SOLUTION: The perimeter of the rectangle 38ft. The length is 5ft longer than the width. Find the dimensions. Write a system of linear equations and solve the resulting system. Let x be the l

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The perimeter of the rectangle 38ft. The length is 5ft longer than the width. Find the dimensions. Write a system of linear equations and solve the resulting system. Let x be the l      Log On


   

Question 648577: The perimeter of the rectangle 38ft. The length is 5ft longer than the width. Find the dimensions. Write a system of linear equations and solve the resulting system. Let x be the length y be the width.
2x+2y=_
x=y+__
What is the Length? __ft
What is the Width? __Ft

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Let x be the length y be the width.
The perimeter of the rectangle P=2x%2B2y
given:
P=+38ft
so,
2x%2B2y=38ft...solve for x
2x=38ft-2y
x=38ft%2F2-2y%2F2
x=19ft-y.......1

since given that the length x is 5ft longer than the width y
x=y%2B5ft......2
equations 1 and two have equal left sides, so right sides must be equal too; than we have
19ft-y=y%2B5ft.........solve for y

19ft-5ft=y%2By
14ft=2y
7ft=y
highlight%28y=7ft%29.........the Width
now find x
x=y%2B5ft......2
x=7ft%2B5ft
highlight%28x=12ft%29.......the Length