4 unknown 4 equetions

clc;clear all;close all;clear vars;

syms a b c d

eq1=a+b+c+d==-9.3;

eq2=(b*a+2)+(2+d*c)+(c+d)*(b+a)==+7;

eq3=(b+a)+(d+c)*(b*a+2)+(a+b)*(2+d*c)+(c+d)==-4.507;

eq4=(b+a)*(d+c)+(2+d*c)*(b*a+2)+(c+d)*(a+b)==-11.181;

S=solve(eq1,eq2,eq3,eq4);

I couldnt find values pls help me

Answer 1

[Star Strider]

syms a b c d
eq1=a+b+c+d==-9.3;
eq2=(b*a+2)+(2+d*c)+(c+d)*(b+a)==+7;
eq3=(b+a)+(d+c)*(b*a+2)+(a+b)*(2+d*c)+(c+d)==-4.507;
eq4=(b+a)*(d+c)+(2+d*c)*(b*a+2)+(c+d)*(a+b)==-11.181;
S=solve(eq1,eq2,eq3,eq4);

a = vpa(S.a, 10)

a = 

b = vpa(S.b, 10)

b = 

c = vpa(S.c, 10)

c = 

d = vpa(S.d, 10)

d = 

If you want to convert them to double precision variables to use in numeric (rather than symbolic) calculations, use double instead of vpa.

.

[Samil Ekrem Salli]

Why this numbers have 6 solutions

[Star Strider]

Apparently because they generate 6-degree polynomials.

[Samil Ekrem Salli]

Thank you very much

[John D'Errico]

At least, the equivalent to a 6th degree polynomial remains, once you eliminate 3 of the 4 unknowns.

[Star Strider]

@Samil Ekrem Salli — As always, my pleasure!