The perimeter of a rectangle is 44 m. if the width were doubled and the length were increased by 8 m, the perimeter would be 76 m. what is the length of the rectangle

Let us assume the length of the first rectangle = x meter
let us assume the width of the first rectangle = y meter
Perimeter of the first rectangle = 44 meter
we already know
Perimeter of a rectangle = 2 (Length + Width)
Then in the case of the first rectangle we get
44 = 2(x + y)
44/2 = x + y
x + y = 22
y = 22 - x
Now coming to the case of the second rectangle
The length of the second rectangle = x + 8
The width of the second rectangle = 2y
Perimeter of the second rectangle = 76 meter
then
76 = 2[(x + 8) + 2y]
76/2 = x + 8 + 2y
38 = x + 2y + 8
x + 2y = 38 - 8
x + 2y = 30
Now we replace the value of y that we found from the first equation. Then
x + 2(22 - x) = 30
x - 2x + 44 = 30
-x = 30 - 44
-x = -14
x = 14
So the length of the first rectangle is 14 meters.

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm