1. **State the problem:** Solve the system of equations: $$y = -2x + 3$$ $$y = x^2 + x - 1$$ 2. **Set the equations equal to find intersection points:** Since both expressions equal $y$, set them equal: $$-2x + 3 = x^2 + x - 1$$ 3. **Rearrange to form a quadratic equation:** Move all terms to one side: $$0 = x^2 + x - 1 + 2x - 3$$ $$0 = x^2 + 3x - 4$$ 4. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=3$, and $c=-4$. Calculate the discriminant: $$\Delta = 3^2 - 4 \times 1 \times (-4) = 9 + 16 = 25$$ Calculate the roots: $$x = \frac{-3 \pm \sqrt{25}}{2} = \frac{-3 \pm 5}{2}$$ 5. **Find the two solutions for $x$:** - For $x = \frac{-3 + 5}{2} = \frac{2}{2} = 1$ - For $x = \frac{-3 - 5}{2} = \frac{-8}{2} = -4$ 6. **Find corresponding $y$ values:** Substitute back into $y = -2x + 3$: - For $x=1$: $y = -2(1) + 3 = -2 + 3 = 1$ - For $x=-4$: $y = -2(-4) + 3 = 8 + 3 = 11$ 7. **Final solutions:** $$\boxed{(1, 1) \text{ and } (-4, 11)}$$ These points are where the parabola and the line intersect, solving the system.