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
$x-3 < 0$


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.
