## 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: 58

#### Answer

$-7\lt t \lt 1$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $5|t+3|\lt20 ,$ 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} 5|t+3|\lt20 \\\\ |t+3|\lt\dfrac{20}{5} \\\\ |t+3|\lt4 .\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} -4\lt t+3 \lt4 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} -4\lt t+3 \lt4 \\\\ -4-3\lt t+3-3 \lt4-3 \\\\ -7\lt t \lt 1 .\end{array} Hence, the solution set is $-7\lt t \lt 1 .$ 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.

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