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: 35

Answer

$\left( -\infty,-\dfrac{2}{3} \right)\cup \left( 4,\infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \left|5-3x \right|\gt7 ,$ 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} 5-3x\gt7 \\\\\text{OR}\\\\ 5-3x\lt-7 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 5-3x\gt7 \\\\ -3x\gt7-5 \\\\ -3x\gt2 \\\\\text{OR}\\\\ 5-3x\lt-7 \\\\ -3x\lt-7-5 \\\\ -3x\lt-12 .\end{array} Dividing by a negative number and consequently reversing the sign results to \begin{array}{l}\require{cancel} x\lt\dfrac{2}{-3} \\\\ x\lt-\dfrac{2}{3} \\\\\text{OR}\\\\ x\gt\dfrac{-12}{-3} \\\\ x\gt4 .\end{array} Hence, the solution is the interval $ \left( -\infty,-\dfrac{2}{3} \right)\cup \left( 4,\infty \right) .$
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