SOLUTION: The length of a rectangle is 12 cm more than its width. A second rectangle is 3cm shorter in width and 5cm longer in length than the first rectangle, and has a perimeter of 68cm.
Algebra ->
Systems-of-equations
-> SOLUTION: The length of a rectangle is 12 cm more than its width. A second rectangle is 3cm shorter in width and 5cm longer in length than the first rectangle, and has a perimeter of 68cm.
Log On
Question 58273: The length of a rectangle is 12 cm more than its width. A second rectangle is 3cm shorter in width and 5cm longer in length than the first rectangle, and has a perimeter of 68cm. What are the dimensions of this second rectangle.
So far, I got first rectangle is L=w+12 and then got confused. I know the second rectangle 2L+2W=68. This question is not from a textbook, but from a worksheet. Answer by faceoff57(108) (Show Source):
You can put this solution on YOUR website! L=w+12
2L+2w=68
2(w+17)+2(w-3)=68
2w+34+2w-6=68
4w+28=68
4w=40
w=10
THE SECOND RECTANGLE HAS A LENGTH OF 27CM AND A WIDTH OF 7CM.