College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Summary, Review, and Test - Test - Page 206: 16

Answer

$x=6$ or $x=12$

Work Step by Step

$|\frac{2}{3}x-6|-2=0$ Applying absolute rule: $|\frac{2}{3}x-6|=2$ or $|\frac{2}{3}x-6|=-2$ So: $\frac{2}{3}x-6=2$ $\frac{2}{3}x=8$ $\frac{\frac{2}{3}}{\frac{2}{3}}x=\frac{8}{\frac{2}{3}}$ $x=12$ or $\frac{2}{3}x-6=-2$ $\frac{2}{3}x= 4$ $\frac{\frac{2}{3}}{\frac{2}{3}}x=\frac{4}{\frac{2}{3}}$ $x=6$
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