# How do you solve x/(x+20)>2/(x+8) using a sign chart?

May 29, 2017

The solution is $x \in \left(- \infty , - 20\right) \cup \left(- 10 , - 8\right) \cup \left(4 + \infty\right)$

#### Explanation:

Let's rearrange the equation, we cannot do crossing over

$\frac{x}{x + 20} > \frac{2}{x + 8}$

$\frac{x}{x + 20} - \frac{2}{x + 8} > 0$

The LCD is $\left(x + 20\right) \left(x + 8\right)$

So,

$\frac{x \left(x + 8\right) - 2 \left(x + 20\right)}{\left(x + 20\right) \left(x + 8\right)} > 0$

$\frac{{x}^{2} + 8 x - 2 x - 40}{\left(x + 20\right) \left(x + 8\right)} > 0$

$\frac{{x}^{2} + 6 x - 40}{\left(x + 20\right) \left(x + 8\right)} > 0$

$\frac{\left(x + 10\right) \left(x - 4\right)}{\left(x + 20\right) \left(x + 8\right)} > 0$

Let $f \left(x\right) = \frac{\left(x + 10\right) \left(x - 4\right)}{\left(x + 20\right) \left(x + 8\right)}$

We can build the 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}$$- 20$$\textcolor{w h i t e}{a a a a}$$- 10$$\textcolor{w h i t e}{a a a a}$$- 8$$\textcolor{w h i t e}{a a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 20$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{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}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 10$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$$| |$$\textcolor{w h i t e}{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}$$+$

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

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

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

Therefore,

$f \left(x\right) > 0$ when $x \in \left(- \infty , - 20\right) \cup \left(- 10 , - 8\right) \cup \left(4 + \infty\right)$