College Algebra (11th Edition)

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

Chapter 1 - Review Exercises: 122

Answer

$\left( -\infty,-\dfrac{11}{7} \right)\cup\left( -\dfrac{5}{7},\infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ |7x+8|-6\gt-3 ,$ use the properties of inequality to isolate the absolute value expression. Then use the definition of absolute value inequalities. $\bf{\text{Solution Details:}}$ Using the properties of inequality, the given expression is equivalent to \begin{array}{l}\require{cancel} |7x+8|\gt-3+6 \\\\ |7x+8|\gt3 .\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} 7x+8\gt3 \\\\\text{OR}\\\\ 7x+8\lt-3 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 7x+8\gt3 \\\\ 7x\gt3-8 \\\\ 7x\gt-5 \\\\ x\gt\dfrac{-5}{7} \\\\ x\gt-\dfrac{5}{7} \\\\\text{OR}\\\\ 7x+8\lt-3 \\\\ 7x\lt-3-8 \\\\ 7x\lt-11 \\\\ x\lt-\dfrac{11}{7} .\end{array} Hence, the solution set is the interval $ \left( -\infty,-\dfrac{11}{7} \right)\cup\left( -\dfrac{5}{7},\infty \right) .$
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