If $5P+7Q= 109$ for primes $P$ and $Q$, find $P$. Suppose $P$ and $Q$ are both prime numbers
  $$5P+7Q= 109$$
  Find the value of $P$.


I thought about examining all terms for parity, and I approached it by this:

Suppose $5P$ was an even number. However, the only even prime number is $2$, so that would make $5P$ = $10$ and $7Q =99$. However, since $7$ doesn't divide fully into $99$, this means that $7Q$ must be even. But how do I approach it from there?

P.S. I'm only an Year 7 and this is Year 9 work. Please explain clearly how you found the solution.
You have already deduced that one of $(P,Q)$ must be $2$.

Also, you’ve tried putting $P=2$ and correctly found that did not give a solution.

The only other option is $Q=2$, so $7Q=14$ and the equation becomes,
$$5P+14=109$$
$$5P=95$$
$$P=19$$
$$5(-2)+7(17)=109$$

Since (-2) is not a prime number, the problem should be:

Suppose $P$ and $Q$ are both prime numbers

$$5P= 109-7Q$$

Find the value of $P$

The answer: $P=2$