54 (1182). Solve the equation with two variables

$$
x^{2}+2 \sqrt{3} x+y-4 \sqrt{y}+7=0
$$
Solution. We have:

$$
x^{2}+2 \sqrt{3} x+3+y-4 \sqrt{y}+4=0
$$

or

$$
(x+\sqrt{3})^{2}+(\sqrt{y}-2)^{2}=0
$$

from which

$$
x=-\sqrt{3}, \sqrt{y}=2, y=4
$$

Answer: $(-\sqrt{3} ; 4)$.

55(1183). Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}+y^{2}-2 z^{2}=0 \\
x+y+z \\
x y=-z^{2}
\end{array}\right.
$$

Solution. Substitute the value $-z^{2}=x y$ from the third equation into the first. We get $(x+y)^{2}=0$, i.e., $x=-y$, so $z=8$ (from the second equation). Substituting the values $x=-y$ and $z=8$ into the third equation of the system, we find $x^{2}=64$, i.e., $x_{1}=8, x_{2}=-8$.

Answer: $\{(8 ;-8 ; 8),(-8 ; 8 ; 8)\}$.

$\square$ 56(1184). Solve the system of equations in natural numbers:

$$
\left\{\begin{array}{l}
x+y+z=14 \\
x+y z=19
\end{array}\right.
$$

Solution. Subtract the first equation of the system from the second. We get $y z-y-z=5$, from which $y(z-1)=5+z$,

$$
y=\frac{5+z}{z-1}=\frac{z-1+6}{z-1}=1+\frac{6}{z-1}
$$

Since $y \in \mathbb{N}, z \in \mathbb{N}$, it follows from (1) that

$$
z_{1}=2, y_{1}=7 ; z_{2}=3, y_{2}=4 ; z_{3}=4, y_{3}=3 ; z_{4}=7, y_{4}=2 \text {. }
$$

Substituting the values of $y$ and $z$ into one of the equations of the given system, we find the corresponding values of $x$.

Answer: $\{(5 ; 7 ; 2),(7 ; 4 ; 3),(7 ; 3 ; 4),(5 ; 2 ; 7)\}$.