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 $1<x\leq\frac{3}{2}$

Why quadratic? Linear is sufficient. 
If $x > 1$, then $(x-1)$ is positive so we can multiply by that to get: $2(x-1) \lt x \le 3(x-1)$. Left-hand gives us $x < 2$ and right-hand gives us $x \ge \frac32$. Thus: $\frac32 \le x < 2$.
If $x < 1$, then we have to flip the signs and get $2(x-1) > x \ge 3(x-1)$. But there, the left-hand gives us $x > 2$, so there's no solution here.
