3. Solve the system of equations:

$$
\left\{\begin{aligned}
x+y-2018 & =(x-2019) \cdot y \\
x+z-2014 & =(x-2019) \cdot z \\
y+z+2 & =y \cdot z
\end{aligned}\right.
$$
Answer: (2022, 2, 4), (2018, 0, -2).

Solution: Introduce the substitution $\tilde{x}=x-2019$. Then the system becomes $\left\{\begin{array}{l}\tilde{x}+y+1=\tilde{x} \cdot y, \\ \tilde{x}+z+5=\tilde{x} \cdot z, \\ y+z+2=y \cdot z .\end{array} \quad\right.$ Transform the system to the form: $\left\{\begin{array}{l}(\tilde{x}-1)(y-1)=2, \\ (\tilde{x}-1)(z-1)=6, \\ (y-1)(z-1)=3 .\end{array} \quad\right.$ Make the substitution $a=\tilde{x}-1, b=y-1, c=z-1$. Then the system becomes:
$\left\{\begin{array}{l}a b=2, \\ a c=6, \\ b c=3 .\end{array}\right.$

$a b c= \pm 6$. Using the last equality, we get that the system has two solutions: $\left\{\begin{array}{l}a=2, \\ b=1, \\ c=3 .\end{array}\left\{\begin{array}{l}a=-2, \\ b=-1, \\ c=-3 .\end{array}\right.\right.$ Then $\left\{\begin{array}{l}\tilde{x}-1=2, \\ y-1=1, \\ z-1=3 .\end{array}\right.$ or $\left\{\begin{array}{l}\tilde{x}-1=-2, \\ y-1=-1, \\ z-1=-3 .\end{array}\right.$ Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_4de35ed8ca5599d2cb91g-2.jpg?height=160&width=1628&top_left_y=520&top_left_x=180)

Remark: 1 point for each correct solution found by trial and error.