A mistake made is a lesson learned.

So to show you how I've learned from my mistake I'll create a sample problem that is similar to the one below.


The width of a rectangle is 2 meters less than twice it's length. The perimeter is 32 meters. Find the dimensions of the rectangle.


What we know: Width (w) = 2l - 2
                          Perimeter  (p) = 32 m

What we don't know: Length ( l )

First, find the length:   
 p = 2w + 2l

32 = 2(2l - 2) + 2l
32 = 4l - 4 + 2l 
32 = 6l - 4
+4        +4
36 = 6l
/6     /6
6 =  l (The length of the rectangle is 6m)

Now, find the width:
w = 2l - 2

w = 2(6) - 2
w = 12 - 2
w = 10 (The width of the rectangle is 10m)

Then, check!!
p = 2w + 2l

32 = 2(6) + 2(10)
32 = 12 + 20
32 = 32 (Therefore the answer is correct)

My final answer (in a full sentence) : 

The dimensions of the rectangle are 6 x 10 m.

Booyah. :)