1. The problem is to analyze and solve the first equation: $2x^3 + y^2 = 1 - 4y$. 2. Start by rewriting the equation to isolate terms and simplify: $$2x^3 + y^2 + 4y = 1$$ 3. Notice that the $y$ terms form a quadratic expression $y^2 + 4y$. We can complete the square to better understand it: $$y^2 + 4y = (y + 2)^2 - 4$$ 4. Substitute back: $$2x^3 + (y + 2)^2 - 4 = 1$$ 5. Add 4 to both sides: $$2x^3 + (y + 2)^2 = 5$$ 6. This equation relates $x$ and $y$ implicitly. For given $x$, we can solve for $y$: $$(y + 2)^2 = 5 - 2x^3$$ 7. Taking square roots: $$y + 2 = \pm \sqrt{5 - 2x^3}$$ 8. Finally, solve for $y$: $$y = -2 \pm \sqrt{5 - 2x^3}$$ 9. Important note: The expression under the square root must be non-negative for real $y$: $$5 - 2x^3 \geq 0 \implies 2x^3 \leq 5 \implies x^3 \leq \frac{5}{2}$$ 10. This means $x \leq \sqrt[3]{\frac{5}{2}}$ for real solutions. Final answer: $$y = -2 \pm \sqrt{5 - 2x^3}$$ with domain $x \leq \sqrt[3]{\frac{5}{2}}$ for real $y$ values.