Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 9 - Inequalities and Problem Solving - 9.3 Absolute-Value Equations and Inequalities - 9.3 Exercise Set - Page 600: 75


$x\lt-\dfrac{1}{2} \text{ or } x\gt\dfrac{7}{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inqeuality, $ 6+|3-2x|\gt10 ,$ use the properties of inequality to isolate the absolute value expression. Then use the definition of a greater than absolute value inequality and solve each resulting inequality. Finally, graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of inequality to isolate the absolute value expression, the given is equivalent to \begin{array}{l}\require{cancel} 6+|3-2x|\gt10 \\\\ |3-2x|\gt10-6 \\\\ |3-2x|\gt4 .\end{array} Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} 3-2x\gt4 \\\\\text{OR}\\\\ 3-2x\lt-4 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 3-2x\gt4 \\\\ -2x\gt4-3 \\\\ -2x\gt1 \\\\\text{OR}\\\\ -2x\lt-4-3 \\\\ -2x\lt-7 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol) results to \begin{array}{l}\require{cancel} -2x\gt1 \\\\ x\lt\dfrac{1}{-2} \\\\ x\lt-\dfrac{1}{2} \\\\\text{OR}\\\\ -2x\lt-7 \\\\ x\gt\dfrac{-7}{-2} \\\\ x\gt\dfrac{7}{2} .\end{array} Hence, the solution set is $ x\lt-\dfrac{1}{2} \text{ or } x\gt\dfrac{7}{2} .$
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