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)}#