# How do you solve x/(x-2)> -1/(x+3) using a sign chart?

Dec 10, 2017

The solution is $x \in \left(- \infty , - 4.45\right] \cup \left(- 3 , 0.45\right] \cup \left(2 , + \infty\right)$

#### Explanation:

Simplify the inequality, we cannot do crossing over.

$\frac{x}{x - 2} > - \frac{1}{x + 3}$

$\frac{x}{x - 2} + \frac{1}{x + 3} > 0$

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

$\frac{{x}^{2} + 3 x + x - 2}{\left(x - 2\right) \left(x + 3\right)} > 0$

$\frac{{x}^{2} + 4 x - 2}{\left(x - 2\right) \left(x + 3\right)} > 0$

The roots of the numerator

${x}^{2} + 4 x - 2 = 0$

are

$x = \frac{- 4 \pm \sqrt{16 - 4 \left(1\right) \left(- 2\right)}}{2}$

$= - 2 \pm \sqrt{6}$

${x}_{1} = - 2 - \sqrt{6} = - 4.45$

${x}_{2} = - 2 + \sqrt{6} = 0.45$

Let

$f \left(x\right) = \frac{\left(x - {x}_{1}\right) \left(x - {x}_{2}\right)}{\left(x - 2\right) \left(x + 3\right)}$

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}$${x}_{1}$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a}$${x}_{2}$$\textcolor{w h i t e}{a a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{1}$$\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 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}$$x - 3$$\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}$$+$$\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}$$x - {x}_{2}$$\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}$color(white)(aa)-$\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}$$x - 2$$\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}$color(white)(aa)-$\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}$$+$

$\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}$$+$$\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}$$+$$\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}$$+$

Therefore,

$f \left(x\right) > 0$ when $x \in \left(- \infty , - 4.45\right] \cup \left(- 3 , 0.45\right] \cup \left(2 , + \infty\right)$

graph{(x^2+4x-2)/((x-2)(x+3)) [-10, 10, -5, 5]}