# How do you solve and write the following in interval notation: 12/ (x-5) > 10 /(x+1)?

Jun 8, 2017

The solution is $x \in \left(- 31 , - 1\right) \cup \left(5 , + \infty\right)$

#### Explanation:

We cannot do crossing over.

Rearrange the inequality

$\frac{12}{x - 5} > \frac{10}{x + 1}$

$\frac{12}{x - 5} - \frac{10}{x + 1} > 0$

$\frac{12 \left(x + 1\right) - 10 \left(x - 5\right)}{\left(x - 5\right) \left(x + 1\right)} > 0$

$\frac{12 x + 12 - 10 x + 50}{\left(x - 5\right) \left(x + 1\right)} > 0$

$\frac{2 x + 62}{\left(x - 5\right) \left(x + 1\right)} > 0$

$\frac{2 \left(x + 31\right)}{\left(x - 5\right) \left(x + 1\right)} > 0$

Let $f \left(x\right) = \frac{2 \left(x + 31\right)}{\left(x - 5\right) \left(x + 1\right)}$

We can 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}$$- 31$$\textcolor{w h i t e}{a a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a a a a}$$5$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 31$$\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}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\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}$$+$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

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

Therefore,

$f \left(x\right) > 0$, when $x \in \left(- 31 , - 1\right) \cup \left(5 , + \infty\right)$