Find all prime numbers $p$ and $q$, such that $7p+q$ and $pq+11$ are also prime numbers. 
Find all prime numbers $p$ and $q$, such that $7p+q$ and $pq+11$ are
  also prime numbers.

Based on the fact that all primes, besides 2, are odd, I found that either $p$ or $q$ must be $2$ in order for $pq+11$ to be a prime number. From here, I found several pairs of $p$ and $q$ that work, but I don't know how to find all $p$ and $q$. I tried letting 
$7p+q=r$
$pq+11=s$ 
and then adding the equations and using SFFT to get:
$(p+1)(q+7)=r+s-4$
but it doesn't really help.
 A: So, say $p=2$.  Then your expressions are $14+q$ and $2q+11$. Working $\pmod 3$ we see that these are $q-1$ and $2-q$  Easy to see that one of these is divisible by $3$ unless $q=3$ which is a valid example.
Now say $q=2$,  Then your expressions are $7p+2$ and $2p+11$.  Working $\pmod 3$ we see that these are $p+2$ and $2(p+1)$ and again one of these terms must be divisible by $3$ unless $p=3$, which is again a valid example.
A: We must have either $p=2$ or $q=2$ , Otherwise $7p+q$ is even .
If $p=2$ , then let  $14+q=x$ and $2q+11 = y$. Adding these equation and taking $\text{modulo } 3$ , we get :
$$x+y \equiv 1\mod 3 \implies x,y \equiv 2\mod 3$$
Substituting this back into the original equation gives $$q\equiv 0\mod 3 \implies q = 3$$
If $q = 2$ , then let $7p + 2 = x$ and $2p + 11 = y$ . Again adding and taking $\text{ modulo } 3$ , we get :
$$x+y\equiv 1\mod 3\implies x,y \equiv2  \mod 3$$
Substituting this back into the original equation gives 
$$p\equiv 0\mod 3 \implies p = 3$$
A: As you have already pointed out that $p=2$ or $q=2$ 
Take cases , 
$Case$ $1$. Let, $p=2$ , Since we know every prime greater than $3$ is of the form $6k+1$ or $6k-1$. Taking $q=6k+1$ implies $7p+q$ divisible by $3$ and taking $q=6k-1$ implies $pq+11$ is divisible by $3$. Thus only possible $q=3$. In that case both expressions are equal to $17$.
$Case$ $2$. Let, $q=2$ if $p=6k+1$ then $7p+q$ is divisible by $3$ and for $p=6k-1$ then $pq+11$ is divisible by $3$. Thus only possible $p=3$. Then clearly first expression is $23$ and second expression is $17$ and both of these are primes. So only solution. $(p,q)=(3,2) , (2,3)$
