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: 36


$\left( -\infty,1 \right)\cup \left( \dfrac{11}{3},\infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \left|7-3x \right|\gt4 ,$ 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} 7-3x\gt4 \\\\\text{OR}\\\\ 7-3x\lt-4 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 7-3x\gt4 \\\\ -3x\gt4-7 \\\\ -3x\gt-3 \\\\\text{OR}\\\\ 7-3x\lt-4 \\\\ -3x\lt-4-7 \\\\ -3x\lt-11 .\end{array} Dividing by a negative number and consequently reversing the sign results to \begin{array}{l}\require{cancel} x\lt\dfrac{-3}{-3} \\\\ x\lt1 \\\\\text{OR}\\\\ -3x\lt-11 \\\\ x\gt\dfrac{-11}{-3} \\\\ x\gt\dfrac{11}{3} .\end{array} Hence, the solution is the interval $ \left( -\infty,1 \right)\cup \left( \dfrac{11}{3},\infty \right) .$
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