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 - Page 168: 49


The solution is $\Big(-\dfrac{3}{2},\dfrac{13}{10}\Big)$

Work Step by Step

$\Big|5x+\dfrac{1}{2}\Big|-2\lt5$ Take $2$ to the right side of the inequality: $\Big|5x+\dfrac{1}{2}\Big|\lt5+2$ $\Big|5x+\dfrac{1}{2}\Big|\lt7$ Solving this absolute value inequality is equivalent to solving the following inequality: $-7\lt5x+\dfrac{1}{2}\lt7$ $\textbf{Solve the inequality shown above:}$ $-7\lt5x+\dfrac{1}{2}\lt7$ Multiply the whole equation by $2$: $2\Big(-7\lt5x+\dfrac{1}{2}\lt7\Big)$ $-14\lt10x+1\lt14$ Subtract $1$ from all three parts of the inequality: $-14-1\lt10x+1-1\lt14-1$ $-15\lt10x\lt13$ Divide all three parts of the inequality by $10$: $-\dfrac{15}{10}\lt\dfrac{10x}{10}\lt\dfrac{13}{10}$ $-\dfrac{3}{2}\lt x\lt\dfrac{13}{10}$ Expressing the solution in interval notation: $\Big(-\dfrac{3}{2},\dfrac{13}{10}\Big)$
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