Algebra 2 (1st Edition)

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

Chapter 3 Linear Systems and Matrices - 3.2 Solve Linear Systems Algebraically - 3.2 Exercises - Skill Practice - Page 164: 32

Answer

$$y=5,\:x=1$$

Work Step by Step

Solving the equation using substitution, we find: $$5\cdot \frac{-2+y}{3}+2y=15 \\ y=5$$ Thus: $$x=\frac{-2+5}{3}=1$$
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