SOLUTION: Suppose the length of a certain rectangle is 5 meters more than twice its width and the perimeter is 22 meters. The width must be ___ meters?

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Question 831506: Suppose the length of a certain rectangle is 5 meters more than twice its width and the perimeter is 22 meters. The width must be ___ meters?
Answer by JulietG(1812) About Me  (Show Source):
You can put this solution on YOUR website!
Remember that the perimeter is a fence. You'll need two pieces of length and two pieces of width. 2L + 2W = P
The problem gives you the following:
P = 22
L = 2W + 5 ("length of a certain rectangle is 5 meters more than twice its width")
.
Substitute the value of L into the root equation
2(2W + 5) + 2W = P
Distribute
4W +10 + 2W = 22
Add the Ws
6W + 10 = 22
Subtract 10 from each side
6W = 12
Divide each side by 6
W = 2 <<--
.
Let's plug it back into the original question to make certain.
"Suppose the length of a certain rectangle is 5 meters more than twice its width [(2*2)+5 = 9] and the perimeter is 22 meters"
Then we'd have 2 pieces of 9' and 2 pieces of 2'. 9+9+2+2 = 22
Success!