Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

Chapter 2 - Systems of Linear Equations and Inequalities - 2.5 Absolute Value Equations and Inequalities - 2.5 Exercises - Page 188: 42

Answer

$-6\lt p \lt0$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|p+3|\lt3 ,$ use the definition of absolute value inequality. Then use the properties of inequality to isolate the variable. Graph the solution. 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 $c\gt0$, $|x|\lt c$ 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} -3\lt p+3 \lt3 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} -3\lt p+3 \lt3 \\\\ -3-3\lt p+3-3 \lt3-3 \\\\ -6\lt p \lt0 .\end{array} The graph above confirms the solution set of the inequality.

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