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?

2. 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...

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

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

$C(BA)+C=A$

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.

3. Would that be

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

Thanks Skeeter

4. 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 ...

$B(AC)=A-C
$

$BAC + C = A$

$C(BA + 1) = A$

$C = \frac{A}{BA+1}$

5. 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.

6. Edit: Seems question had already been answered but only the original post was showing up for me! Apols..