Question 1066116
 Suppose that a cyclist began a 442 mi ride across a state,
 at the western edge of the​ state, at the same time that a car traveling toward it leaves the eastern end of the state.
 If the bicycle and car met after 6.5 hr and the car traveled 31.4 mph faster than the​ bicycle, 
 find the average rate of each.
:
When the two meet, they will have traveled a total of 442 mi
let s = speed of the bike
then
(s+31.4) = speed of the car
:
Write distance equation; dist = time * speed
:
Bike dist + Car dist = 442 mi
6.5s + 6.5(s+31.4) = 442
6.5s + 6.5s + 204.1 = 442
13s = 442 - 204.1
13s = 237.9
s = 237.9/13
s = 18.3 mph is the speed of the bike
and
18.3 + 31.4 = 49.7 mph is the speed of the car
:
:
Check this by finding the actual dist each traveled:
18.3*6.5 = 118.95 mi
49.7*6.5 = 323.05
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total mi:  442.0 mi