SOLUTION: The perimeter of a rectangle is 16 feet. If twice its length is 4 times its width, find the width and the length of a rectangle. My Work: Equation 1: 2L=4W Equation 2: L+W=16

Algebra ->  Linear-equations -> SOLUTION: The perimeter of a rectangle is 16 feet. If twice its length is 4 times its width, find the width and the length of a rectangle. My Work: Equation 1: 2L=4W Equation 2: L+W=16       Log On


   

Question 1177440: The perimeter of a rectangle is 16 feet. If twice its length is 4 times its width, find the width and the length of a rectangle.
My Work:
Equation 1: 2L=4W
Equation 2: L+W=16
W=16/3
L=10/6
Found 3 solutions by Boreal, mananth, MathLover1:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
perimeter is 2L +2W=16; L+W=8, the half-perimeter, not 16
2L=4W, so L=2W
6W=16
W=8/3 feet
4W=32/3 feet, and 2L=32/3 feet so L=16/3 feet
width=8/3 feet
Half-perimeter is 24/3 feet (8)

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!

My Work:
Equation 1: 2L=4W
Equation 2: 2L+2W=16
Therefore 4w+2w=16
6w=16
w=16/6 =8/3
l=2*8/3=16/3

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

The perimeter of a rectangle: P=2%28L%2BW%29

if the perimeter of a rectangle is 16 feet, then
2%28L%2BW%29=16............eq.1
if twice its length is 4 times its width, then
2L=4W.......solve for L
L=2W.......eq.2, substitute in eq.1

2%282W%2BW%29=16............eq.1
%283W%29=16%2F2
3W=8
W=8%2F3

go to
L=2W.......eq.2, substitute W
L=2%288%2F3%29
L=16%2F3

the width is 8%2F3 ft and the length 16%2F3 ft