Wayne is hanging a string of lights 75 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is 3 feet longer than twice its width. Find the length and width of the patio.

Width of the patio is 
 feet
Length of the patio is 
 feet

Let's denote the width of the patio as \( w \) feet. According to the problem, the length of the patio is 3 feet longer than twice its width. Therefore, we can express the length of the patio as \( l = 2w + 3 \) feet.

Wayne is hanging a string of lights around three sides of the patio. Since one side is adjacent to the house, the lights are only needed for the other three sides: the two widths and one length.

The total length of the string of lights is 75 feet. Therefore, we have:

\[
2w + l = 75
\]

Substituting the expression for \( l \) from the above:

\[
2w + (2w + 3) = 75
\]

Simplify and solve for \( w \):

\[
2w + 2w + 3 = 75
\]
\[
4w + 3 = 75
\]
\[
4w = 75 - 3
\]
\[
4w = 72
\]
\[
w = \frac{72}{4} = 18 \text{ feet}
\]

Now that we have the width, we can find the length:

\[
l = 2w + 3 = 2(18) + 3 = 36 + 3 = 39 \text{ feet}
\]

So, the width of the patio is \( 18 \) feet, and the length of the patio is \( 39 \) feet.

Would you like any further details or explanations?

Here are 5 related questions:
1. How would the solution change if the string of lights were 80 feet long?
2. What if the length of the patio was 4 feet longer than twice its width?
3. How do you calculate the area of the patio with the dimensions found?
4. Can you calculate the perimeter of the patio, including the side adjacent to the house?
5. How would you solve for the dimensions if the patio was a square?

**Tip:** When setting up an equation from a word problem, clearly define your variables and write out the relationships described in the problem. This helps prevent errors when solving the equations.

Wayne is hanging a string of lights 75 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is 3 feet longer than twice its width. Find the length and width of the patio.

Solve a real-world algebra problem involving the perimeter of a rectangular patio. Given that the length is 3 feet longer than twice its width, learn how to calculate the exact dimensions of the patio. Perfect for students in Grades 7-9.

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