# How to solve linear and non linear equation system?

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Hello

Can anyone tell me how I can solve the following system of equation

1) a1v1-a2v2-a3v3-a4v4=0

2) a3v3-a5v5-a6v6=0

3) a2v2+a6v6-a7v7=0

4) a7v7+a5v5+a4v4-a8v8=0

5) 500v1^2+p1-p2-500v2^2=0

6) 500v1^2+p1-p3-500v3^2=0

7) 500v1^2+p1-p4-500v4^2=0

8) 500v3^2+p3-p5-500v5^2=0

9) 500v3^2+p3-p6-500v6^2=0

10) 500v7^2+p7-p2-500v2^2=0

11) 500v7^2+p7-p6-500v6^2=0

12) 500v8^2+p8-p4-500v4^2=0

13) 500v8^2+p8-p5-500v5^2=0

14) 500v8^2+p8-p7-500v7^2=0

How to solve the above equation system for v's and p's where v1 and p1 are known

Please guide!!!

Regards

### Accepted Answer

George Papazafeiropoulos
on 4 Jun 2014

Edited: George Papazafeiropoulos
on 4 Jun 2014

##### 7 Comments

### More Answers (1)

John D'Errico
on 4 Jun 2014

Edited: John D'Errico
on 4 Jun 2014

You have 14 equations. And it looks like 16 variables, {v1,...,v8,p1,...,p8}.

But you tell us that v1 and p1 are known, so effectively only 14 unknowns. With as many equations as unknowns, there will OFTEN be exact solution(s), but there is no assurance thereof. In fact, for nonlinear problems, there will often be multiple solutions. (See below.)

Worse, the equations are mostly quadratic in the variables. So the system would correspond to a quite high order polynomial equation, on the order of 25. So no exact solution is available in any event, since the effective degree is greater than 5.

At best you can solve for numerical solutions, using a tool like lsqnonlin or fsolve, ASSUMING that you will provide values for all known variables. If they are left as symbolic variables, you will fail.

There will be many such solutions, so different starting values will yield different solutions. I describe the idea of a basin of attraction in my optimization tips and tricks doc on the file exchange. Effectively, if you start an optimizer anywhere in the basin of a given solution, it will converge to that solution. Such basins will not in general be convex sets, since there will be many solutions.

Edit: I talked about an effective degree before. With polynomial problems as we have here, suppose we had TWO quadratic equations in TWO unknowns. A "simple" version of your problem, sort of. The idea would be to use the quadratic formula to solve for one variable in one of the equations. Then substitute back into the other equation. This will generally yield a FOURTH order polynomial problem to solve. So a system of two quadratic equations in two unknowns is equivalent to an effective total order of 4.

A system with a dozen quadratic equations, plus some linear equations, will be equivalent to at least an effective order of 25 or so. This tells me there will be no symbolic solution to be found, unless you were very lucky. In real life, you never get lucky on these things. (Were it homework, you often get lucky.)

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