College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.8 - Absolute Value Equations and Inequalities - 1.8 Exercises - Page 154: 65


$\left( -\infty, -\dfrac{2}{3} \right) \cup \left( -\dfrac{2}{3}, \infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ |3x+2|\gt0 ,$ exclude the solution to the equation $3x+2=0.$ $\bf{\text{Solution Details:}}$ The absolute value of $x$, written as $|x|,$ is the distance of $x$ to $0,$ and hence, is always a nonnegative number. In the same reason, for any $x,$ the left side of the given equation is always a nonnegative number. This is always greater than or equal to $0$ (the right-hand expression.) But since the solution required should be greater than zero, then exclude the solution to the equation $3x+2=0.$ That is, \begin{array}{l}\require{cancel} 3x+2\ne0 \\\\ 3x\ne-2 \\\\ \dfrac{3x}{3}\ne-\dfrac{2}{3} \\\\ x\ne-\dfrac{2}{3} .\end{array} Hence, the solution is the interval $ \left( -\infty, -\dfrac{2}{3} \right) \cup \left( -\dfrac{2}{3}, \infty \right) .$
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