# How do you solve | 3 + 2x | >| 4 - x |?

Jul 22, 2017

$x < - 7 \mathmr{and} x > \frac{1}{3}$

#### Explanation:

$\left\mid 3 + 2 x \right\mid > \left\mid 4 - x \right\mid$
Since both sides are positive
You can do the square at both sides to get rid of the abs sign
$\because {\left(\left\mid x \right\mid\right)}^{2} = {\left(x\right)}^{2}$

${\left(\left\mid 3 + 2 x \right\mid\right)}^{2} > {\left(\left\mid 4 - x \right\mid\right)}^{2}$
${\left(3 + 2 x\right)}^{2} > {\left(4 - x\right)}^{2}$
$9 + 12 x + 4 {x}^{2} > 16 - 8 x + {x}^{2}$
$3 {x}^{2} + 20 x - 7 > 0$
$\left(3 x - 1\right) \left(x + 7\right) > 0$

$\textcolor{w h i t e}{\ldots .}$+$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .}$$-$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots}$$+$
<--------$\circ$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots .}$$\circ$-------------->
----------/--------------------------/-----------------
$\textcolor{w h i t e}{\ldots . .}$$- 7$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots . .}$$\frac{1}{3}$

$\therefore x < - 7 \mathmr{and} x > \frac{1}{3}$