# How to solve this simple system of 2 equation in MATLAB

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Abdulaziz Abutunis on 4 May 2018
Commented: John BG on 10 May 2018
Hi All;
I wonder if Matlab can solve this two equations to find the Fx and Fy as a function in the other constants. I know I can do it by hand just want to validate. Please see the initial code below
syms D L COST SINT Fx Fy
Fx * COST + Fy *SINT - D =0;
Fx * SINT + Fy * COST -L =0;
solve ( text: the two equations for Fx and Fy)
Thank you for your valuable suggestion
Aziz

John BG on 6 May 2018
Edited: John BG on 6 May 2018
Hi Abdulaziz
syms v Fx Fy D L
b=[D ;L]
b =
D
L
A=[(1-v^2)^.5 v;v (1-v^2)^.5]
A =
[ (1 - v^2)^(1/2), v]
[ v, (1 - v^2)^(1/2)]
F=A\b
=
(D*v^2 - D + L*v*(1 - v^2)^(1/2))/((1 - v^2)^(1/2)*(2*v^2 - 1))
(D*v - L*(1 - v^2)^(1/2))/(2*v^2 - 1)
.
these are the expressions
.
Fx=F(1)
=
(D*v^2 - D + L*v*(1 - v^2)^(1/2))/((1 - v^2)^(1/2)*(2*v^2 - 1))
Fy=F(2)
Fy =
(D*v - L*(1 - v^2)^(1/2))/(2*v^2 - 1)
.
with
v=sin(t)
.
thanks in advance for time and attention
John BG

Stephan on 4 May 2018
Edited: Stephan on 4 May 2018
Hi,
you could do so:
% declare syms
syms D L t;
% Coefficient Matrix
A = [cos(t) sin(t); sin(t) cos(t)];
% RHS
b = [D ; L];
% Unknown: Fx, Fy
F = A\b;
% create Matlab function
fun = matlabFunction(F);
% Test for D = 0, L = -1 and t = pi()
D = 0;
L = -1;
t = pi();
[F] = fun(D, L, t)
this gives you a vector F containing Fx and Fy:
F =
0.0000
1.0000
or you do the same in a live script with symbolic toolbox and get the same result but nice: Best regards
Stephan

Show 1 older comment
Jan on 7 May 2018
@Abdulaziz Abutunis: The shown output for F is the x and y component already in dependence to the other variables. So what do you call "not the final result"?
John BG on 7 May 2018
Abdulaziz may not need the anonymous function, just the expressions of the results, have a look at my answer.
Abdulaziz Abutunis on 7 May 2018
That what I meant John. Sorry, Stephan if my question statement was not clear enough.

Abdulaziz Abutunis on 7 May 2018
Thank you John and all of you who have valuable suggestions. John, that was exactly what I want

#### 1 Comment

John BG on 10 May 2018
Thanks Abdulaziz
feel free to ask me about having a look at any other particular question that you may consider I would be able to assist with.
Regards
John BG