# Thread: Rearranging an algebraic equation

1. ## Rearranging an algebraic equation

Having some difficulty making q the subject:

((a-c)/b(2+vj)-(q/2+vj) = ((a-c)/b)-q(2+vi)

Any help would be appreciated.

2. ## Re: Rearranging an algebraic equation

Originally Posted by SergioYaya
Having some difficulty making q the subject:

((a-c)/b(2+vj)-(q/2+vj) = ((a-c)/b)-q(2+vi)

Any help would be appreciated.
Separate everything involving q from every thing NOT involving q:
[(a-c)/(b(2+vj))]- q/2- [vj]= [(a- c)/b]- (2+vi)q

Get everything involving q on the left side, everything not involving q on the right side:
-q/2- (2+vi)q= (a-c)b+ vj- (a-c)/(b(2+vj))

Factor q out on the left:
(-1/2- 2+vi)q= (a-c)b+ vj- (a-c)/(b(2+vj))

Divide both sides by -1/2- 2+ vi:
q= [(a-c)b+ vj-(a-c)/(b(2+vj))]/(-1/2- 2+ vi)

3. ## Re: Rearranging an algebraic equation

Thank you for the reply, however, I still do not fully understand how you came to this conclusion. My notes tell me that a simpler version of this problem:
((a-c)/b)-q(2+v)=((a-c)/b(2+v))-(q/2+V) equates to q*= (a-c)/b(3+v) and I just don't understand how this is possible.

4. ## Re: Rearranging an algebraic equation

Originally Posted by SergioYaya
Having some difficulty making q the subject:

((a-c)/b(2+vj)-(q/2+vj) = ((a-c)/b)-q(2+vi)

Any help would be appreciated.
what are a,b,c,i,j and q?

5. ## Re: Rearranging an algebraic equation

q is quantity, c is costs, a and b form part of pricing, vi and vj are conjectural variation terms so they are all simply letters which must be rearranged.