1. **State the problem:** Solve the system of equations algebraically: $$y = x^2 - 2x - 3$$ $$x + y = 3$$ 2. **Use substitution:** From the second equation, express $y$ as: $$y = 3 - x$$ 3. **Set the two expressions for $y$ equal:** $$x^2 - 2x - 3 = 3 - x$$ 4. **Bring all terms to one side to form a quadratic equation:** $$x^2 - 2x - 3 - 3 + x = 0$$ Simplify: $$x^2 - x - 6 = 0$$ 5. **Factor the quadratic:** $$x^2 - x - 6 = (x - 3)(x + 2) = 0$$ 6. **Solve for $x$:** $$x - 3 = 0 \Rightarrow x = 3$$ $$x + 2 = 0 \Rightarrow x = -2$$ 7. **Find corresponding $y$ values using $y = 3 - x$:** For $x = 3$: $$y = 3 - 3 = 0$$ For $x = -2$: $$y = 3 - (-2) = 3 + 2 = 5$$ 8. **Final solutions:** $$\boxed{(3, 0) \text{ and } (-2, 5)}$$ These points are where the parabola and the line intersect.