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: 49


$\left( -\dfrac{3}{2},\dfrac{13}{10} \right)$

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+\dfrac{1}{2} \right|-2\lt5 ,$ is \begin{array}{l}\require{cancel} \left| 5x+\dfrac{1}{2} \right|\lt5+2 \\\\ \left| 5x+\dfrac{1}{2} \right|\lt7 \\\\ -7\lt 5x+\dfrac{1}{2} \lt7 \\\\ -7-\dfrac{1}{2}\lt 5x+\dfrac{1}{2}-\dfrac{1}{2} \lt7-\dfrac{1}{2} \\\\ -\dfrac{15}{2}\lt 5x \lt \dfrac{13}{2} \\\\ -\dfrac{\dfrac{15}{2}}{5}\lt \dfrac{5x}{5} \lt \dfrac{\dfrac{13}{2}}{5} \\\\ -\dfrac{15}{10} \lt x \lt \dfrac{13}{10} \\\\ -\dfrac{3}{2} \lt x \lt \dfrac{13}{10} .\end{array} Hence, the solution set is the interval $ \left( -\dfrac{3}{2},\dfrac{13}{10} \right) .$
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