1. **State the problem:** Solve the equation $$n\sqrt{1 + y} + y\sqrt{1 + n} = 0$$ for $y$ in terms of $n$. 2. **Rewrite the equation:** $$n(1 + y)^{\frac{1}{2}} + y(1 + n)^{\frac{1}{2}} = 0$$ 3. **Isolate one term:** Move the second term to the other side: $$n(1 + y)^{\frac{1}{2}} = -y(1 + n)^{\frac{1}{2}}$$ 4. **Square both sides** to eliminate the square roots: $$\left(n(1 + y)^{\frac{1}{2}}\right)^2 = \left(-y(1 + n)^{\frac{1}{2}}\right)^2$$ $$n^2(1 + y) = y^2(1 + n)$$ 5. **Expand and rearrange:** $$n^2 + n^2 y = y^2 + n y^2$$ 6. **Bring all terms to one side:** $$n^2 + n^2 y - y^2 - n y^2 = 0$$ 7. **Group terms:** $$n^2 + y(n^2 - y - n y) = 0$$ Rewrite the $y$ terms: $$n^2 + y(n^2 - y(1 + n)) = 0$$ 8. **Rewrite as a quadratic in $y$:** $$-y^2(1 + n) + n^2 y + n^2 = 0$$ Multiply both sides by $-1$ for clarity: $$y^2(1 + n) - n^2 y - n^2 = 0$$ 9. **Use quadratic formula:** For equation $$a y^2 + b y + c = 0$$ with $$a = 1 + n, \quad b = -n^2, \quad c = -n^2$$ The solutions are: $$y = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a} = \frac{n^2 \pm \sqrt{(-n^2)^2 - 4(1 + n)(-n^2)}}{2(1 + n)}$$ 10. **Simplify under the square root:** $$b^2 - 4 a c = n^4 + 4 n^2 (1 + n) = n^4 + 4 n^2 + 4 n^3 = n^2 (n^2 + 4 + 4 n) = n^2 (n^2 + 4 n + 4)$$ Note that $$n^2 + 4 n + 4 = (n + 2)^2$$ So: $$\sqrt{b^2 - 4 a c} = \sqrt{n^2 (n + 2)^2} = |n| |n + 2|$$ 11. **Write the final solutions:** $$y = \frac{n^2 \pm |n| |n + 2|}{2 (1 + n)}$$ 12. **Interpretation:** The solution depends on the sign of $n$ and $n + 2$ due to absolute values. **Final answer:** $$\boxed{y = \frac{n^2 \pm |n| |n + 2|}{2 (1 + n)}}$$