The perimeter of a rectangle is 12 ft. the length is 2 ft longer than the width. find the dimensions. write a system of linear equations and solve the resulting system. let x be the length and y be the width.

Length = x

Width = y

Given,

Perimeter = 12 ft

2(x + y) = 12

x + y = 6

x = 6 - y

Given,

x = 2 + y

6 - y = 2 + y

2y = 4

--- y = 2

--- x = 6 - y = 6 - 2 = 4

Thus, the length of the rectangle is 4 ft and the width is 2 ft

P=2(L+W)
12=2(L+W)
if L=x and y=W
12=2(x+y)
divide both sides by 2
6=x+y

length is 2 ft longer than width
x=2+y

subsitute 2+y for x

6=x+y
6=2+y+y
6=2+2y
minus 2 both sides
4=2y
divide by 2
2=y

sub back
x=2+y
x=2+2
x=4

the length is 4ft
width is 2ft

The answer is lol i dont know