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Exact solution? (solving equations)

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dennislau

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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
 

v_c

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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,
v_c
 

kalyanram

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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.

~Kalyan
 

v_c

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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 :D and what he is saying is right.
 

dennislau

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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).
 

zajbanlik

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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)
 

neils_arm_strong

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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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