The area of a is 99yd and the length of the rectangle is 7yd more than twice the width. Find the dimensions of the rectangle

Question

The area of a rectangle is 99yd and the length of the rectangle is 7yd more than twice the width. Find the dimensions of the rectangle.

Answer 1

99 = w(2w+7)

99 = 2w^2 + 7w

0 = 2w^2 + 7w - 99

0 = ( 2w - 11)(w + 9 )

2w-11 = 0 ---> w = 11/2

w+9=0 results in negative measures

Length is 2(11/2)+ 7 = 18

Area is 18 * 11/2 = 9 * 11 = 99