Solve the following system of equations

$$
\begin{aligned}
& 3 x^{2}+2 y^{2}+2 x+3 y=0 \\
& 4 x^{2}-3 y^{2}-3 x+4 y=0
\end{aligned}
$$
Add to the first equation three times the second equation, and divide the resulting equation by 17. Then the equation

$$
x^{2}+y=0
$$

is obtained. Now subtract from four times the first equation of (1) three times the second equation. Then we get

$$
y^{2}+x=0
$$

From this it follows that $x=-y^{2}$. Using this in equation (2) yields

$$
y^{4}+y=0, \quad \text { or } \quad y\left(y^{3}+1\right)=0
$$

The solutions to this equation are: $y=0$ and $y=-1$. The corresponding values of $x$ can be obtained from equation (3): if $y=0$, then $x=0$, and if $y=-1$, then $x=-1$. Consequently, the solutions to the system of equations (1) can only be the pairs $x=y=0$ or $x=y=-1$. Substituting these into the system of equations, we can verify that they are indeed solutions.