Question

What is the solution set for `abs(4x – 3) – 1 > 12`?

Answer 1

`x in (-oo, -5/2) uu (4, +oo)`

Explanation:

Start by isolating the modulus on one side of the inequality. This can be done by adding `1` to both sides

`|4x-3| - color(red)(cancel(color(black)(1))) + color(red)(cancel(color(black)(1))) > 12 + 1`

`|4x-3| > 13`

Since you're dealing with absolute value, you need to take into account the fact that the expression inside the modulus can be negative or positive

• `4x-3>0 implies |4x-3| = 4x-3`

This means that your inequality can be written as

`4x - 3 > 13`

`4x > 16 implies x > 4`

• `4x-3<0 implies |4x-3| = -(4x-3)`

This time, you have

`-(4x-3) > 13`

`-4x + 3 > 13`

`-4x > 10 implies x < -5/2`

What this tells you is that for any value of `x` that is bigger than `4` or smaller than `-5/2`, this inequality will be true. This means that the solution set will will `x in (-oo, -5/2) uu (4, +oo)`.