1. Find all triples of real numbers $(x, y, z)$ that satisfy the system of equations

$$
\begin{aligned}
x^{2}+y^{2}+4 z^{2} & =6 y-4 \\
2 x y-4 x z+4 y z & =y^{2}+5
\end{aligned}
$$
II/1. We observe that if we subtract the second equation from the first, we get a square of a trinomial on the left side. We get

$$
(x-y+2 z)^{2}=-y^{2}+6 y-9
$$

which can be further transformed into

$$
(x-y+2 z)^{2}+(y-3)^{2}=0
$$

Since \(x, y\), and \(z\) are real numbers, both parentheses in the above expression must be equal to 0, so \(y=3\) and \(x=y-2 z=3-2 z\). Substituting this into the first equation of the system and after rearranging, we get \(2 z^{2}-3 z+1=0\) or \((2 z-1)(z-1)=0\). The same result is obtained if we substitute \(y=3\) and \(x=3-2 z\) into the second equation. Therefore, \(z=1\) or \(z=\frac{1}{2}\). The solutions to the problem are thus the triples \((1,3,1)\) and \(\left(2,3, \frac{1}{2}\right)\).

Subtracting the equations .1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-10.jpg?height=62&width=1633&top_left_y=957&top_left_x=220)

Writing the square \((x-y+2 z)^{2}\)..................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-10.jpg?height=57&width=1633&top_left_y=1057&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-10.jpg?height=51&width=1633&top_left_y=1111&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-10.jpg?height=54&width=1631&top_left_y=1161&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_2324ea98e20bf8c655d1g-10.jpg?height=68&width=1633&top_left_y=1208&top_left_x=220)

The solutions must be clearly justified to satisfy both equations. If this justification is not good, but both solutions are written, 1 point is awarded.