SOLUTION: CAN SOMEONE PLEASE ASSIST: THE LENGTH OF A RECTANGULAR GARDEN IS 8FT LESS THEN 3 TIMES ITS WIDTH. IF THE PERIMETER OF THE GARDEN IS 48FT, FIND THE DIMENSIONS OF THE GARDEN..

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Question 56903: CAN SOMEONE PLEASE ASSIST:
THE LENGTH OF A RECTANGULAR GARDEN IS 8FT LESS THEN 3 TIMES ITS WIDTH. IF THE PERIMETER OF THE GARDEN IS 48FT, FIND THE DIMENSIONS OF THE GARDEN..
Found 2 solutions by Scriptor, tutorcecilia:
Answer by Scriptor(36) About Me  (Show Source):
You can put this solution on YOUR website!
Call the length = x
Call the width = y
You know that x*y = 48
You also know that x = 3y - 8
Substitute this x in x*y=48 to obtain:
y * (3y-8) = 48
3y² - 8y - 48 = 0
Now solve this for y (delete the negative root)
The rest of the excercise is for you :-)
(hint: y = 4/3 + 4*sqrt(10)/3 )

Answer by tutorcecilia(2152) About Me  (Show Source):
You can put this solution on YOUR website!
CAN SOMEONE PLEASE ASSIST:
THE LENGTH OF A RECTANGULAR GARDEN IS 8FT LESS THEN 3 TIMES ITS WIDTH. IF THE PERIMETER OF THE GARDEN IS 48FT, FIND THE DIMENSIONS OF THE GARDEN..
P=2(length+width) [Use the formula for the perimeter of a rectangle.]
P=48
Length = (3)(width)-8=(3w-8)
Width = width = w
.
48=2[(3w-8)+w] [Plug-in the values]
48=2(3w-8+w] [Simplify]
48=2(4w-8]
48=8w-16 [Solve for w]
48+16=8w
64=8w
8=w
So, the width is 8 ft.
.
To find the length:
length=(3w-8)=(3)(8)-8=24-8=16 feet
.
Check by plugging all of the values back into the original equation.
48=2(3(8)-8+8)
48=2(24)
48=48 [Checks out]