Example 9 Let $x, y$ be non-negative integers, $x+2y$ is a multiple of 5, $x+y$ is a multiple of 3, and $2x+y \geqslant 99$. Then the minimum value of $7x+5y$ is $\qquad$
(15th "Five Sheep Forest" Junior High School Mathematics Competition (Initial $\exists)$)
Explanation: Let $x+2y=5A, x+y=3B, A, B$ be integers.
Since $x, y \geqslant 0$, hence $A, B \geqslant 0$, solving we get
$$
\begin{array}{l}
x=6B-5A, y=5A-3B, \\
2x+y=9B-5A \geqslant 99, \\
S=7x+5y=27B-10A .
\end{array}
$$

The problem is transformed to: integers $A, B \geqslant 0, 6B \geqslant 5A \geqslant$ $3B, 9B \geqslant 5A+99$, find the minimum value of $S$.
Since $15A \geqslant 9B \geqslant 99+5A$, hence
$$
10A \geqslant 99, A \geqslant 10.
$$

Thus, $9B \geqslant 50+99=149, B \geqslant 17$.
Then, $5A \geqslant 51, A \geqslant 11$.
Further, $9B \geqslant 55+99=154, B \geqslant 18$.
If $B \geqslant 19$, then
$$
\begin{array}{l}
S=9B+2(9B-5A) \\
\geqslant 9 \times 19+2 \times 99=369.
\end{array}
$$

If $B=18$, then
$$
5A \leqslant 9B-99=63, A \leqslant 12.
$$

Therefore, $A$ can only be 12 or 11.
When $A=12$, $S=27 \times 18-10 \times 12=366$ reaches the minimum value, and this is the only case, at this time $x=48, y=6$.