in the table. 200x + 30y + 40z = 1470 1 250x + 60y + 50z = 1950 2 Saturday’s sales. 150x + 30y = 1050 3 Sunday’s sales. Friday’s sales. A variable is “missing” in the last equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Example 2 Continued Step 2 Eliminate z. Multiply equation 1 2 1 by 5 and equation 5(200x + 30y + 40z = 1470) –4(250x + 60y + 50z = 1950) 2 by –4 and add. 1000x + 150y + 200z = 7350 –1000x – 240y – 200z = –7800 y =5 By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y. Holt Algebra 2 Solving Linear Systems in Three Variables 3-6 Example 2 Continued Step 3 Use equation 150x + 30y = 1050 150x + 30(5) = 1050 3 x=6 Holt Algebra 2 3 to solve for x. Substitute 5 for y. Solve for x. 3-6 Solving Linear Systems in Three Variables Example 2 Continued Step 4 Use equations 1 or 2 to solve for z. 1 1 200x + 30y + 40z = 1470 200(6) + 30(5) + 40z = 1470 Substitute 6 for x and 5 for y. Solve for x. z=3 The solution to the system is (6, 5, 3). So, the cost of an orchestra seat is $6, the cost of a mezzanine seat is $5, and the cost of a balcony seat is $3. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 2 Jada’s chili won first place at the winter fair. The table shows the results of the voting. How many points are first-, second-, and third-place votes worth? Winter Fair Chili Cook-off Name 1st Place 2nd Place 3rd Place Total Points Jada 3 1 4 15 Maria 2 4 0 14 Al 2 2 3 13 Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 2 Continued Step 1 Let x represent first-place points, y represent second-place points, and z represent thirdplace points. Write a system of equations to represent the data in the table. 3x + y + 4z = 15 1 Jada’s points. 2x + 4y = 14 2 Maria’s points. 2x + 2y + 3z = 13 3 Al’s points. A variable is “missing” in one equation; however, the same solution methods apply. Elimination is a good choice because eliminating z is straightforward. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 2 Continued Step 2 Eliminate z. Multiply equation 1 by 3 and equation 3 by –4 and add. 1 3(3x + y + 4z = 15) 9x + 3y + 12z = 45 3 –4(2x + 2y + 3z = 13) –8x – 8y – 12z = –52 x – 5y Multiply equation 4 2 4 –2(x – 5y = –7) 2x + 4y = 14 Holt Algebra 2 = –7 by –2 and add to equation 2 4 . –2x + 10y = 14 2x + 4y = 14 y = 2 Solve for y. Solving Linear Systems in Three Variables 3-6 Check It Out! Example 2 Continued Step 3 Use equation 2 2x + 4y = 14 2x + 4(2) = 14 x=3 Holt Algebra 2 2 to solve for x. Substitute 2 for y. Solve for x. 3-6 Solving Linear Systems in Three Variables Check It Out! Example 2 Continued Step 4 Substitute for x and y in one of the original equations to solve for z. 3 2x + 2y + 3z = 13 2(3) + 2(2) + 3z = 13 6 + 4 + 3z = 13 z=1 Solve for z. The solution to the system is (3, 2, 1). The points for first-place is 3, the points for second-place is 2, and 1 point for third-place. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables The systems in Examples 1 and 2 have unique solutions. However, 3-by-3 systems may have no solution or an infinite number of solutions. Remember! Consistent means that the system of equations has at least one solution. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Example 3: Classifying Systems with Infinite Many Solutions or No Solutions Classify the system as consistent or inconsistent, and determine the number of solutions. 2x – 6y + 4z = 2 –3x + 9y – 6z = –3 5x – 15y + 10z = 5 Holt Algebra 2 1 2 3 3-6 Solving Linear Systems in Three Variables Example 3 Continued The elimination method is convenient because the numbers you need to multiply the equations are small. First, eliminate x. Multiply equation 1 2 1 by 3 and equation 3(2x – 6y + 4z = 2) 2(–3x + 9y – 6z = –3) 2 by 2 and add. 6x – 18y + 12z = 6 –6x + 18y – 12z = –6 0=0 Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Example 3 Continued 1 3 Multiply equation 1 by 5 and equation 3 by –2 and add. 10x – 30y + 20z = 10 5(2x – 6y + 4z = 2) –10x + 30y – 20z = –10 –2(5x – 15y + 10z = 5) 0 = 0 Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent and has an infinite number of solutions. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3a Classify the system, and determine the number of solutions. 3x – y + 2z = 4 1 2x – y + 3z = 7 2 –9x + 3y – 6z = –12 Holt Algebra 2 3 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3a Continued The elimination method is convenient because the numbers you need to multiply the equations by are small. First, eliminate y. Multiply equation 1 3 2 by –1 and add to equation 3x – y + 2z = 4 –1(2x – y + 3z = 7) . 3x – y + 2z = 4 –2x + y – 3z = –7 x Holt Algebra 2 1 – z = –3 4 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3a Continued Multiply equation 2 3 2 by 3 and add to equation 3(2x – y + 3z = 7) –9x + 3y – 6z = –12 –9x + 3y – 6z = –12 Now you have a 2-by-2 system. –3x + 3z = 9 Holt Algebra 2 4 5 . 6x – 3y + 9z = 21 –3x x – z = –3 3 + 3z = 9 5 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3a Continued Eliminate x. 4 5 3(x – z = –3) –3x + 3z = 9 3x – 3z = –9 –3x + 3z = 9 0=0 Because 0 is always equal to 0, the equation is an identity. Therefore, the system is consistent, dependent, and has an infinite number of solutions. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3b Classify the system, and determine the number of solutions. Holt Algebra 2 2x – y + 3z = 6 1 2x – 4y + 6z = 10 2 y – z = –2 3 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3b Continued Use the substitution method. Solve for y in equation 3. 3 y – z = –2 y=z–2 Solve for y. 4 Substitute equation 4 in for y in equation 2x – y + 3z = 6 2x – (z – 2) + 3z = 6 2x – z + 2 + 3z = 6 2x + 2z = 4 Holt Algebra 2 5 1 . 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3b Continued Substitute equation 4 in for y in equation 2x – 4y + 6z = 10 2x – 4(z – 2) + 6z = 10 2x – 4z + 8 + 6z = 10 2x + 2z = 2 6 Now you have a 2-by-2 system. 2x + 2z = 4 2x + 2z = 2 Holt Algebra 2 5 6 2 . 3-6 Solving Linear Systems in Three Variables Check It Out! Example 3b Continued Eliminate z. 5 6 2x + 2z = 4 –1(2x + 2z = 2) 02 Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions. Holt Algebra 2 3-6 Solving Linear Systems in Three Variables Lesson Quiz: Part I 1. At the library book sale, each type of book is priced differently. The table shows the number of books Joy and her friends each bought, and the amount each person spent. Find the price of each type of book. Hardcover Paper- Audio Total back Books Spent Hal 3 4 1 $17 Ina 2 5 1 $15 Joy 3 3 2 $20 Holt Algebra 2 hardcover: $3; paperback: $1; audio books: $4 3-6 Solving Linear Systems in Three Variables Lesson Quiz: Part II Classify each system and determine the number of solutions. 2x – y + 2z = 5 2. –3x +y – z = –1 inconsistent; none x – y + 3z = 2 9x – 3y + 6z = 3 3. 12x – 4y + 8z = 4 –6x + 2y – 4z = 5 Holt Algebra 2 consistent; dependent; infinite

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