# How do you solve (-x-3)/(x+2)<=0?

Jul 27, 2016

$\textcolor{g r e e n}{x > - 2} \mathmr{and} \textcolor{g r e e n}{x \le - 3}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \frac{- x - 3}{x + 2} \le 0$

Provided $x \ne - 2$ we need to consider two possibilities:

Case 1. If $\textcolor{b l a c k}{x < - 2}$
Since $\left(x + 2\right) < 0$ we need to reverse the inequality when dividing by $\left(x + 2\right)$
$\textcolor{w h i t e}{\text{XXX}} - x - 3 \ge 0$

$\textcolor{w h i t e}{\text{XXX}} - x \ge 3$

$\textcolor{w h i t e}{\text{XXX}} x \le - 3$
$\textcolor{w h i t e}{\text{XXX}}$Considering the two restrictions $x < - 2$ and $x \le - 3$
$\textcolor{w h i t e}{\text{XXX}}$we employ the more restrictive:
$\textcolor{w h i t e}{\text{XXXXXX}} \textcolor{g r e e n}{x \le - 3}$

Case 2. If $\textcolor{b l a c k}{x > - 2}$
Since $\left(x + 2\right) > 0$ we can divide by $\left(x + 2\right)$ without effecting the inequality
$\textcolor{w h i t e}{\text{XXX}}$then -x-3 <= 0#

$\textcolor{w h i t e}{\text{XXX}} - x \le 3$

$\textcolor{w h i t e}{\text{XXX}} x \ge - 3$
$\textcolor{w h i t e}{\text{XXX}}$Again considering the two restrictions $x > - 2$ and $x \ge - 3$
$\textcolor{w h i t e}{\text{XXX}}$we employ the more restrictive
$\textcolor{w h i t e}{\text{XXXXXX}} \textcolor{g r e e n}{x > - 2}$

Here is a graph of the $\frac{- x - 3}{x + 2}$ to help verify this result: