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

##### 1 Answer

Sep 5, 2015

#### 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| = -(x-2)#

This time you have

#-(x-2) > 3#

#-x + 2 > 3#

#-x > 1 implies x < -1#

So, for any value of **greater** than **smaller** than