Question 261619
"One train leaves city A heading for city B which is 390 miles away. At the same time a second train leaves city B heading for city A, Going 15 mph faster then the first train. If they meet in 3 hours and 20 minutes how fast were the trains traveling?
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Train A DATA:
rate = x mph ; time = 3 1/3 = 10/3 hrs ; distance = rt = (10/3)x miles
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Train B DATA:
rate = x+15 mph ; time = (10/3) hrs ; distance = rt = (10/3)(x+15) miles
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Equation:
A distance + B distance = 390 miles
(10/3)x + (10/3)(x+15) = 390
10x + 10(x+15) = 1170
20x + 150 = 1170
20x = 1020
x = 51 mph (Train A rate)
x+15 = 66 mph (Train B rate)
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Cheers,
Stan H.









 
I keep trying to put it in the formula 3.2(x)+ 3.2(x-15)=390 from there I get (simplified) 6.4x= 438 
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