# How do you solve m/(m-5)+3/(m-1)>0 using a sign chart?

Apr 11, 2017

The solution is $m \in \left(- \infty , - 5\right) \cup \left(1 , 3\right) \cup \left(5 , + \infty\right)$

#### Explanation:

Let's do some simplification

$\frac{m}{m - 5} + \frac{3}{m - 1} > 0$

$\frac{{m}^{2} - m + 3 m - 15}{\left(m - 5\right) \left(m - 1\right)} > 0$

$\frac{{m}^{2} + 2 m - 15}{\left(m - 5\right) \left(m - 1\right)} > 0$

$\frac{\left(m + 5\right) \left(m - 3\right)}{\left(m - 5\right) \left(m - 1\right)} > 0$

Let $f \left(m\right) = \frac{\left(m + 5\right) \left(m - 3\right)}{\left(m - 5\right) \left(m - 1\right)}$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$m$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 5$$\textcolor{w h i t e}{a a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$3$$\textcolor{w h i t e}{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}$$m + 5$$\textcolor{w h i t e}{a a a a}$$-$$\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}$$+$$\textcolor{w h i t e}{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 a a}$$m - 1$$\textcolor{w h i t e}{a a a a}$$-$$\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}$$+$$\textcolor{w h i t e}{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 a a}$$m - 3$$\textcolor{w h i t e}{a a a a}$$-$$\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}$$-$$\textcolor{w h i t e}{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 a a}$$m - 5$$\textcolor{w h i t e}{a a a a}$$-$$\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}$$-$$\textcolor{w h i t e}{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 a a}$$f \left(m\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\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}$$+$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

So,

$f \left(m\right) > 0$ when $m \in \left(- \infty , - 5\right) \cup \left(1 , 3\right) \cup \left(5 , + \infty\right)$