SOLUTION: The length of a rectangle is 2 centimeters less than twice its width. The perimeter of the rectangle is 32 cm. What are the dimensions of the rectangle?

Question 256885: The length of a rectangle is 2 centimeters less than twice its width. The perimeter of the rectangle is 32 cm. What are the dimensions of the rectangle?
Found 2 solutions by EMStelley, CharlesG2:
Answer by EMStelley(208) (Show Source): You can put this solution on YOUR website!
Let's start by naming variables. Let's call the length L and the width W. Then the first sentence tell us that:
L = 2W - 2
Remember that the perimeter of a rectangle is P = 2L + 2W. So the second sentence tells us:
2L + 2W = 32
Now, in order to solve these two equations, notice that we can use the first equation to replace L with 2W - 2. This gives us:
2(2W - 2) + 2W = 32
4W - 4 + 2W = 32
6W - 4 = 32
6W = 36
W = 6
So the width is 6 cm. To finish the problem, we need to find the length, so let's substitute 6 in for W in the first equation:
L = 2(6) - 2
L = 12 - 2
L = 10
So the dimensions of the rectangle are 10 cm by 6 cm.
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 2 centimeters less than twice its width. The perimeter of the rectangle is 32 cm. What are the dimensions of the rectangle?
P perimeter = 2 * (L length + W width)
P = 2 * (2*W -2 + W)
32 = 2 * (2*W -2 + W)
16 = 2*W - 2 + W
16 = 3*W - 2
18 = 3*W
6 = W
L=2*W -2
L=2*6 -2
L=12-2
L=10

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