SOLUTION: I am not sure how to set this up so I can solve it. Your help is greatly appreciated. 62. a 60 ft by 80 ft parking lot is torn up to install a sidewalk of uniform width around

Question 71134: I am not sure how to set this up so I can solve it. Your help is greatly appreciated.
62. a 60 ft by 80 ft parking lot is torn up to install a sidewalk of uniform width around it's perimeters. The new area of the parking lot is 2/3 of the old area. How wide is the sidewalk?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=width of sidewalk
Then (60-2x) and (80-2x) would be new dimensions of parking lot
And (60-2x)(80-2x) would be new area of parking lot
Also (60)(80)=4800 sq ft= old area

Now we are told that the new area is 2/3 that of the old area. So our equation to solve is:
(60-2x)(80-2x)=(2/3)(4800) get rid of parens
4800-280x+4x^2=3200 subtract 3200 from both sides
4800-3200-280x+4x^2=3200-3200 collect like terms:
4x^2-280x+1600=0 divide all terms by 4
x^2-70x+400=0
We'll solve using the quadratic formula:

x =6.277 ft------------------------------correct ans
and
x=63.723 ft--------------------------------impossible ans

CK

2x=2(6.277)=12.554
(60-12.554)(80-12.554)=3200
(47.446)(67.446)=3200
3200=3200

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