4. Solve the system of equations $\left\{\begin{array}{l}2 x+\sqrt{2 x+3 y}-3 y=5, \\ 4 x^{2}+2 x+3 y-9 y^{2}=32 .\end{array}\right.$
Answer: $\left(\frac{17}{4} ; \frac{5}{2}\right)$.

Solution. Let $\sqrt{2 x+3 y}=u, 2 x-3 y=v$. Then the system becomes

$$
\left\{\begin{array} { l } 
{ u + v = 5 , } \\
{ u ^ { 2 } v + u ^ { 2 } = 3 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=5-u \\
u^{2}(5-u)+u^{2}=32
\end{array}\right.\right.
$$

From the second equation of the last system, we get $u^{3}-6 u^{2}+32=0$. By selecting the integer root $u=-2$ and factoring out $(u+2)$ from the left side of the last equation, we obtain $(u+2)\left(u^{2}-8 u+16\right)=0$, hence $u=-2$ or $u=4$. The value $u=-2$ is not suitable. For $u=4$, we get $v=1$. Then

$$
\left\{\begin{array} { l } 
{ 2 x + 3 y = 1 6 } \\
{ 2 x - 3 y = 1 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 4 x = 1 7 , } \\
{ 6 y = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{17}{4} \\
y=\frac{5}{2}
\end{array}\right.\right.\right.
$$