Solve the following systems of equations:
a.) $x+y=4$
$x z+y u=7$
$x z^{2}+y u^{2}=12$
b.) $x^{3}+y^{3}=1$
$x z^{3}+y u^{3}=21$
$x^{2} z+y^{2} u=0$
$x z^{2}+y u^{2}=-2$
$x z^{2}+y u^{2}=-2$
$z^{3}+u^{3}=22$
65. a.)

\[
\begin{array}{r}
x+y=4 \\
x z+y u=7 \\
x z^{2}+y u^{2}=12 \\
x z^{3}+y u^{3}=21
\end{array}
\]

I. solution: Eliminate \( x \) from equations (1) and (2), (2) and (3), and finally (3) and (4).

\[
\begin{aligned}
y z - y u & = 4 z - 7 \\
y z u - y u^{2} & = 7 z - 12 \\
y z u^{2} - y u^{3} & = 12 z - 21
\end{aligned}
\]

Divide (6) by (5) and (7) by (6):

\( u = \frac{7 z - 12}{4 z - 7} = \frac{12 z - 21}{7 z - 12} \). This gives a quadratic equation in \( z \). \( z^{2} - 3 = 0, z = \pm \sqrt{3} \); \( u = -\sqrt{3} \) if \( z = \sqrt{3} \) and \( u = \sqrt{3} \) if \( z = -\sqrt{3} \). For \( x \) and \( y \), equations (1) and (2) give \( \frac{12 + 7 \sqrt{2}}{6}, \frac{12 - 7 \sqrt{3}}{6} \), or \( \frac{12 - 7 \sqrt{3}}{6}, \frac{12 + 7 \sqrt{3}}{6} \), depending on whether \( z = +\sqrt{3} \) or \( z = -\sqrt{3} \).

Solved by: Bánó Klára, Bécsy Cecília, Boda I., Csordás L., Csuhássy Edit, Fülöp M., Gaál E., ifj. Gacsályi S., Gehér L., Hosszú M., Izsák I., Kemény Judit, Kővári T., Magyar Á. Sz., Pál L., Párkány M., Róna P., Sós Vera, Spitz Vera, Szathmári D., Szépfalussy P., Szűcs L., Váczy Irén, Vékony Mária, Vörös M., Zsulán J.

II. solution:

\[
\begin{gathered}
(x z + y u) \cdot (z + u) = x z^{2} + y u^{2} + x u z + y u z = \\
= x z^{2} + y u^{2} + u z (x + y)
\end{gathered}
\]

and

\[
\begin{gathered}
\left(x z^{2} + y u^{2}\right)(z + u) = x z^{3} + y u^{3} + x z^{3} u + y z u^{2} = \\
= x z^{3} + y u^{3} + u z (x z + y u)
\end{gathered}
\]

From this, using equations (1)-(4), we get the following system of linear equations for \( z + u \) and \( z u \):

\[
\begin{aligned}
7(z + u) & = 12 + 4 \cdot z u \\
12(z + u) & = 21 + 7 \cdot z u
\end{aligned}
\]

From which we get:

\[
z + u = 0, \quad z u = -3
\]

Thus, \( z \) and \( u \) are the roots of the following equation:

\[
X^{2} - 3 = 0, \quad \text{so} \quad z = \sqrt{3}, \quad u = -\sqrt{3}
\]

From equations (1) and (2):

\[
x + y = 4, \quad x - y = \frac{7 \sqrt{3}}{3}, \quad x = 2 + \frac{7 \sqrt{3}}{6}, \quad y = 2 - \frac{7 \sqrt{3}}{6}
\]

If we swap the values of \( z \) and \( u \), \( x \) and \( y \) will also swap, which is evident from their symmetric roles.

Silfen Péter (Budapest, Bolyai reálisk. VII. o.)
65. b.

\[
\begin{array}{r}
x^{3} + y^{3} = 1 \\
x^{2} z + y^{2} u = 0 \\
x z^{2} + y u^{2} = -2 \\
z^{3} + u^{3} = 22
\end{array}
\]

Solution: By substituting \( x^{3} = X, \quad y^{3} = Y, \quad \frac{z}{x} = Z, \quad \frac{u}{y} = U \), we get the system of equations \( X + Y = 1, \quad X Z + Y U = 0, \quad X Z^{2} + Y U^{2} = -2, \quad X Z^{3} + Y U^{3} = 22 \), which can be solved similarly to the problem (a) and leads to the result:

\[
\begin{aligned}
X = \frac{1}{2} - \frac{11}{2 \sqrt{113}}, \quad Y & = \frac{1}{2} + \frac{11}{2 \sqrt{113}}, \quad Z = \frac{-11 - \sqrt{113}}{2} \\
U & = \frac{-11 - \sqrt{113}}{2}
\end{aligned}
\]

By symmetry, swapping \( X \) with \( Z \) and \( Y \) with \( U \), we get the other solution of the system of equations.

Solved by: Boda I., Fülöp M., Gaál E., Gehér L., Izsák I., Kővári T., Magyar Á. Sz., Róna P., Ungár P.