To solve the equation:

$\sqrt{x-2} = 2\sqrt{y+1}$

First, let's square both sides to eliminate the square roots:

$(\sqrt{x-2})^2 = (2\sqrt{y+1})^2$

This simplifies to:

$x - 2 = 4(y + 1)$

Next, distribute the 4 on the right-hand side:

$x - 2 = 4y + 4$

Now, add 2 to both sides to isolate $x$:

$x = 4y + 6$

So the relationship between $x$ and $y$ is:

$x = 4y + 6$

This equation shows that $x$ is directly related to $y$.

Would you like to explore more details or have any questions about this solution? Here are some related questions:

1. What would happen if the equation involved cubes instead of square roots?
2. How does the domain of $x$ and $y$ affect the solution?
3. Can we solve for $y$ in terms of $x$ instead?
4. How would the solution change if the equation had an additional constant term on either side?
5. What is the graphical interpretation of this equation?

Tip: Always check the domain restrictions when dealing with square roots. The expressions inside the square roots must be non-negative to have real solutions.