4 unknown 4 equetions
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Samil Ekrem Salli
on 8 Jun 2021
Commented: Star Strider
on 8 Jun 2021
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
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Star Strider
on 8 Jun 2021
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)
b = vpa(S.b, 10)
c = vpa(S.c, 10)
d = vpa(S.d, 10)
If you want to convert them to double precision variables to use in numeric (rather than symbolic) calculations, use double instead of vpa.
.
5 Comments
John D'Errico
on 8 Jun 2021
Edited: John D'Errico
on 8 Jun 2021
At least, the equivalent to a 6th degree polynomial remains, once you eliminate 3 of the 4 unknowns.
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