5. Solve the system of equations $\left\{\begin{array}{l}y+\sqrt{y-3 x}+3 x=12, \\ y^{2}+y-3 x-9 x^{2}=144\end{array}\right.$.
Answer: $(-24 ; 72),\left(-\frac{4}{3} ; 12\right)$.

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

$$
\left\{\begin{array} { l } 
{ u + v = 1 2 } \\
{ u ^ { 2 } v + u ^ { 2 } = 1 4 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=12-u \\
u^{2}(12-u)+u^{2}=144
\end{array}\right.\right.
$$

From the second equation of the last system, it follows that $u^{3}-13 u^{2}+144=0$. By selecting the integer root $u=4$ and factoring out the factor $(u-4)$ from the left side of the last equation, we get $(u-4)\left(u^{2}-9 u-36\right)=0$, from which $u=-3, u=4$ or $u=12$. The value $u=-3$ is not suitable. For $u=4$, we get $v=8$. Then

$$
\left\{\begin{array} { l } 
{ y - 3 x = 1 6 } \\
{ y + 3 x = 8 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 y = 2 4 , } \\
{ 6 x = - 8 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=-\frac{4}{3} \\
y=12
\end{array}\right.\right.\right.
$$

For $u=12$, we get $v=0$. Then

$$
\left\{\begin{array} { l } 
{ y - 3 x = 1 4 4 , } \\
{ y + 3 x = 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 y = 1 4 4 } \\
{ 6 x = - 1 4 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=-24 \\
y=72
\end{array}\right.\right.\right.
$$