5. Solve the system of equations $\left\{\begin{array}{l}3 x+\sqrt{3 x-y}+y=6, \\ 9 x^{2}+3 x-y-y^{2}=36 .\end{array}\right.$
Answer: $(2, -3), (6, -18)$.

Solution. Let $\sqrt{3 x-y}=u, 3 x+y=v$. Then the system takes the form

“Phystech-2016”, mathematics, solution to ticket 4

$$
\left\{\begin{array}{l}
u + v = 6, \\
u^2 v + u^2 = 36
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=6-u, \\
u^2(6-u)+u^2=36
\end{array}\right.\right.
$$

From the second equation of the last system, it follows that $u^3 - 7 u^2 + 36 = 0$. By selecting the integer root $u=-2$ and factoring out the factor $(u+4)$ from the left side of the last equation, we get $(u+2)(u^2 - 9 u + 18) = 0$, from which $u=-2, u=3$ or $u=6$. The value $u=-2$ is not suitable. For $u=3$, we get $v=3$. Then

$$
\left\{\begin{array}{l}
3 x - y = 9, \\
3 x + y = 3
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 y = -6, \\
6 x = 12
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=2 \\
y=-3
\end{array}\right.\right.\right.
$$

For $u=6$, we get $v=0$. Then

$$
\left\{\begin{array}{l}
3 x - y = 36 \\
3 x + y = 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 y = -36 \\
6 x = 36
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=6 \\
y=-18
\end{array}\right.\right.\right.
$$