Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 3 Linear Systems and Matrices - 3.4 Solve Systems of Linear Equations in Three Variables - 3.4 Exercises - Skill Practice - Page 182: 20

Answer

$(9, -4, -5)$ is a solution of the system

Work Step by Step

The second equation: $x+y+z=0$ Solve for x: $x=-y-z$ Substitute for x in the first equation: $3x + 5y - z = 12 $ $3(-y-z) + 5y - z = 12 $ $-3y-3z+5y-z=12$ $2y-4z=12$ (*) Substitute for x in the third equation: $-x + 2y + 2z = -27$ $-(-y-z) + 2y + 2z = -27$ $y+z+2y+2z=-27$ $3y+3z=-27$ $\rightarrow 3y=-3z-27$ $\rightarrow y=-z-9$ Substitute for y: $2y-4z=12$ $2(-z-9)-4z=12$ $-2z-18-4z=12$ $-6z=30$ $z=-5$ Solve for y: $y=-(-5)-9=-4$ Solve for x: $z=-(-4)-(-5)=9$
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