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

Aug 14, 2015

$x \in \left(- \infty , - \frac{5}{2}\right) \cup \left(4 , + \infty\right)$

#### Explanation:

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

$| 4 x - 3 | - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} > 12 + 1$

$| 4 x - 3 | > 13$

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

• $4 x - 3 > 0 \implies | 4 x - 3 | = 4 x - 3$

This means that your inequality can be written as

$4 x - 3 > 13$

$4 x > 16 \implies x > 4$

• $4 x - 3 < 0 \implies | 4 x - 3 | = - \left(4 x - 3\right)$

This time, you have

$- \left(4 x - 3\right) > 13$

$- 4 x + 3 > 13$

$- 4 x > 10 \implies x < - \frac{5}{2}$

What this tells you is that for any value of $x$ that is bigger than $4$ or smaller than $- \frac{5}{2}$, this inequality will be true. This means that the solution set will will $x \in \left(- \infty , - \frac{5}{2}\right) \cup \left(4 , + \infty\right)$.