## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 1 - Section 1.5 - Linear Inequalities in One Variable - 1.5 Exercises - Page 100: 16

#### Answer

$[-18,\infty)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $-\dfrac{2}{3}x\le12 ,$ 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} 3\left( -\dfrac{2}{3}x\right)\le3(12) \\\\ -2x\le36 .\end{array} Dividing both sides by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} x\ge\dfrac{36}{-2} \\\\ x\ge-18 .\end{array} The red graph is the graph of the solution set. In interval notation, the solution set is $[-18,\infty) .$

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