Originally Posted by

**HyperKaehler** A lot of people have problems with this. I think it is because it is not taught too clearly in school - they do not differentiate multiplication enough from addition, and so when variables are involved and the expressions are complicated people tend to confuse the two. I can't say I can explain it more clearly in a few sentences, but I can offer a few examples.

Say you are given the following algebraic fraction:

$\displaystyle \frac{(x)(x)(a)}{(x)(1)}$

Now in this case it is very simple, because everything is written very neatly and there is no addition confusing things for you. So it should be easy to simplify this one: all you do is cancel an x on the top and an x on the bottom; they cancel each other out. So you are left with:

$\displaystyle \frac{(x)(a)}{(1)}$

and as you probably know this can just be written as

$\displaystyle xa$

(If you didn't understand that step, tell me and I'll explain in more detail.)

It turns out that the difficult part is not the canceling; it's getting the fraction into a form which makes it easy for you to cancel. Let's look at a harder one:

$\displaystyle \frac{x^2a-x^3b}{x}$

For this one, to get it in an easier form, we have to factor. First, we expand the $\displaystyle x^2$ and $\displaystyle x^3$ in the numerator. Since $\displaystyle x^2$ is just shorthand for x*x, and since $\displaystyle x^3$ is just shorthand for x*x*x, we can write

$\displaystyle \frac{xxa - xxxb}{x}$

Now, the important thing to remember is that you cannot cancel a variable on the top unless it is present in *all* of the things being added. That is, if you want to cancel an x on the top, you have to make sure there is an x in (x)(x)(a) *and* an x in (x)(x)(x)(b). Obviously there is! The first has two x's and the second has 3 x's. But we only have one x on the bottom, so we only need to take away one x from each term on the top.

A simpler way to say this is we just need to factor the numerator some more. We see that both the terms on the top have two x's in them. So we can use the distributive property to write the fraction like this:

$\displaystyle \frac{xx(a-xb)}{x}$

To make things really clear, it might help you to add an imaginary one to the denominator. Since x*1 is the same as x, we can change the denominator to x*1 like this:

$\displaystyle \frac{xx(a-xb)}{x1}$

Now we have the fraction in an easy form! Just like you did above, you can cancel one x on top and one x on bottom, and you are left with

$\displaystyle \frac{x(a-xb)}{1}$

which of course is the same thing as just

$\displaystyle x(a-xb)$

I hope this made sense. Please ask if you have any more questions!