simple 2 sides inequality 
$$2<\frac{x}{x-1}\leq 3$$

Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and  $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$

you can rewrite $$2 < \frac x{x-1} = 1 + \frac 1{x-1} \le 3 \to 1 < \frac1 {x-1} \le 2 \tag 1$$  from the graph of $y = \frac1{x-1},$ we see that there are no solutions $(-\infty, 1)$ and $x - 1 > 0$ is necessary so $(1)$ is equivalent to $$\frac 12 \le x - 1< 1\to \frac32 \le x < 2. $$ 
