What is another way to approach this inequality? I was given this inequality to solve today and then somebody showed me a very clever and simple way to solve it. They made two assumptions and with both of those assumptions, they deduced the proper interval of the inequality. I tried to replicate their method, but I was not successful. 
This is how I solved it, but I would like to see that "assumptions" method again. It seemed more efficient. 
Steps I took:
$$\frac { 2x }{ x-3 } <1\\ $$
Finding the critical points: $$\frac { 2x }{ x-3 } =1\\ $$
$$2x=x-3$$
Critical points are: $$x=3\quad or\quad x=-3$$
Testing around the critical points shows that the interval is $(-3,3)$
Can someone show me the "two assumptions" method?
 A: Maybe the assumptions were on whether $x<3$ or $x>3$.


*

*If $x<3$, then $x-3<0$, meaning that $$\frac{2x}{x-3}<1 \iff 2x > x-3\iff x>-3$$
meaning that $x\in(-3,3)$.

*If $x>3$, then $x-3 > 0$, meaning that $$\frac{2x}{x-3} < 1 \iff 2x < x-3 \iff x<-3$$ which is impossible.

A: Maybe it was this?
$$
\begin{cases}
x-3>0\\
2x<x-3
\end{cases}
\qquad \lor \qquad 
\begin{cases}
x-3<0\\
2x>x-3
\end{cases}
$$
so that:
$$
\left(\begin{cases}
x>3\\
x<-3
\end{cases}
\qquad \lor \qquad 
\begin{cases}
x<3\\
x>-3
\end{cases}\right)
\iff -3<x<3
$$
A: Divide it into two cases (clearly $x=3$ doesn't make sense).
Case 1: $x > 3$
Multiply both sides of the equality by $x-3$ to get
$$2x < x-3,$$
which is equivalent to
$$x < -3.$$
This is a contradiction, so the first case brings no solutions with it.
Case 2: $x < 3$
Multiply both sides with the same, but flip the inequality to get
$$2x > x-3,$$
or
$x > -3$.
In conclusion, the set of solutions is $\{x\ |-3<x<3\}$.
A: $$\frac { 2x }{ x-3 } <1\Rightarrow \frac { 2x }{ x-3 } -1=\frac { 2x-x+3 }{ x-3 }= \frac { x+3 }{ x-3 }<0 $$
$\frac { x+3 }{ x-3 }<0\iff{ (x+3 )}{ (x-3 )}<0$
Note that: $(x-a)(x-b)<0$ with $a<b\implies a<x<b$
