SOLUTION: george built a rectangular pen for his rabbit such that the length in 7 ft. less than twice the width. If the perimeter is 40 ft., what are the dimensions of the pen

Question 212505: george built a rectangular pen for his rabbit such that the length in 7 ft. less than twice the width. If the perimeter is 40 ft., what are the dimensions of the pen
Found 2 solutions by drj, checkley77:
Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
George built a rectangular pen for his rabbit such that the length is 7 ft. less than twice the width. If the perimeter is 40 ft., what are the dimensions of the pen

Step 1. Let l = w-7 length of rectangle and let w be the width

Step 2. Let P = 40 ft be the perimeter. Perimeter means adding the 4 sides of a rectangle. So,

Step 3. Add 14 to both sides of equation to get 4w by itself

Step 4. Divide 4 to both sides of equation

Step 5. w = 13.5 is the width of the rectangle.

Check P=4w-14+2w=4(13.5)-14=54-14=40 So w = 13.5 and l= 6.5 is the solution.

Hope the above steps were helpful. Good luck in your homework and studies!

Respectfully
Dr J

Hope you understood and followed the steps. Good luck in your studies. Dr J

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
L=2W-7
2L+2W=40
2(2W-7)+2W=40
4W-14+2W=40
6W=40+14
6W=54
W=54/6
W=9 ANS. FOR THE WIDTH.
L=2*9-7
L=18-7
L=11 ANS. FOR THE LENGTH.
PROOF:
2*11+2*9=40
22+18=40
40=40
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