Question 400472
a rectangular garden is 9ft 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

let x=width of first garden
then, x+9=length of first garden
then, x+6=width of second garden
then,2(x+9)=length of second garden
area = length*width

x(x+9)+(x+6)(2)(x+9)=528
x^2+9x+2(x^2+15x+54)=528
x^2+9x+2x^2+30x+108=528
3x^2+39x+108=528
3x^2+39x-420=0
x^2+13x-140=0
(x+20)(x-7)=0
x=-20(reject) or x=7
ans:area of first garden=x(x+9)=7*16=112sqft