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: 31


$\left( -\dfrac{3}{2},\dfrac{5}{2}\right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \left|\dfrac{1}{2}-x \right|\lt2 ,$ remove the absolute value sign using the properties of absolute value inequality. Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ For any $a\gt0,$ $|x|\le a$ implies $-a\le x\le a.$ (Note that the symbol $\le$ may be replaced with $\lt.$) Hence, the inequality above is equivalent to \begin{array}{l}\require{cancel} -2\lt \dfrac{1}{2}-x \lt2 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 2(-2)\lt 2\left(\dfrac{1}{2}-x\right) \lt2(2) \\\\ -4\lt 1-2x\lt4 \\\\ -4-1\lt 1-1-2x\lt4-1 \\\\ -5\lt -2x\lt3 .\end{array} Dividing both sides by $ -2 $ and reversing the inequality symbols, the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-5}{-2}\gt \dfrac{-2x}{-2}\gt\dfrac{3}{-2} \\\\ \dfrac{5}{2}\gt x\gt-\dfrac{3}{2} \\\\ -\dfrac{3}{2}\lt x\lt\dfrac{5}{2} .\end{array} Hence, the solution is the interval $ \left( -\dfrac{3}{2},\dfrac{5}{2}\right) .$
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