# How do you solve compound inequalities p + 4 > 6 and 3p < -18?

Jul 16, 2015

$p + 4 > 6 \mathmr{and} 3 p < - 18$
$\textcolor{w h i t e}{\text{XXXX}}$$\Rightarrow 2 < p < - 6$ impossible
There are no values of $p$ that satisfy this pair of compound inequalities

#### Explanation:

Evaluate the two inequalities separately

$p + 4 > 6$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$Subtract 4 from each side
$\textcolor{w h i t e}{\text{XXXX}}$$p > 2$

$3 p < - 6$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$Divide each side by 3
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(this is valid without modifying the orientation of the inequality, since 3 > 0)
$\textcolor{w h i t e}{\text{XXXX}}$$p < - 6$

$p + 4 > 6$ and $3 p < - 18$
are those values of $p$ for which both conditions are true:

As we can see from the above number line,
there are no values for which both conditions are true.