1. Find all $x$ for which $\left[\frac{8 x+19}{7}\right]=\frac{16(x+1)}{11}$, where $[t]$ is the integer part of $t$. Answer: $\left\{1 \frac{1}{16} ; 1 \frac{3}{4} ; 2 \frac{7}{16} ; 3 \frac{1}{8} ; 3 \frac{13}{16}\right\}$.
# Solution:

Let $t=\frac{16(x+1)}{11}$, where $t \in Z$ and express $x=\frac{11 t-16}{16}$. Then the equation becomes: $\left[\frac{8 \frac{11 t-16}{16}+19}{7}\right]=t$ or $\left[\frac{11 t+22}{14}\right]=t$. Therefore, $0 \leq \frac{11 t+22}{14}-t\frac{8}{3}, \\ t \leq \frac{22}{3},\end{array} \stackrel{t \in Z}{\Rightarrow} t_{1}=3, t_{2}=4, t_{3}=5, t_{4}=6, t_{5}=7\right.\right.$.

Considering that $x=\frac{11 t-16}{16}$, we get $x_{1}=1 \frac{1}{16}, x_{2}=1 \frac{3}{4}, x_{3}=2 \frac{7}{16}, x_{4}=3 \frac{1}{8}, x_{5}=3 \frac{13}{16}$.