how to solve simultaneous equations?

Dear sir/madam,

I need to solve two simultaneous linear equations. How could I do this in matlab? Looking forward to hearing from you soon.

Thanking you, BSD

[MUYIDEEN MOMOH]

Question

3x+2y=12
4x+6y=18

Matlab code

A=[3, 2 ; 4, 6];
B=[12; 18];
sol=linsolve(A,B)

Answer 1

[Paulo Silva]

Equations:

1x + 2y = 0
2x + 2y = 0

MATLAB code:

A = [1 2;2 2]
B = [0;0]
X = A\B

see examples from this pdf

[KELVIN]

Mhhhh!

Answer 2

[Ishika Shivahre]

x1+x2=1

0.718+y2 = 1

x1*P"= 0.718*86.8

x2*P2" = y2* 86.8

[Walter Roberson]

syms x1 x2 y2 P_dprime P2_dprime
eqn1 = x1 + x2 == 1

eqn1 = 

eqn2 = 0.718 + y2 == 1

eqn2 = 

eqn3 = x1 * P_dprime == 0.718*86.8

eqn3 = 

eqn4 = x2 * P2_dprime == y2 * 86.8

eqn4 = 

sol = solve([eqn1, eqn2, eqn3, eqn4], [x1, x2, y2, P_dprime])

sol = struct with fields:
          x1: (1250*P2_dprime - 30597)/(1250*P2_dprime)
          x2: 30597/(1250*P2_dprime)
          y2: 141/500
    P_dprime: (77903*P2_dprime)/(1250*P2_dprime - 30597)

That is as far as you can get. You have 4 equations in 5 variables, so you cannot solve for all of them simultaneously.

Answer 3

[Yaavendra Ramsaroop]

A=[3, 2 ; 4, 6];

B=[12; 18];

sol=linsolve(A,B)

Answer 4

[KELVIN]

Step 1: Express your equations into an Augmented Matrix where each equation represents a row of that matrix (excluding the answers/ the value beyond "=" sign.), assign the matrix to a variable. Let say A.

Step 2: Form a column matrix of the answers/ values beyond the "=" sign. Assign the column matrix to another variable B.

Step 3: Compute the solution by 'linsolve()' function OR sipmly A\B=inverse(A)*B

Solution=linsolve(A,B)

Answer 5

[SHRDRACK]

A=[3, 2 ; 4, 6];
B=[12; 18];
sol=inv(A)*B

enter in comand window

[Walter Roberson]

It is not recommended that you use inv() for this purpose; it is less precise then some of the alternatives.

A=[3, 2 ; 4, 6];
B=[12; 18];
sol=A\B

sol = 2x1

3.6000 0.6000

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

[Sam Chak]

Hi @SHRDRACK, In my Grade 10 Math class, my teacher taught me this method of solving a system of linear equations. It's natural to use the inv() command when searching online for how to compute the inverse of a square matrix. Interestingly, even the professor who taught my Numerical Methods course never showed me the trick of Left Matrix Division using "A\b" like @Walter Roberson did.

A    = [3, 2; 4, 6];
b    = [12; 18];
% step 1
detA = det(A)

detA = 10

% step 2
adjA = adjoint(A)

adjA = 2x2

6.0000 -2.0000 -4.0000 3.0000

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

% step 3
invA = (1/detA)*adjA

invA = 2x2

0.6000 -0.2000 -0.4000 0.3000

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

% step 4
x    = invA*b

x = 2x1

3.6000 0.6000

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>