1。 Party A and Party B set out from A and B at the same time, face to face, and meet at a distance of 800 meters .. After meeting, they go on, and immediately return along the original road after reaching the starting point of the other party, and meet at a distance of 300 meters. How long is the public road between A and B?

Solution: Let the distance between the two places be s meters, the speed of A is X (m/h), and the speed of B is Y (m/h).

When we first met, A walked 800 meters and B walked (S-800) meters. Because they started at the same time, the time spent was equal. Therefore,

There is an equation: 800/x = (s-800)/y ............ (1).

From the departure to the second meeting, A walked (S+300) meters, and B walked (2S-300) meters. It took the same time for both of them, so there was another time.

Equation: (s+300)/x = (2s-300)/y...(2)

( 1)÷(2)800/(S+300)=(S-800)/(2S-300)

800(2S-300)=(S+300)(S-800) excluding the denominator.

Extended 1600S-240000=S? -500-240000

So you got an s? -2100s = (s-2100) S=0, so S = 2 100m or s = 0 (omitted).

That is, the distance between a and b is 2 100 meters.

2。 After school. Xiaoming and Swallow walked out of the school gate together and went home. Xiaoming walks 45 meters per minute, and the swallow walks 50 meters per minute. At the same time, my father picked up Xiaoming at a speed of 55 meters per minute. Two minutes after meeting Swallow, he met Xiaoming. How far is it from school to Xiaoming's home?

Solution: Suppose Xiaoming's father meets Xiaoyan in t minutes, then Xiaoming's father meets Xiaoming in (t+2) minutes, so there is an equation;

(50+55)t=(45+55)(t+2), that is, 105t= 100t+200, ∴5t=200, t=40 points.

So the distance from school to Xiaoming's home = (50+55) × 40 =105 × 40 = 4200m.