1. Let's consider a challenging algebra problem involving solving a system of nonlinear equations. 2. Problem: Solve the system of equations: $$\begin{cases} x^2 + y^2 = 25 \\ xy = 12 \end{cases}$$ 3. Formula and rules: We have two equations, one representing a circle of radius 5 centered at the origin, and the other representing the product of $x$ and $y$ equal to 12. 4. Step 1: From the second equation, express $y$ in terms of $x$: $$y = \frac{12}{x}$$ 5. Step 2: Substitute $y$ into the first equation: $$x^2 + \left(\frac{12}{x}\right)^2 = 25$$ 6. Step 3: Simplify the equation: $$x^2 + \frac{144}{x^2} = 25$$ 7. Step 4: Multiply both sides by $x^2$ to clear the denominator: $$x^4 + 144 = 25x^2$$ 8. Step 5: Rearrange to form a quadratic in $x^2$: $$x^4 - 25x^2 + 144 = 0$$ 9. Step 6: Let $z = x^2$, then: $$z^2 - 25z + 144 = 0$$ 10. Step 7: Solve the quadratic equation for $z$ using the quadratic formula: $$z = \frac{25 \pm \sqrt{25^2 - 4 \cdot 1 \cdot 144}}{2} = \frac{25 \pm \sqrt{625 - 576}}{2} = \frac{25 \pm \sqrt{49}}{2}$$ 11. Step 8: Calculate the roots: $$z_1 = \frac{25 + 7}{2} = 16, \quad z_2 = \frac{25 - 7}{2} = 9$$ 12. Step 9: Recall $z = x^2$, so: $$x^2 = 16 \Rightarrow x = \pm 4$$ $$x^2 = 9 \Rightarrow x = \pm 3$$ 13. Step 10: Find corresponding $y$ values using $y = \frac{12}{x}$: - For $x = 4$, $y = \frac{12}{4} = 3$ - For $x = -4$, $y = \frac{12}{-4} = -3$ - For $x = 3$, $y = \frac{12}{3} = 4$ - For $x = -3$, $y = \frac{12}{-3} = -4$ 14. Final answer: The solutions to the system are: $$(4, 3), (-4, -3), (3, 4), (-3, -4)$$ This problem exercises substitution, quadratic solving, and understanding of nonlinear systems.