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 $(4+1)=(2^2+1^2)=(3^2-2^2)=(3-2)(3+2)=5$ 
 A: 
...is there a method to find all the solutions[?]

Yes, there is: draw the graphs of the functions $x\mapsto3-\sqrt{x}$ and $x\mapsto(5-x)^2$ on $[0,5]$, these intersect at $(4,1)$ and there only hence the unique solution of your system is $(x,y)=(4,1)$.
A: For the question in the title, you can move the $\sqrt y$ to the other side, square, and get a quartic which yields to the rational root theorem.
A: If you make the change of variables 
$$\begin{equation*}
u=\sqrt{x}\geq 0,\qquad v=\sqrt{y}\geq 0,
\end{equation*}$$
then you need to solve
$$\begin{equation*}
\left\{ 
\begin{array}{c}
u+v^{2}=3 \\ 
u^{2}+v=5
\end{array}
\right. 
\end{equation*}$$
or
$$\begin{eqnarray*}
&&\left\{ 
\begin{array}{c}
u=3-v^{2} \\ 
v^{4}-6v^{2}+v+4=0
\end{array}
\right.  \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
u=3-v^{2} \\ 
\left( v-1\right) \left( v^{3}+v^{2}-5v-4\right) =0.
\end{array}
\right. 
\end{eqnarray*}$$
So $u=2,v=1$ is a solution, i.e. $x=4,y=1$ in the original variables. And
since $v^{3}+v^{2}-5v-4=0$ has 3 real solutions, one positive $v_{1}\approx
2.164\,2\gt \sqrt{3}$ and two negative $v_{2}\approx -2.391\,4,v_{3}\approx
-0.772\,87$, there are no other solutions because $u=3-v_{1}^{2}<0$. 
A: $(3-y)^2+\sqrt{y}=5$
$y^2-6y+\sqrt{y}+4=0$
$(\sqrt{y}-1)(y\sqrt{y}+y-5\sqrt{y}-4)=0$
i.e. $\sqrt{y}=1$, and $x=4$ is a solution
