Let's solve the equation:

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

Step 1: Square both sides to eliminate the square roots.

Squaring both sides, we get:

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

This simplifies to:

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

Step 2: Expand and simplify the equation.

Expand the right-hand side:

$x - 2 = 4y + 4$

Now, isolate $x$ by adding 2 to both sides:

$x = 4y + 6$

Step 3: Consider the domain restrictions.

Since we are dealing with square roots, we must ensure that the expressions inside the square roots are non-negative:

1. $x - 2 \geq 0$ implies $x \geq 2$.
2. $y + 1 \geq 0$ implies $y \geq -1$.

These restrictions are essential because they ensure the values of $x$ and $y$ lead to real numbers under the square roots.

Final Answer:

The relationship between $x$ and $y$ is given by:

$x = 4y + 6$

with the constraints $x \geq 2$ and $y \geq -1$.

Would you like further details or have any questions? Here are some related topics:

1. How would the solution be affected if the equation included cube roots?
2. Can we express $y$ in terms of $x$ and explore its implications?
3. What does the graph of this equation look like in the $xy$-plane?
4. How do the domain restrictions impact the solution set?
5. What if the equation had a different constant factor, such as $3$ instead of $2$?

Tip: When solving equations involving square roots, always remember to check for extraneous solutions, which may arise from squaring both sides.