MATLAB: How do i solve a system of equasion with mulitple variables

algebrasystem of equations

Im trying to solve this for c1 c3 and c4

equ1: m1*c1=m2*c2
equ2: v = c3*(h1+h2)+c4
equ3: c1*h1=c3*h1+c4

im not sure if this code i wrote is even close to correct any help is appreciated.

syms c1 c3 c4 m1 m2 h1 h2 v
equ1 = m2*c3/m1;
equ2 = (v-c4)/(h1+h2);
equ3 = c1*h1-c3*h1;
sol = solve(equ1,equ2,equ3)

Best Answer

syms c1 c3 c4 m1 m2 h1 h2 v
equ1 = c1==m2*c3/m1;
equ2 = c3==(v-c4)/(h1+h2);
equ3 = c4==c1*h1-c3*h1;
sol = solve([equ1,equ2,equ3],[c1,c3,c4]);
c1=sol.c1,c3=sol.c3,c4=sol.c4
c1 = 
c3 = 
c4 =

Related Solutions

MATLAB: Solve symbol four equations for c1 c2 c3 c4

Thanks, I solve it,

clc
clear
syms a b x y c1 c2 c3 c4 T1 T2 T3 T4  
T=c1+c2*x+c3*y+c4*x*y;
%%
r1=c1+c2*a+c3*b+c4*a*b-T1;
r2=c1-c2*a+c3*b-c4*a*b-T2;
r3=c1-c2*a-c3*b+c4*a*b-T3;
r4=c1+c2*a-c3*b-c4*a*b-T4;
%%Find C1 and C4
p13=r1+r3;
p24=r2+r4;
n1=solve(p13==0,c1);
n2=solve(p24==0,c1);
C4=solve(n1==n2,c4)
w1=solve(p13==0,c4);
w2=solve(p24==0,c4);
C1=solve(w1==w2,c1)
r1=subs(r1,c1,C1);
r1=subs(r1,c4,C4);
r2=subs(r2,c1,C1);
r2=subs(r2,c4,C4);
r3=subs(r3,c1,C1);
r3=subs(r3,c4,C4);
r4=subs(r4,c1,C1);
r4=subs(r4,c4,C4);
%%Find C3
q12=r1+r2;
q34=r3+r4;
C3=solve(q12==0,c3)
r1=subs(r1,c3,C3);
r2=subs(r2,c3,C3);
r3=subs(r3,c3,C3);
r4=subs(r4,c3,C3);
%% find C2
C2=solve(r1==0,c2)
r1=subs(r1,c2,C2);
r2=subs(r2,c2,C2);
r3=subs(r3,c2,C2);
r4=subs(r4,c2,C2);
C1
C2
C3
C4

MATLAB: How to solve 4 linear equations with 4 unknowns

A = [-1 -1 -1.644 8.938
     -1 -1 -2.263 1.761
     -1 -1 -12.62 0.63
     -1 -1 -1     -1];
b = [93.386; 73.02; -141.48; 31.725];
A\b

However, it is obviously rank deficient. R1 and R2 are only ever used in the form R1+R2, so it is not possible to distinguish the two of them. You can rewrite all the R1+R2 as a temporary variable R1plusR2 and then the first three equations would be three equations in three variables, which would be enough to find all of the values, R1plusR2, R3, and R4. If you were then to substitute those results into the fourth equation, since it has no new forms of R1 or R2, then the left and right sides are completely determined, and the equation would either be false (inconsistent with the first 3) or true (consistent with the first 3) -- but if it were true then you still would not be able to figure out R1 separately from R2.

Best Answer

Related Solutions

MATLAB: Solve symbol four equations for c1 c2 c3 c4

MATLAB: How to solve 4 linear equations with 4 unknowns

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