Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 2 - Section 2.3 - Solving Linear Equations - Exercise Set - Page 142: 76



Work Step by Step

$\frac{2}{3}$x=$\frac{1}{4}$x-8 12($\frac{2}{3}$x)=12($\frac{1}{4}$x-8) $\frac{24}{3}$x=$\frac{12}{4}$x-12(8) 8x=3x-96 8x-3x=-96 5x=-96 x=-$\frac{96}{5}$
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