Example 10 Let real numbers $s, t$ satisfy $19 s^{2}+99 s+1=0, t^{2}+99 t+19=0$, and $s t \neq 1$. Find the value of $\frac{s t+4 s+1}{t}$.
Solve: Given $19 s^{2}+99 s+1=0, t^{2}+99 t+19=0$, we have
$$
19 s^{2}+99 s+1=0, 19\left(\frac{1}{t}\right)^{2}+99 \cdot\left(\frac{1}{t}\right)+1=0 \text {. }
$$

Since $s t \neq 1$, we know $s \neq \frac{1}{t}$. Therefore, $s, \frac{1}{t}$ are the two distinct real roots of the equation $19 x^{2}+99 x+1=0$.
By the relationship between roots and coefficients, we have $s+\frac{1}{t}=-\frac{99}{19}, s \cdot \frac{1}{t}=\frac{1}{19}$.
$$
\text { Thus } \frac{s t+4 s+1}{t}=s+4\left(\frac{s}{t}\right)+\frac{1}{t}=\left(s+\frac{1}{t}\right)+4 \cdot s \cdot \frac{1}{t}=-\frac{99}{19}+\frac{4}{19}=-5 \text {. }
$$