Cancelling Rules

**CSG18** · April 29th 2010, 02:22 PM

Hi,

I'm quite hesitant when it comes to cancelling terms in algebraic fractions. So how do you know when not to cancel?

Does the term in the denominator have to be present outside another term in the numerator, in order for it to be cancelled? Do you know what I mean? I've found that sometimes you have to factorise the top of an algebraic fraction in order to cancel it down further - why is that?

I know it depends on the type of question and sorry for the vagueness.

I would be grateful for any help,

Thanks.

**shenanigans87** · April 29th 2010, 02:44 PM

Think of fractions as always being factored, always. If you can't factor it any further, and you can't see a similar, factored term in both the numerator and denominator, it can't be canceled.

For example

$\frac{x^2-7x+10}{(x-5)(x-2)}=\frac{(x-5)(x-2)}{(x-5)(x-2)}=\frac{1}{1}=1$

and

$\frac{3(x^2+5)}{3y+9z}=\frac{3(x^2+5)}{3(y+3z)}=\f rac{x^2+5}{y+3z}$

**Quacky** · April 29th 2010, 02:56 PM

Someone will easily be able to explain better than I can.

Hey, let's take a basic, factorized example such as:

If I have $\frac{(x+1)(x+2)}{(x-1)(x+1)}$ ,

You'll immediately notice that there is a common term of $x+1$ which can be cancelled. The reason this can be cancelled is that you are timesing everything by $(x+1)$ and then dividing everything by $(x+1)$ . You're just doing the inverse.But if you take this example:

$\frac{4x+2}{x}$ You can not cancel the x because you have not timsed everything by x (the 2 has not been timsed by x). Can you see the difference? Above, the whole fraction was timsed by (x+1) and then the whole fraction was divided by (x+1) which clearly will just give you 1. Here, the WHOLE fraction has been divided by x, but only one PART of the top fraction has been timsed by x. You have not got the inverse.

If you are not sure when to cancel, the easiest thing to do is to factorize first. Why? Because this shows you exactly WHAT has been timsed together. For example:

$\frac{x^3+2x^2-5x-6}{x^3-3x^2-25x-21}$

How would you know what you're timesing together? By looking at it, you would have no clue.

If you factorize, you learn that this is equal to:
$<br /> \frac{(x+1)(x-2)(x+3)}{(x+3)(x+1)(x-7)}<br />$

And now you can quite easily spot the common factors of $x+3$ and $x+1$ Which cancel to give

$\frac{x-2}{x-7}$

**HyperKaehler** · April 29th 2010, 03:05 PM

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:

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

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:

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

and as you probably know this can just be written as

$xa$

(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:

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

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

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

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:

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

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:

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

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

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

which of course is the same thing as just

$x(a-xb)$

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

**CSG18** · April 30th 2010, 10:45 AM

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:

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

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:

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

and as you probably know this can just be written as

$xa$

(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:

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

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

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

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:

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

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:

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

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

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

which of course is the same thing as just

$x(a-xb)$

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

Thank you very much for taking the time to explain, this has definitely made me understand it much better. I understood everything you said. Thanks again.

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Cancelling Rules

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