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wallace and vernonville are 112 miles part. A car leaves wallace traveling towards vernonville and another car leaves vernonville at the same time traveling towards wallace. The car leaving wallace averages 10 miles per hour more the than the other and they meet after 1 hour and 36 minutes. what are the average speeds of the cars?
the car leaving wallace averages?
the car leaving vernoville averages?

Hi Jhnaesha;
distance...112 miles
Wallace to Vernonville...(x+10) miles/hour
Vernonville to Wallace...x miles/hour
time...1 hour 36 minutes
Our first priority concerns the fact that time is represented in hours and minutes.  Speed is represented in miles/hour.  Henceforth, we must eliminate the unit of minutes...
(36 minutes)[(1 hour)/(60 minutes)]
The unit of minutes is in the numerator and denominator.  It cancels...
(36 minutes)[(1 hour)/(60 minutes)]
[(36)(hours)]/60
3/5 hours
0.6 hours
time=1.6 hours
(112 miles)/(1.6 hours)=[(x+10) miles/hour]+(x miles/hour)
The unit of miles/hour is in the numerator and denominators of both sides of the equation.  These cancel...
(112 miles)/(1.6 hours)=[(x+10) miles/hour]+(x miles/hour)
112/1.6=2x+10
70=2x+10
Let's subtract 10 from both sides...
70-10=2x+10-10
60=2x
Let's divide both sides by 2...
60/2=(2x)/2
30=x

Vernonville to Wallace...30 miles/hour
Wallace to Vernonville...(30 +10) miles/hour=40 miles/hour