Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.8 Absolute Value Equations and Inequalities - 1.8 Exercises: 40

Answer

The solution is $\Big(-\infty,\dfrac{26}{9}\Big)\cup\Big(\dfrac{34}{9},\infty\Big)$

Work Step by Step

$\Big|\dfrac{5}{3}-\dfrac{1}{2}x\Big|\gt\dfrac{2}{9}$ Solving this absolute value inequality is equivalent to solving two separate inequalities, which are: $\dfrac{5}{3}-\dfrac{1}{2}x\gt\dfrac{2}{9}$ and $\dfrac{5}{3}-\dfrac{1}{2}x\lt-\dfrac{2}{9}$ $\textbf{Solve the first inequality:}$ $\dfrac{5}{3}-\dfrac{1}{2}x\gt\dfrac{2}{9}$ Multiply the whole inequality by $18$: $18\Big(\dfrac{5}{3}-\dfrac{1}{2}x\gt\dfrac{2}{9}\Big)$ $30-9x\gt4$ Take $30$ to the right side: $-9x\gt4-30$ $-9x\gt-26$ Take $-9$ to divide the right side and reverse the direction of the inequality sign: $x\lt\dfrac{-26}{-9}$ $x\lt\dfrac{26}{9}$ $\textbf{Solve the second inequality:}$ $\dfrac{5}{3}-\dfrac{1}{2}x\lt-\dfrac{2}{9}$ Multiply the whole inequality by $18$: $18\Big(\dfrac{5}{3}-\dfrac{1}{2}x\lt-\dfrac{2}{9}\Big)$ $30-9x\lt-4$ Take $30$ to the right side: $-9x\lt-4-30$ $-9x\lt-34$ Take $-9$ to divide the right side and reverse the direction of the inequality sign: $x\gt\dfrac{-34}{-9}$ $x\gt\dfrac{34}{9}$ Expressing the solution in interval notation: $\Big(-\infty,\dfrac{26}{9}\Big)\cup\Big(\dfrac{34}{9},\infty\Big)$
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