Solve the following system of equations:

$$
x y+x z=10, \quad y z+y u=6, \quad z u+z x=2, \quad u x+u y=-2
$$
$$
x y+x z=10, \quad y z+y u=6, \quad z u+z x=2, \quad u x+u y=-2
$$

Solution I. By subtracting the sum of the two middle equations from the sum of the two extreme equations, we get $(x-z)(y+u)=0$. Thus, either $x=z$ or $y=-u$. Furthermore, if any of these and the first three equations of system (1) are satisfied, then the fourth equation is also satisfied.

In the first case, substitute $z$ with $x$ in the first three equations. We immediately factorize:

$$
x(x+y)=10, \quad y(u+x)=6, \quad x(u+x)=2
$$

From the second and third equations, we get $y=3x$ (since $u+x$ cannot be 0), so substituting into the first equation of (2) gives $x^2=10/4$, from which we get the roots (with $u=-x+2/x$ and $z=x$):

| $x$ | $y$ | $z$ | $u$ |
| :---: | :---: | :---: | :---: |
| $\frac{\sqrt{10}}{2}$ | $\frac{3 \sqrt{10}}{2}$ | $\frac{\sqrt{10}}{2}$ | $-\frac{1}{\sqrt{10}}$ |
| $-\frac{\sqrt{10}}{2}$ | $-\frac{3 \sqrt{10}}{2}$ | $-\frac{\sqrt{10}}{2}$ | $\frac{1}{\sqrt{10}}$ |

These are solutions to (1) because all three equations in (2) are satisfied.

In the second case, $y=-u$. So, write $(-y)$ instead of $u$ in (1):

$$
x y+x z=10 ; \quad y z-y^{2}=6, \quad -y z+x z=2
$$

From the second equation of (3), express $z$ and from the first, express $x$ in terms of $y$:

$$
z=\frac{6+y^{2}}{y}, \quad x=\frac{10}{y+z}=\frac{10}{y+\left(6+y^{2}\right) / y}=\frac{5 y}{3+y^{2}} .
$$

If we choose $z$ and $x$ this way, the first two equations in (3) are definitely satisfied. For the third equation to be satisfied, $y$ must be chosen such that

$$
-y \cdot \frac{6+y^{2}}{y}+\frac{5 y}{3+y^{2}} \cdot \frac{6+y^{2}}{y}=2
$$

which simplifies to $y^{4}+6 y^{2}-6=0$. Solving this (considering only real solutions), we get the additional roots, which, as can be seen similarly to the previous ones, are also solutions to (1):

| $x$ | $y$ | $z$ | $u$ |
| :---: | :---: | :---: | :---: |
| $\frac{5 \sqrt{-3+\sqrt{15}}}{\sqrt{15}}$ | $\sqrt{-3+\sqrt{15}}$ | $\frac{3+\sqrt{15}}{\sqrt{-3+\sqrt{15}}}$ | $-\sqrt{-3+\sqrt{15}}$ |
| $-\frac{5 \sqrt{-3+\sqrt{15}}}{\sqrt{15}}$ | $-\sqrt{-3+\sqrt{15}}$ | $-\frac{3+\sqrt{15}}{\sqrt{-3+\sqrt{15}}}$ | $\sqrt{-3+\sqrt{15}}$ |

Thus, the system of equations has the four solutions listed above.

Gábor Megyesi (Szeged, Ságvári E. Gyak. Gimn., I. o. t.)

Solution II. If the right-hand sides of the equations in (1) are denoted by $a, b, c, d$, and we introduce the notations $x z=A, u y=B$, the system of equations can be written as:

$$
x y=a-A, \quad y z=b-B, \quad z u=c-A, \quad u x=d-B
$$

From this,

$$
(a-A)(c-A)=A B, \quad(b-B)(d-B)=A B
$$

If we solve for $B$ from the first equation of (5) and substitute it into the second, after rearranging, we get:

$$
\begin{gathered}
(a+b+c+d) A^{3}-\left(a b+a c+a d+b c+b d+c d+(a+c)^{2}\right) A^{2}+ \\
+a c(2 a+b+2 c+d) A-a^{2} c^{2}=0
\end{gathered}
$$

Knowing the solutions of this, the original system (1) can be easily solved: from the first equation of (5), we get the value of $B$, and thus the values of the double products $x y, y z, x u$, etc. For example, $x^2$ can be obtained as follows:

$$
x^{2}=\frac{x y \cdot x z}{y z}=\frac{(a-A) A}{b-B}
$$

In our case, $a=10, b=6, c=2, d=-2$, so the equation (6) becomes:

$$
16 A^{3}-200 A^{2}+560 A-400=0
$$

Fortunately, the left-hand side of this can be factored as:

$$
8(2 A-5)\left(A^{2}-10 A+10\right)=0
$$

from which the possible values of $A$ are $5/2, 5-\sqrt{15}, 5+\sqrt{15}$. From these, we get the solutions already written.