Answer
$x\geq\frac{c-b}{x}$ or $x\leq\frac{-c-b}{x}$.
Work Step by Step
$|ax+b|\geq c$ implies $ax+b\geq c\\ax\geq c-b\\x\geq\frac{c-b}{x}$ or $ax+b\leq -c\\ax\leq -c-b\\x\leq\frac{-c-b}{x}$.
Thus $x\geq\frac{c-b}{a}$ or $x\leq\frac{-c-b}{a}$.
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 1, Equations and Inequalities - 1.7 Solve Absolute Value Equations and Inequalities - 1.7 Exercises - Skill Practice - Page 56: 71
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