SOLUTION: Please help me solve this problem. It says suppose that the lenght of a certain rectangle is five inches more than two times its width. The perimeter of the rectangle is 40 inches.

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Question 115554: Please help me solve this problem. It says suppose that the lenght of a certain rectangle is five inches more than two times its width. The perimeter of the rectangle is 40 inches. Find the length and width of the rectangle. This problem must be solved algebracially.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
The length is 5 inches more than 2 times the width. 2 times the width is 2w, and 5 inches more than that is 2w + 5, so now we can say that:

l=2w%2B5

The perimeter of a rectangle is given by P=2l%2B2w and we know that the perimeter is 40 inches, so 2l%2B2w=40.

Now substitute the expression we developed for l into the perimeter equation:

2%282w%2B5%29%2B2w=40

Distribute
4w%2B10%2B2w=40

Collect like terms
6w%2B10=40

Add -10 to both sides
6w=30

Divide both sides by 6
w=5

Now we know the width is 5 inches, so put that into our expression for the length:
l=2%285%29%2B5=10%2B5=15

And the length is 15 inches.

Check:
2%2815%29%2B2%285%29=30%2B10=40, so those measurements give the correct perimeter.
2%285%29%2B5=10%2B5=15, so the length and width have the correct relationship. The answer checks.