# How do you solve and write the following in interval notation: 2 + | x/3 - 1 | > 6?

Apr 27, 2017

$\left(- \setminus \infty , - 9\right) \setminus \cup \left(15 , \setminus \infty\right)$

#### Explanation:

First, subtract two from both sides to get
$\setminus \left\mid \setminus \frac{x}{3} - 1 \right\mid > 4$

The absolute value can be split into
$\setminus \frac{x}{3} - 1 > 4$
$- \left(\setminus \frac{x}{3} - 1\right) > 4$

Simplifying for the first equation we get the following
$\setminus \frac{x}{3} > 5$
$x > 15$

For the second equation
$1 - \setminus \frac{x}{3} > 4$
$- \setminus \frac{x}{3} > 3$
$\setminus \frac{x}{3} < - 3$
$x < - 9$

So all values less than -9 and greater than 15 satisfy the inequality. This can be written as $\left(- \setminus \infty , - 9\right) \setminus \cup \left(15 , \setminus \infty\right)$.

Open brackets are used since infinity is not included as a point and 9 and 15 do not satisfy the inequality.