SOLUTION: solve and show the equation used for the solution. The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 34 in., find the dimension

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Question 122147: solve and show the equation used for the solution.
The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 34 in., find the dimensions of the rectangle.
Answer by praseenakos@yahoo.com(507) (Show Source):
You can put this solution on YOUR website!
Question:
The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 34 in., find the dimensions of the rectangle
Answer:
The perimeter of the rectangle is given by the formula,

P = 2(length +width)

Assume that, width = x inches

Then its length = 2x+2 inches( because it is given that more than twice its width)

So P = 2( 2x+2 +x) that is substitute the values of length and width.

But P = 34 inches, given

==> 34 = 2( 3x + 2)

==> 34 = 6x + 4

Subtract 4 from both sides

==> 34- 4 = 6x + 4 -4

==> 30 = 6x

Divide by 6 on both sides..

==>

==> 5 = x

That is width = 5 inches

So length = 2(5) + 2

==> length = 12 inches.

Hope ypou found the explanation useful.

Regards.
Praseena.