From the second equation:
3x-y = 2
=> -y = 2-3x
=> y = 3x-2
Substituting this into the first equation:
8x^2-y^2 = 16
=> 8x^2-(3x-2)^2 = 16
=> 8x^2-(9x^2-12x+4) = 16
=> -x^2+12x-4 = 16
=> x^2 - 12x + 20 = 0
=> (x-2)(x-10) = 0
=> x-2 = 0 or x-10 = 0
=> x = 2 or x = 10
Substituting these back into y = 3x-2:
y = 3(2) - 2 or y = 3(10)-2
=> y = 4 or y = 28
So we get (x,y) = (2, 4) or (x, y) = (10, 28).
Checking our results, we find that both (x, y) pairs fulfill the given system of equations, and thus we have the solution set
(x, y) in {(2, 4), (10, 28)}