SOLUTION: The length of a rectangle is 6 ft. longer than twice the width. If the perimeter is 90 ft. find the length and width of the rectangle.

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Question 884457: The length of a rectangle is 6 ft. longer than twice the width. If the perimeter is 90 ft. find the length and width of the rectangle.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
A dimension of a rectangle is x ft. longer than k the other dimension, y. If the perimeter is p ft. find the dimensions x and y of the rectangle.

x=k+y and 2x+2y=p.
The unknown variables are x and y.
Solve for x and y.







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Return again to the equation from the description for how x and y are related.





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If you use your given values from the beginning, there could be fewer steps needed. If you chose to solve the problem all symbolically, then you would finish by substituting the given values from the problem description into the solved formulas to compute the x and y values.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 6 ft. longer than twice the width. If the perimeter is 90 ft. find the length and width of the rectangle.


Let width be W, and length, L

Then L = 2W + 6 ------- eq (i)

Since perimeter = 2L + 2W, then 2L + 2W = 90

2(L + W) = 2(45) ------ Factoring out GCF, 2

L + W = 45 ------------ eq (ii)

2W + 6 + W = 45 ------- Substituting  2W + 6 for L in eq (ii)

3W = 39

W, or width = , or  ft

Use this info to determine the value of L, or the length.

Then do a check!! 



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