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.

Learn how to solve the equation sqrt(x-2) = 2sqrt(y+1) with a detailed step-by-step solution. This problem covers equations with square roots, including domain restrictions to ensure real solutions. Suitable for advanced high school students and beyond.

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