Question

How do you solve `|3r - 1| \leq 11`?

Answer 1

`-10/3 <= r <= 4`

Explanation:

Using the property of absolute values that

`absx <= a => -a <= x <= a`

we get:

`-11 <= 3r - 1 <= 11 => -10 <= 3r <= 12 => -10/3 <= r <= 4`