# How do solve (3-x)/(x+5)<=0 and write the answer as a inequality and interval notation?

##### 1 Answer
May 30, 2017

The solution is $x \in \left(- \infty , - 5\right) \cup \left[3 , + \infty\right)$ or
$x < - 5$ and $x \ge 3$

#### Explanation:

Let $f \left(x\right) = \frac{3 - x}{x + 5}$

We build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$- 5$$\textcolor{w h i t e}{a a a a a a a}$$3$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 5$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$3 - x$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left(- \infty , - 5\right) \cup \left[3 , + \infty\right)$ in interval notation or

$x < - 5$ and $x \ge 3$ as inequality