Simplifying Algebraic FractionsDate: 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. I really appreciate any help you can give, thanks, Rebecca 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 additions, 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/