If the length of a rectangle is 11 yds more than twice the width and the area of the rectangle is 63yds^2 what is the dimension of the rectangle?

Width = w

length = 2w + 11

area = length * width = (2w+11) * w = 63

w = 3 gives area = 3*17 = 51. w=4 gives area = 4 * 19 = 96. So w is somewhere between 3 and 4.

2w^2 + 11w - 63 = 0

That's a quadratic equation. We can test and find out if it's easily solved by factoring. Otherwise the quadratic formula can solve it.

(2w - a) (w + b) = 0

a*b = 63

2b - a = 11

(or, a - 2b = 11)

Let's factor 63.

63 = 3*21 = 3*3*7

1 and 63 (that doesn't work).

7 and 9: 14-9=5 nope, 18-7=11 yes. So b=9, a=7 will work.

(2w - a) (w + b) = 0 becomes

(2w - 7) (w + 9) = 0

That means w can be 7/2, or - 9. - 9 is obviously wrong, so let's check 7/2. The area of the rectangle was w * (2w+11) That's 3.5 * (7+11) = 3.5*18 = 63. So the rectangle's width is 3.5, and its length is 18.