solve two equations for two unknown variables

19 visualizaciones (últimos 30 días)
yasser
yasser el 20 de Dic. de 2011
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.

Respuesta aceptada

Dr. Seis
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);

Más respuestas (5)

Walter Roberson
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
yasser el 20 de Dic. de 2011
now how can i solv it?
  1 comentario
Walter Roberson
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

Iniciar sesión para comentar.


Brian
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!

yasser
yasser el 20 de Dic. de 2011
sorry, my bad in my case , i would say z1=ax+by z2=cx+dy a,b,c,d constants i want to put z1=const1 & z2=const2 then solve and find x & y ?
  1 comentario
Walter Roberson
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)

Iniciar sesión para comentar.


yasser
yasser el 20 de Dic. de 2011
@elige grant thx alot

Categorías

Más información sobre Systems of Nonlinear Equations en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by