Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 1, Equations and Inequalities - 1.7 Solve Absolute Value Equations and Inequalities - 1.7 Exercises - Problem Solving - Page 57: 79

Answer

$|x-45|\lt15$

Work Step by Step

If $a\lt x\lt b$, we can turn this into an absolute value inequality in the following way: Take the mean of $a$ and $b$: $\frac{a+b}{2}$. This will be equidistant from $a$ and $b$. Thus if we take x's difference from this, it must be within $|a-\frac{a+b}{2}|=|b-\frac{a+b}{2}|$. Thus the inequality will be: $|x-\frac{a+b}{2}|\lt|a-\frac{a+b}{2}|$. Hence here the inequality is: $|x-\frac{30+60}{2}|\lt|30-\frac{30+60}{2}|\\|x-45|\lt15$
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.