# What is the solution set for abs(x – 2) > 3?

Sep 5, 2015

$x \in \left(- \infty , - 1\right) \cup \left(5 , + \infty\right)$

#### Explanation:

When you're dealing with absolute value inequalities, you need to take into account the fact that, for real numbers, the absolute value function returns a positive value regardless of the sign of the number that's inside the modulus.

This means that you have two cases to examine, one in which the expression inside the modulus is positive, and the other in which the expression inside the modulus would be negative.

• $x - 2 > 0 \implies | x - 2 | = x - 2$

The inequality becomes

$x - 2 > 3 \implies x > 5$

• $x - 2 < 0 \implies | x - 2 | = - \left(x - 2\right)$

This time you have

$- \left(x - 2\right) > 3$

$- x + 2 > 3$

$- x > 1 \implies x < - 1$

So, for any value of $x$ that is greater than $5$ or smaller than $\left(- 1\right)$, the inequality will be satisfied. This means that the solution set will be $\left(- \infty , - 1\right) \cup \left(5 , + \infty\right)$.