Question 82862
Sally drives 720 miles to work and 720 miles back. Her average rate going to work is 12 miles per hour greater than returning. If the entire road trip took 35 hours, what was her average speed going and returning?
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To-work DATA:
distance = 720 miles ; Rate = x+12 mph ; Time = d/r = 720/(x+12) hrs
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From-work DAA:
distance = 720 miles ; Rate = x mph ; Time = d/r = 720/x hrs
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EQUATION:
time + time = 35 hrs
720/(x+12) + 720/x = 35
Multiply thru by x(x+12) to get:
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720x + 720(x+12) = 35x(x+12)
1440x + 8640 = 35x^2 + 420x
35x^2-1020x-8640
7x^2-204x-1728=0
x = [204+-sqrt90000]/14
x = [204+-300]/14
x = 504/14 or -96/14
Positive solution: x = 36 (Rate driving from work)
x+12 = 48 (Rate driving to work)
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Cheers,
Stan H>