Solve system of nonlinear equation

[abotaha]

Hello guys,

I am not good in advance mathematics.
I have system of nonlinear equation and I want to solve it analytically, but I face some difficulties.

Homework Statement

The system is :

(a-b)y^2-exy^2-2fx=-c
2fx^2-2cx-2fy^2=-d

Homework Equations

where: a,b,c,d,e,and f are constants.

The Attempt at a Solution

any suggestion and help would be appreciated.

 

[HallsofIvy]

That looks rather messy, especially for general coefficients like that. I think what I would try would be to solve one of the equations, the first, say, since it is linear in x, for x as a function of y, then put that value back into the second equation:
(\epsilon y^2+ 2f)x= (a- b)y^2+ c
x= \frac{(a-b)y^2+ c}{y^2+ 2f}

Replacing x by that in the second equation will result, eventually, in a sixth degree equation for y. There are no general methods of solving such an equation.

 

[jackmell]

When you can't (practially) solve it, then aren't you just curious what the answer might be? A CAS is such a wonderful tool in mathematics I believe and extends one's "reach" significantly in my opinion. Now you said you needed an "analytic" solution to it right? And since that can't be done "exactly", why not just try to solve it in Mathematica?

You could do:

Clear[a, b, c, d, e, f, x, y]
mysol = Solve[{(a - b) y^2 - e x y^2 - 2 f x == -c, 
   2 f x^2 - 2 c x - 2 f y^2 == -d}, {x, y}]

Surprisingly, Mathematica returns a "solution expression" in mysol. It's the six solutions indicated by Hallsofivy above and is too large for Mathemtica to display on the screen conveniently. In some ways, the expression returned by Mathematica can be considered the "analytic" solution although I assume it contains "root" objects which are then numerically computed when an explicit result is desired. Just for a test, I tried back-substituting the first solution into the left side of the first equation with some simple values for the constants. The result is the expected negative of the value of c which is -4.

In[19]:= 
N[(a - b) y^2 - e x y^2 - 2 f x //. {a -> 1, b -> 2, e -> 3, c -> 4, 
   d -> 5, e -> 6, f -> 7, mysol[[1, 1]], mysol[[1, 2]]}]

Out[19]= -4.

Now, if you wanted to study this further, you could then study the particular format Mathematica is using to represent the solutions.

Work like this offers quite a unique and new perspective into Mathematics education which is beneficial to the student's success in the field in my opinion. They should do it regularly.

 

[abotaha]

Thanks guys,
I also tried to simplify the system such as:
(a-b)(2fx^2 - 2cx + d) - ex(2fx^2 - 2cx + d) - 4f^2x = -2fc

I searched for the cubic formula http://en.wikipedia.org/wiki/Cubic_formula#General_formula_of_roots". I found that there are some restriction (limitation or may say conditions) that gives different solution which I really did not understand it.
could you explain to me little bit please or any suggestion about solution using cubic formula.

 

[vela]

Since y only appears as y², you could change variables to u=y², which leaves you with equations for two conic sections. Perhaps that might help you understand what the solutions are.