# How do you solve and graph the compound inequality 3x + 8 < 2 or x + 12 > 2 - x?

Jul 9, 2015

$3 x + 8 < 2$ or $x + 12 > 2 - x$
for all possible Real values of $x$.

Draw a solid number line or shade the entire Cartesian plane for the solution set of $x$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXXX}}$$3 x + 8 < 2$
or
[2]$\textcolor{w h i t e}{\text{XXXX}}$$x + 12 > 2 - x$

From [1]
$\textcolor{w h i t e}{\text{XXXX}}$$3 x < - 6$
$\textcolor{w h i t e}{\text{XXXX}}$$x < - 2$

From [2]
$\textcolor{w h i t e}{\text{XXXX}}$$2 x > - 10$
$\textcolor{w h i t e}{\text{XXXX}}$$x > - 5$

So the solution set for $x$ all all values for which
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{red}{x < - 2}$ or $\textcolor{b l u e}{x > - 5}$

Any value that is not $< - 2$
will be $> - 5$
(and visa versa).