SOLUTION: THE LENGTH OF A RECTANGULAR GARDEN IS 8FT LESS THAN 3 TIMES IT'S WIDTH. IF THE PERIMETER OF THE GARDEN IS 48FT, FIND THE DIMENSIONS OF THE GARDEN.
IF SOMEONE CAN PLEASE HELP I W
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IF SOMEONE CAN PLEASE HELP I W
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Question 57084: THE LENGTH OF A RECTANGULAR GARDEN IS 8FT LESS THAN 3 TIMES IT'S WIDTH. IF THE PERIMETER OF THE GARDEN IS 48FT, FIND THE DIMENSIONS OF THE GARDEN.
IF SOMEONE CAN PLEASE HELP I WOULD APPRECIATE IT Found 2 solutions by Scriptor, tutorcecilia:Answer by Scriptor(36) (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)
Solution: y = 4/3 + 4*sqrt(10)/3
x = 48 / (4/3 + 4*sqrt(10)/3)
You can put this solution on YOUR website! P=2(length + width) [Use the formula for a perimeter]
P=48ft
Length= 3w-8
Width=w
.
48=2[(3w-8)+w]
48=2(3w-8+w)
48=2(4w-8)
48=8w-16
48+16=8w
64=8w
64/8=w
8ft=width
.
Length:
l=3w-8
l=3(8)-8
l=24-8
Length = 16 ft
.
Check by plugging the values of the perimeter, length and width into the formula for the perimeter of a rectangle.
48=2(16+8)
48=2(24)
48=48 [checks out]
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