1. **State the problem:** Solve the simultaneous equations: $$\sqrt{x} - \sqrt{y} = 1$$ $$\sqrt{xy} = 20$$ where $x, y \in \mathbb{N}$ and $x > y$. 2. **Introduce variables:** Let $a = \sqrt{x}$ and $b = \sqrt{y}$. Then the equations become: $$a - b = 1$$ $$ab = 20$$ 3. **Use substitution:** From the first equation, express $a$ as: $$a = b + 1$$ 4. **Substitute into the second equation:** $$ab = (b + 1) b = 20$$ which simplifies to: $$b^2 + b = 20$$ 5. **Rewrite as a quadratic equation:** $$b^2 + b - 20 = 0$$ 6. **Solve the quadratic equation using the quadratic formula:** $$b = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-20)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm \sqrt{81}}{2}$$ 7. **Calculate the roots:** $$b = \frac{-1 + 9}{2} = 4 \quad \text{or} \quad b = \frac{-1 - 9}{2} = -5$$ Since $b = \sqrt{y} \geq 0$, discard $b = -5$. 8. **Find $a$:** $$a = b + 1 = 4 + 1 = 5$$ 9. **Find $x$ and $y$ by squaring $a$ and $b$:** $$x = a^2 = 5^2 = 25$$ $$y = b^2 = 4^2 = 16$$ 10. **Check the solution:** $$\sqrt{25} - \sqrt{16} = 5 - 4 = 1$$ $$\sqrt{25 \times 16} = \sqrt{400} = 20$$ Both equations are satisfied. **Final answer:** $$x = 25, \quad y = 16$$