Question 322244: Edward leaves home at 9AM to hike to another town 12 miles away. He rests for 1 hour, and then hikes back on the same path and reaches home at 5PM. When coming back, his speed was 1mph < his speed going to the other town.
-What was his speed going both ways.
NOTE: I was able to solve some of this problem:
Distance = 12 miles (one way)
Rate = (x+1) going and (x) coming back
Time = Total 8 hours, but 1 hour spent resing so 7 hours. PLEASE HELP!!!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Edward leaves home at 9AM to hike to another town 12 miles away. He rests for 1 hour, and then hikes back on the same path and reaches home at 5PM. When coming back, his speed was 1mph < his speed going to the other town.
-What was his speed going both ways.
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To town DATA:
distance = 12 miles ; rate = x mph ; time = 12/x hrs
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From town DATA:
distance = 12 miles ; rate = (x-1) mph ; time = 12/(x-1) hrs
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Note: Total time moving was (3+5-1) = 7 hrs
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Equation:
12/x + 12/(x-1) = 7
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12(x-1) + 12x = 7x(x-1)
24x -12 = 7x^2 - 7x
7x^2 -31x + 12 = 0
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Factor:
(x-4)(7x-3) = 0
x = 4 mph or x = 3/7
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Reasonable answer:
Only x = 4 mph (rate to town)
x - 1 = 3 mph (rate from town)
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Cheers,
Stan H.
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