Find the set of values of x for which:

$\displaystyle \frac{x}{x-3} > \frac{1}{x-2}$.

I rearranged it to form:

$\displaystyle \frac{x^2 - 3x + 3}{(x-3)(x-2)} > 0$

Therefore critical points of the denominator are x = 2, 3.

It says that the numerator is always positive, could you explain why exactly this is, or is this just something that you "know".

I am not sure then how to work out a set of values for x, they have x<2 and x>3 in the book.

Thanks in advance for the help :)