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 154: 48


$\left( -\dfrac{7}{5},\dfrac{3}{5} \right)$

Work Step by Step

Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a,$ then the solution to the given equation, $ \left| 5x+2 \right|-2\lt3 ,$ is \begin{array}{l}\require{cancel} \left| 5x+2 \right|\lt3+2 \\\\ \left| 5x+2 \right|\lt5 \\\\ -5\lt 5x+2\lt5 \\\\ -5-2\lt 5x+2-2\lt5-2 \\\\ -7\lt 5x\lt3 \\\\ -\dfrac{7}{5}\lt \dfrac{5x}{5}\lt\dfrac{3}{5} \\\\ -\dfrac{7}{5}\lt x\lt\dfrac{3}{5} .\end{array} Hence, the solution set is the interval $ \left( -\dfrac{7}{5},\dfrac{3}{5} \right) .$
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