SOLUTION: Tom is building a rectangular dog pen that we wants to enclose. The width of the pen is 2 yards less than the length. If the area of dog pen is 15 square yards, hpw many yards of f

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: Tom is building a rectangular dog pen that we wants to enclose. The width of the pen is 2 yards less than the length. If the area of dog pen is 15 square yards, hpw many yards of f      Log On

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Question 129612: Tom is building a rectangular dog pen that we wants to enclose. The width of the pen is 2 yards less than the length. If the area of dog pen is 15 square yards, hpw many yards of fence would he need to completely enclose the pen? Please and thank you :)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Width =wyds and length=lyds
l+=+w+%2B+2
Area = 15 yds2
Area = w%2Al
15 = w%28w+%2B+2%29
w%5E2+%2B+2w+=+15
w%5E2+%2B+2w+-+15+=+0
%28w+-+3%29%28w+%2B+5%29+=+0
w+=+3
w+=+-5can't used a negative, reject
The perimeter would be
w+%2B+w+%2B+%28w+%2B+2%29+%2B+%28w+%2B+2%29
2w+%2B+2w+%2B+4
4w+%2B+4
4%2A3+%2B+4
16yds
He will need 16 yds of fence