# Relatively basic fraction multplication!

• November 11th 2011, 01:17 AM
ktharmer
Relatively basic fraction multplication!
Hi all,

The below expression is part of a longer equation which I have simplified for this example. I would like to remove the r from the very bottom denominator. What do I multiply the remaining variables by?

a/b * (1-cd)/(1-(cd/r)) + e/f

Many thanks in advance!
• November 11th 2011, 04:49 AM
SpringFan25
Re: Relatively basic fraction multplication!
$\frac{a}{b} \cdot \frac{1-cd}{1-\frac{cd}{r}} + \frac{e}{f}$

i assume that by "very bottom denominator" you mean the $\frac{cd}{r}$ part.

multiply everything by 1/r

$\frac{a}{b} \cdot \frac{1-cd}{r \left(1-\frac{cd}{r} \right)} + \frac{e}{fr}$

$= \frac{a}{b} \cdot \frac{1-cd}{r-cd} + \frac{e}{fr}$
• November 11th 2011, 05:02 AM
ktharmer
Re: Relatively basic fraction multplication!
Thanks!
• November 12th 2011, 01:04 PM
SpringFan25
Re: Relatively basic fraction multplication!
oh dear, how silly of me, if you wanted a fraction that was equivalent to he first one, that wont be it....as the above is multiplied by (1/r)

hopefully you made appropriate adjustments to ther est of your equation.
• November 12th 2011, 08:31 PM
Wilmer
Re: Relatively basic fraction multplication!
Quote:

Originally Posted by ktharmer
a/b * (1-cd)/(1-(cd/r)) + e/f

1 - cd/r = (r - cd) / r
So:
a/b * [r(1 - cd) / (r - cd)] + e/f