At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist heading north is riding 4 km/hour

One rider's rate is x....the other is x+4

Once they pass each other, they travel for 1.5 hrs...and the combined distance they travel = 39 km

So....Rate x Time = Distance....and we have

x(1.5) + (x+4)(1.5) = 39     simplify

1.5x + 1.5x + 6 = 39

3x + 6 = 39        subtract 6 from both sides

3x = 33              divide by 3 on both sides

x = 11  km/hr     this si the slower rider's rate

x + 4 =   11 + 4 = 15 km/hr   and this is the faster rider's rate  

Let the speed of the bike going south be s km/hr.  Then the speed of the bike going north is (s+4) km/hr

The speed of the bike going north is (2s + 4) km/hr relative to the one going south.  Therefore In 1.5 hours they will be a distance (2s + 4)81.5 km apart.  This, we are told, ios 39 km.

(2s + 4)*1.5 = 39

3s + 6 = 39

3s = 33

s = 11

So the south going bike is travelling at 11 km/hr and the north going bike is travelling at 15 km/hr

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Chris beat me to it!

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I'll still give you some points.......