The length of a rectangular garden is 8 ft less than 2 times its width. If the perimeter of the garden is 44 ft, find the dimensions of the garden. Define variables, set up equation and solve.
2007-01-21 16:44:12 · 3 answers · asked by sittinncat 1 in Science & Mathematics ➔ Mathematics
l=2w-8 or 2w=l+8
p=44=2l+2w=2l+l+8
44=3l+8
3l=36
l=12
2w=l+8=12+8=20
w=10
the garden is 10' X 12'
2007-01-21 16:49:17 · answer #1 · answered by yupchagee 7 · 13⤊ 0⤋
Let the width be W ft and the length L ft.
L = 2W - 8
2L + 2W = 44
Substituting for L in the second equation gives
2(2W - 8) + 2W = 44
=> 4W - 16 + 2W = 44
=> 6W = 60
=> W = 10
Then L = 2W - 8 = 20 - 8 = 12.
So the garden is 10 ft wide and 12 ft long.
2007-01-22 00:49:26 · answer #2 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
let width = x
Then the length = 2x-8
So the perimeter
44 = x+x+2x-8+2x-8
44 = 6x-16
60 = 6x
x = 10
Width = 10 Length = 12
2007-01-22 00:49:24 · answer #3 · answered by trichbopper 4 · 0⤊ 0⤋