Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.5 - Linear Inequalities in One Variable - 1.5 Exercises: 37

Answer

$\left( -\infty,\infty \right)$
1512800616

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ 3(2x-4)-4x\lt2x+3 ,$ 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} 3(2x)+3(-4)-4x\lt2x+3 \\\\ 6x-12-4x\lt2x+3 \\\\ 6x-4x-2x\lt3+12 \\\\ 0\lt15 \text{ (TRUE)} .\end{array} Since the solution above ended with a TRUE statement, then the solution set is the set of all real numbers. In interval notation, the solution set is $ \left( -\infty,\infty \right) .$ The graph of the solution set is the number line itself.
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