SOLUTION: the perimeter of a rectangle lot is 350ft. the length of the lot is 10ft. more than twice the width. find the deminsion of the lot?

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Question 814330: the perimeter of a rectangle lot is 350ft. the length of the lot is 10ft. more than twice the width. find the deminsion of the lot?
Found 2 solutions by TimothyLamb, ewatrrr:
Answer by TimothyLamb(4379)   (Show Source): You can put this solution on YOUR website!
2L + 2w = 350
L = 2w + 10
---
2L + 2w = 350
2(2w + 10) + 2w = 350
4w + 20 + 2w = 350
6w = 330
---
w = 55
L = 2w + 10
L = 110 + 10
L = 120
---
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Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

P = 350ft

the length of the lot is 10ft more than twice the width 

Let x and (2x+10) represent the width and length



 2(2x+10) + 2x = 350ft

  6x = 330ft

   x = 55ft, the width.  The length is 120ft   (2x+10)

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