3. Find $x$ and $y$ for which $\left\{\begin{array}{l}x-2 y+[x]+3\{x\}-\{y\}+3[y]=2.5 \\ 2 x+y-[x]+2\{x\}+3\{y\}-4[y]=12\end{array}\right.$, where $[x],[y]$ and $\{x\},\{y\}$ are the integer and fractional parts of the numbers $x$ and $y$. The integer part of a number $a$ is the greatest integer not exceeding $a$, and $\{a\}=a-[a]$.
Solution. Given that $x=[x]+\{x\}$ and $y=[y]+\{y\}$, we can rewrite the system as:

$$
\left\{\begin{array}{l}
2[x]+4\{x\}+[y]-3\{y\}=2.5 \\
{[x]+4\{x\}-3[y]+4\{y\}=12}
\end{array}\right.
$$

Introduce the notations: $\{x\}=u \in[0 ; 1),\{y\}=v \in[0 ; 1)$.

The system becomes: $\left\{\begin{array}{l}2[x]+[y]=2.5-4 u+3 v \\ {[x]-3[y]=12-4 u-4 v}\end{array}\right.$.

Express $[x]$ and $[y]$ in terms of $u$ and $v:\left\{\begin{array}{l}{[x]=\frac{19.5-16 u+5 v}{7}} \\ {[y]=\frac{-21.5+4 u+11 v}{7}}\end{array}\right.$.

Estimate the expression $\frac{19.5-16 u+5 v}{7}: 0.5=\frac{3.5}{7}<\frac{19.5-16 u+5 v}{7}<\frac{24.5}{7}=3.5$.

Thus, for $[x]$ only three values are possible: $[x]=1 ; 2 ; 3$.

Case 1. $[x]=1$. We have $\frac{19.5-16 u+5 v}{7}=1$. From this we get

$$
16 u-5 v=12.5 \rightarrow u=\frac{12.5+5 v}{16} \in[0 ; 1) \rightarrow v \in[-2.5 ; 0.7) \rightarrow v \in[0 ; 0.7)
$$

Then $[y]=\frac{49 v-73.5}{28} \in(-2.625 ;-1.4)$ for $v \in[0 ; 0.7)$ and with the condition of integer $[y]$ we get $\quad[y]=-2 . \quad$ This $\quad$ value $\quad$ corresponds to $\quad v_{1}=\frac{5}{14}, u_{1}=\frac{25}{28} \quad$ and $x_{1}=1+\frac{25}{28}=\frac{53}{28}, y_{1}=-2+\frac{5}{14}=-\frac{23}{14}$.

Case 2. $[x]=2$. We have $\frac{19.5-16 u+5 v}{7}=2$. From this we get

$$
16 u-5 v=5.5 \rightarrow u=\frac{5 v+5.5}{16} \in[0 ; 1) \rightarrow v \in[-1.1 ; 2.1) \rightarrow v \in[0 ; 1) .
$$

Then $[y]=\frac{49 v-80.5}{28} \in[-2.875 ;-1.125)$ for $v \in[0 ; 1)$ and with the condition of integer $[y]$ we get $[y]=-2$. This value corresponds to $v_{2}=0.5, u_{2}=0.5$ and $x_{2}=2+0.5=2.5, y_{2}=-2+0.5=-1.5$.

Case 3. $[x]=3$. We have $\frac{19.5-16 u+5 v}{7}=3$. From this we get

$$
16 u-5 v=-1.5 \rightarrow u=\frac{5 v-1.5}{16} \in[0 ; 1) \rightarrow v \in[0.3 ; 3.5) \rightarrow v \in[0.3 ; 1)
$$

Then $[y]=\frac{49 v-87.5}{28} \in[-2.6 ;-1.375)$ for $v \in[0 ; 1)$ and with the condition of integer $[y]$ we get $[y]=-2 . \quad$ This $\quad$ value $\quad$ corresponds to $\quad v_{3}=\frac{9}{14}, u_{3}=\frac{3}{28} \quad$ and $x_{3}=3+\frac{3}{28}=\frac{87}{28}, y_{3}=-2+\frac{9}{14}=-\frac{19}{14}$.

Answer: $\left(\frac{53}{28} ;-\frac{23}{14}\right) ;\left(\frac{5}{2} ;-\frac{3}{2}\right) ;\left(\frac{87}{28} ;-\frac{19}{14}\right)$.