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 - 2.5 Absolute Value Equations and Inequalities - 2.5 Exercises - Page 188: 70

Answer

$-47 \le k \le25$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ -|k+11|\ge-36 ,$ 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 $\ge.$ $\bf{\text{Solution Details:}}$ Dividing both sides by a negative number (and consequently reversing the inequality symbol) results to \begin{array}{l}\require{cancel} -|k+11|\ge-36 \\\\ \dfrac{-|k+11|}{-1}\le\dfrac{-36}{-1} \\\\ |k+11|\le36 .\end{array} Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -36 \le k+11 \le36 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} -36 \le k+11 \le36 \\\\ -36-11 \le k+11-11 \le36-11 \\\\ -47 \le k \le25 .\end{array} Upon checking, any value of the variable in the solution set satisfies the original inequality and any value not in the solution set does not satisfy the original inequality. Hence, the solution set is $ -47 \le k \le25 .$
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