B2. Find all real numbers $x, y$, and $z$ that solve the system of equations

$$
\frac{3 x y}{x-y}=2, \quad \frac{2 y z}{y+2 z}=3, \quad \frac{x z}{z-4 x}=3
$$
B2. In the equation, we first eliminate the fractions and obtain

$$
\begin{aligned}
3 x y & =2 x-2 y \\
2 y z & =3 y+6 z \\
x z & =3 z-12 x
\end{aligned}
$$

We transform the first equation into $\mathrm{v}(3 y-2) x=-2 y$. If $3 y-2=0$ or $y=\frac{2}{3}$, then it follows that $y=0$, which is a contradiction. Therefore, we can divide the equation by $3 y-2$ and express $x=\frac{-2 y}{3 y-2}$. We transform the second equation into $(2 y-6) z=3 y$ and similarly express $z=\frac{3 y}{2 y-6}$. We now substitute both into the last equation, to get

$$
\frac{-2 y}{3 y-2} \cdot \frac{3 y}{2 y-6}=\frac{9 y}{2 y-6}+\frac{24 y}{3 y-2}
$$

We multiply the equation by $(3 y-2)(2 y-6)$, to get

$$
-6 y^{2}=9 y(3 y-2)+24 y(2 y-6)
$$

which can be simplified to $81 y^{2}-162 y=0$. We factor the left side and get $81 y(y-2)=0$. If $y=0$, it follows from the above relations that $x=0$ and $z=0$. However, this is not a solution to the system of equations, as the fractions in the equations are not defined in this case. Therefore, $y=2$ and consequently $x=\frac{-4}{4}=-1$ and $z=\frac{6}{-2}=-3$. The only solution to the system is thus $x=-1, y=2$ and $z=-3$.

Expressing two variables as functions of the third, where special cases where division by zero might occur (e.g., $x, z$ as functions of $y$) ................................................................................................. Substituting the expressions into the remaining equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 point Transforming the equation into a quadratic polynomial in the third variable (y) ............. 1 point Both solutions of the quadratic polynomial .......................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-20.jpg?height=57&width=1742&top_left_y=1122&top_left_x=160)

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-20.jpg?height=54&width=1739&top_left_y=1172&top_left_x=164)

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-20.jpg?height=57&width=1676&top_left_y=1274&top_left_x=230)

Substituting $p, q, r$ or equivalent variables ......................................................................

Transforming the equations and determining that it is a system of linear equations .................... 2 points

Solving the system of linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-20.jpg?height=54&width=1753&top_left_y=1475&top_left_x=160)