A room has a perimeter of 52 ft. The length *x* is 6 ft greater than the width *y*. Which shows the correct system and the dimensions of the room?

A.

2*x* + 2*y* = 52; *x* + 6 = *y* + 6;  *x* = 10 ft, *y* = 16 ft

B.

2*x* + 2*y* = 52; *x* + 6 = *y*;  *x* = 16 ft, *y* = 10 ft

C.

*x* + *y* = 52; *x* = *y* + 6;  *x* = 22 ft, *y* = 28 ft

D.

2*x* + 2*y* = 52; *x* = *y* + 6; *x* = 16 ft, *y* = 10 ft

To solve this problem, we need to set up and solve a system of equations based on the given information.

1. The perimeter of the room is 52 feet:
   \[
   2x + 2y = 52
   \]

2. The length \(x\) is 6 feet greater than the width \(y\):
   \[
   x = y + 6
   \]

We can now solve this system of equations.

First, let's simplify the perimeter equation:
\[
2x + 2y = 52 \implies x + y = 26
\]

Now, we substitute \(x = y + 6\) into the simplified perimeter equation:
\[
(y + 6) + y = 26
\]
\[
2y + 6 = 26
\]
\[
2y = 20
\]
\[
y = 10
\]

Now, substitute \(y = 10\) back into the equation \(x = y + 6\):
\[
x = 10 + 6
\]
\[
x = 16
\]

Thus, the dimensions of the room are \(x = 16\) feet and \(y = 10\) feet.

Now let's identify the correct option from the given choices:

A. 
\[
2x + 2y = 52; x + 6 = y + 6; x = 10 \text{ ft}, y = 16 \text{ ft}
\]
This is incorrect because it swaps the dimensions and has an incorrect second equation.

B.
\[
2x + 2y = 52; x + 6 = y; x = 16 \text{ ft}, y = 10 \text{ ft}
\]
This matches our solution.

C.
\[
x + y = 52; x = y + 6; x = 22 \text{ ft}, y = 28 \text{ ft}
\]
This is incorrect because it has the wrong perimeter equation and incorrect dimensions.

D.
\[
2x + 2y = 52; x = y + 6; x = 16 \text{ ft}, y = 10 \text{ ft}
\]
This is also correct.

Thus, the correct answer is:
\[
\boxed{D. \ 2x + 2y = 52; \ x = y + 6; \ x = 16 \text{ ft}, \ y = 10 \text{ ft}}
\]

Would you like further details or have any questions?

Here are 8 related questions that expand on this information:
1. What would be the perimeter if both dimensions were doubled?
2. How would the equations change if the length was 8 feet greater than the width?
3. If the perimeter was increased to 60 feet, what would the new dimensions be?
4. How would you find the area of the room given its dimensions?
5. If the width was increased by 2 feet, what would be the new perimeter?
6. What is the relationship between perimeter and area for a rectangular room?
7. How do the equations change if the room is square?
8. What is the ratio of the length to the width of the room?

**Tip**: Solving systems of equations is a fundamental skill in algebra that helps in finding unknown values that satisfy multiple conditions simultaneously.

Learn how to solve a perimeter problem involving a room with specific dimensions using algebraic equations. This problem covers key algebraic concepts, including linear equations and the calculation of perimeter for rectangles. Suitable for students in Grades 7-9.

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