Algebra 2 (1st Edition)

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

Chapter 3 Linear Systems and Matrices - Chapter Review - Page 224: 15

Answer

The solution is $(-3, -8,4)$.

Work Step by Step

Rewrite the system as a linear system in two variables. Add Equation 1 and $6$ times Equation 2. Then add $2$ times Equation 2 to Equation 3: $6x - y + 4z =6$ $-6x -18y +6z = 186$ ______________________ $-19y+10z=192$ $-2x -6y + 2z = 62$ $2x + 2y- 5z =-42$ _________________________ $-4y-3z=20$ Solve the new linear system for both of its variables. $-19y+10z=192$ $-4y-3z=20$ Now, we can eliminate z: $-57y+30z=576$ $-40y-30z=200$ ____________________ $-97y=776$ $y=-8$ Solve for z: $-4(-8)-3z=20$ $12=3z$ $z=4$ Solve for y: $-x-3(-8)+4=31$ $x=-3$ The solution is $(-3, -8,4)$
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