SOLUTION: A rectangular garden is 9 ft. longer than it is wide. A second rectangular garden is planned so that it will be 6 ft. wider and twice as long as the first garden. Find the area of
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-> SOLUTION: A rectangular garden is 9 ft. longer than it is wide. A second rectangular garden is planned so that it will be 6 ft. wider and twice as long as the first garden. Find the area of
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Question 313344: A rectangular garden is 9 ft. longer than it is wide. A second rectangular garden is planned so that it will be 6 ft. wider and twice as long as the first garden. Find the area of the first garden if the sum of the areas of both gardens will be 528 ft squared Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! let the width of the first garden be x
the length will be x+9
The area of this garden = x(x+9)
..
The width of the second garden is 6 more than the first
so width will be x+6
The length of the second garden = 2(x+9)
so the area of this garden = (x+6)*2(x+9)
..
The sum of their areas = 528ft^2
x(x+9)+2(x+6)(x+9)= 528
x^2+9x +2(x^2+15x+54)=528
x^2 +9x +2x^2+30x+108=528
3x^2+39x-420= 0
3(x^2+13x-140)=0
x^2+13x-140=0
x^2+20x-7x-140=0
x(x+20)-7(x+20)=0
(x+20(x-7)=0
x=7 the width of the first garden.
the length will be x+9 = 7+9
=16
Area of the garden = 16*7
= 112 sq. ft
