Solving an Algebra Formula
I need to solve this equation. The end goal is to be able to create a worksheet where I can set a goal profitability for a project and have the spreadsheet kick out the suggested price. Here is the equation:
(Price - Labor - Database - Commission)/Price = % profitability
Labor is assumed to be a constant.
Commission has the following equation: Commission = (Price - Labor - Database)*.14
So basically I need to solve for Price and I can't remember for the life of me how to get rid of the parentheses.
Simple Version: (P - L - D - ((P - L - D)*.14))/P = X (Solve for P)
Any help would be incredibly appreciated.
Re: Solving an Algebra Formula
Quote:
Originally Posted by
noeltroy
I need to solve this equation. The end goal is to be able to create a worksheet where I can set a goal profitability for a project and have the spreadsheet kick out the suggested price. Here is the equation:
(Price - Labor - Database - Commission)/Price = % profitability
Labor is assumed to be a constant.
Commission has the following equation: Commission = (Price - Labor - Database)*.14
So basically I need to solve for Price and I can't remember for the life of me how to get rid of the parentheses.
Simple Version: (P - L - D - ((P - L - D)*.14))/P = X (Solve for P)
Any help would be incredibly appreciated.
 }{P} &= X \\ \frac{(1 - 0.14)\left( P - L - D \right)}{P} &= X \\ \frac{0.86\left( P - L - D \right)}{P} &= X \\ 0.86\left( P - L - D \right) &= X\,P \\ 0.86\,P - 0.86\,L - 0.86\,D &= X\,P \\ 0.86\,P - X\,P &= 0.86\,L + 0.86\,D \\ \left(0.86 - X\right) P &= 0.86\left(L + D\right) \\ P &= \frac{0.86\left( L + D \right) }{0.86 - X} \end{align*})
Re: Solving an Algebra Formula
Brilliant! Thanks so much! There were a couple of moves in there I'd forgotten you could do. Really, really helpful. Thanks!
Re: Solving an Algebra Formula
Being lazy, I'd set L + D = K and rewrite:
P - K - .14(P - K) = XP
P - K - .14P + .14K = XP
.86P - XP = K - .14K
P(.86 - X) = K(1 - .14)
P = .86K / (.86 - X)
Re: Solving an Algebra Formula
That is really helpful as well. Thanks!