# How do you solve 3abs(x+2)-2>7?

Jul 5, 2015

$x < - 5$
or
$x > 1$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXXX}}$$3 \left\mid x + 2 \right\mid - 2 > 7$
Adding 2 to both sides maintains the inequality, so
$\textcolor{w h i t e}{\text{XXXX}}$$3 \left\mid x + 2 \right\mid > 9$
Dividing both sides by a positive amount also maintains the inequality, so
$\textcolor{w h i t e}{\text{XXXX}}$$\left\mid x + 2 \right\mid > 3$

There are two possibilities.

Possibility 1: $\left(x + 2\right) < 0 \rightarrow x < - 2$
In this case
$\textcolor{w h i t e}{\text{XXXX}}$$\left\mid x + 2 \right\mid = - \left(x + 2\right)$
So
$\textcolor{w h i t e}{\text{XXXX}}$$\left\mid x + 2 \right\mid > 3$
becomes
$\textcolor{w h i t e}{\text{XXXX}}$$- x - 2 > 3$
Adding 2 to both sides (which maintains the inequality)
$\textcolor{w h i t e}{\text{XXXX}}$$- x > 5$
Multiplying by $\left(- 1\right)$ (remembering that multiplication by a negative reverses the inequality)
$\textcolor{w h i t e}{\text{XXXX}}$$x < - 5$
The two conditions for Possibility 1: $x < - 2$ and $x < - 5$ simplify to
$\textcolor{w h i t e}{\text{XXXX}}$$x < - 5$

Possibility 2: $\left(x + 2\right) \ge 0 \rightarrow x \ge - 2$
In this case
$\textcolor{w h i t e}{\text{XXXX}}$$\left\mid x + 2 \right\mid = x + 2$
So
$\textcolor{w h i t e}{\text{XXXX}}$$\left\mid x + 2 \right\mid > 3$
becomes
$\textcolor{w h i t e}{\text{XXXX}}$$x + 2 > 3$
Subtracting 2 from both sides (inequality maintained)
$\textcolor{w h i t e}{\text{XXXX}}$$x > 1$
The two conditions for Possibility 2: $x \ge - 2$ and $x > 1$ simplify to
$\textcolor{w h i t e}{\text{XXXX}}$$x > 1$

Combining
Either Possibility 1 or Possibility 2
so
$\textcolor{w h i t e}{\text{XXXX}}$$x < - 5$ or $x > 1$