Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapters R-5 - Cumulative Review Exercises - Page 363: 8

Answer

$(2,3)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ 2x+4\lt10 \text{ and } 3x-1\gt5 ,$ solve each inequality separately. Since the conjunction "AND" is used, the solution set is the set of numbers common to both inequalities. $\bf{\text{Solution Details:}}$ Using the properties of inequality to solve each inequality separately results to \begin{array}{l}\require{cancel} 2x+4\lt10 \\\\ 2x\lt10-4 \\\\ 2x\lt6 \\\\ x\lt\dfrac{6}{2} \\\\ x\lt3 \\\\\text{AND}\\\\ 3x-1\gt5 \\\\ 3x\gt5+1 \\\\ 3x\gt6 \\\\ x\gt\dfrac{6}{3} \\\\ x\gt2 .\end{array} Since "AND" is used, then the solution set is the set of numbers common to both inequalities. Hence, the solution set is the interval $ (2,3) .$
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