11. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+\sqrt{x y}=28 \\
x^{2}+y^{2}+x y=336 .
\end{array}\right.
$$
11. $(4,16)$ and $(16,4)$
After rewriting the first equation into $x+y=28-\sqrt{x y}$, we square to get
$$
x^{2}+x y+y^{2}=784-56 \sqrt{x y} \text {. }
$$

Using the second given equation, the last equation becomes
$$
336=784-56 \sqrt{x y} \text { or } \sqrt{x y}=8 .
$$

Substituting this last equation to the first given equation, we get $y=$ $20-x$, and the second given equation then becomes
$$
x^{2}+(20-x)^{2}+64=336,
$$
which yields $x=4$ and $x=16$. Knowing that $x y=64$, the solutions are the ordered pairs $(4,16)$ and $(16,4)$.