Intermediate Algebra (12th Edition)

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

Chapter 1 - Review Exercises: 39

Answer

$\left( \dfrac{3}{2},\infty \right]$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \dfrac{6x+3}{-4} \lt -3 ,$ 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 a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} -4\left(\dfrac{6x+3}{-4}\right) \gt -4(-3) \\\\ 6x+3 \gt 12 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 6x \gt 12-3 \\\\ 6x \gt 9 \\\\ x \gt \dfrac{9}{6} \\\\ x \gt \dfrac{3}{2} .\end{array} In interval notation, the solution set is $ \left( \dfrac{3}{2},\infty \right] .$
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