Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1-2 - Cumulative Review: 50

Answer

$x\lt\dfrac{2}{3} \text{ OR } x\gt 4$
1514191267

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ |3x-7|+10\gt15 ,$ isolate first the absolute value expression. Then use the definition of absolute value inequality. Use the properties of inequality to isolate the variable. 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, the given is equivalent to \begin{array}{l}\require{cancel} |3x-7|+10\gt15 \\\\ |3x-7|\gt15-10 \\\\ |3x-7|\gt5 .\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} 3x-7\gt5 \\\\\text{OR}\\\\ 3x-7\lt-5 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 3x-7\gt5 \\\\ 3x\gt5+7 \\\\ 3x\gt12 \\\\ x\gt\dfrac{12}{3} \\\\ x\gt4 \\\\\text{OR}\\\\ 3x-7\lt-5 \\\\ 3x\lt-5+7 \\\\ 3x\lt2 \\\\ x\lt\dfrac{2}{3} .\end{array} Hence, the solution set is $ x\lt\dfrac{2}{3} \text{ OR } x\gt4 .$
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