My approach is the following:

We need the greatest prime number, so we have to maximize x-2y, how? y has to be negative

So then:

\(x^{2}<9^{2}\) ---> x = 9

\(y^{2}<(-5)^{2}\) ---> y = -5

Limit = 9 + 10 = 19 prime but we can't take it, we need a prime less than 19

If we take one digit less we have that:

\(x^{2}=8^{2}\) ---> x = 8

\(y^{2}=(-4)^{2}\) ---> y = -4

So then 8 + 8 =16 not prime

There are two numbers between 16 and 19 --> 17 (prime) and 18 (not prime)

Answer E
_________________

Kudos please if you liked my post

Thanks!