SOLUTION: Find the dimensions of a rectangle with a perimeter of 54 meters, if its length is 3 meters less than twice its width. I thought it was: 54=2w-3 54-2w=2w-2w-3 54-2w=-3 54-54

Algebra ->  Rectangles -> SOLUTION: Find the dimensions of a rectangle with a perimeter of 54 meters, if its length is 3 meters less than twice its width. I thought it was: 54=2w-3 54-2w=2w-2w-3 54-2w=-3 54-54      Log On


   

Question 217694: Find the dimensions of a rectangle with a perimeter of 54 meters, if its length is 3 meters less than twice its width.
I thought it was:
54=2w-3
54-2w=2w-2w-3
54-2w=-3
54-54-2w=-3+-54
-2w=-57
-2w/-2=-57/-2
w=28.5
but it's not right with the answer in the back of the book
what am I not doing right?
Please help
Thanks Leona
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let L=length, W=width


Since "length is 3 meters less than twice its width", we know that L=2W-3.


Also, because "the dimensions of a rectangle with a perimeter of 54 meters", this means that 2L%2B2W=54. In other words, you add the length twice (there are 2 'lengths') with the width twice to get the perimeter 54 m.


2L%2B2W=54 Start with the second equation.


2%282W-3%29%2B2W=54 Plug in L=2W-3.


4W-6%2B2W=54 Distribute.


6W-6=54 Combine like terms on the left side.


6W=54%2B6 Add 6 to both sides.


6W=60 Combine like terms on the right side.


W=%2860%29%2F%286%29 Divide both sides by 6 to isolate W.


W=10 Reduce.


So the width is 10 m.


L=2W-3 Go back to the first equation.


L=2%2A10-3 Plug in W=10


L=20-3 Multiply


L=17 Subtract


So the length is 17 m.