To solve this system of equations, we can use substitution or elimination method.

Let's first subtract the second equation from the first equation:

4x^2 + 5y^2 - (x^2 + 5y^2) = 16 - 25
3x^2 = -9
x^2 = -3

Now we can substitute this value of x^2 into either of the equations to solve for y:

x^2 + 5y^2 = 25
-3 + 5y^2 = 25
5y^2 = 28
y^2 = 28/5
y = ±√(28/5)

Hence, the solutions for this system of equations are x = ±√(-3), y = ±√(28/5).