Solving a system of equations without "syms"

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Daniel Miller
Daniel Miller 2019-10-11
回答： GAGANDEEP KAUR 2020-11-2
Hello!
I have been given the following system of equations that I should solve:
2x1 + 4x2 + 7x3 = 64
3x1 + x2 + 8x3 = 71
-2x = -4
Now, the problem is that I'm on the MatLab Grader platform and it doesn't seem to have this Symbolic Math Tool (i.e. "syms") in it. It only returns the error "Undefined function 'syms' for input arguments of type 'char'."
My code looks like this:
syms x1 x2 x3
equation1 = 2*x1 + 4*x2 + 7*x3 == 64;
equation2 = 2*x1 + 1*x2 + 8*x3 == 71;
equation3 = -2*x1 == -4;
solutionX = solve([equation1, equation2, equation3], [x1, x2, x3]);
SolutionX1 = solution.x1
SolutionX2 = solution.x2
SolutionX3 = solution.x3
Is there any other method I could use instead of using "syms"?
Thank you in advance!

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采纳的回答

jeewan atwal
jeewan atwal 2019-10-11
A*x = b;
for your case
A = [2 4 7; 2 1 8; -2 0 0];
b = [64;71;-4];
where x = [x1;x2;x3]
solution x can be found using either of two methods as follows:
x = inv(A)*b;
or
x = linsolve(A,b)
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jeewan atwal
jeewan atwal 2019-10-11
If you have any doubt, you are free to ask. Happy to help.
Thankyou Steven Lord for the info.
Daniel Miller
Daniel Miller 2019-10-11
Alright, thanks for the additional heads-up. Don't know how I would've solved this without your help so big appreciations for it!

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更多回答（2 个）

Stephan
Stephan 2019-10-11
编辑：Stephan 2019-10-11
GAGANDEEP KAUR
GAGANDEEP KAUR 2020-11-2
I also need to determine some variables using syms with solve command but find some issue with syms itself.
Code is like this:
for i=1:9
syms a b c d e ;
%calculating mole fractions of ionic species
x1=[0.5096 0.5092 0.5087 0.4852 0.4847 0.4834 0.4804 0.4805 0.4803];
x2=[0.0963 0.0964 0.0965 0.1163 0.1161 0.1158 0.1275 0.1266 0.1253];
x3=[0.3941 0.3944 0.3948 0.3985 0.3992 0.4008 0.3921 0.3929 0.3943];
T=[394.15 399.15 404.15 375.15 390.15 405.15 374.15 392.15 406.15];
%Equilibrium constant for reaction 1 (Solvation reaction)
K1=exp((-8.549)+(6692/T(i)));
%Equilibrium constant for reaction 2(Ionization of water)
K2=10^(-14);
%Equilibrium constant for reaction 3(Dissociation of HI)
K3=exp((16.93565)+((1250)/T(i))+(-2.575*log(T(i))));
%Equilibrium constant for reaction 4(Polyiodide formation a)
K4=exp((-936.28)+((40216.27)/T(i))+(151.983*(log(T(i))))+(-0.1675*(T(i))));
%Equilibrium constant for reaction 5(Polyiodide formation b)
K5=exp((1044.78)+(-45171.42/T(i))+(-165.20*log(T(i)))+(0.1511*(T(i))));
eqns=[((d*(c-e))/((x1(i)-a-b-c-d)^5)*(x2(i)-d-b-c))==K1,((a+b+c)*a)/((x1(i)-a-b-c-d)^2)==K2,(((a+b+c)*(d+b))/((x1(i)-a-b-c-d)*(x2(i)-d-b-c-e)))==K3,(((a+b+c)*(c-e))/((x1(i)-a-b-c-d)^4*(x2(i)-d-b-c-e)))==K4,(((e)*(x1(i)-a-b-c-d)^3)/((c-e)*(x3(i)-e)))==K5];
S=solve(eqns,a, b, c, d, e)
S.a(S.a<0)=[];
S.b(S.b<0)=[];
S.c(S.c<0)=[];
S.d(S.d<0)=[];
S.e(S.e<0)=[];
S.a=double(S.a);
S.b=double(S.b);
S.c=double(S.c);
S.d=double(S.d);
S.e=double(S.e);
end
A positive response is awaited

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