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Exact solution? (solving equations)

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Dec 16, 2005
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Could anyone can help to solve the exact solution if possible or the ways to find the values of A, B and C? Thanks.

Given the values of x, y and z (where x+y+z=1), find the values of A, B and C and their relationship is:

x=A/(A+B+C); y=B/(A+B+C), z=C/(A+B+C)

for example: x=0.310, y=0.338, z=0.352

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.

Best regards,

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,

Added after 6 minutes:

Looks like kalyanram beat me to it :D and what he is saying is right.

I get it. Thanks a lot kalyanram and v_c :D

So, it seems that the solutions for A, B and C are:

A=kx, B=ky, C=kz for any positivie integers k with the condition A+B+C=k(x+y+z).

dennislau said:
A=kx, B=ky, C=kz for any positivie integers k with the condition A+B+C=k(x+y+z).
It doesn't have to be integer, can be any number except 0. (For example 0.5)

4 equations are given and 3 unknowns are to be found.Hence the system of equations will have infinite number of solutions.

dennislau is right.

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