How do I solve this algebra problem The question goes, solve in real number.
$x^5 - 5 x^3y - 5x^2 + 5xy^2 + 5y = 16 \tag{1}$ 
$ y^5 + 5xy^3 + 5y^2 + 5x^2y + 5x = -57 \tag{2}$ 
I tried simplifying the first equation to,
$$ x^5 + 5\left[ \left(xy+1 \right) \left( y - x^2 \right) \right] = 16 $$
and second equation to, 
$$ y^5 + 5 \left[ \left(xy+1 \right) \left( y^2 + x \right) \right] = -57$$
I know not much efforts shown, but this is where I'm stuck. Any hints on where do I go from here? 
Thanks
 A: COMMENT.- It seems that Batominovski's comment about $(1,-2)$ is true. One has $$x^5+5(xy+1)(y-x^2)=16\\y^5+5(xy+1)(y^2+x)=-57$$ From which 
$$\frac{y^2+x}{y-x^2}=\frac{y^5+57}{x^5-16}\qquad(1)$$
 A necessary condition for solutions $(x,y)$ is equality $(1)$. However it is not suffisant because if not it would be an infinity of solutions. Making separately
$$\frac{y^2+x}{y-x^2}=a\qquad(2)$$ $$\frac{y^5+57}{x^5-16}=a\qquad(3)$$ we have in $(2)$ a conic, hyperbola, ellipse, circle ($a=1$) and two lines ($a=-1$) while in $(3)$ there is a quintic.The corresponding graphics are suggestive in order to search solutions. In any case the graphics of the given equations (in the attached figure green and black respectively) also give $(1,-2)$ as probably only real solution. 

A: By adding these two equations I ended up in a quite symmetric term 
$$x^5+y^5+5[(1+xy)(y+x)(y-x+1)]~=~-41$$
I would also recommend to use the fact that you can write any $(x+y)^n$ only in terms of $xy$ and $(x+y)$ and the two $x^n,y^n$ like
$$(x+y)^3~=~x^3+y^3+3xy(x+y)$$
Maybe someone can proceed form there.
