Solving a distance, rate, time problem using a system of linear equations

Question

Two buses leave towns 1260 kilometers apart at the same time and travel toward each other. One bus travels 16km/h slower than the other. If they meet in 6 hours, what is the rate of each bus? Rate of slower bus _ km/hr. Rate of faster bus _ km/hr

Answer 1

Ley "s" be the speed of the faster bus. Then the speed of the slower bus will be "s - 16".

At the time they meet, the distance that the faster bus travels added to the distance the slower bus traveled will equal the whole 1260 km.

The distance the faster bus travels is 6s (6 hrs times "s" km/hr)

The distance the slower bus travels is 6(s - 16) or 6s - 96

So 6s + 6s - 96 = 1260

12s - 96 = 1260

12s = 1356

s = 113 km/hr

s - 16 = 97 km/hr

So the faster bus traveled at 113 km/hr while the slower bus traveled at 97 km/hr

Answer 2

As usual, you need two equations, one for distance and one for speed comparison.

D = rt D = distance, r is speed or rate, t is time. (they both travel for 6 hrs)

A = speed of Bus A; B = speed of bus B.

What do we know about their speeds? A = B - 16 (A is defined here as the slower bus)

What about distance? 6A + 6B = 1260 (total distance - they start from either end and meet somewhere in between after 6 hrs)

Now can substitute: 6(B - 16) + 6B = 1260

6B - 96 + 6B = 1260

12B = 1356; B = 113 km/hr

Then A = 113 - 16 = 97 km/hr

Check: 6(113) + 6(97) = 678 + 582 = 1260 km

(Thought they were speeding until I realized they are going 70.2 and 60.3 mph, respectively!)