# Express as a single fraction

• Oct 29th 2012, 05:26 AM
lsmcal1984
Express as a single fraction
Hi all,

This is my first post so thank you for reading it. I have started advanced level maths after 12 years of no studying of the subject. I have learned most of the stuff we have studied so far but it seems like I'm not alone in having issues with fractions. I've been asked to express the following as a single fraction:

 (m+n)^2 + M^2+mn m^2-n^2 n^2-mn

I've tried many videos and pages online but to no avail. I'd be eternally grateful if someone could explain this step by step and that way I could practice it with other equations. I feel like I have missed how to do these with no numbers somewhere along the line.

Many thanks
• Oct 29th 2012, 06:22 AM
earboth
Re: Express as a single fraction
Quote:

Originally Posted by lsmcal1984
Hi all,

This is my first post so thank you for reading it. I have started advanced level maths after 12 years of no studying of the subject. I have learned most of the stuff we have studied so far but it seems like I'm not alone in having issues with fractions. I've been asked to express the following as a single fraction:

 (m+n)^2 + M^2+mn m^2-n^2 n^2-mn

I've tried many videos and pages online but to no avail. I'd be eternally grateful if someone could explain this step by step and that way I could practice it with other equations. I feel like I have missed how to do these with no numbers somewhere along the line.

Many thanks

1. I assume that you want to simplify

$\displaystyle \frac{(m+n)^2}{m^2-n^2} + \frac{m^2+mn}{n^2-mn}$

2. If so factor as much as possible:

$\displaystyle \frac{(m+n)^2}{m^2-n^2} + \frac{m^2+mn}{n^2-mn} = \frac{(m+n)(m+n)}{(m+n)(m-n)} + \frac{m(m+n)}{n(n-m)}$

3. For $\displaystyle m+n\ne 0$ you can cancel one factor in the 1st summand. Because $\displaystyle n-m=-(m-n)$ you'll get:

$\displaystyle \frac{(m+n)}{(m-n)} - \frac{m(m+n)}{n(m-n)}$

4. Obviously the common denominator is $\displaystyle n(m-n)$. Therefore:

$\displaystyle \frac{(m+n)}{(m-n)} - \frac{m(m+n)}{n(m-n)} = \frac{n(m+n)-m(m+n)}{n(m-n)}\\ = \frac{(n-m)(m+n)}{n(m-n)} = - \frac{m+n}n$
• Oct 29th 2012, 07:13 AM
lsmcal1984
Re: Express as a single fraction
Thank you very much indeed - this puts together various things I've read on the Internet and I now know how to tackle algebraic fractions.
Much appreciated.
Leo