solve two equations for two unknown variables

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yasser
yasser 2011 年 12 月 20 日
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.

採用された回答

Dr. Seis
Dr. Seis 2011 年 12 月 20 日
G*m = d
G = [f1,f2;f3,f4];
d = [z1;z2];
m = G\d;
x = m(1);
y = m(2);

その他の回答 (5 件)

Walter Roberson
Walter Roberson 2011 年 12 月 20 日
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
  1 件のコメント
yasser
yasser 2011 年 12 月 20 日
yes, f1,f2,f3&f4 are constants.

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yasser
yasser 2011 年 12 月 20 日
now how can i solv it?
  1 件のコメント
Walter Roberson
Walter Roberson 2011 年 12 月 20 日
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

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Brian
Brian 2011 年 12 月 20 日
you can solve a 2x2 by substitution or you can use rref.m. substitution is probably easier in this case!

yasser
yasser 2011 年 12 月 20 日
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 件のコメント
Walter Roberson
Walter Roberson 2011 年 12 月 20 日
x = (f4*z1-f2*z2)/(f1*f4-f3*f2)
y = (f1*z2-f3*z1)/(f1*f4-f3*f2)

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yasser
yasser 2011 年 12 月 20 日
@elige grant thx alot

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