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: 12

Answer

$$y=\frac{10}{3},\:x=\frac{40}{9}$$

Work Step by Step

We solve the system of equations using substitution: $$ \frac{1}{2}\cdot \frac{4}{3}y-\frac{1}{15}y=2\\ y=\frac{10}{3} \\ x=\frac{4}{3}\cdot \frac{10}{3}\\ x=\frac{40}{9}$$
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