Example 2 Let real numbers $s, t$ satisfy
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, \\
t^{2}+99 t+19=0(s t \neq 1) . \\
\text { Find the value of } \frac{s t+4 s+1}{t} \text { . }
\end{array}
$$
(1999, National Junior High School Mathematics Competition)
【Analysis】Transform the first equation of the known equations, and combine it with the second equation to find that $\frac{1}{s} 、 t(s t \neq 1)$ are the two roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$

Since $s \neq 0$, the first equation can be transformed into
$$
\left(\frac{1}{s}\right)^{2}+99\left(\frac{1}{s}\right)+19=0 \text {. }
$$

Also, $t^{2}+99 t+19=0$, and $s t \neq 1$, so $\frac{1}{s} 、 t$ are the two distinct real roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$

Thus, $\frac{1}{s}+t=-99, \frac{1}{s} \cdot t=19$, which means
$$
s t+1=-99 s, t=19 s . \\
\text { Therefore, } \frac{s t+4 s+1}{t}=\frac{-99 s+4 s}{19 s}=-5 .
$$