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Solving a system of equations without "syms"

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Daniel Miller
Daniel Miller on 11 Oct 2019
Answered: GAGANDEEP KAUR on 2 Nov 2020 at 7:31
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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Accepted Answer

jeewan atwal
jeewan atwal on 11 Oct 2019
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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Steven Lord
Steven Lord on 11 Oct 2019
DON'T use inv. I know that may be how you were told to solve a linear system in school. In theory, that may be fine. In practice, use the backslash operator (\) instead.
jeewan atwal
jeewan atwal on 11 Oct 2019
If you have any doubt, you are free to ask. Happy to help.
Thankyou Steven Lord for the info.
Daniel Miller
Daniel Miller on 11 Oct 2019
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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More Answers (2)


GAGANDEEP KAUR
GAGANDEEP KAUR on 2 Nov 2020 at 7:31
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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