Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Review - Page 113: 78

Answer

$\left( -\infty, -4 \right]\cup\left[ 1,\infty \right)$

Work Step by Step

Since for any $a\gt0$, $|x|\ge a$ implies $x\ge a\text{ or } x\le-a$, then the expression, $ \left| 6+4x \right|\ge10 ,$ is equivalent to \begin{array}{l}\require{cancel} 6+4x\ge10 \\\\ 4x\ge10-6 \\\\ 4x\ge4 \\\\ x\ge\dfrac{4}{4} \\\\ x\ge1 ,\\\\\text{ OR }\\\\ 6+4x\le-10 \\\\ 4x\le-10-6 \\\\ 4x\le-16 \\\\ x\le-\dfrac{16}{4} \\\\ x\le-4 .\end{array} Hence, the solution set is $ \left( -\infty, -4 \right]\cup\left[ 1,\infty \right) .$
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