There are four; the first equation is an ellipse, and the second is a hyperbola that intercepts the ellipse at (±3, ±5). Try substitution as follows:
y^{2} = 16 + x^{2}, so 5x^{2} + 3(16 + x^{2}) = 120 and solve as
5x^{2} + 48 + 3x^{2} = 120
8x^{2} = 120  48
x^{2} = 72/8 = 9 so x = ±3
either of these values plugged into the second equation yields y^{2} = 16 + 9 = 25 so y = ±5
the full set would be (3, 5), (3, 5), (3, 5), (3, 5)
Good Luck!
May 6

Ron S.