From city A to city B, the distance between which is 120 km, two cyclists left at the same time.

From city A to city B, the distance between which is 120 km, two cyclists left at the same time. The speed of the first one is 3 km / h more than the second, so he arrived in city B 2 hours earlier. Determine the speeds of the cyclists.

Decision:

1. Let’s designate: x km / h – speed of the second cyclist. This means that the speed of the former is x + 3 km / h.

2. According to the condition of the problem, an equation was drawn up:

120 / x = 120 / (x + 3) + 2;

120 / x – 120 / (x + 3) = 2;

(120 * (x + 3) – 120x) / (x ^ 2 + 3x) = 2;

(120x + 360 – 120x) / (x ^ 2 + 3x) = 2;

360 / (x ^ 2 + 3x) – 2 = 0;

(360 – 2 * (x ^ 2 + 3x)) / (x ^ 2 + 3x) = 0;

(-2x ^ 2 – 6x + 360) / (x ^ 2 + 3x) = 0;

3. The fraction is 0 when the numerator is 0 and the denominator is not equal to it:

-2x ^ 2 – 6x + 360 = 0;

x ^ 2 + 3x – 180 = 0;

Discriminant = 3 * 3 + 4 * 1 * 180 = 729 (the root of 729 is 27);

x = (-3 + 27) / 2 or x = (-3 – 27) / 2;

x = 12 or x = -15;

The speed cannot be negative, so it is 12 km / h.

If x = 12, then x + 3 = 12 + 3 = 15.

Answer: the speed of the first cyclist is 15 km / h, the speed of the second cyclist is 12 km / h.