solve two equations for two unknown variables
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how can i solve two equations for two unknown variables?
like these two:
z=f1(x)+f2(y)
z=f3(x)+f4(y)
for z=0
please help, thx in advance.
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Respuesta aceptada
Dr. Seis
el 20 de Dic. de 2011
G*m = d
G = [f1,f2;f3,f4];
d = [z1;z2];
m = G\d;
x = m(1);
y = m(2);
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Más respuestas (5)
Walter Roberson
el 20 de Dic. de 2011
That cannot be solved without knowing f1(x), f2(y), f3(x), f4(y)
Are f1, f2, f3, and f4 perhaps constants, as in f1*x + f2*y ? If so then the solution is x = 0, y = 0, unless f1*f4 = f2*f3
yasser
el 20 de Dic. de 2011
1 comentario
Walter Roberson
el 20 de Dic. de 2011
If your equations are
0=f1*x+f2*y
0=f3*x+f4*y
then the only solution is x = 0 and y = 0, unless it happens that f1*f4 = f2*f3 is exactly zero.
If f1*x + f2*y = 0 then y = -f1/f2 * x . Substitute this in to f3*x + f4*y = 0, and you end up with -x*(f1*f4-f3*f2)/f2 = 0 . That has solutions only if x = 0 or f1*f4-f3*f2 = 0 . If you try x = 0 then because f1*x + f2*y = 0, then f1 * 0 + f2*y = 0, then f2 * y = 0, which has a solution only if y = 0 or f2 = 0. But if f2 = 0 then the solution for x would have had a division by 0 so that possibility is out. This leaves only x = 0 and y = 0, or f1*f4 = f2*f3
Brian
el 20 de Dic. de 2011
you can solve a 2x2 by substitution or you can use rref.m. substitution is probably easier in this case!
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yasser
el 20 de Dic. de 2011
1 comentario
Walter Roberson
el 20 de Dic. de 2011
x = (f4*z1-f2*z2)/(f1*f4-f3*f2)
y = (f1*z2-f3*z1)/(f1*f4-f3*f2)
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