Example 9 Let real numbers $s, t$ satisfy the equations
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, t^{2}+99 t+19=0, \text { and } s t \neq 1 . \\
\text { Then } \frac{s t+4 s+1}{t}=
\end{array}
$$
(1999, "Mathematics Weekly Cup" National Junior High School Mathematics Competition)
Solution: Clearly, $t \neq 0$.
Thus, $19\left(\frac{1}{t}\right)^{2}+99 \cdot \frac{1}{t}+1=0$.
By comparison, $s$ and $\frac{1}{t}$ are the two distinct roots of the equation $19 x^{2}+99 x+1=0$ (since $s t \neq 1$, i.e., $s \neq \frac{1}{t}$). Therefore, by Vieta's formulas, we have
$$
\begin{array}{l}
s+\frac{1}{t}=-\frac{99}{19}, \frac{s}{t}=\frac{1}{19} . \\
\text { Hence } \frac{s t+4 s+1}{t}=\left(s+\frac{1}{t}\right)+4 \cdot \frac{s}{t} \\
=-\frac{99}{19}+\frac{4}{19}=-5 .
\end{array}
$$