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 Test - Page 227: 13

Answer

The solution is $(5,1,-3)$.

Work Step by Step

Use elimination for equation 1 and equation 2: $x + y + z = 3 $ $-x + 3y + 2z= -8$ __________________________ $4y+3z=-5$ The third equation: $5y+z=2$ $\rightarrow z=2-5y$ Substitute for z: $4y+3z=-5$ $4y+3(2-5y)=-5$ $-11y=-11$ $y=1$ Solve for z: $z=2-5(1)=-3$ Solve for x: $x+1+(-3)=3$ $x=5$ The solution is $(5,1,-3)$.
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