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: 47


$\left( -\dfrac{4}{3},\dfrac{2}{3} \right)$

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| 3x+1 \right|-1\lt2 ,$ is \begin{array}{l}\require{cancel} \left| 3x+1 \right|\lt2+1 \\\\ \left| 3x+1 \right|\lt3 \\\\ -3\lt 3x+1\lt3 \\\\ -3-1\lt 3x+1-1\lt3-1 \\\\ -4\lt 3x\lt2 \\\\ -\dfrac{4}{3}\lt \dfrac{3x}{3}\lt\dfrac{2}{3} \\\\ -\dfrac{4}{3}\lt x\lt\dfrac{2}{3} .\end{array} Hence, the solution set is the interval $ \left( -\dfrac{4}{3},\dfrac{2}{3} \right) .$
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