1. **State the problem:** Solve the system of equations: $$x^2 + y^2 = 9$$ $$x - y = 1$$ 2. **Express one variable in terms of the other:** From the second equation, we have: $$x = 1 + y$$ 3. **Substitute into the first equation:** Replace $x$ with $1 + y$ in the first equation: $$ (1 + y)^2 + y^2 = 9 $$ 4. **Expand and simplify:** $$ 1 + 2y + y^2 + y^2 = 9 $$ $$ 1 + 2y + 2y^2 = 9 $$ 5. **Bring all terms to one side:** $$ 2y^2 + 2y + 1 - 9 = 0 $$ $$ 2y^2 + 2y - 8 = 0 $$ 6. **Simplify by dividing all terms by 2:** $$ \cancel{2}y^2 + \cancel{2}y - \cancel{8} = 0 $$ $$ y^2 + y - 4 = 0 $$ 7. **Solve the quadratic equation:** Use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=1$, $c=-4$: $$ y = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-4)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2} $$ 8. **Find corresponding $x$ values:** Recall $x = 1 + y$, so: $$ x = 1 + \frac{-1 + \sqrt{17}}{2} = \frac{2 - 1 + \sqrt{17}}{2} = \frac{1 + \sqrt{17}}{2} $$ and $$ x = 1 + \frac{-1 - \sqrt{17}}{2} = \frac{2 - 1 - \sqrt{17}}{2} = \frac{1 - \sqrt{17}}{2} $$ 9. **Final solutions:** $$ \left( \frac{1 + \sqrt{17}}{2}, \frac{-1 + \sqrt{17}}{2} \right) \quad \text{and} \quad \left( \frac{1 - \sqrt{17}}{2}, \frac{-1 - \sqrt{17}}{2} \right) $$