# How do you solve p/(p-16)+2/(p-6)<=0 using a sign chart?

May 7, 2017

The solution is $p \in \left[- 4 , 6\right) \cup \left[8 , 16\right)$

#### Explanation:

Let 's simplify the $L H S$ of the inequality

$\frac{p}{p - 16} + \frac{2}{p - 6} \le 0$

$\frac{p \left(p - 6\right) + 2 \left(p - 16\right)}{\left(p - 16\right) \left(p - 6\right)} \le 0$

$\frac{{p}^{2} - 6 p + 2 p - 32}{\left(p - 16\right) \left(p - 6\right)} \le 0$

$\frac{{p}^{2} - 4 p - 32}{\left(p - 16\right) \left(p - 6\right)} \le 0$

$\frac{\left(p + 4\right) \left(p - 8\right)}{\left(p - 16\right) \left(p - 6\right)} \le 0$

Let $f \left(p\right) = \frac{\left(p + 4\right) \left(p - 8\right)}{\left(p - 16\right) \left(p - 6\right)}$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$p$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a a a}$$6$$\textcolor{w h i t e}{a a a a a a}$$8$$\textcolor{w h i t e}{a a a a a a a}$$16$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$p + 4$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$p - 6$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$p - 8$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$p - 16$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

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

Therefore,

$f \left(p\right) \le 0$ when $p \in \left[- 4 , 6\right) \cup \left[8 , 16\right)$