# Question e97c3

Mar 14, 2017

The solution is x in ]-oo,1[ uu [4,+oo[

#### Explanation:

Let's simplify the equation

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

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

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

$\frac{4 - x}{x - 1} \le 0$

Let $f \left(x\right) = \frac{4 - x}{x - 1}$

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}$$1$$\textcolor{w h i t e}{a a a a a a a a}$$4$$\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}$$4 - x$$\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}$$f \left(x\right)$$\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,

$f \left(x\right) \le 0$, when x in ]-oo,1[ uu [4,+oo[#