The two places of AB are 18 kilometers apart, and there are two people from place A

It is most convenient to use equations to solve, and the quantitative relations can be clearly represented.

The walking distance of Set B is x km.

Then the distance of ac can be expressed as [(x 4) 60+x] 2=8xThe total time taken by A is: 8x 60+(18-8x) 4The total time taken by B is: x 4+(18-x) 60

The total time is the same, and the equation can be listed.

8x 60+(18-8x) 4=x 4+(18-x) 60 is multiplied by 60 on both sides of the equation

8x+15(18-8x)=15x+18-x8x+270-120x=14x+18

270-18=14x+120x-8x

126x=252

x = 28x = 8 2 = 16 km.

18-8x = 18-8 2 = 2 km.

The distance of AC is 16 km, and both A and B walked 2 km.

Assuming that B's walking distance is x km, then AC's distance can be expressed as [(x 4) 60+x] 2=8x A time is: 8x 60+(18-8x) 4

Time B is: x 4+(18-x) 60

The total time is the same, available.

8x 60+(18-8x) 4=x 4+(18-x) 60 solves this unary equation to get x=2

8x=8 2=16km.

18-8x = 18-8 2 = 2 km.

The distance of AC is 16 km, and both A and B walked 2 km.

A velocity v

B speed 2v+

Traveling in the opposite direction, the combined speed of the two is 3V+

After four hours, the total distance was 12v+2

v is less than or equal to 89 6, then the maximum distance is 12 * 89 6 + 2 = 180

So the fastest is just the right to meet, and the slower is within 180 km, depending on how slow.

If it is greater than 89 6 km h, the two meet each other and walk against each other, and the two people will distance themselves again, and the specific distance is 12*v+2-18010. It is known that the two places of AB are 180 kilometers apart, and A and B set off from the two places of AB at the same time and traveled in the opposite direction, more than twice as many kilometers. , when the speed of A is v kilometers, V is less than or equal to 89/6, after four hours, how many kilometers are A and B apart? What if v is greater than 89/6.

20 minutes = 1 3 hours 1 hour 40 minutes of the wheel clock = 5 3 hours set: B travels x kilometers per hour, then A travels x + 3 kilometers per hour.

5/3(x+3)+(5/3+1/3)x=605/3x+5+2x=60

11/3x=55

x=15 So, B is 15 kilometers per hour, and A is 15 + 3 = 18 (kilometers) per hour

Let the velocity of A be x, then the velocity of B is (20-2x) 2=10-x, after the encounter, the velocity of A returns unchanged and still x, then the time taken is also 2 hours, and the unary equation is unary equation: (10-x)*4+2=20 The velocity of solving x= B is.

Actually, this problem doesn't need to be an equation, 20 kilometers, B walks for 4 hours, and there is still 2 kilometers left, so in fact, B walks 18 kilometers in 4 hours, so the speed is 18 4=

Solution: Let the velocity of A be x km-h, and the velocity of B be ykm-h;

Then: (x+y)*2=20

x-y)*2=1

Solution: x=, y=

You are not 1 in 7 years 10

Remember the difficult scriptures that you have to remember in class.

The speed of the armor is x

Then B's velocity is (20-2x) 2

When A returns to point A, B is still moving towards place A, and when A returns to place A, B is still two kilometers away from place A, and A and B each walk for four hours. ）

B's distance = speed (20-2x) 2 x time4 = 20-2 A's speed is: B's speed is:

4x+2=20 (let B be x).

then A is 20 2-x

The speed of B is: The speed of A is:

Solution: Let car A travel x kilometers per hour, and vehicle B travel y kilometers per hour.

6（x-y）=120①

2/3（x+y）=120-40②

Solution: x=70, y=50

Vehicle A travels 70 kilometers per hour, and vehicle B travels 50 kilometers per hour.

1. Set up a hidden to do B to set off x hours after meeting and teasing, 6 (solution, x=17 Shan Zheng 14 hours.

2. Set X hours after B catches up with A, solve with relative velocity, 8x-6x=20x=10 hours.

Set time X and stroke Y

then y=10(x-8)chang(x8).

Let time B be x and the auspicious segment of the trip be y

then y=40 (Sincerely.