Intermediate Algebra: Connecting Concepts through Application

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

Chapter 2 - Systems of Linear Equations and Inequalities - Chapter Review Exercises - Page 211: 40


$x\le-15 \text{ OR } x\ge9$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ |x+3|\ge12 ,$ 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 $\ge.$ $\bf{\text{Solution Details:}}$ 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} x+3\ge12 \\\\\text{OR}\\\\ x+3\le-12 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} x+3\ge12 \\\\ x\ge12-3 \\\\ x\ge9 \\\\\text{OR}\\\\ x+3\le-12 \\\\ x\le-12-3 \\\\ x\le-15 .\end{array} Hence, the solution set is $ x\le-15 \text{ OR } x\ge9 .$
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