1. Let's create a challenging algebra problem involving quadratic equations and systems. 2. Problem: Solve the system of equations: $$\begin{cases} x^2 + y^2 = 25 \\ x + y = 7 \end{cases}$$ 3. Formula and rules: - The first equation represents a circle with radius 5. - The second is a linear equation. - To solve, substitute $y = 7 - x$ into the first equation. 4. Substitute and simplify: $$x^2 + (7 - x)^2 = 25$$ $$x^2 + 49 - 14x + x^2 = 25$$ $$2x^2 - 14x + 49 = 25$$ 5. Rearrange: $$2x^2 - 14x + 24 = 0$$ 6. Divide entire equation by 2: $$x^2 - 7x + 12 = 0$$ 7. Factor the quadratic: $$(x - 3)(x - 4) = 0$$ 8. Solutions for $x$: $$x = 3 \quad \text{or} \quad x = 4$$ 9. Find corresponding $y$ values: - If $x=3$, then $y = 7 - 3 = 4$ - If $x=4$, then $y = 7 - 4 = 3$ 10. Final solutions: $$(x, y) = (3, 4) \quad \text{or} \quad (4, 3)$$ This problem combines substitution, quadratic factoring, and understanding geometric shapes.