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 105: 24

Answer

Interval Notation: $(-∞,\frac{-11}{3})∪(-3,∞)$ Graph:

Work Step by Step

$|10+3x|+1 \gt 2$ By adding $-1$ to both sides of the inequality, isolate the absolute value expression. $|10+3x|+1-1 \gt 2-1$ $|10+3x| \gt 1$ Using absolute value inequality property, it is equivalent to $10+3x \lt -1$ or $ 10+3x \gt 1$ Add $-10$ to both sides of the inequalities, $10+3x-10 \lt -1-10 $ or $ 10+3x-10 \gt 1-10$ $3x \lt -11 $ or $ 3x \gt -9$ Divide both sides of the inequalities by $3$ $x \lt \frac{-11}{3} $ or $ x \gt \frac{-9}{3}$ $x \lt \frac{-11}{3} $ or $ x \gt -3$ Interval Notation: $(-∞,\frac{-11}{3})∪(-3,∞)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.