1. **State the problem:** Solve the system of equations: $$y = x^2 - 7x - 5$$ $$y = -x^2 - 3x + 11$$ We want to find the values of $x$ and $y$ where these two parabolas intersect. 2. **Set the equations equal to each other:** Since both equal $y$, set the right sides equal: $$x^2 - 7x - 5 = -x^2 - 3x + 11$$ 3. **Bring all terms to one side:** $$x^2 - 7x - 5 + x^2 + 3x - 11 = 0$$ Simplify: $$2x^2 - 4x - 16 = 0$$ 4. **Simplify the equation by dividing all terms by 2:** $$\cancel{2}x^2 - \cancel{2} \cdot 2x - \cancel{2} \cdot 8 = 0 \implies x^2 - 2x - 8 = 0$$ 5. **Factor the quadratic:** $$x^2 - 2x - 8 = (x - 4)(x + 2) = 0$$ 6. **Solve for $x$:** $$x - 4 = 0 \implies x = 4$$ $$x + 2 = 0 \implies x = -2$$ 7. **Find corresponding $y$ values by substituting back into one of the original equations, e.g., $y = x^2 - 7x - 5$:** For $x=4$: $$y = 4^2 - 7 \cdot 4 - 5 = 16 - 28 - 5 = -17$$ For $x=-2$: $$y = (-2)^2 - 7 \cdot (-2) - 5 = 4 + 14 - 5 = 13$$ 8. **Write the solution as ordered pairs:** $$(4, -17), (-2, 13)$$ **Final answer:** The solutions to the system are $(4, -17)$ and $(-2, 13)$.