## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 9 - Inequalities and Problem Solving - 9.3 Absolute-Value Equations and Inequalities - 9.3 Exercise Set - Page 600: 103

#### Answer

$t\le \dfrac{5}{2}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $t-2\le|t-3| ,$ use the definition of a greater than (greater than or equal to) absolute value equality. $\bf{\text{Solution Details:}}$ Reading the original expression from right to left, the given expression is equivalent to \begin{array}{l}\require{cancel} |t-3|\ge t-2 .\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} t-3\ge t-2 \\\\\text{OR}\\\\ t-3\le -(t-2) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} t-3\ge t-2 \\\\ t-t\ge -2+3 \\\\ 0\ge 1 \text{ (FALSE)} \\\\\text{OR}\\\\ t-3\le -(t-2) \\\\ t-3\le -t+2 \\\\ t+t\le 2+3 \\\\ 2t\le 5 \\\\ t\le \dfrac{5}{2} .\end{array} Hence, $t\le \dfrac{5}{2} .$

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