Can a fraction be simplified like this? Ridiculously embarrassing question, but can $\frac{x^2-x}{x^2-25}$ be simplified to simply $\frac{1-x}{1-25}$?
Full thought process here is that this is essentially $\frac{x*x-x}{x*x-25}$ so the $x$s should cancel.  The full problem is:$$\frac{x^2-x-30}{x^2-25}$$
sorry
I'm used to programming forums where a simplest-case example of an error is the way to ask about it.  I should have made the full problem clearer earlier as $-$ unfortunately $-$ this lead to someone who gave more information being wrong at the final problem and I can't mark both answers right.
 A: If you are unsure, then one way to check whether things like this might be true is to plug in a value for $x$.  Let $x = 2$.  We get:
$$\frac{x^2-x}{x^2 - 25} = \frac{2}{-21} \neq \frac{1-x}{1-25} = \frac{-1}{-24} = \frac{1}{24}$$
So in this case, you made a mistake somewhere.  
Of course, if you plug in a value and equality does hold, then that doesn't imply it always holds.  E.g. $2x \neq x^2$ in general even though it holds when $x = 2$.
A: No.  If you want to divide numerator and denominator by $x^2$, you will get $\dfrac{1-\frac1x}{1-\frac{25}{x^2}}$, which isn't really simpler.  If you really want to do something to simplify it, you can rewrite it as
$$\frac{x^2-25-x+25}{x^2-25}=\frac{x^2-25}{x^2-25}+\frac{-x+25}{x^2-25}=1+\frac{25-x}{x^2-25}$$
A: To simplify a fraction, you want to factor the numerator and denominator and see what cancels.  The denominator $x^2-25$ is a difference of squares:  $x^2-25=(x+5)(x-5)$, so see if either of those factors divides the numerator.  When you say $x$s should cancel you should understand that "cancel" means "divide by".  The $25$ term does not have a factor of $x$, so you can't cancel it.
A: $\displaystyle \frac{x^2-x-30}{x^2-25}=\frac{(x-6)(x+5)}{(x-5)(x+5)}=\frac{x-6}{x-5}$
