# How do you solve x/(x-3)<=-8/(x-6) using a sign chart?

Mar 15, 2017

The solution is  x in [-6,3[ uu [4,6 [

#### Explanation:

We cannot do crossing over

$\frac{x}{x - 3} \le - \frac{8}{x - 6}$

We rearrange and factorise the inequality

$\frac{x}{x - 3} + \frac{8}{x - 6} \le 0$

$\frac{x \left(x - 6\right) + 8 \left(x - 3\right)}{\left(x - 3\right) \left(x - 6\right)} \le 0$

$\frac{{x}^{2} - 6 x + 8 x - 24}{\left(x - 3\right) \left(x - 6\right)} \le 0$

$\frac{{x}^{2} + 2 x - 24}{\left(x - 3\right) \left(x - 6\right)} \le 0$

$\frac{\left(x - 4\right) \left(x + 6\right)}{\left(x - 3\right) \left(x - 6\right)} \le 0$

Let $f \left(x\right) = \frac{\left(x - 4\right) \left(x + 6\right)}{\left(x - 3\right) \left(x - 6\right)}$

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}$$- 6$$\textcolor{w h i t e}{a a a a a a a}$$3$$\textcolor{w h i t e}{a 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}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 6$$\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 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}$$x - 3$$\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 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}$$x - 4$$\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 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}$$x - 6$$\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 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}$$f \left(x\right)$$\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 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}$$+$

Therefore,

$f \left(x\right) \le 0$ when  x in [-6,3[ uu [4,6 [