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
$x-3 < 0$
 A: $$\frac{2x-5}{x-2}<1$$
it is alwys better that you have 0 on  the RHS. Then
$$\frac{2x-5}{x-2}-1<0 \implies \frac{x-3}{x-2} <0$$
$A/B <0 \implies AB<0 \implies (1):$ $A<0$ and $B>0$ or (2): $A>0$ and $B<0$
So we get (1): $x-3<0$ and $x-2>0 \implies 2<x<3.$ or $(ii) x>3$ and $x<2 \implies$ no solution (as the oprverlap is null).
So the the solution is $2<x<3$
A: 
How did the author of the book reached to the conclusion that we will need to evaluate such cases?

Because we don't know the sign of $x-2.$ All we know is that it may either be positive ($>0$) or negative ($<0$); the case $x-2=0$ not arising since then the fraction is not a real number.

why did the orientation of the inequalites change?

I don't know which of the inequalities you're talking about, but when you're considering the case when $x-2<0,$ you have to change the original inequality from $\text{LHS}<\text{RHS}$ to $\text{LHS}>\text{RHS}$ because whenever you multiply both sides of an inequality by a negative number, the order is reversed.
A way to proceed by not evaluating cases is simply to multiply both sides by $(x-2)^2$ without bothering about signs, since a square can never be negative.
A: Because $$\frac{a}{b}<1\iff\begin{cases}a<b\\ b>0\end{cases}\quad \text{or}\quad \begin{cases}a>b\\ b<0\end{cases}.$$
