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

The simplest, as we have a homographic function, is to write it in canonical form:
\begin{align*}
2<\frac{x}{x-1}\leq 3&\iff 2 < 1+\frac1{x-1}\leq 3 \iff 1< \frac1{x-1}\leq 2\\
&\iff \frac12\le x-1 <1 \iff \frac32 \le x <2
\end{align*}
The third equivalence is valid because all numbers at the end of the first line have the same sign.
