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 - Page 100: 34


$\left( -\infty,\lt\dfrac{71}{150} \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \dfrac{7}{5}(10x-1)\lt\dfrac{2}{3}(6x+5) ,$ 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, the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{7}{5}(10x)+\dfrac{7}{5}(-1)\lt\dfrac{2}{3}(6x)+\dfrac{2}{3}(5) \\\\ 14x-\dfrac{7}{5}\lt4x+\dfrac{10}{3} .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 15\left( 14x-\dfrac{7}{5} \right) \lt15\left( 4x+\dfrac{10}{3} \right) \\\\ 210x-21\lt60x+50 \\\\ 210x-60x\lt50+21 \\\\ 150x\lt71 \\\\ x\lt\dfrac{71}{150} .\end{array} The red graph is the graph of the solution set. In interval notation, the solution set is $ \left( -\infty,\lt\dfrac{71}{150} \right) .$
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