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: 52


$\left( -\infty, 1 \right] \cup \left[ 3, \infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ |12-6x|+3\ge9 ,$ use the properties of equality to isolate the absolute value expression. Then use the properties of absolute value inequality. $\bf{\text{Solution Details:}}$ Using the properties of inequality, the statement above is equivalent to \begin{array}{l}\require{cancel} |12-6x|\ge9-3 \\\\ |12-6x|\ge6 .\end{array} Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} 12-6x\ge6 \\\\\text{OR}\\\\ 12-6x\le-6 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 12-6x\ge6 \\\\ -6x\ge6-12 \\\\ -6x\ge-6 \\\\ x\le\dfrac{-6}{-6} \text{ (reverse the sign)}\\\\ x\le1 \\\\\text{OR}\\\\ 12-6x\le-6 \\\\ -6x\le-6-12 \\\\ -6x\le-18 \\\\ x\ge\dfrac{-18}{-6} \text{ (reverse the sign)}\\\\ x\ge3 .\end{array} Hence, the solution set is the interval $ \left( -\infty, 1 \right] \cup \left[ 3, \infty \right) .$
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