Example 8: A survey of 50 students on their attitudes towards events $A$ and $B$ yielded the following results: The number of students who support $A$ accounts for three-fifths of the total, with the rest not supporting; The number of students who support $B$ is 3 more than those who support $A$, with the rest not supporting; Additionally, the number of students who do not support both $A$ and $B$ is 1 more than one-third of the number of students who support both $A$ and $B$. How many students support both $A$ and $B$, and how many do not support either?
The number of people who support $A$ is $50 \times \frac{3}{5}=30$, the number of people who support $B$ is $30+3=33$. As shown in the figure, let the set of 50 students be $U$, the set of all students who support event $A$ be $A$; the set of all students who support event $B$ be $B$.
Let the number of students who support both events $A$ and $B$ be $x$, then the number of students who do not support either $A$ or $B$ is $\frac{x}{3}+1$, the number of students who support $A$ but not $B$ is $30-x$, and the number of students who support $B$ but not $A$ is $33-x$. According to the problem, $(30-x)+(33-x)+x+\left(\frac{x}{3}+1\right)=50$, solving this gives $x=21$. Therefore, the number of students who support both $A$ and $B$ is 21, and the number of students who do not support either is 8.
Solution notes Generally, using $|A|$ to represent the number of elements in set $A$, then $|A \cup B|=|A|+$
$$
\begin{array}{l}
|B|-|A \cap B| ;|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|B \cap C|- \\
|A \cap C|+|A \cap B \cap C| .
\end{array}
$$