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: 52


$\left( -\dfrac{5}{2},-\dfrac{1}{2} \right]$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ 4 \le -2x+3 \lt 8 ,$ 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 $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 4-3 \le -2x+3-3 \lt 8-3 \\\\ 1 \le -2x \lt 5 .\end{array} Dividing all sides by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{-2} \ge \dfrac{-2x}{-2} \gt \dfrac{5}{-2} \\\\ -\dfrac{1}{2} \ge x \gt -\dfrac{5}{2} \\\\ -\dfrac{5}{2} \lt x \le -\dfrac{1}{2} .\end{array} In interval notation, the solution set is $ \left( -\dfrac{5}{2},-\dfrac{1}{2} \right] .$ The red graph is the graph of the solution set.
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