# How do you solve and write the following in interval notation: |x + 2| >16?

Feb 1, 2018

Solution: $x > 14 \mathmr{and} x < - 18 \mathmr{and} x | \left(- \infty , - 18\right) \cup \left(14 , \infty\right)$

#### Explanation:

$| x + 2 | > 16 \mathmr{and} x + 2 > 16 \mathmr{and} x > 14$ OR

$| x + 2 | > 16 \mathmr{and} x + 2 < - 16 \mathmr{and} x < - 18$

Solution: $x < - 18 \mathmr{and} x > 14$

In interval notation : $x | x | \left(- \infty , - 18\right) \cup \left(14 , \infty\right)$ [Ans]