SOLUTION: If 1/x+1/y =-1 , x^3 + y^3 = 4 find x & y

1.   +  = -1  ====>  x + y = - xy     (1)


2.   +  = 4  ====>

     = 4.                  (2)


        In (2), replace (x+y) by -xy   and  replace  (x^2 - xy + y^2)  by   =  - ,   based on  (1). 
        You will get instead of (2)


      = 4.              (3)

     Let t = xy  be new variable.  Then  (3)  becomes 

     -t*(t^2 -3t) = 4,

     t^3 - 3t^2 + 4 = 0.               (4)


     It is easy to check that t= -1 and t= 2  are the roots to (4).

     So, from the two original equations we have  two and only two possibilities for xy

         xy = -1   OR   xy = 2.


3.  Thus the original system of two equations (one of which is of the degree 3 !)
    deploys in two systems of much simpler equations:

    a)  x + y =  1
        xy    = -1

        which implies x*(1-x) = -1  ====>  x*(x-1) = 1  ====>  x^2 -x - 1 = 0  ====>   =  = 

OR
    b)  x + y = -2
        xy    =  2

        which implies x*(-2-x) = 2  ====>  x*(x+2) = -2  ====>  x^2 +2x + 2 = 0  ====>   =  =  = .


Answer.  There are 4 solutions:  two real  (x,y) = ( ,  ),  (x,y) = ( ,  ),  

                           and two complex  (x,y) = ( ,  ),  (x,y) = ( ,  ).