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: 13



Work Step by Step

Solving the system of equations using substitution, we find: $$ \frac{-1-\left(-\frac{-14z+8}{5}\right)+z}{3}-\left(-\frac{-14z+8}{5}\right)+3z=3\\ z=-3 \\ y=-\frac{-14\left(-3\right)+8}{5}=-10 \\ x=\frac{-1-\left(-10\right)-3}{3}=2$$
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