4. 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 becomes

$$
\left\{\begin{array} { l } 
{ u + v = 6 , } \\
{ u ^ { 2 } v + u ^ { 2 } = 3 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=6-u, \\
u^{2}(6-u)+u^{2}=36 .
\end{array}\right.\right.
$$

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

$$
\left\{\begin{array} { l } 
{ 3 x - y = 9 , } \\
{ 3 x + y = 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 y = - 6 , } \\
{ 6 x = 1 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=2 \\
y=-3
\end{array}\right.\right.\right.
$$

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

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