1. **State the problem:** Solve the simultaneous equations: $$y = -x + 2$$ $$y = x^2 + 2x - 1$$ We need to find the values of $x$ and $y$ that satisfy both equations simultaneously. 2. **Set the equations equal:** Since both expressions equal $y$, set them equal to each other: $$-x + 2 = x^2 + 2x - 1$$ 3. **Rearrange to form a quadratic equation:** Move all terms to one side: $$0 = x^2 + 2x - 1 + x - 2$$ $$0 = x^2 + 3x - 3$$ 4. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=3$, and $c=-3$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 3^2 - 4 \times 1 \times (-3) = 9 + 12 = 21$$ So, $$x = \frac{-3 \pm \sqrt{21}}{2}$$ 5. **Find corresponding $y$ values:** Substitute each $x$ back into the linear equation $y = -x + 2$. For $$x_1 = \frac{-3 + \sqrt{21}}{2}$$ $$y_1 = -\left(\frac{-3 + \sqrt{21}}{2}\right) + 2 = \frac{3 - \sqrt{21}}{2} + 2 = \frac{3 - \sqrt{21} + 4}{2} = \frac{7 - \sqrt{21}}{2}$$ For $$x_2 = \frac{-3 - \sqrt{21}}{2}$$ $$y_2 = -\left(\frac{-3 - \sqrt{21}}{2}\right) + 2 = \frac{3 + \sqrt{21}}{2} + 2 = \frac{3 + \sqrt{21} + 4}{2} = \frac{7 + \sqrt{21}}{2}$$ **Final answers:** $$x = \frac{-3 + \sqrt{21}}{2}, \quad y = \frac{7 - \sqrt{21}}{2}$$ $$x = \frac{-3 - \sqrt{21}}{2}, \quad y = \frac{7 + \sqrt{21}}{2}$$