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 - 8.2 Solving by Substitution or Elimination - 8.2 Exercise Set - Page 517: 36

Answer

1,3

Work Step by Step

Solving the two equations using substitution, we find: $$ 2\cdot \frac{-16+7y}{5}+8y=26\\ y=3 \\ x=\frac{-16+7\cdot \:3}{5}\\ x=1$$
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