The problem states that the square and rectangle have equal perimeters, and we are given the following information:

• The side length of the square is $4x - 1$.
• The length of the rectangle is $2x + 2$, and the width of the rectangle is $2x$.

Step 1: Find the perimeter of the square.

The perimeter of a square is given by: $\text{Perimeter of square} = 4 \times (\text{side length})$ Substituting the side length: $\text{Perimeter of square} = 4 \times (4x - 1) = 16x - 4$

Step 2: Find the perimeter of the rectangle.

The perimeter of a rectangle is given by: $\text{Perimeter of rectangle} = 2 \times (\text{length} + \text{width})$ Substituting the given length and width: $\text{Perimeter of rectangle} = 2 \times ((2x + 2) + 2x) = 2 \times (4x + 2) = 8x + 4$

Step 3: Set up the equation for equal perimeters.

Since the square and rectangle have equal perimeters, we can set the two expressions equal to each other: $16x - 4 = 8x + 4$

Step 4: Solve the equation.

1. Subtract $8x$ from both sides: $16x - 8x - 4 = 4$ $8x - 4 = 4$
2. Add 4 to both sides: $8x = 8$
3. Divide both sides by 8: $x = 1$

Conclusion:

The value of $x$ is $1$.

Do you want more details on any step, or do you have any further questions?

Here are 5 related questions to deepen your understanding:

1. What would happen to the perimeters if the side of the square was doubled?
2. How does changing the width of the rectangle affect its perimeter?
3. Can you find the perimeters of both shapes when $x = 1$?
4. What if the side of the square was $3x - 2$ instead? How would that change the solution?
5. How would you write a similar problem involving a circle and a triangle?

Tip: Always double-check your setup when equating formulas for geometric shapes to avoid simple mistakes.