Question 1072074: The perimeter of a rectangle is 50m. If the width were doubled and the length increased by 19, the perimeter would be 108m. What would the width of the rectangle. I need a formula please.
Found 2 solutions by math_helper, MathTherapy:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! The perimeter of a rectangle is 50m. If the width were doubled and the length increased by 19, the perimeter would be 108m. What would the width of the rectangle. I need a formula please.
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There are two formulas:
2L+2W = 50 (1)
and
2(L+19) + 2(2W) = 108 (2)
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(2) simplifies to 2L+38 + 4W = 108
2L + 4W = 70 (2')
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You can solve for L and W if you need to get their individual values.
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangle is 50m. If the width were doubled and the length increased by 19, the perimeter would be 108m. What would the width of the rectangle. I need a formula please.
Let width and length be W and L, respectively
Twice length, plus twice width = Perimeter of rectangle, OR
2L + 2W = 50_____2(L + W) = 2(25)______L + W = 25______L = 25 - W
Doubling the width makes new width: 2W
Increasing length by 19 makes new length: L + 19, or 25 - W + 19, or 44 - W
We then get: 2(2W) + 2(44 - W) = 108
2(2W) + 2(44 - W) = 2(54)
2W + 44 - W = 54
2W - W = 54 - 44
W, or original width = 
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