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: 33

Answer

$\left( -\infty,\dfrac{23}{6} \right]$
1512797647

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \dfrac{2}{3}(3x-1)\ge\dfrac{3}{2}(2x-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} \dfrac{2}{3}(3x)+\dfrac{2}{3}(-1)\ge\dfrac{3}{2}(2x)+\dfrac{3}{2}(-3) \\\\ 2x-\dfrac{2}{3}\ge3x-\dfrac{9}{2} .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 6\left( 2x-\dfrac{2}{3} \right)\ge6\left( 3x-\dfrac{9}{2} \right) \\\\ 12x-4\ge18x-27 \\\\ 12x-18x\ge-27+4 \\\\ -6x\ge-23 .\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\le\dfrac{-23}{-6} \\\\ x\le\dfrac{23}{6} .\end{array} The red graph is the graph of the solution set. In interval notation, the solution set is $ \left( -\infty,\dfrac{23}{6} \right] .$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.