1. **State the problem:** We are given the system of equations: $$x + y = 3$$ $$xy = 9$$ We need to find the values of $x$ and $y$ that satisfy both equations. 2. **Use the substitution or quadratic approach:** From the first equation, express $y$ in terms of $x$: $$y = 3 - x$$ 3. **Substitute into the second equation:** $$x(3 - x) = 9$$ 4. **Expand and rearrange:** $$3x - x^2 = 9$$ 5. **Rewrite as a quadratic equation:** $$-x^2 + 3x - 9 = 0$$ Multiply both sides by $-1$ to simplify: $$\cancel{-}x^2 + \cancel{3x} - \cancel{9} = \cancel{0} \Rightarrow x^2 - 3x + 9 = 0$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-3$, and $c=9$. Calculate the discriminant: $$\Delta = (-3)^2 - 4(1)(9) = 9 - 36 = -27$$ Since the discriminant is negative, there are no real solutions for $x$. 7. **Conclusion:** The system has no real solutions for $x$ and $y$ because the product $xy=9$ and sum $x+y=3$ cannot be satisfied simultaneously with real numbers. **Final answer:** No real solutions exist for the system.