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:$$\frac{x^2-x-30}{x^2-25}$$


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.

  • $\begingroup$ Did you try plugging in numbers, like $x = 2$, to see if those two expressions are equal for all $x$? $\endgroup$ – JimmyK4542 Aug 15 '14 at 2:18
  • $\begingroup$ No but perhaps if you tell us what your thought process was we could further clarify $\endgroup$ – illysial Aug 15 '14 at 2:19
  • $\begingroup$ @illysial The OP undoubtedly "canceled" the factors of $x^2$. $\endgroup$ – Gahawar Aug 15 '14 at 2:21
  • $\begingroup$ Was in the process of adding that, thank you. $\endgroup$ – theStandard Aug 15 '14 at 2:21
  • $\begingroup$ @Gahawar Ah I see that now ..i should've noticed earlier! $\endgroup$ – illysial Aug 15 '14 at 2:29

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 $x = 2$.

  • 6
    $\begingroup$ A good rule of thumb to avoid the issue of "fortuitous equality" is to use a transcendental constant like $\pi$ rather than "nice" numbers like $2$. When dealing with a polynomial/rational equation, no transcendental constant will ever satisfy it unless the equation is identically true. $\endgroup$ – Deepak Aug 15 '14 at 2:29
  • $\begingroup$ @Deepak, great point! $\endgroup$ – Kaj Hansen Aug 15 '14 at 2:32
  • $\begingroup$ @Deepak that can be generalized even further to: given a function $f(x)$ and a set of values $Q$ that can be produced by plugging in an algebraic number $x$ into $f$ and/or $w$ that produces an algebraic number when plugged into $f$, one should opt to test with a number $u$ that exists outside this group. That way even with funky sines and cosines where certain transcendental constants lead to "fortuitous equality" we know to avoid them $\endgroup$ – frogeyedpeas Aug 15 '14 at 7:43
  • $\begingroup$ @frogeyedpeas Thank you that's a good point. I was thinking of transcendental equations and how the complementary approach would work but decided to just focus on the problem at hand. $\endgroup$ – Deepak Aug 15 '14 at 22:48

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



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.


$\displaystyle \frac{x^2-x-30}{x^2-25}=\frac{(x-6)(x+5)}{(x-5)(x+5)}=\frac{x-6}{x-5}$

  • 3
    $\begingroup$ I would suggest that this will not contribute to understanding by OP (even though correct as long as $x \neq -5$) $\endgroup$ – Ross Millikan Aug 15 '14 at 2:30
  • $\begingroup$ He seemed to have thought about the solution after the discussion about improper cancellation. I thought about just saying he should factor the polynomials, but see Gahawars post calling the $x^2$ terms "factors". It seemed that this would confuse the issue. Perhaps I was wrong. $\endgroup$ – Paul Sundheim Aug 15 '14 at 2:36
  • $\begingroup$ (changed which answer was marked as correct, although this was helpful in a secondary sort of way) $\endgroup$ – theStandard Aug 15 '14 at 2:37

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