Question 814080
x = 2y + 4
xy = 16


Substitute the first equation into the second.


(2y + 4)(y) = 16
2y^2 + 4y - 16 = 0
y^2 + 2y - 8 = 0
(y + 4)(y - 2) = 0
y = -4 or y = 2


solutions: (x, y) = (-4, -4) or (8, 2)