Output Answer by solve
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Hello
In the output the values give this result. I want to get the values:
syms X1 X2 X3 X4
eq=[X1^2+X2^2==1949.947, (X1*X3)+(X2*X4)==-28.338, eq3=(X3*X1)+(X4*X2)==-28.338, eq4=X3^2=X4^2-81.892]
[x1,x2,x3,x4]=solve([eq],[X1,X2,X3,X4])
x1 =
Empty sym: 0-by-1
x2 =
Empty sym: 0-by-1
x3 =
Empty sym: 0-by-1
x4 =
Empty sym: 0-by-1
Thank you for helping me
5 个评论
Walter Roberson
2022-6-1
syms X1 X2 X3 X4
eq=[X1^2+X2^2==1949.947, (X1*X3)+(X2*X4)==-28.338, eq3=(X3*X1)+(X4*X2)==-28.338, eq4=X3^2=X4^2-81.892]
[x1,x2,x3,x4]=solve([eq],[X1,X2,X3,X4])
You have eq3= and eq4= in the middle of the []. These days that is equivalent to
eq = [ X1^2+X2^2==1949.947,...
(X1*X3)+(X2*X4)==-28.338, ...
'eq3', ...
(X3*X1)+(X4*X2)==-28.338, ...
'eq4', ...
X3^2=X4^2-81.892]
[x1,x2,x3,x4]=solve([eq],[X1,X2,X3,X4])
Farshid R
2022-6-1
Torsten
2022-6-1
Is there another way?
Yes: copy KSSV's solution.
Maybe he will be so kind to replace the numbers by symbolic variables so that you can play with the right-hand side of your equations.
Farshid R
2022-6-1
Walter Roberson
2022-6-1
编辑：Walter Roberson
2022-6-1
Your second and third equations are the same. There would not be a unique solution .
回答（1 个）
syms X1 X2 X3 X4
eq1 = X1^2+X2^2==1949.947 ;
eq2 = (X1*X3)+(X2*X4)==-28.338;
eq3 = (X3*X1)+(X4*X2)==-28.338;
eq4 = X3^2==X4^2-81.892 ;
eqs = [eq1, eq2, eq3, eq4] ;
s=solve(eqs,[X1,X2,X3,X4]) ;
s.X1
s.X2
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2022-6-1
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