1. **State the problem:** We are given the system of equations: $$x + y = 6$$ $$xy = 36$$ We need to find the values of $x$ and $y$ that satisfy both equations. 2. **Use the formulas and rules:** From the first equation, $y = 6 - x$. Substitute this into the second equation: $$x(6 - x) = 36$$ 3. **Simplify and solve the quadratic equation:** $$6x - x^2 = 36$$ Rearranged: $$-x^2 + 6x - 36 = 0$$ Multiply both sides by $-1$ to simplify: $$\cancel{-}x^2 + \cancel{6x} - \cancel{36} = \cancel{0}$$ becomes $$x^2 - 6x + 36 = 0$$ 4. **Calculate the discriminant to check for real solutions:** $$\Delta = b^2 - 4ac = (-6)^2 - 4(1)(36) = 36 - 144 = -108$$ Since $\Delta < 0$, there are no real solutions for $x$ and $y$. 5. **Find complex solutions:** $$x = \frac{6 \pm \sqrt{-108}}{2} = \frac{6 \pm \sqrt{108}i}{2}$$ Simplify $\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}$: $$x = \frac{6 \pm 6\sqrt{3}i}{2} = 3 \pm 3\sqrt{3}i$$ 6. **Find corresponding $y$ values:** $$y = 6 - x = 6 - (3 \pm 3\sqrt{3}i) = 3 \mp 3\sqrt{3}i$$ **Final answer:** $$\boxed{(x, y) = (3 + 3\sqrt{3}i, 3 - 3\sqrt{3}i) \text{ or } (3 - 3\sqrt{3}i, 3 + 3\sqrt{3}i)}$$