# Canceling With Variables

• Aug 14th 2008, 09:17 AM
cmf0106
Canceling With Variables
The Author of this book does not really explain the process very thoroughly, consequently I am lost.

$\frac{72x}{18xy}$

to solve:
$\frac{4}{y} * \frac{18x}{18x} = \frac{4}{y}$

I am completely stumped as to how the author gets $\frac{4}{y}$ and $\frac{18x}{18x}$, if someone could please clarify and tell me the process, and why its implemented it would be greatly appreciated.

When I look at it I dont see why you cant just cancel out the x in the numerator and denominator and get $\frac{72}{18y} = \frac{4}{y}$ I dont understand how she sets it up at all.
• Aug 14th 2008, 10:47 AM
Twig
Hi
The author has done nothing special.

Since you have an expression $\frac{72x}{18xy}$ with just factors in it, you CAN definetely "just cancel out" the x.
The author has simply broken out the GCF (greatest common factor) from both the $72x \ \mbox{and the} \ 18xy$

So $\mbox{GCF}(72x,18xy) \ = 18x$

Now the $18x$ gets cancelled out, and we´re left with just
$\frac{4}{y}$
• Aug 14th 2008, 11:37 AM
cmf0106
Ah many thanks. I learned GCF many years ago but this book, which is meant for beginners, jumps around from topic to topic. Even in the introduction of the book the author states she will build on previous material from chapter to chapter; however, GCF has not even been included this far. But at any rate many thanks for clarifying.

Also I know the same process must be going on here but how is the GCF of this expression 3x?

$\frac{6xy(2x-1)}{3x} = \frac{2y(2x-1)}{1} * \frac{3x}{3x} = 2y(2x-1)$

Looks like the 3 divides the 6 and the x just divides into the other x on the numerator.
• Aug 15th 2008, 01:47 AM
Twig
hi
The GCF here is definetely $3x$.

You always try to "cancel out" as much as possible when simplifying expressions. The key thing not to forget is that you can only cancel out factors!