Oops, my mistake... I mean, mistakes.

The length of a rectangle is 6 meters less that twice it's width. The perimeter is 48 meters. find the dimensions of the rectangle.

I got this lovely multi-step word problem wrong on my math Common Assessment on October 14th 2010.  I spoted some mistakes as you can see above. I...

In finding the width:
1) Didn't distribute the 2 to EVERYTHING in the parentheses. I should have gotten 2w+4w-12
2) After doing that I added 48 and 6 and got 42. What was I thinking? The world may never know. I should have added 12 to 48 and have gotten 60.

In finding the length:
1) I substituted 7 in for 'W' but 7 did not satisfy the equation 2w+2(2w-6)=48. Therefore my answer will automatically be wrong.  

But never fear, I do it correctly here!

and here!

To check we substitute the "w" and the "l" in an equation that translates to the perimeter of the rectangle.

P=(2l)(2w)

And so...

48=2(14)+2(10)

48=28+20

48=48

Therefore the correct dimensions are 14x10

Not too shabby...

It's great to be able to write a correct answer in bold letters, it means you know what you're doing. Talk about smarticle particles.