# How do you solve |8r - 3| <67?

Dec 6, 2016

The solution to this problem is the interval:

$- 8 < r < \frac{35}{4}$

#### Explanation:

Because this is an absolute value inequality we must convert this to an inequality set and solve the inequality set while keeping the set balanced:

$- 67 < 8 r - 3 < 67$

$- 67 + 3 < 8 r - 3 + 3 < 67 + 3$

$- 64 < 8 r < 70$

$- \frac{64}{8} < \frac{8 r}{8} < \frac{70}{8}$

$- 8 < r < \left(\frac{2}{2}\right) \left(\frac{35}{4}\right)$

$- 8 < r < \frac{35}{4}$