4. Solve the system of equations $\left\{\begin{array}{l}x+\sqrt{x+2 y}-2 y=\frac{7}{2}, \\ x^{2}+x+2 y-4 y^{2}=\frac{27}{2}\end{array}\right.$.
Answer: $\left(\frac{19}{4} ; \frac{17}{8}\right)$.

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

$$
\left\{\begin{array} { l } 
{ u + v = \frac { 7 } { 2 } } \\
{ u ^ { 2 } v + u ^ { 2 } = \frac { 2 7 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=\frac{7}{2}-u \\
u^{2}\left(\frac{7}{2}-u\right)+u^{2}=\frac{27}{2}
\end{array}\right.\right.
$$

From the second equation of the last system, it follows that $2 u^{3}-9 u^{2}+27=0$. By selecting the integer root $u=3$ and factoring out the factor $(u-3)$ from the left side of the last equation, we get $(u-3)\left(2 u^{2}-3 u-9\right)=0$, hence $u=3$ or $u=-\frac{9}{2}$. The value $u=-\frac{9}{2}$ is not suitable. For $u=3$, we get $v=\frac{1}{2}$. Then

$$
\left\{\begin{array} { l } 
{ x + 2 y = 9 , } \\
{ x - 2 y = \frac { 1 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 x = \frac { 1 9 } { 2 } } \\
{ 4 y = \frac { 1 7 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{19}{4} \\
y=\frac{17}{8}
\end{array}\right.\right.\right.
$$