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How to solve this system of equations?

  1. Sep 29, 2012 #1
    I don't know if I'm posting in the wright section, but i need to solve this system of equations:

    {5x^4 - 4x + 5y = 0
    5y^4 + 5x - 4y = 0}

    I know one answer is (0;0), but i don't know how to show it, also there is another point.
    Maybe someone can help me?
  2. jcsd
  3. Sep 29, 2012 #2


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    Dearly Missed

    If you subtract the equations from each other, you get the following:


    Now, you have:
    x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3), and thus, you may factorize the above equation as:
  4. Sep 29, 2012 #3
    Thank you very much :)
  5. Oct 13, 2012 #4

    Ray Vickson

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    One can use Groebner basis methods to get at all the solutions. Using the Groebner package in Maple 11, I get:
    sys:=[5*x^4-4*x+5*y,5*y^4+5*x-4*y]:lprint(sys); <== input
    [5*x^4-4*x+5*y, 5*y^4+5*x-4*y]
    with(Groebner): <=== load the Groebner package
    [225*y+756*y^4-1280*y^7+2400*y^10-2000*y^13+625*y^16, 5*y^4+5*x-4*y]

    This B is the output. The two original equations are equivalent to setting the two components of B to zero (although, of course, there may be some extraneous roots, so those must be checked in the original system). Note that the first component of B is a polynomial in y alone, so we can find roots of the two-equation system by first solving the polynomial in y, then solving for x by setting the second expression in B to zero.

    225*y+756*y^4-1280*y^7+2400*y^10-2000*y^13+625*y^16 <--- set to zero

    Let's try to factor the polynomial.

    y*(5*y^3+1)*(125*y^12-425*y^9+565*y^6-369*y^3+225) <=== the factors

    So, y = 0 or 5*y^3 + 1 = 0, or
    125*y^12 -425*y^9 + 565*y^6 -360*y^3 + 225 = 0, which is a 4th degree polynomial in z = y^3.

    After solving for y, we get x from 5*x = 4*y - 5*y^4.

    Note: doing anything like this by hand would take months or years of work and require hundreds of pages of algebraic work. Use of a computer algebra system is essential.

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