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.

Learn how to solve the equation sqrt(x-2) = 2sqrt(y+1) step-by-step. This problem involves equations with square roots, suitable for advanced high school students. Explore related mathematical concepts and detailed solutions.

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