Intermediate Algebra (12th Edition)

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

Chapter 1 - Section 1.7 - Absolute Value Equations and Inequalities - 1.7 Exercises: 68

Answer

$\left( -\infty, \dfrac{3}{5} \right] \cup \left[ 1,\infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ |8-10x| \ge 2 ,$ use the definition of absolute value inequalities. Then use the properties of inequality to isolate the variable. 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 $\ge.$ $\bf{\text{Solution Details:}}$ 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} 8-10x \ge 2 \\\\\text{OR}\\\\ 8-10x \le -2 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} 8-10x \ge 2 \\\\ -10x \ge 2-8 \\\\ -10x \ge -6 \\\\\text{OR}\\\\ 8-10x \le -2 \\\\ -10x \le -2-8 \\\\ -10x \le -10 .\end{array} Dividing by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} -10x \ge -6 \\\\ x \le \dfrac{-6}{-10} \\\\ x \le \dfrac{3}{5} \\\\\text{OR}\\\\ -10x \ge -10 \\\\ x \ge \dfrac{-10}{-10} \\\\ x \ge 1 .\end{array} In interval notation, the solution set is $ \left( -\infty, \dfrac{3}{5} \right] \cup \left[ 1,\infty \right) .$
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