College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.8 - Absolute Value Equations and Inequalities - 1.8 Exercises - Page 153: 17

Answer

$x= \left\{ -\dfrac{4}{3},\dfrac{2}{9} \right\}$

Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ or $x=-a$, then the solution/s to the given equation, $ \left| \dfrac{6x+1}{x-1} \right|=3 ,$ is/are \begin{array}{l}\require{cancel} \dfrac{6x+1}{x-1}=3 \\\\ 6x+1=3(x-1) \\\\ 6x+1=3x-3 \\\\ 6x-3x=-3-1 \\\\ 3x=-4 \\\\ x=-\dfrac{4}{3} \\\\\text{OR}\\\\ \dfrac{6x+1}{x-1}=-3 \\\\ 6x+1=-3(x-1) \\\\ 6x+1=-3x+3 \\\\ 6x+3x=3-1 \\\\ 9x=2 \\\\ x=\dfrac{2}{9} .\end{array} Hence, the solutions are $ x= \left\{ -\dfrac{4}{3},\dfrac{2}{9} \right\} .$
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