A and B set out from a to B at the same time, 26km away. A rode a bicycle and B walked. It is known that a's speed is faster than B's A and B set out from place a to place B 26km away at the same time. A rides a bicycle and B walks. It is known that a's speed is 1km / h faster than twice that of B. after a arrives at place B, he immediately returns from place B. on the way, he meets B. The time from their departure is 4 hours. Q: what are their respective speeds? How far is the meeting point from place B?

A and B set out from a to B at the same time, 26km away. A rode a bicycle and B walked. It is known that a's speed is faster than B's A and B set out from place a to place B 26km away at the same time. A rides a bicycle and B walks. It is known that a's speed is 1km / h faster than twice that of B. after a arrives at place B, he immediately returns from place B. on the way, he meets B. The time from their departure is 4 hours. Q: what are their respective speeds? How far is the meeting point from place B?

Let B be x, then a = 2x + 1
From the problem, we can get the equation 4 (2x + 1) + 4x = 26 × 2
We can get x = 4
B's speed is 4km / h
4×2+1=9
A's speed is 9km / h
The distance between the meeting point and B is the total distance minus the distance b has traveled
26-4×4=10KM

A and B go from a to B at the same time. A rides a bicycle and B walks. The speed of a is 2 km / h faster than that of B. A stops at B for half an hour, and then returns to a from B. on the way, he meets B. at this time, it is 3 hours after their departure. The distance between a and B is 18.5 km, so the speed of a is
Complete with equation of degree one variable

A's distance + B's distance = 2 times of AB's distance
Let B be XKM / h, then a be (2x + 2) km / h
2.5 (2x + 2) + 3x = 18.5 * 2
x=4
A: if B's speed is 4km / h, then a's speed is 10km / h
This kind of problem needs a sketch, so it is clearer~

Both Party A and Party B go from place a to place B, which is 51 kilometers away from each other. Party A rides and Party B walks. The speed of Party A is more than three times faster than that of Party B, and it is one kilometer per hour
Stop for 1 hour, then return from B to a, and meet B on the way. At this time, according to their departure time, it's exactly 6 hours. What's their speed?

Let B be x km / h, then a be 3x + 1 km / h
6x+(6-1)(3x+1)=51×3
6x+5(3x+1)=153
6x+15x+5=153
21x=147
x=7
3x+1=22
Speed a is 22 km / h, speed B is 7 km / h

A and B travel from a to B 25 (1 / 2) km apart at the same time. A rides a bicycle while B walks. A's speed is 2 km / h faster than B's speed. A
After arriving at B first, return from B immediately, and meet B on the way. The time from their departure is 3 hours

Let the velocity be x, so the velocity of a is 2x + 2
According to the question 3x + 3 (2x + 2) = 25, so
9x=19 x=19/9
A's speed: 38 / 9 + 2