is there a method to find all the solutions to, $y^2-6y+\sqrt{y}+4=0$ is there a method to find all the solutions to the following set of irrational equations,
$\sqrt{x}+y=3$
$x+\sqrt{y}=5$
NOTE: $(4-1)=(2-1)(2+1)=3$ and $(4+1)=(2^2+1^2)=(3^2-2^2)=(3-2)(3+2)=5$ 

$(3-y)^2+\sqrt{y}=5$
$y^2-6y+\sqrt{y}+4=0$
$(\sqrt{y}-1)(y\sqrt{y}+y-5\sqrt{y}-4)=0$
i.e. $\sqrt{y}=1$, and $x=4$ is a solution
