# Solve the following

Question:

Solve $\left|\frac{2 x-1}{x-1}\right|>2$

Solution:

As, $\left|\frac{2 x-1}{x-1}\right|>2$

$\Rightarrow \frac{2 x-1}{x-1}<-2$ or $\frac{2 x-1}{x-1}>2 \quad($ As, $|x|>2 \Rightarrow x<-2$ or $x>2)$

$\Rightarrow \frac{2 x-1}{x-1}+2<0$ or $\frac{2 x-1}{x-1}-2>0$

$\Rightarrow \frac{2 x-1+2 x-2}{x-1}<0$ or $\frac{2 x-1-2 x+2}{x-1}>0$

$\Rightarrow \frac{4 x-3}{x-1}<0$ or $\frac{1}{x-1}>0$

$\Rightarrow \frac{4 x-3}{x-1}<0$ or $x-1>0$

$\Rightarrow[(4 x-3>0$ and $x-1<0)$ or $(4 x-2<0$ and $x-1>0)]$ or $[x-1>0]$

$\Rightarrow\left[\left(x>\frac{3}{4}\right.\right.$ and $\left.x<1\right)$ or $\left(x<\frac{3}{4}\right.$ and $\left.\left.x>1\right)\right]$ or $[x>1]$

$\Rightarrow\left[\left(\frac{3}{4}1]$

$\Rightarrow\left[\frac{3}{4}$\Rightarrow \frac{3}{4}1\therefore x \in\left(\frac{3}{4}, 1\right) \cup(1, \infty)\$