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: 89


$\left[ -\dfrac{1}{2},\dfrac{3}{2} \right]$

Work Step by Step

Using the properties of inequality, the given expression, $ 8-|2x-1|\ge6 ,$ is equivalent to \begin{array}{l}\require{cancel} -|2x-1|\ge6-8 \\\\ -|2x-1|\ge-2 \\\\ |2x-1|\le2 .\end{array} 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, $ -2\le 2x-1\le2 ,$ is \begin{array}{l}\require{cancel} -2+1\le 2x\le2+1 \\\\ -1\le 2x\le3 \\\\ -\dfrac{1}{2}\le x\le\dfrac{3}{2} .\end{array} In interval notation, the solution set is $ \left[ -\dfrac{1}{2},\dfrac{3}{2} \right] .$
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