SOLUTION: In a rectangle, the perimeter is 82 inches. The length of the rectangle is 8 inches more than twice the width. What is the length and width of the rectangle?

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Question 30526: In a rectangle, the perimeter is 82 inches. The length of the rectangle is 8 inches more than twice the width. What is the length and width of the rectangle?
Found 2 solutions by checkley71, sdmmadam@yahoo.com:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
L=8+2W W=WIDTH SO 2TIMES THE WIDTH IS 2W & 2 TIMES THE LENGTH IS 2(8+2W)
ADDING THE 2 WIDTHS AND THE 2 LENGTHS SHOULD EQUAL 82 AS FOLLOWS:
82=2W+2(8+2W) OR 82=2W+16+4W OR 82=6W+16 OR 66=6W OR W=11
LENGTH=8+2W OR L=8+2X11 OR L=8+22 OR L=30.
11+11+30+30=82.

Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length be L inches
and let the width be B inches
the perimeter is 82 inches
Therefore 2(L+B) = 82
dividing by 2
That is (L+B) = 41 ----(1)
The length of the rectangle is 8 inches more than twice the width
That is L = B+8 ----(2)
putting (2) in (1)
(B+8)+B = 41
2B +8 = 41
2B = 41-8
2B = 33
B = 33/2 = 16.5
Putting B = 16.5 in (2)
L = B +8
L= 16.5+8
L= 24.5
Answer: Length of the rectangle = 24.5 inches
and width of the rectangle = 16.5 inches
Verification:Perimeter = 2(L+B) = 2X(24.5+16.5) = 2X41 = 82 which is correct