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: 40


$\left( -\infty,\dfrac{26}{9} \right)\cup \left( \dfrac{34}{9},\infty \right) $

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \left|\dfrac{5}{3}-\dfrac{1}{2}x \right|\gt\dfrac{2}{9} ,$ 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 $a\gt0,$ $|x|\ge a$ implies $x\ge a$ OR $x\le -a$ (Note that the symbols $\ge$ may be replaced with $\gt$ and the symbol $\le$ with $\lt.$) Hence, the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5}{3}-\dfrac{1}{2}x\gt\dfrac{2}{9} \\\\\text{OR}\\\\ \dfrac{5}{3}-\dfrac{1}{2}x\lt-\dfrac{2}{9} .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5}{3}-\dfrac{1}{2}x\gt\dfrac{2}{9} \\\\ 18\left(\dfrac{5}{3}-\dfrac{1}{2}x\right)\gt\left(\dfrac{2}{9}\right)18 \\\\ 30-9x\gt4 \\\\ -9x\gt4-30 \\\\ -9x\gt-26 \\\\ x\lt\dfrac{-26}{-9} \\\\ x\lt\dfrac{26}{9} \\\\\text{OR}\\\\ \dfrac{5}{3}-\dfrac{1}{2}x\lt-\dfrac{2}{9} \\\\ 18\left(\dfrac{5}{3}-\dfrac{1}{2}x\right)\lt\left(-\dfrac{2}{9}\right)18 \\\\ 30-9x\lt-4 \\\\ -9x\lt-4-30 \\\\ -9x\lt-34 \\\\ x\gt\dfrac{-34}{-9} \\\\ x\gt\dfrac{34}{9} .\end{array} Note that division by a negative number entails reversing the inequality symbol. Hence, the solution is the interval $ \left( -\infty,\dfrac{26}{9} \right)\cup \left( \dfrac{34}{9},\infty \right) .$
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