# What is the solution set of |4x – 3| – 1 > 12?

### 2 Answers | Add Yours

l 4x-3 l -1 > 12

First we will add 1 to both sides.

==> l 4x-3 l > 13

Now by definition we will rewrite:

==> 4x-3 > 13 OR 4x-3 < -13

We will solve both cases:

==> 4x-3 > 13

==> 4x > 16

==> x > 4 ............(1)

==> 4x-3 < -13

==> 4x < -10

==> x < -10/4

==> x < -5/2 ............(2)

Then the solution is :

**x belongs to ( -inf, -5/2) U (4, inf)**

|4x – 3| – 1 > 12

First add 1 on both sides

By adding, you should get

**|4x - 3| > 13 **Since there are absolute value signs change your equation into

**4x - 3 > 13 and 4x - 3 < -13 **now add 3 on both sides to both equation

By adding, you should get

**4x > 16 and 4x < -10** now divide 4 on both sides

By dividing, you should get

**x > 4 and x < -10/4 **