• Jul 10th 2009, 06:08 AM
vorsilva
I was given an equation and asked to solve for C. I've been able to simplify it to

B(AC)=A-C

I can't seem to get any further. Can anyone offer some suggestions?
• Jul 10th 2009, 06:12 AM
colby2152
Quote:

Originally Posted by vorsilva
I was given an equation and asked to solve for C. I've been able to simplify it to

B(AC)=A-C

I can't seem to get any further. Can anyone offer some suggestions?

You need to break the variable out of the multiple term on the LHS...

$\displaystyle BAC=A-C$
$\displaystyle C(BA)=A-C$

Now, group the terms with C's on one side, and the terms without C's on the other side.

$\displaystyle C(BA)+C=A$

Factor out C on the LHS, and divide the equality by the other factor, and you will get your answer. Practice this with many other forms of multiples, divisions, additions, and subtractions.
• Jul 10th 2009, 06:15 AM
allyourbass2212
Would that be

$\displaystyle C(BA-1)=A$
$\displaystyle C=\frac{A}{BA+1}$

Thanks Skeeter
• Jul 10th 2009, 06:58 AM
skeeter
Quote:

Originally Posted by vorsilva
I was given an equation and asked to solve for C. I've been able to simplify it to

B(AC)=A-C

I can't seem to get any further. Can anyone offer some suggestions?

folks need to pay attention to signs ...

$\displaystyle B(AC)=A-C$

$\displaystyle BAC + C = A$

$\displaystyle C(BA + 1) = A$

$\displaystyle C = \frac{A}{BA+1}$
• Jul 10th 2009, 12:10 PM
colby2152
Quote:

Originally Posted by skeeter
folks need to pay attention to signs ...

Thanks for spotting that typo. It's been a while since I posted here at MHF.(Rofl)
• Jul 10th 2009, 12:14 PM
Unenlightened
Edit: Seems question had already been answered but only the original post was showing up for me! Apols..