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: 50


$\left( -\dfrac{5}{3},\dfrac{4}{3} \right)$

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| 2x+\dfrac{1}{3} \right|+1\lt4 ,$ is \begin{array}{l}\require{cancel} \left| 2x+\dfrac{1}{3} \right|\lt4-1 \\\\ \left| 2x+\dfrac{1}{3} \right|\lt3 \\\\ -3\lt 2x+\dfrac{1}{3} \lt3 \\\\ -3-\dfrac{1}{3}\lt 2x+\dfrac{1}{3}-\dfrac{1}{3} \lt3-\dfrac{1}{3} \\\\ -\dfrac{10}{3}\lt 2x \lt \dfrac{8}{3} \\\\ -\dfrac{\dfrac{10}{3}}{2}\lt \dfrac{2x}{2} \lt \dfrac{\dfrac{8}{3}}{2} \\\\ -\dfrac{10}{6}\lt x \lt \dfrac{8}{6} \\\\ -\dfrac{5}{3}\lt x \lt \dfrac{4}{3} .\end{array} Hence, the solution set is the interval $ \left( -\dfrac{5}{3},\dfrac{4}{3} \right) .$
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