# How do you solve abs(x-a)>b?

Jan 25, 2017

$x > a + b$, when $b \ge 0.$
$x < a - b$, when $b < 0$

#### Explanation:

Despite that $| x - a | \ge 0$, b could be negative.

So, I allow negative b also.

The inequality breaks up to

$x - a > b$, giving $x > a + b$, for $x \ge a \to b \ge 0$ and

$- \left(x - a\right) > b$, giving $x < a - b$, for $x \le a \to b < 0$