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### Simplifying Algebraic Fractions

Date: 05/01/2008 at 08:28:16
From: Rebecca
Subject: simplify the following:

Simplify:

(2az+6bz)/(6az+3bz) x (8a+4b)/(az+bz) x (2az+4bz)/(3a+9b)

The answer should be 16(a+2b)/9(a+b), and although I have tried
different things, I cannot arrive at this answer and don't see how
they got it.

I started by canceling out the az's etc, and then got to a point
where I had 8/9 x 12/2z x 6z/12 = 576z/216z and the z's could then
cancel out leaving 576/216 which would then reduce to 16/6.

Date: 05/01/2008 at 11:29:40
From: Doctor Ian
Subject: Re: simplify the following:

Hi Rebecca,

A good way to get started is by looking for common factors in the

2az + 6bz    8a + 4b    2az + 4bz
---------- * ------- *  ---------
6az + 3bz    az + bz     3a + 9b

2(az + 3bz)   4(2a + b)    2(az + 2bz)
= ----------- * --------- *  -----------
3(2az + bz)    az + bz     3(a + 3b)

2z(a + 3b)   4(2a + b)    2z(a + 2b)
= ---------- * --------- *  -----------
3z(2a + b)   z(a + b)     3(a + 3b)

So far, so good?  Now, since we're multiplying, anything that appears
in both a numerator and a denominator can be canceled.  Do you see
why?  Let's look for some of those.  In each step, note what's been
removed, and make sure you understand WHY it's been removed:

2(a + 3b)   4(2a + b)    2z(a + 2b)
= --------- * --------- *  -----------
3(2a + b)   z(a + b)     3(a + 3b)

2(a + 3b)   4(2a + b)    2(a + 2b)
= --------- * --------- *  -----------
3(2a + b)    (a + b)     3(a + 3b)

2           4(2a + b)    2(a + 2b)
= --------- * --------- *  -----------
3(2a + b)    (a + b)     3

2           4            2(a + 2b)
= --------- * --------- *  -----------
3            (a + b)     3

Is there anything else that can be canceled?  There isn't.  And what
remains combines to the answer you expected.

My guess is that you tried to cancel things that were being added; but
that doesn't work.  That is, we can do this,

3 * 4   4
----- = -              This is okay.
3 * 5   5

but not this,

3 + 4   4
----- = -              This is NOT okay.
3 + 5   5

It's important--actually, crucial--to understand why the first works
and the second doesn't, so if you're not absolutely clear on that,
please write back and we can discuss it further.

Does this help?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

Date: 05/01/2008 at 13:41:23
From: Rebecca
Subject: Thank you (simplify the following:)

Hi Dr Ian, thank you so much for the help, at first I was a little bit
confused, but I have been through it all stage by stage and I actually
understand it now and how we arrived at that answer.  I was, as you
said, trying to cancel out terms that were being added, and this as I
found out didn't work.  Many thanks again, it is so nice to know
what I am doing now!  Rebecca
Associated Topics:
High School Polynomials

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