Re: difficult system of equation

leonid.uk@xxxxxxxxx a écrit : 
oh..I am sorry, last equation must be 8*y*z+y^2+2*z^2=1


Hi,

let e1=4*x*y+y^2+2*z^2=-3
    e2=4*x*z+x^2+2*z^2=1
    e3=8*y*z+y^2+2*z^2=1

consider e3-e1

                       8 y z - 4 x y = 4

solve(e3-e1,y);

                                1

                       y = - --------

                             -2 z + x

supposing that -2z+x\neq 0

substitute in e2, y by 1/(x-2*z)

                              2      2

                     4 x z + x  + 2 z  = 1

equation of 2° degree in x for example

x=-2*z+(2*z^2+1)^(1/2) ou x=-2*z-(2*z^2+1)^(1/2)

....
.