# How do you solve 3/(x-2)<5/(x+2) using a sign chart?

Mar 6, 2017

The solution is x in ]-2,2 [ uu ]8, +oo [

#### Explanation:

We cannot do crossing over

Let's simplify the inequality

$\frac{3}{x - 2} < \frac{5}{x + 2}$

$\frac{3}{x - 2} - \frac{5}{x + 2} < 0$

$\frac{3 \left(x + 2\right) - 5 \left(x - 2\right)}{\left(x - 2\right) \left(x + 2\right)} < 0$

$\frac{3 x + 6 - 5 x + 10}{\left(x - 2\right) \left(x + 2\right)} < 0$

$\frac{16 - 2 x}{\left(x - 2\right) \left(x + 2\right)} < 0$

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

Let $f \left(x\right) = \frac{2 \left(8 - x\right)}{\left(x - 2\right) \left(x + 2\right)}$

We, now, 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}$$- 2$$\textcolor{w h i t e}{a a a a a a a a a}$$2$$\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}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a 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 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 a a}$$x - 2$$\textcolor{w h i t e}{a a 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 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 a a}$$8 - x$$\textcolor{w h i t e}{a a 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 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 a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a 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 a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

Therefore,

$f \left(x\right) < 0$ when x in ]-2,2 [ uu ]8, +oo [