Intermediate Algebra (12th Edition)

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

Chapter 1 - Review Exercises - Page 131: 42


$\left( \dfrac{59}{31},\infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \dfrac{5}{3}(x-2)+\dfrac{2}{5}(x+1)\gt1 ,$ 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 $\ge.$ $\bf{\text{Solution Details:}}$ Multiplying both sides by the $LCD=15,$ the inequality above is equivalent to \begin{array}{l}\require{cancel} 15\left( \dfrac{5}{3}(x-2)+\dfrac{2}{5}(x+1) \right) \gt 15(1) \\\\ 25(x-2)+6(x+1)\gt 15(1) .\end{array} Using the Distributive Property and the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 25(x)+25(-2)+6(x)+6(1)\gt 15(1) \\\\ 25x-50+6x+6\gt 15 \\\\ 25x+6x\gt 15+50-6 \\\\ 31x\gt 59 \\\\ x\gt \dfrac{59}{31} .\end{array} In interval notation, the solution set is $ \left( \dfrac{59}{31},\infty \right) .$
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