Question 179311
Kim starts to walk 3 mi to school at
7:30 A.M. with a temperature of 0°F. 
Her brother Bryan starts at 7:45 A.M. on his bicycle, 
traveling 10 mph faster than Kim. 
If they get to school at the same time, then how
fast is each one traveling?
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Kim DATA:
distance = 3 miles ; rate = x mph ; time = d/r = 3/x hrs
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Bryan DATA:
distance = 3 miles ; rate = x+10 mph ; time = d/r = 3/(x+10) hrs
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Equation:
Kim time - Bryan time = (1/4) hr
3/x - 3/(x+10) = 1/4

Divide thru by 3 to get:

1/x - 1/(x+10) = 1/12

Simplify:
12(x+10) - 12(x) = x(x+10)

12*10 = x^2 + 10x

x^2 + 10x - 120 = 0
x = [-10 +- sqrt(100 - 4*-120]/2

x = [-10 +- sqrt(580)]/2

Positive solution:
x = [-10 + 24.083]/2

x = 14.083/2

x = 7.0415 mph (Kim's rate)
x+10 = 17.0415 mph (Bryan's rate)
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Cheers,
Stan H.