1. $[x]$ - the integer part of the number $x$ (the greatest integer not exceeding $x$), $\{x\}=x-[x]-$ the fractional part of the number $x$.

For which $a$ does the system $\left\{\begin{array}{c}2 x-[x]=4 a+1 \\ 4[x]-3\{x\}=5 a+15\end{array}\right.$ have a solution? Find such $x$.
Solution. Let $\left\{\begin{array}{l}u=[x] \\ v=\{x\}\end{array} \rightarrow u+v=x, 0 \leq v<1\right.$.

Then the system will take the form: $\left\{\begin{array}{c}u+2 v=4 a+1 \\ 4 u-3 v=5 a+15\end{array} \rightarrow\left\{\begin{array}{c}u=2 a+3 \\ v=a-1\end{array}\right.\right.$.

Since $0 \leq v<1$, then $0 \leq a-1<1 \rightarrow a \in[1 ; 2)$. Therefore, $u=2 a+3 \in[5 ; 7)$.

Due to the integrality of $u$, we get $: u=[x]=5$ and $u=[x]=6$.

$$
u=5: a=1 \rightarrow v=0 \rightarrow x=5 ; \quad u=6: a=\frac{3}{2} \rightarrow v=\frac{1}{2} \rightarrow x=\frac{13}{2} .
$$

Answer: 1) when $a=1, x=5 ; 2)$ when $a=\frac{3}{2}, x=\frac{13}{2}$.