11.3. Solve the equation $20[x]-14\{x\}=2014$ ([x] - the integer part of the number $x$, i.e., the greatest integer not exceeding $x,\{x\}$ - the fractional part of the number $x$ : $\{x\}=x-[x]$).
Answer: $x=101 \frac{3}{7}$.

From the inequality $0 \leq\{x\}<1$ it follows that $0 \leq 14\{x\}<14,0 \leq 20[x]-2014<14$. Adding 2014 and dividing by 20, we get $\frac{2014}{20} \leq[x]<\frac{2028}{20}$ or $100.7 \leq[x]<101.4$. Thus, $[x]=101,\{x\}=\frac{20 \cdot 101-2014}{14}=\frac{3}{7}$ and $x=101+\frac{3}{7}$.

Second solution. From the condition, it follows that the number $14\{x\}$ must be an integer, so it is one of the numbers $0,1,2, \ldots, 13$. At the same time, its sum with 2014 must be divisible by 20. Therefore, $14\{x\}=6,[x]=\frac{2014+6}{20}=101$.