SOLUTION: The length of a rectangle is 3 yards less than the twice the width, and the area of the rectangle is 27 yards to the power of two. Find the dimensions of the rectangle.

Question 1028653: The length of a rectangle is 3 yards less than the twice the width, and the area of the rectangle is 27 yards to the power of two. Find the dimensions of the rectangle.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x equals the length
y equals the width
the length is 3 yards less than twice the width.
x = 2y-3
area is 27^2
length * width = area.
x*y = 27^2
since x = 2y-3, the formula becomes (2y-3)*y = 27^2
simplify to get 2y^2 - 3y = 729
subtract 27^2 from both sides of the equation to get 2y^2 - 3y - 729 = 0
use the quadratic formula to get:
y = 19.856608804285 or y = -18.something.
y can't be negative, so it has to be 19.8566.....
x = 2y-3, so x = 36.71321761
the area of the rectangle is x*y = 19.856608804285 * 36.71321761 = 729.
729 is equal to 27^2, so the requirements of the problem are satisfied.

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