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I'm trying to simplify: $$\frac{x^2}{x^2-4}-\frac{x+1}{x+2}$$ but I can't get to the answer: $$\frac{1}{x-2}$$

How to do it?

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It may be worth noting that as complicated as it may seem, the same technique you would use to simplify $\frac{13}{15}-\frac{2}{3}$ to $\frac{1}{5}$ applies. However, $15=3\cdot 5$ is probably more obvious than $x^2-4=(x-2)(x+2)$. –  Jonas Meyer Feb 4 '11 at 0:24
@Jonas: true, though "difference of squares" tends to be the one that most students who went through/suffered through algebra seem to remember. –  Arturo Magidin Feb 4 '11 at 0:26
@Arturo: I agree, which is why I commented on the similarity. I have occasionally found that when a student is struggling with "complicated" algebra with rational functions, pointing out an analogous arithmetic problem clears up what is going on. Incidentally, the example I gave is just a special case of $\frac{x^2-3}{x^2-1}-\frac{x-2}{x-1}=\frac{1}{x+1}$. –  Jonas Meyer Feb 4 '11 at 0:30

2 Answers 2

up vote 3 down vote accepted

First, do the operation by finding the least common denominator. Since $x^2-4 = (x-2)(x+2)$, the least common denominator is already $x^2-4$. Then do some simple algebra: \begin{align*} \frac{x^2}{x^2-4} - \frac{x+1}{x+2} &= \frac{x^2 - (x+1)(x-2)}{x^2-4} = \frac{x^2-(x^2-x-2)}{x^2-4}\\ &= \frac{x+2}{x^2-4} = \frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}. \end{align*}

If you didn't realize that $x+2$ already divides $x^2-4$, you probably would get \begin{align*} \frac{x^2}{x^2-4} - \frac{x+1}{x+2} &= \frac{x^2(x+2) - (x+1)(x^2-4)}{(x^2-4)(x+2)}\\ &= \frac{x^3 + 2x^2 - (x^3 +x^2 - 4x - 4)}{(x^2-4)(x+2)}\\ &= \frac{x^2 +4x + 4}{(x^2-4)(x+2)} = \frac{(x+2)^2}{(x^2-4)(x+2)}\\ &= \frac{x+2}{x^2-4} = \frac{1}{x-2}, \end{align*} with the extra work for not noticing.

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you type it or use a software? that's too fast.. –  Tom Brito Feb 4 '11 at 0:22
I type. I've been touch-typing since 1982, and TeX-ing since 1987. –  Arturo Magidin Feb 4 '11 at 0:23
I used the second approach (multiplying the denominators), but I was trying to simplify directly, instead of multiplying all. Actually, I symplified even the denominators before multipling.. A big mental loop.. –  Tom Brito Feb 4 '11 at 0:36

HINT $\ \ $ Taking fractional parts (by subtracting $\:1\:$ from both terms) it reduces to the following

$$\rm \frac{4}{x^2-4}\ +\ \frac{1}{x+2}\ =\ \frac{1}{x-2}$$

which you'll probably find easier to derive.

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