# Solve for $\frac{b+c}{a+b+c+d}$ given $\frac{b}{a+b}, \frac{c}{c+d}, \frac{c}{c+a},\frac{b}{b+d}$

Given $\frac{b}{a+b}, \frac{c}{c+d}, \frac{c}{c+a},\frac{b}{b+d}$, where $a,b,c,d \geq 1$, can we solve for the value $\frac{b+c}{a+b+c+d}$? While it seems that we have 4 equations and 4 unknowns, due to the non-linearity I haven't been able to find a solution. I was thinking about inverting the equations, by saying $\frac{b}{a+b} = k \implies \frac{b+a}{b} = k \implies 1+\frac{a}{b} = k \implies \frac{a}{b}=k-1$. However, after doing this for all four equations, and trying to substitute, I haven't made any progress. I'd appreciate any help (or, if this isn't possible, I'd love to understand why)!

'Motivation': In my statistics class earlier today, we were talking about confusion matrices (https://en.wikipedia.org/wiki/Confusion_matrix), and I was curious to see if algebraically we can compute the 'total error' in a model from the error values. There really isn't any statistical importance of this question though (as far as I can tell)

• what is here to solve, an equation is missing – Dr. Sonnhard Graubner Feb 19 '18 at 20:02

Let $$\frac b{a+b}=A\implies b=\frac{Aa}{1-A}$$ $$\frac c{c+d}=B\implies d=\frac{1-B}Bc$$ $$\frac c{a+c}=C\implies c=\frac{Ca}{1-C}$$ so $$d=\frac{1-B}B\cdot\frac{Ca}{1-C}=\frac{C(1-B)}{B(1-C)}a$$ Hence \begin{align}\frac{b+c}{a+b+c+d}&=\frac{\frac{A}{1-A}+\frac{C}{1-C}}{1+\frac{A}{1-A}+\frac{C}{1-C}+\frac{C(1-B)}{B(1-C)}}\quad\text{cancel}\,a\\&=\frac{A(1-C)+C(1-A)}{(1-A)(1-C)+A(1-C)+C(1-A)+\frac CB(1-A)(1-B)}\\&=\frac{B(A+C-2AC)}{B(1-AC)+C(1-A)(1-B)}\end{align}