Intermediate Algebra (6th Edition)

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

Chapter 2 - Section 2.7 - Absolute Value Inequalities - Exercise Set - Page 106: 68

Answer

$x=\left\{ -4,\dfrac{17}{3} \right\}$

Work Step by Step

Since $|x|=a$ implies $x=a \text{ OR } x=-a$, then the statement $ |5-6x|=29 $ evaluates to \begin{array}{l} 5-6x=29\\ -6x=24\\ x=-4 ,\\\\\text{OR}\\\\ 5-6x=-29\\ -6x=-34\\\\ x=\dfrac{34}{6}\\\\ x=\dfrac{17}{3} .\end{array} Hence, the solution set is $ x=\left\{ -4,\dfrac{17}{3} \right\} $.
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