1. **State the problem:** Solve the system of equations: $$x + y = 4$$ $$xy = 16$$ 2. **Use the formulas:** We want to find values of $x$ and $y$ that satisfy both equations simultaneously. 3. **Express $y$ from the first equation:** $$y = 4 - x$$ 4. **Substitute $y$ into the second equation:** $$x(4 - x) = 16$$ 5. **Expand and rearrange:** $$4x - x^2 = 16$$ $$-x^2 + 4x - 16 = 0$$ Multiply both sides by $-1$ to simplify: $$x^2 - 4x + 16 = 0$$ 6. **Calculate the discriminant to check for real solutions:** $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 1 \times 16 = 16 - 64 = -48$$ 7. **Interpret the discriminant:** Since $\Delta < 0$, there are no real solutions for $x$ and $y$. 8. **Find complex solutions using the quadratic formula:** $$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{4 \pm \sqrt{-48}}{2} = \frac{4 \pm 4i\sqrt{3}}{2} = 2 \pm 2i\sqrt{3}$$ 9. **Find corresponding $y$ values:** $$y = 4 - x = 4 - (2 \pm 2i\sqrt{3}) = 2 \mp 2i\sqrt{3}$$ **Final answer:** $$\boxed{(x, y) = (2 + 2i\sqrt{3}, 2 - 2i\sqrt{3}) \text{ or } (2 - 2i\sqrt{3}, 2 + 2i\sqrt{3})}$$