How can I simplify this fraction problem?

I have the problem $$\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$$ which should simplify to $$\frac{1}{x-2}$$

I have simplified $$x^2-4$$, which becomes:

$$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$$

However, if I combine the fractions I get, $$x^2-x-1$$ for the numerator, which can't be factored. That's where I get stuck.

How can I get $$\frac{1}{x-2}$$ out of this problem?

You need to put them over a common denominator, $$\frac {x+1}{x+2}=\frac {(x+1)(x-2)}{(x+2)(x-2)}=\frac {x^2-x-2}{x^2-4}$$ Now you can subtract the numerators $$x^2-(x^2-x-2)=x+2$$ and finally divide out the $$x+2$$ from numerator and denominator

Your $$-1$$ should be $$-2$$. You didn't show your work, so I can't see why it happened.

• Regarding your last line, I wonder if OP just added the two numerators together without finding a common denominator. – Matthew Leingang Jan 30 at 18:54
• @MatthewLeingang That's right. It's been awhile since I worked with fractions, and I'm very rusty. – LuminousNutria Jan 30 at 18:56
• @LuminousNutria It's a very common error. But if you think about it with familiar fractions you'll remember it can't be true. For instance, $\frac{1}{2} + \frac{1}{2}$ is $1$, not $\frac{2}{4}$ (which is $\frac{1}{2}$ again). – Matthew Leingang Jan 30 at 19:00

\begin{align} \frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}&=\frac1{x+2}\left(\frac{x^2}{x-2} - (x+1)\right)\\ &=\frac1{x+2}\left(\frac{x^2-(x-2)(x+1)}{x-2}\right)\\ &=\frac1{x+2}\left(\frac{x^2-(x^2-x-2)}{x-2}\right)\\ &=\frac1{x+2}\left(\frac{x+2}{x-2}\right)\\ &=\frac1{x-2} \end{align}

Write $$\frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{(x+1)(x-2)}{(x+2)(x-2)}=…$$ Note that it must be $$x\ne 2,-2$$

The numerator should be $$x^2-(x-2)(x+1) = x+2$$ which simplifies with a factor of the denominator.

You probably just forgot a term when subtracting the two fractions.