The perimeter of a rectangle is 44ft the length is 4ft longer than the width. Find the demensions. Write a system of linear equations and solve the resulting system. Let x be the length and y be the width.

perimeter=2L+2W

L-4=W

To solve for the dimensions of the rectangle, we need to set up a system of linear equations based on the given information.

Let's start by assigning variables to the dimensions of the rectangle. We'll let x represent the length and y represent the width.

Given:
The perimeter of a rectangle is 44ft, which can be expressed as:
2(length + width) = 44
2(x + y) = 44

We are also given the length is 4ft longer than the width, so we can write:
x = y + 4

Now we have a system of two equations:
1. 2(x + y) = 44
2. x = y + 4

To solve this system, we can use substitution or elimination method.

Let's solve it using substitution:
We rearrange equation 2 to be x - y = 4.

Now we substitute x in equation 1 with (y + 4):
2((y + 4) + y) = 44
2(2y + 4) = 44
4y + 8 = 44
4y = 44 - 8
4y = 36
y = 36/4
y = 9

Now we substitute this value of y back into equation 2 to find x:
x = y + 4
x = 9 + 4
x = 13

Therefore, the dimensions of the rectangle are 13ft (length) and 9ft (width).