The problem is as follows, the method adds the answer, wants the formula Divide two numbers, the quotient is 22, the remainder is 8, the divisor, the divisor, the quotient, the sum of the remainder is 866, how much is the divisor and the divisor? There's no point. Have pity on me

The problem is as follows, the method adds the answer, wants the formula Divide two numbers, the quotient is 22, the remainder is 8, the divisor, the divisor, the quotient, the sum of the remainder is 866, how much is the divisor and the divisor? There's no point. Have pity on me

Let two numbers be x and Y respectively
22y = X-8; X + y + 22 + 8 = 866; just solve X and Y. however, this seems to be a primary school problem. Is this solution a junior high school solution? I don't know
x=22y+8;
22y+8+y+22+8=866;23y=828;y=36;
x=22*36+8=800
The answer is divided by 800 and 36

To explain the method, not only the formula, say about ideas and skills!
1. Divide two numbers, the quotient is 12, the remainder is 26, the sum of divisor, divisor, quotient and remainder is 454, please write the division formula
2. It is known that the sum of a factor plus another factor and the product is 13. One of the factors is 2.5. What is the other factor?

1. Divisor = (454-12-26-26) / (12 + 1) = 30
Divisor = 30 × 12 + 26 = 386
386÷30=12…… twenty-six
2. Another factor = 13 △ 2 △ 2.5 = 2.6

Want to say to solve a problem train of thought, list a formula, can do a few questions, do a few questions!
1. A car drove from place a to place B at a speed of 40 km. After 3 hours, due to rain, the average speed of the car was reduced by 10 km per hour. As a result, it was 45 minutes later than expected
2. Xiao Zhang and his father took the bus at home to visit their grandfather in their hometown. Halfway through the journey, Xiao Zhang asked the driver when he arrived at the railway station. The driver estimated that the train would just leave when he continued to take the bus to the railway station. According to the driver's suggestion, Xiao Zhang and his father immediately got off the bus and took a taxi. The speed doubled. As a result, they arrived at the railway station 15 minutes before the train started, It is known that the average speed of the bus is 30 km. How far is it from Xiaozhang to the railway station?
3. A good horse takes 20 days to walk. A bad horse takes 24 days to walk. Now a bad horse takes 2 days to walk from a to B. ask if a good horse can catch up with a bad horse in a few days
4. The distance between a and B is 20 kilometers. A and B set out from a and B at the same time. After two hours, they meet. After meeting, a returns to the original place. B is still moving towards A. when a arrives at a, B is 2000 meters away from A. ask for the speed of a and B
5. Party A and Party B go to the East Village, and Party C go to the West Village. Party A walks 4 kilometers per hour, already 4.5 kilometers per hour, and Party C walks 5 kilometers per hour
6. The track of a sports ground is 400 meters long. If a takes a bicycle, the average minute is 490 meters, and has run, the average minute is 250 meters
(1) How long did it take them to meet for the first time
(2) How long does it take for two people to meet each other if they are traveling in the same place at the same time
7. On a mountain road, someone has to walk from the bottom of the mountain to the top of the mountain for 3 hours, but it is still one kilometer short to reach the top of the mountain. If it takes 150 minutes to walk from the top of the mountain to the bottom of the mountain, it is known that the downhill speed is 1.5 times of the uphill speed, then how many kilometers per minute is the uphill speed?

Set up equations to solve application problems, as for checking calculation, it doesn't matter. Some problems don't need so many unknowns, just for the convenience of understanding
1: Assuming that the distance is y and the scheduled time is x, then 40x = y (estimated) 40 * 3 + 30 * (x-3 + 0.75) = y (actual)
The solution is 10x = 52.5, x = 5.25, y = 210
2: Suppose the same as above, the distance is y and the scheduled time is x, then 30x = y (bus)
(Y / 2) / 30 + (Y / 2) / 60 = x-0.25 (the actual time spent is the sum of bus and taxi time)
The solution is: Y / 120 = 0.25, y = 30 (km) x = 1 (H)
3: Suppose that the speed of a good horse is y, and that of a bad horse is X. if a good horse can catch up with a bad horse after Z days, then 24x = 20Y, YZ = x (Z + 2)
The solution is: z = 2x / (Y-X) = 2x / (24 / 20x-x) = 10
PS: there are three unknowns and two equations in this problem, so we can't calculate the specific X and y, but Z is only related to the ratio of X and y, and has nothing to do with the specific values of X and y
A simple algorithm. A good horse takes 20 days to walk, and a bad horse takes 24 days to walk. A constant proportion formula 20:24:4 (difference days) is obtained. To make the end value become 2, there is only 10:12:2, that is, a good horse takes 10 days to walk, and a bad horse takes 12 days to walk
4: Suppose that the speed of a and B is x and Y respectively, and the distance between the meeting place and a is Z
2 * (x + y) = 20 (before meeting) z = 2x (distance of a)
2Y = Z + 2
The solution is y = 18 / 4 = 4.5 x = 5.5 z = 11
5: Suppose that C meets B after X hours, then it meets a after (1 / 6 + x) hours
(4.5 + 5) x = y (encounter of B and C) (4 + 5) * (1 / 6 + x) = y (encounter of a and C)
The solution is 0.5x = 9 / 6, x = 3, y = 28.5
6: Suppose two people start at the same place and direction at the same time and meet for the first time in X minutes. If two people walk at the same place and face each other at the same time and meet in Y minutes, then (490-250) * x = 400
(490 + 250) * y = 400
The solution is: x = 5 / 3, y = 20 / 37
7: Assuming that the uphill speed is x (km / h) and the distance is y (km), then the downhill speed is 1.5x and 150 minutes is 2.5 hours
3x = Y-1 1.5x * 2.5 = y solution: x = 4 / 3, y = 5
I'm so tired of typing so many words that I have to switch input methods all the time