Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 8 - Systems of Linear Equations and Problem Solving - Mid-Chapter Review - Mixed Review - Page 540: 14

Answer

$$z=\frac{1}{3},\:y=-4,\:x=\frac{1}{2}$$

Work Step by Step

Solving the system of equations using substitution, we find: $$ 2\cdot \frac{-4-\left(9z-7\right)+3z}{2}-\left(9z-7\right)+6z=7\\ z=\frac{1}{3} \\ y=9\cdot \frac{1}{3}-7=-4 \\ x=\frac{-4-\left(-4\right)+3\cdot \frac{1}{3}}{2}=\frac{1}{2}$$
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