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: 17

Answer

$x=-6$ $y=4$ $z=-4$

Work Step by Step

The third equation: $x-y-2z=-2$ Solve for x: $\rightarrow x=y+2z-2$ Substitute for z in the first equation: $4x + y +5z = -40$ $4(y+2z-2)+ y+ 5z = -40$ $4y+8z-8+ y+ 5z= -40$ $5y+13z=-32$ (1) Substitute for z in the second equation: $-3x + 2y + 4z = 10$ $-3(y+2z-2)+ 2y + 4z = 10$ $-y-2z=4$ $\rightarrow y=-2z-4$ Substitute for y in equation (1): $5(-2z-4)+13z=-32$ $3z=-12$ $z=-4$ Solve for y: $y=-2z-4$ $y=-2(-4)-4$ $y=4$ Solve for x: $x=y+2z-2$ $x=4+2(-4)-2$ $x=-6$
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