# How do you solve |3x + 6| >15?

Mar 28, 2018

$\left(- \infty , - 7\right) \cup \left(3 , \infty\right)$

#### Explanation:

The absolute value is defined as:

$| a | = a \textcolor{w h i t e}{888888}$ if and only if $\textcolor{w h i t e}{88888} a \ge 0$

$| a | = - a \textcolor{w h i t e}{.888}$ if and only if $\textcolor{w h i t e}{888888} a < 0$

Because of this we need to solve both:

$3 x + 6 > 15$ and $- \left(3 x + 6\right) > 15$

I

$3 x + 6 > 15$

$3 x > 9$

$x > 3$

II

$- \left(3 x + 6\right) > 15$

$3 x + 6 < - 15$

$3 x < - 21$

$x < - 7$

This gives us:

$- \infty < x < - 7$

and

$3 < x < \infty$

We can express this as a union of intervals:

$\left(- \infty , - 7\right) \cup \left(3 , \infty\right)$