Example 8 Given that positive integers $p, q$ are both prime numbers, and $7p+q, pq+11$ are also prime numbers. Find $p+q$.

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When $p \equiv q \equiv 1(\bmod 2)$,
$$
7 p+q \equiv 7+1 \equiv 0(\bmod 2) \text{. }
$$

But $7 p+q \geqslant 7>2$, which contradicts the fact that $7 p+q$ is a prime number. Therefore, one of $p, q$ must be even.
Since $p, q$ are both primes, then $p=2$ or $q=2$.
(1) When $p=2$, $14+q$ and $2 q+11$ are both primes.
If $q \equiv 1(\bmod 3)$, then
$$
14+q \equiv 14+1 \equiv 0(\bmod 3) ; \quad \div i
$$

Contradiction.
If $q \equiv-1(\bmod 3)$, then
$$
2 q+11 \equiv-2+11 \equiv 0(\bmod 3),
$$
Contradiction.
Therefore, $q \equiv 0(\bmod 3)$.
But $q$ is a prime, so $q=3$.
(2) When $q=2$, $7 p+2$ and $2 p+11$ are both primes.
If $p \equiv 1(\bmod 3)$, then
$$
7 p+2 \equiv 7+2 \equiv 0(\bmod 3),
$$

Contradiction.
If $p \equiv-1(\bmod 3)$, then
$$
2 p+11 \equiv-2+11 \equiv 0(\bmod 3),
$$

Contradiction.
Therefore, $p \equiv 0(\bmod 3)$.
But $p$ is a prime, so $p=3$.
Upon verification, $p=2, q=3$ or $p=3, q=2$ satisfy the conditions. Hence $p+q=5$.