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