Solve. Please show all of your work.

Find the length and width of a rectangular lot with a perimeter of 108 meters if the length is 8 meters more than the width.

Perimeter is

2L + 2W = 108
L = W + 8

Solve by substitution.

2(W+8) + 2W = 108
4W = 92
Finish it.

The rest of your posts have been removed.

Once you write up YOUR THOUGHTS, please re-post, and someone here will be happy to comment.

To find the length and width of a rectangular lot with a given perimeter, we need to set up equations based on the information given in the problem.

Let's assume that the width of the lot is represented by the variable "x" meters. Since the length of the lot is 8 meters more than the width, we can represent the length as "x + 8" meters.

The perimeter of a rectangle is calculated by adding the lengths of all four sides. For a rectangle, opposite sides have the same length. So, for this problem, the perimeter is twice the sum of the length and width.

Using this information, we can set up the equation:
2 * (length + width) = perimeter

Substituting the values from the problem into the equation, we have:
2 * (x + (x + 8)) = 108

Simplifying the equation, we get:
2 * (2x + 8) = 108

Expanding the brackets, we have:
4x + 16 = 108

Next, we'll isolate the variable on one side of the equation. Subtracting 16 from both sides, we get:
4x = 108 - 16
4x = 92

To solve for x, we divide both sides of the equation by 4:
x = 92 / 4
x = 23

Now that we have determined the value of x, we can substitute it back into the equation to find the length:
Length = x + 8 = 23 + 8 = 31

So, the width of the rectangular lot is 23 meters, and the length is 31 meters.