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Hurkyl
#5
Jan12-06, 06:01 PM
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Linear Algebra can solve for a,b,d, and e if a solution exists. It doesn't do anything regarding x and y.

You can not solve u and v through that at all.
Let me try explaining it a different way:

Suppose the system of equations

au + bv = c
du + ev = f

has a unique solution (u,v) = (m,n). (As is typical with 2 equations and 2 unknowns)


Then, if you were asked to solve the system of equations

a*f(x) + b*g(y) = c
d*f(x) + e*g(y) = f

for x and y.

The previous work allows you to immediately reduce this problem to solving

f(x) = m
g(y) = n