Question 652547: The Perimeter of a rectangle is 46 ft. The width is 3ft less than twice the lenth. Find the lenth and width of the rectangle.
Answer by mralgebra(3)   (Show Source):
You can put this solution on YOUR website! The Perimeter of a rectangle is 46 ft. The width is 3ft less than twice the length. Find the length and width of the rectangle.
With problems like this, you'll just want to set up basic equations that use the information you're given in a pretty basic way.
"The Perimeter of a rectangle is 46 ft."
If we call our length L and our width W, then we know this means that:
2L + 2W = 46
"The width is 3ft less than twice the length."
Again, with the length L and width W:
W = 2L - 3
Now we have a simple system of two equations.
2L + 2W = 46
W = 2L - 3
Let's substitute in 2L-3 for W in the first equation:
2L + 2W = 46
2L + 2(2L - 3) = 46
From here, we'll simplify the equation to find the length L.
2L + 2(2L - 3) = 46
2L + 4L - 6 = 46
6L - 6 = 46
6L - 6 + 6 = 46 + 6
6L = 52
L = 52/6 = 26/3
Since we know L, we can substitute it back into one of the equations to get W.
W = 2L - 3
W = 2(26/3) - 3
W = 52/3 - 3
We need a common denominator to do the subtraction.
W = 52/3 - 9/3
W = 43/3
So our final answer is that the length is 26/3 ft and the width is 43/3 ft.
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