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6 equation with 6 unknowns

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enis çoko 2020-4-9

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编辑： Stephen23 2021-3-6

在 MATLAB Online 中打开

Hello everyone ,

I have 6 equations with 6 unknowns and ı dont know how to implement in matlab .Could you please help me ?

967.85 = L+V
611.6812=L*x(1) + V*y(1)
265.1909=L*x(2) +V*y(2)
x(1)*256.98=y(1)*1350.49
x(2)*1350.49=y(2)*556.29
(1-x(1)-x(2))*556.29=(1-y(1)-y(2))*256.98

Unknowns ; L ,V,x(1),x(2),y(1),y(2)

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Torsten 2020-4-9

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Try "fsolve".

enis çoko 2020-4-9

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在 MATLAB Online 中打开

syms L V x1 y1 x2 y2;
eqns = [L+V==967.85, L*x1+V*y1==611.6812, L*x2+V*y2==265.1909, x1/y1==1350.49/256.98, x2/y2==556.29/1350.49, (1-x1-x2)/(1-y1-y2)==256.98/556.29];
S=solve(eqns, [L V x1 y1 x2 y2]);
a=S.L;
b=S.V;
c=S.x1;
d=S.x2;
e=S.y1;
f=S.y2;
fprintf('%.2f\n',a)
fprintf('\n')
fprintf('%.2f\n',b)
fprintf('\n')
fprintf('%.2f\n',c)
fprintf('\n')
fprintf('%.2f\n',d)
fprintf('\n')
fprintf('%.2f\n',e)
fprintf('\n')
fprintf('%.2f\n',f)

I tried this method but one of the solution has minus sign.It must be positive but I can't see where I made a mistake

Torsten 2020-4-9

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Rounding the coefficients to two decimal places might cause an unexpected solution.

Mathematica gives two solutions, but both have negative components.

The equations look correctly implemented.

enis çoko 2020-4-9

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If I follow a new path for example I want to use ''fsolve'' what should ı do ?

Torsten 2020-4-9

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编辑：Torsten 2020-4-9

fsolve won't help - in my opinion, the rounding is the problem.

What you could do is search for a positive solution that satisfies the equations best possible. For this, try "lsqnonlin" and set the lower bounds of the variables to 0.

Ameer Hamza 2020-4-9

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编辑：Ameer Hamza 2020-4-9

enis, you mentioned "I tried this method but one of the solution has minus sign.It must be positive"

Do you know for sure that all the solutions of these equations are positive and solutions given by the solve() function are wrong? Or is this a requirement that the solutions should be positive?

enis çoko 2020-4-9

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Ameer, this is separation process problem that means x and y unknowns are the concentration and we know that the concentration must be positive.All in all, yes ı am sure about that.

Torsten 2020-4-9

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x and y are positive for the mathematica solution, but V is negative.

enis çoko 2020-4-9

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Could you send me mathematica codes? May be I can rearrange that.I don't think of any other method at this stage.

Ameer Hamza 2020-4-9

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enis, yes, from the physical interpretation, it makes sense that the solution must be positive? Can you show us the equations in mathematical form? Because if you put the solution given by MATLAB into your equations, it will satisfy all the equations. I suspect there might be an issue in writing the equations in MATLAB syntax.

Torsten 2020-4-9

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在 MATLAB Online 中打开

I can give you the result:

L = 977.53
V = -9.68
x(1) = 0.6269
x(2) = 0.2779
y(1) = 0.1192
y(2) = 0.6747

enis çoko 2020-4-9

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enis çoko 2020-4-9

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exactly this

Alex Sha 2020-4-10

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There is one more solution:

No.	2
l	1763.32683383609
v	-795.476833836095
x1	0.379464645385751
x2	-1.58011154773944
y1	0.0722069949212733
y2	-3.83599353593743

enis çoko 2020-4-10

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Alex , thanks for your attention but all values must be positive value thats is not enough.

Ameer Hamza 2020-4-10

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enis, please recheck your hand derivations. The negative solution does satisfy these equations. There may be a mistake in these equations, or there is some error in the numeric coefficients.