## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 1 - Section 1.5 - Linear Inequalities in One Variable - 1.5 Exercises: 39

no solution

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $8\left(\dfrac{1}{2}x+3 \right)\lt8\left( \dfrac{1}{2}x-1 \right) ,$ use the Distributive Property and 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 Distributive Property and the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 8\left(\dfrac{1}{2}x\right)+8(3)\lt8\left( \dfrac{1}{2}x\right)-8(1) \\\\ 4x+24\lt4x-8 \\\\ 4x-4x\lt-8-24 \\\\ 0\lt-32 \text{ (FALSE)} .\end{array} Since the solution above ended with a FALSE statement, then there is $\text{ no solution .}$ There is no graph since there is no solution set.

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