## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.9 - Linear Inequalities and Absolute Value Inequalities - Exercise Set - Page 138: 65

#### Answer

$(-6,0)$

#### Work Step by Step

The solutions of $|u| \lt c$ are the numbers that satisfy $-c \lt u \lt c.$ The idea is to isolate $x$ in the middle. Adding, subtracting or multiplying/dividing with a positive number preserves order (the inequality symbols remain). Multiplying/dividing with a negative number inverts the order (the inequality symbols changes). --- $|\displaystyle \frac{2y+6}{3}| \lt 2\qquad$... is equivalent to ... $-2 \lt \displaystyle \frac{2y+6}{3} \lt 2\qquad$ ... multiply all parts with $3$ $-6 \lt 2y+6 \lt 6\qquad$ ... add $-6$ to all parts $-12 \lt 2y \lt 0 \qquad$ ... divide with $2$ $-6 \lt y \lt 0$ The interval borders are excluded from the interval. The solution set is $(-6,0)$

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