Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 89: 85


$\left( -4,8 \right)$

Work Step by Step

For any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$. (The symbol $\lt$ may be replaced with $\le$.) Using the concept above, the solutions to the given inequality, $ \left| \dfrac{x-2}{3} \right|\lt2 ,$ is \begin{array}{l}\require{cancel} -2\lt\dfrac{x-2}{3}\lt2 \\\\ -6\lt x-2\lt6 \\\\ -6+2\lt x\lt6+2 \\\\ -4\lt x\lt8 .\end{array} In interval notation, the solution set is $ \left( -4,8 \right) .$
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