Solving inequality with fraction in one side confusion In the book I'm using an example is given as follow:
$\frac{2x - 5}{x-2}< 1$
then it proceeds to say that we could multiple both sides by $x-2$ to get rid of the denominator in the left hand side (I understand that). But then it goes on to say that this method would require to consider the following cases,
$x-2 > 0$ and $x-2<0$ separately.
How did the author of the book reached to the conclusion that we will need to evaluate such cases? and why did the orientation of the inequalites change? 
I tried doing some algebraic manipulation myself but I could reach a conclusion. I tried:
multiplying both sides by $x-2$
$2x-5<x-2$
subtract x from both sides
$x-5 < -2 $
add 2 to both sides
$x-3 < 0$

Because $$\frac{a}{b}<1\iff\begin{cases}a<b\\ b>0\end{cases}\quad \text{or}\quad \begin{cases}a>b\\ b<0\end{cases}.$$
