# Subtracting rational functions

I'm trying to find out how to solve this:

$$\frac{x-2}{x^2+2x} - \frac{x+2}{x^2-2x} - \frac{4x}{x^2-4}$$

The answer is $\displaystyle \frac{-4}{x-2}$

What is this called? And is there any video of it on http://www.khanacademy.org ? Thanks!

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Think about this equation: $$\frac{\alpha}{ab} - \frac{\beta}{ac} - \frac{\gamma}{bc} = \frac{\alpha c - \beta b - \gamma a}{abc}.$$ Now, try to figure out what are the values of $\alpha, \beta, \gamma, a, b,$ and $c$ in your case. – user2468 May 28 '11 at 0:33
"solve" is for equations. What you want to do is called "simplify", not "solve". – Gerry Myerson May 28 '11 at 5:58
@J.M. I really don't know when the [algebra] tag can be used. I have added it, but apparently this question is not to be tagged as so, because you removed it. I thought a [algebra-precalculus] question might also have the [algebra] tag in most cases. But apparently I am wrong. – Américo Tavares Sep 15 '11 at 14:31
@Américo: from my current understanding, we use algebra-precalculus only for those questions regarding things one might encounter in "secondary school" or the freshman year of a college course. From the tag blurb: "use the algebra-precalculus tag for topics relating to symbolic manipulation, basic functions, and other algebra/advanced-algebra/precalculus topics." – J. M. Sep 15 '11 at 14:34
@J.M. I think I understand the use of the [algebra-precalculus] tag. What I do not understand is the use of the [algebra] tag. – Américo Tavares Sep 15 '11 at 14:38

Added: The method is to transform the sum of the given rational fractions (the numerator and denominator consists of polynomials) into a single equivalent fraction.

The properties used are

1. $$\frac{A(x)}{B(x)}=\frac{A(x)P(x)}{B(x)P(x)}\qquad\text{for }P(x)\neq 0.$$

2. $$\frac{A(x)}{B(x)}\pm \frac{C(x)}{D(x)}=\frac{A(x)D(x)\pm B(x)C(x)}{B(x)D(x)}.$$

I would calculate as follows, starting with the factorization of denominators

$$\begin{eqnarray*} &&\frac{x-2}{x^{2}+2x}-\frac{x+2}{x^{2}-2x}-\frac{4x}{x^{2}-4} \\ &=&\frac{x-2}{x(x+2)}-\frac{x+2}{x\left( x-2\right) }-\frac{4x}{\left( x-2\right) \left( x+2\right) } \\ &=&\frac{\left( x-2\right) ^{2}}{x(x+2)\left( x-2\right) }-\frac{\left( x+2\right) ^{2}}{x\left( x-2\right) \left( x+2\right) }-\frac{4x^{2}}{% x\left( x-2\right) \left( x+2\right) } \\ &=&\frac{\left( x-2\right) ^{2}-\left( x+2\right) ^{2}-4x^{2}}{x\left( x-2\right) \left( x+2\right) } \\ &=&\frac{x^{2}-4x+4-x^{2}-4x-4-4x^{2}}{x\left( x-2\right) \left( x+2\right) } \\ &=&\frac{-4x^{2}-8x}{x\left( x-2\right) \left( x+2\right) } \\ &=&-\frac{4x\left( x+2\right) }{x\left( x-2\right) \left( x+2\right) } \\ &=&-\frac{4}{x-2}. \end{eqnarray*}$$

so that I reduce the rational fractions to a common denominator first to allow me to add (subtract) them, and simplify the numerator thereafter. Finally I divide by the common factors to both the numerator and denominator. This is valid iff $x\ne -2$ and $x\ne 0$, because you cannot divide by zero. Also for $x=2$ the fraction is not defined.

See "Simplifying Rational Expressions 1, 2 and 3", "Adding and Subtracting Rational Expressions 1, 2 and 3" on http://www.khanacademy.org.

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I think that the main idea used is to decompose every denominator into simplest factors, and then use the basic knowledge in fraction addition. – Beni Bogosel May 28 '11 at 10:23
@Beni Bogosel: That's right. – Américo Tavares May 28 '11 at 10:24
Américo Tavares@ Thanks a lot! – Muazam May 28 '11 at 14:21
@Muazam: You are welcome! – Américo Tavares May 28 '11 at 22:41