# Question #c6156

May 28, 2017

The solution is $x \in \left(2 , \frac{7}{3}\right) \cup \left(3. + \infty\right)$

#### Explanation:

${x}^{2} - 5 x + 6 = \left(x - 2\right) \left(x - 3\right)$

Let's simplify the inequality, without crosing over

$\frac{3}{{x}^{2} - 5 x + 6} + \frac{4 - x}{3 - x} > \frac{6 - x}{2 - x}$

$\frac{3}{\left(x - 2\right) \left(x - 3\right)} - \frac{4 - x}{x - 3} + \frac{6 - x}{x - 2} > 0$

$\frac{3 - \left(4 - x\right) \left(x - 2\right) + \left(6 - x\right) \left(x - 3\right)}{\left(x - 2\right) \left(x - 3\right)} > 0$

$\frac{3 - 4 x + 8 + {x}^{2} - 2 x + 6 x - 18 - {x}^{2} + 3 x}{\left(x - 2\right) \left(x - 3\right)} > 0$

$\frac{3 x - 7}{\left(x - 2\right) \left(x - 3\right)} > 0$

Let $f \left(x\right) = \frac{3 x - 7}{\left(x - 2\right) \left(x - 3\right)}$

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 a a}$$2$$\textcolor{w h i t e}{a a a a a a}$$\frac{7}{3}$$\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}$$+ \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 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 a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$3 x - 7$$\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 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}$$-$$\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}$$-$$\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}$$-$$\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}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) > 0$, when $x \in \left(2 , \frac{7}{3}\right) \cup \left(3. + \infty\right)$
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