## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 1 - Section 1.5 - Linear Inequalities in One Variable - 1.5 Exercises: 54

#### Answer

$\left[ -2,\dfrac{5}{2} \right]$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $-12 \le -6x+3 \le 15 ,$ use the properties of inequality. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -12-3 \le -6x+3-3 \le 15-3 \\\\ -15 \le -6x \le 12 .\end{array} Dividing all sides by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-15}{-6} \ge \dfrac{-6x}{-6} \ge \dfrac{12}{-6} \\\\ \dfrac{5}{2} \ge x \ge -2 \\\\ -2 \le x \le \dfrac{5}{2} .\end{array} In interval notation, the solution set is $\left[ -2,\dfrac{5}{2} \right] .$ The red graph is the graph of the solution set.

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