# Question #bfc27

##### 1 Answer
Apr 15, 2017

The solution is $x \in \left(1 , \frac{10}{3}\right)$

#### Explanation:

We cannot do crossing over

So, let's rearrange the inequality

$\frac{2 x + 5}{x - 1} > 5$

$\frac{2 x + 5}{x - 1} - 5 > 0$

$\frac{\left(2 x + 5\right) - 5 \left(x - 1\right)}{x - 1} > 0$

$\frac{2 x + 5 - 5 x + 5}{x - 1} > 0$

$\frac{10 - 3 x}{x - 1} > 0$

Let $f \left(x\right) = \frac{10 - 3 x}{x - 1}$

We can build a 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}$$1$$\textcolor{w h i t e}{a a a a a a a}$$\frac{10}{3}$$\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}$$-$$\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 a a}$$10 - 3 x$$\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 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}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$

Therefore,

$f \left(x\right) > 0$ when $x \in \left(1 , \frac{10}{3}\right)$