1. **Stating the problem:** Solve the system of equations $\{x^2 + y^2 = 10, x + y = -1\}$. 2. **Formula and rules:** We can use substitution or elimination. Here, from the linear equation $x + y = -1$, express $y = -1 - x$. 3. **Substitute into the first equation:** $$x^2 + (-1 - x)^2 = 10$$ 4. **Expand and simplify:** $$x^2 + (1 + 2x + x^2) = 10$$ $$x^2 + 1 + 2x + x^2 = 10$$ $$2x^2 + 2x + 1 = 10$$ 5. **Bring all terms to one side:** $$2x^2 + 2x + 1 - 10 = 0$$ $$2x^2 + 2x - 9 = 0$$ 6. **Divide entire equation by 2 to simplify:** $$\cancel{2}x^2 + \cancel{2}x - \cancel{9} = 0 \Rightarrow x^2 + x - \frac{9}{2} = 0$$ 7. **Use quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=1$, $c=-\frac{9}{2}$. 8. **Calculate discriminant:** $$\Delta = 1^2 - 4 \times 1 \times \left(-\frac{9}{2}\right) = 1 + 18 = 19$$ 9. **Find roots:** $$x = \frac{-1 \pm \sqrt{19}}{2}$$ 10. **Find corresponding $y$ values:** $$y = -1 - x$$ So, $$y = -1 - \frac{-1 + \sqrt{19}}{2} = \frac{-2 + 1 - \sqrt{19}}{2} = \frac{-1 - \sqrt{19}}{2}$$ and $$y = -1 - \frac{-1 - \sqrt{19}}{2} = \frac{-2 + 1 + \sqrt{19}}{2} = \frac{-1 + \sqrt{19}}{2}$$ **Final solutions:** $$\left(\frac{-1 + \sqrt{19}}{2}, \frac{-1 - \sqrt{19}}{2}\right) \quad \text{and} \quad \left(\frac{-1 - \sqrt{19}}{2}, \frac{-1 + \sqrt{19}}{2}\right)$$