College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises - Page 166: 120

Answer

$\left( -\dfrac{22}{3},\dfrac{14}{3} \right)$

Work Step by Step

Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$, then the solution to the given inequality, $ \left| \dfrac{1}{2}x+\dfrac{2}{3} \right|\lt3 ,$ is \begin{array}{l}\require{cancel} -3\lt \dfrac{1}{2}x+\dfrac{2}{3}\lt3 \\\\ 6\cdot(-3)\lt 6\cdot\left( \dfrac{1}{2}x+\dfrac{2}{3} \right)\lt6\cdot(3) \\\\ -18\lt 3x+4\lt18 \\\\ -18-4\lt 3x+4-4\lt18-4 \\\\ -22\lt 3x\lt14 \\\\ -\dfrac{22}{3}\lt \dfrac{3x}{3}\lt\dfrac{14}{3} \\\\ -\dfrac{22}{3}\lt x\lt\dfrac{14}{3} .\end{array} Hence, the solution set is the interval $ \left( -\dfrac{22}{3},\dfrac{14}{3} \right) .$
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