Isn't this just a case of three equations and three unkowns? You can use a number of ways to solve this -- Gaussian elimination, matrix inversion, or just use matlab or mathcad to solve it for you.
Hi Dennis,
The problem stated above has an insufficiency in it. Now givien that x+y+z =1 and having stated that x= A/(A+B+C)--(1) & y = B/(A+B+C)--(2) then we know that z= 1-x-y and can conclude z= C/(A+B+C) so the above problem reduces to solving two equations in three unknowns, Now if we say A+B+C =k(x+y+z)=k
then we have A=kx, B=ky and C=kz so for any given x,y,z satisfying x+y+z =1 we can choose any k and have infinitly many solutions to the above equations.
On another note, I see a problem with uniqueness here ... maybe the determinant of the matrix will be zero in this case. Well let me try a simple solution assuming that I have x=0.310, y=0.338 and z=0.352.
Assume that I have A=0.310 and I let A+B+C=1, then your first equation would be satisified. Then if I have B=0.338, the second equation would be satisfied and picking C=0.352 the third equation would be satisifed.
Now let me pick A=0.620 (twice what I had before) and let A+B+C=2 (twice what I assumed before). I will still satisfy the first equation since I multiplied both the top and the bottom by the same numbers. So if I continue and pick B=0.676 and C=0.704, I will still satisfy the three equations.
All I am trying to say is that the solution to this set of equations is not unique. In fact each looks like a voltage divider calculation to me and you know that to get ratios of resistors you can choose different sets of resistors -- the choice is not unique. But you may have to choose a certain set of resistors so you don't dissipate to much power, draw too much current, etc.
By the way, my choice of a factor of two was arbitrary, you can choose any scaling factor that fits your expressions.
Best regards,
v_c
Added after 6 minutes:
Looks like kalyanram beat me to it and what he is saying is right.