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Let's solve the problem step by step.

We are given:
- A square with an area of \(121 \, \text{in}^2\).
- A rectangle with dimensions \(3x\) and \(x - 2\).
- The perimeters of the square and rectangle are the same.

### Step 1: Find the side length of the square.
The area of the square is given by:
\[
\text{Area of square} = \text{side}^2 = 121 \, \text{in}^2
\]
So, to find the side length:
\[
\text{side} = \sqrt{121} = 11 \, \text{in}
\]

### Step 2: Find the perimeter of the square.
The perimeter of a square is \(4\) times the side length:
\[
\text{Perimeter of square} = 4 \times 11 = 44 \, \text{in}
\]

### Step 3: Write an equation for the perimeter of the rectangle.
The perimeter \(P\) of a rectangle is given by:
\[
P = 2(\text{length} + \text{width})
\]
For this rectangle, the length is \(3x\) and the width is \(x - 2\). So, the perimeter is:
\[
P = 2\left(3x + (x - 2)\right) = 2(3x + x - 2) = 2(4x - 2) = 8x - 4
\]

### Step 4: Set up the equation.
Since the perimeters of the square and the rectangle are the same:
\[
8x - 4 = 44
\]

### Step 5: Solve for \(x\).
\[
8x - 4 = 44
\]
Add \(4\) to both sides:
\[
8x = 48
\]
Divide both sides by \(8\):
\[
x = 6
\]

### Final Answer:
The value of \(x\) is \(6\).

---

Would you like further details on any part of the solution?

Here are some related questions you might consider:
1. What is the perimeter of a square with an area of \(225 \, \text{in}^2\)?
2. How does the formula for perimeter differ for squares and rectangles?
3. If the width of the rectangle were \(x + 3\), how would the equation change?
4. Can you find the area of the rectangle for \(x = 6\)?
5. How would the equation change if the perimeters were not equal?

**Tip:** When solving for unknowns in geometric problems, always start by writing the formula for each shape’s perimeter or area.

The perimeters of the square and the rectangle below are the same. Write and solve an equation to find the value of x.

Learn how to find the value of x by setting up an equation based on the perimeters of a square and rectangle being equal. This problem involves solving using geometry and algebra, suitable for students in Grades 6-8.

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