College Algebra (11th Edition)

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

Chapter 1 - Test: 29

Answer

$\left( -\infty,-6 \right]\cup\left[ 5,\infty \right) $

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ |2x+1|-11\ge0 ,$ isolate first the absolute value expression. Then use the definition of absolute value inequalities. $\bf{\text{Solution Details:}}$ Using the properties of inequality, the given inequality is equivalent to \begin{array}{l}\require{cancel} |2x+1|\ge0+11 \\\\ |2x+1|\ge11 .\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} 2x+1\ge11 \\\\\text{OR}\\\\ 2x+1\le-11 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 2x+1\ge11 \\\\ 2x\ge11-1 \\\\ 2x\ge10 \\\\ x\ge\dfrac{10}{2} \\\\ x\ge5 \\\\\text{OR}\\\\ 2x+1\le-11 \\\\ 2x\le-11-1 \\\\ 2x\le-12 \\\\ x\le-\dfrac{12}{2} \\\\ x\le-6 .\end{array} Hence, the solution set is the interval $ \left( -\infty,-6 \right]\cup\left[ 5,\infty \right) .$
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