Solve the following system of equations:

$$
\begin{aligned}
\frac{1}{x}+\frac{1}{y} & =2 \frac{1}{4} \\
\frac{x^{2}}{y}+\frac{y^{2}}{x} & =32 \frac{1}{16}
\end{aligned}
$$
The two equations can also be written as:

$$
x+y=\frac{9}{4} x y
$$

$$
x^{3}+y^{3}=\frac{513}{16} x y
$$

Divide the second equation by the first:

$$
x^{2}-x y+y^{2}=\frac{57}{4}
$$

Square the first equation:

$$
x^{2}+2 x y+y^{2}=\frac{81}{16} x^{2} y^{2}
$$

Subtract (3) from (4), we get:

$$
\frac{81}{16} x^{2} y^{2}-3 x y=\frac{57}{4}
$$

From this, the product \( x y \) has two values: 2 and \(-\frac{38}{27}\); substituting these values into (1), \( x+y \) has two values: \(\frac{9}{2}\) and \(-\frac{19}{6}\); thus, \( x \) and \( y \) are the roots of the following equations:

$$
z^{2}-\frac{9}{2} z+2=0, u^{2}+\frac{19}{6} u-\frac{38}{7}=0
$$

From these equations:

$$
x_{1}=y_{1}^{\prime}=4, y_{1}=x_{1}^{\prime}=\frac{1}{2} ; x_{2}=y_{2}^{\prime}=\frac{1}{12}\left(-19+\sqrt{\frac{1691}{3}}\right), y_{2}=x_{2}^{\prime}=\frac{1}{12}\left(-19-\sqrt{\frac{1691}{3}}\right)
$$

(Béla Grünhut.)

The problem was also solved by: Bálint B., Devecis M., Freund A., Friedmann B., Geist E., Goldstein K., Goldziher K., Gy. Halász, N. Kántor, L. Kertész, M. Klein, Ö. Kornis, F. Orlowszky, F. Riesz, Ö. Spitzer, K. Szabó.