Question

A rectangular garden is 9ft longer than it is wide. A second rectangular garden is planned so that it will be 6ft wider and twice as long as the first garden. Find the area of the first garden of the sum of the areas of both gardens will be 528 ft^2.

Answer 1

112 sq ft.

Explanation:

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