# What is the solution set for abs(4x – 3) – 2 > 3?

Mar 23, 2018

$\left(- \infty , - \frac{1}{2}\right) \cup \left(2 , \infty\right)$

#### Explanation:

If we look at the definition of absolute value:

$| a | = a \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ if and only if $\setminus \setminus \setminus a \ge 0$

$| a | = - a \setminus \setminus \setminus$ if and only if $\setminus \setminus \setminus a < 0$

It follows from this, that we need to solve both:

$4 x - 3 - 2 > 3$ and $- \left(4 x - 3\right) - 2 > 3$

$4 x - 3 - 2 > 3$

$4 x - 5 > 3$

$x > \frac{8}{4}$

$\textcolor{b l u e}{x > 2}$

$- \left(4 x - 3\right) - 2 > 3$

$- 4 x + 3 - 2 > 3$

$- 4 x > 2$

$\textcolor{b l u e}{x < - \frac{1}{2}}$

This gives us a union of intervals:

$\left(- \infty , - \frac{1}{2}\right) \cup \left(2 , \infty\right)$