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: 23


$\left( -\infty,-\dfrac{15}{2} \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \dfrac{2x-5}{-4}\gt5 ,$ 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 $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Multiplying both sides by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} -4\cdot\left( \dfrac{2x-5}{-4} \right)\lt-4\cdot5 \\\\ 2x-5\lt-20 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 2x\lt-20+5 \\\\ 2x\lt-15 \\\\ x\lt-\dfrac{15}{2} .\end{array} The red graph is the graph of the solution set. In interval notation, the solution set is $ \left( -\infty,-\dfrac{15}{2} \right) .$
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