SOLUTION: The length of a conference room is one and half times its width. A carpet that is twice as long as it is wide is placed on the centre of the room, leaving a 3 m wide border round t

Algebra ->  Equations -> SOLUTION: The length of a conference room is one and half times its width. A carpet that is twice as long as it is wide is placed on the centre of the room, leaving a 3 m wide border round t      Log On


   

Question 772698: The length of a conference room is one and half times its width. A carpet that is twice as long as it is wide is placed on the centre of the room, leaving a 3 m wide border round the carpet. Find the area of the carpet.
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Drawing a picture may be helpufl (but I have not included one here.)

Let the length of the room be x and the width of the room be y.
According to given, highlight%28x=2y%29.

Placement of the carpet is done in order to form a bare area 3 meters around the carpet.
Carpet length is x-2%2A3=x-6
Carpet width is y-2%2A3=y-6
We were also told that carpet length & width are in same ratio as that of the room, so
highlight%28x-6=2%28y-6%29%29

The highlighted equations give a system in the two variables, x and y. The equations are only for lengths, so these are linear equations. Solve the system for x and y and you can compute both the area of the room, and the area of the carpet. You will want to perform a couple of operations on the carpet highlighted equation before the fully handle the two equations as a system; you would want the equations in as much the same format as possible before finishing to find x and y.