Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.7 - Absolute Value Equations and Inequalities - 1.7 Exercises - Page 118: 18

Answer

$x=\left\{ \dfrac{1}{6},\dfrac{3}{2} \right\}$

Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the given equation, $ |-6x+5|=4 ,$ is equivalent to \begin{array}{l}\require{cancel} -6x+5=4 \text{ OR } -6x+5=-4 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} -6x+5=4 \\\\ -6x=4-5 \\\\ -6x=-1 \\\\ x=\dfrac{-1}{-6} \\\\ x=\dfrac{1}{6} \\\\\text{ OR }\\\\ -6x+5=-4 \\\\ -6x=-4-5 \\\\ -6x=-9 \\\\ x=\dfrac{-9}{-6} \\\\ x=\dfrac{3}{2} .\end{array} Hence, the solutions are $ x=\left\{ \dfrac{1}{6},\dfrac{3}{2} \right\} .$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.