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

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) (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 .

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

Start with the second equation.

Plug in .

Distribute.

Combine like terms on the left side.

Add to both sides.

Combine like terms on the right side.

Divide both sides by to isolate .

Reduce.

So the width is 10 m.

Go back to the first equation.

Plug in

Multiply

Subtract

So the length is 17 m.

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