# From point A to point B the distance between which is 27 km at the same time two pedestrians left to meet

**From point A to point B the distance between which is 27 km at the same time two pedestrians left to meet each other and met after 3 hours. The pedestrian who was walking from A arrives at B 1 hour 21 minutes earlier than the other arrives at A. Find the speed of each pedestrian**

1.1 hour 21 minutes = 1 hour + 21/60 hours = 81/60 hours.

2. We take for x (km / h) the speed of the traveler who left point A, for y (km / h) the speed of the traveler walking towards him.

3. Let’s compose two equations:

3x + 3y = 27; x + y = 9; y = 9 – x;

27 / y – 27 / x = 81/60; 1 / y – 1 / x = 1/20; 20x – 20y = xy;

5. Substitute in the second equation y = (9 – x):

20x – 20 (9 – x) = (9 – x) x;

20x – 180 + 20x – 9x + x² = 0;

x² + 31x – 180 = 0;

The first value x = (- 31 + √961 + 180 x 4) / 2 = (- 31 + 41) / 2 = 5 km / h.

The second value x = (- 31 – 41) / 2 = – 36. Not accepted

The speed of the oncoming traveler is 9 – 5 = 4 km / h.

Answer: the speed of the traveler leaving point A is 5 km / h. The speed of the oncoming traveler is 4 km / h.