Let the distance of AB be S and the speed of A be X, then the speed of B is 4x (3 times faster than that of A is 4 times), which can be obtained as follows: s/x=s/4x+4 (A leaves 1 hour earlier than B, while B arrives 3 hours earlier, with a time difference of 4 hours). Because the conditions given by the landlord are incomplete, they can only be listed here ... There are also some equations of travel problems as follows: It can be seen that solving application problems is a major difficulty in middle school mathematics teaching. There are many types of application problems, and each type of problem has its own characteristics. Therefore, the thinking of solving all kinds of problems is different, which increases the teaching difficulty of application problems. But the basic steps to solve all kinds of application problems are the same: first, examine the problems and determine the elements, and express the related quantities with algebra; The second step is to find out the equation relationship that can express the overall meaning of the problem; Step 3, replace the related quantities in the equality relation with algebraic expressions, and list the equations or equations; Fourthly, solving equations or equations, and finding out unknown quantities and values of the unknown quantities; Step five, answer the question according to its meaning. It can be seen that there are rules to follow in solving application problems by using equations, at least in terms of ideas and steps. Among the five steps to solve application problems by using sequence equation, the most crucial step is the second step, that is, to find the equation relationship. In fact, there is a basic relationship in all kinds of application problems, which is the key to find out the equal relationship in the problem. Here, taking the trip problem as an example, I would like to talk about my humble opinion on finding the equation in the teaching process of solving application problems. Travel problem is the most common application problem, and it is also a kind of problem that students find difficult in equations or equations. Therefore, how to teach students to find the equation relationship from travel problems through analysis, and then list equations or equations is a subject that most of our math teachers need to think about for a long time. In fact, all kinds of application problems have their own characteristics. As long as we understand the characteristics of various problems and their basic relationships, the methods to find out the equation relationship are similar. For example, the basic relationship in travel problem is the relationship between distance S, speed V and time T: s=vt, which is not only the soul of travel problem, but also the fundamental basis for finding out the equal relationship in travel problem. As the saying goes, "everything changes without departing from its original religion", as long as this basic relationship is well grasped in various forms of tourism problems, the problem will not be difficult to solve. Let's talk about the role of the relationship s=vt in the process of finding the equality relationship from two aspects. First, find the three basic quantities of the research object in the problem and determine their relationship. For a travel problem, there may be one or two or more research objects. Such as "someone", "Party A and Party B" or "two trains". But for every trip object, there are three basic quantities: distance, speed and time, and there is always a relationship between them: s=vt or v=s/t or t=s/v, some are known and some are unknown. If a known quantity and an unknown quantity are determined, it can be considered that the two basic quantities have been expressed by two algebraic expressions (including constants). At this time, the relationship between the remaining third quantities can be regarded as an equal relational equation. For example, 1: "Party A and Party B are 6 kilometers apart, and both of them set out at the same time and walked in the same direction. Party A caught up with Party B in 3 hours; Walk in opposite directions and meet within 1 hour. What is the average speed of the two? " Here we can assume that the average speeds of A and B are X km/h and Y km/h respectively, then when walking in the same direction, the known quantity is their common time t=3 hours, and the unknown quantity is their speeds V A =x km/h and V B =y km/h, then we use algebra to represent the two quantities A (speed and time) and B (speed and time). Then we can use the remaining quantity, that is, the distance between a and b, as the equality relationship of the equations. A =v, t=3x (km), B =v, t=3y (km). Moreover, the equal relationship between distances is easy to find: when traveling in the same direction, the difference between A and B is the distance when they start, that is, S0 = when traveling in the opposite direction, the sum of the distances between A and B is the distance when they start, that is, s0=s A +s B, and the equation can be obtained as x+y=6. Therefore, the equations 3x-3y=6 and x+y=6. Secondly, by using the determinable quantity between the three basic quantities, we can find that there are three equations with equal relations because of the s = vt of the basic quantity or its variation. If we know the speed and time, we can write the equation by taking the distance as an equivalent. If we know the speed and distance, we can use time as an equivalent to write the equation; For example, if you know time and distance, you can write an equation with speed as the same amount. Secondly, there are differences in the application of different equations or equations. For example, distance is a known quantity and speed is an unknown quantity, then time is the quantity in the equation relationship. But at time t=s/v, the equation obtained is a fractional equation, so the fractional equation is listed. But from the basic relationship, the result is nothing more than three cases: 1, and the two quantities represented by algebra are speed and time, so distance can be regarded as the quantity with equal relationship. Such as the above example is this type. The time and speed in this kind of problems can be known or unknown, and the listed equations are mostly integral equations, that is, linear equations with one variable or linear equations with two variables. 2. The two quantities represented by algebra are speed and distance, so time can be regarded as the quantity of equivalence. For example, Example 2: "An airplane can fly continuously in the air for up to 4 hours with a speed of 950km/h and a speed of 850km/h, so how many kilometers should the airplane fly back?" Suppose that the farthest distance of flying out is X kilometers, there are two quantities expressed by algebra in the process of flying out (namely, flying speed v = 950km/h and distance s=x kilometers), and there are also two quantities expressed by algebra when flying back (namely, speed v = 850km/h and distance s=x kilometers when returning), that is to say, we know two basic quantities, speed and distance, and residual quantity (. Because we know the distance (that is, s=x km) and speed (that is, V-out =950 km/h and V-back =850 km/h), it is easy to get the time of flying out and returning (that is, T-out =x/950 hours and T-back =x/850 hours). Since the total time of flying out and coming back is T =4 hours, it is equal. 3. If these two generations are time and distance, speed can be used as a quantity to find an equal relationship. Example 3: "The correspondent originally planned to travel from place A to place B in five hours. Because the task was urgent, he accelerated at a speed of 3 kilometers per hour and finally arrived within 4 hours to find the distance between A and B. "Here, if the distance between A and B is X kilometers, then two quantities are expressed by algebraic expression: distance (that is, s=x kilometers) and time (that is, the original planned time t0=5 hours, the actual time t=4 hours), and the third quantity (that is, speed v) is to find. Since the distance (that is, s=x km) and time (that is, t0=5 hours and t=4 hours) are already available, it is easy to get the original planned speed (that is, Vyuan =x/5 km/h) and the actual speed (that is, Vreal =x/4 km/h). It is not difficult to see that the equation x/4-x/5=3 can be obtained from the equation relationship (that is, Vreal-Vprimitive = Vpositive). 4. In the application of different equations or equations, the known quantity and the unknown quantity are different among the two quantities expressed by algebraic expressions. (1), as mentioned above, among the three types, when the speed and time are expressed by algebra and the distance is the third quantity, the listed equations are always integral equations. And most of them are unary linear equations or binary linear equations, such as the above example 1. (2) When distance and speed are expressed by algebraic expressions (where speed is a known quantity and distance is an unknown quantity), when time is used as the third quantity to solve the equation, the listed equation is also an integral equation, as in Example 2 above; When distance and time are expressed by algebraic expressions (in which time is a known quantity and distance is an unknown quantity), when velocity is used as the third quantity to solve the equation, the listed equation is also an integral equation, as in Example 3 above. (3) When the distance is a known quantity and the speed (or time) is an unknown quantity among the two quantities expressed by algebra, when the time (or speed) is used to find the equality relation of the third quantity, the listed equations are fractional equations. Example 4: "Workers of agricultural machinery factory go to the production team leaving the factory 15km to overhaul agricultural machinery. Some people rode bicycles for 40 minutes first, and the rest set off by car. As a result, they arrived at the same time. As we all know, the speed of a car is three times that of a bicycle. Find the speed of two cars. " Here, if the speed of bicycle is X km/h and the speed of car is 3xkm/h, then the distance (S = 15km) and speed (V =x km/h, V = 3xkm/h) of two kinds of traveling objects (namely bicycle and car) are expressed by algebraic expressions, and then the third quantity (namely time t). Because of the relationship between distance and speed, it is not difficult to get the travel time of bicycles and cars (that is, T = 15/x hours, T = 15/3x hours). It is easy to see that the time difference between the two trains is 40 minutes, that is, from T to T =40 minutes, and the equation 15/x- 15/3x=40/60 can be listed. In a word, when solving the application problem of the trip problem, we should first analyze the meaning of the problem and find out two quantities in the basic relationship s=vt that can be expressed by known numbers, unknowns or algebraic expressions, and the remaining third quantity is the quantity used to find the equality relationship. In general, it is relatively easy to find the related equation relationship by using the sum, difference, multiple and arithmetic relation of the third quantity. For example, s0=s A -s B and S0 = SA+SB in 1; Ttotal = tout+tback in example 2; In example 3, V is real-V is original =v plus, and in example 4, T is self-T steam = 2/3. I hope it helps you ... Come on!