It is known that the distance between a and B is 40km. A and B both ride from a to B at the same time. A walks 2 hours more than B All the time, the distance between a and B is 40km. Party A and B ride from a to B at the same time. Party A walks 2km more than Party B every hour. Party A walks 4km away from B for some reason, 8km less than the original. As a result, they just arrive at B at the same time. Let party a ride XKM every hour. (fractional equation)

It is known that the distance between a and B is 40km. A and B both ride from a to B at the same time. A walks 2 hours more than B All the time, the distance between a and B is 40km. Party A and B ride from a to B at the same time. Party A walks 2km more than Party B every hour. Party A walks 4km away from B for some reason, 8km less than the original. As a result, they just arrive at B at the same time. Let party a ride XKM every hour. (fractional equation)

36/X+4/（X-8）=40/（X-2）

A and B start from a and B at the same time. A takes the first two to three steps by bike
A and B start from a and B at the same time. A takes the first two-thirds of the distance by bike and the last one-third by car. B takes the first two-thirds of the time by bike and the last one-third by car. If the speed of the bicycle is always a, the speed of the car is always B
1. If a = 18, B = 30, who will arrive first?
2. If a = 1 / 2B, who will arrive first?
3. If B = a + 1, who will arrive first?

(1) Suppose the distance is y, then the time of a is 2-3y-18 + 1-3y-30 = 13-270y. Assuming that they use the same time, comparing the distance, then the distance of B is 13-270yx2-3x18 + 13-270yx1-3x30 = 133-135y. If the distance is less than y, then a arrives first
(2) The same as (1): Party A takes 5-6 years, Party B travels 10-9 years in the same time, and Party B arrives first
(3) The time of a is (3a + 2) \ [(3a + 3) a] y