Answer:
Choice A: 11
Explanation:
We have x² - y² = 19
x² - y² = (x + y)(x-y)
So (x + y)(x-y) = 19
Since x and y are integers as per the first statement (x, y ∈ Z), we also have x + y = integer and x-y = integer
Take (x + y)(x - y) = 19
Since 19 is prime its only factors are 1 and 19
So we have two situations which can satisfy the above
Either x + y = 1 and x -y 19 or x + y = 19 and x -y = 1
1. Let's try x + y = 1 and x - y = 19
Adding both equations together gives us 2x = 20 (y-terms cancel)
==> x = 10 ==> 10 + y = 1 or y = -9
So 2x - 9 => 2(10) - (-9) = 20 + 9 = 29
This is not one of the choices so we can go on to the next case
2 Let's take x + y = 19 and x - y = 1
Adding these two equations gives us
2x = 20, x = 10 as before
Use 10 in the equation x + y = 19
==> 10 + y = 19
==> y = 19 - 10 = 9
2x - y ==> 2(10) - 9 ==> 20 - 9 = 11
So answer is choice A: 11