In fact, an equation is an unknown equation. Solving practical problems with column equations is to express some quantitative relations in practical problems in the form of such equations with unknowns. And each formula in this equation has its own practical significance, which represents the quantity or quantity relationship of a corresponding process in topic setting. Therefore, the key to solving the equation application problem is to "master the basic quantity and find out the equation relationship".
The following comments on several common application problems in the one-dimensional linear equation for students' reference.
1. Travel problems
There are three basic quantities in the travel problem: distance, time and speed. The relationship is: ① distance = speed × time; ② Speed =; ③ Time =.
The equivalent relationships that can be found are: distance relationship, time relationship and speed relationship. In different issues, equal relations are flexible. For example, in encounter problems, distance is often used as an equal relationship, while in sequence problems, time is often used as an equal relationship, and in navigation problems, speed is often used as an equal relationship.
Navigation problem is a special case of travel problem, and its speed will change under different circumstances: ① downstream speed (wind) = still water speed (no wind)+current speed (wind speed); ② Current (wind) speed = still water (windless) speed-current speed (wind speed). From this, we can get an important equivalence relation in navigation problems: downstream (wind) speed-current speed (wind speed) = countercurrent (wind) speed+current speed (wind speed) = still water (windless) speed.
Example 1. A team is 450 meters long and advances at a speed of 90 meters per minute. Someone took something from the end of the line to the head of the line and immediately returned to the end of the line at a speed of 3 meters per second. How long does it take to go back and forth?
Comments: This problem is actually divided into two processes: ① The process from the tail to the head is a catching-up process, which is equivalent to the last person catching up with the previous person; ② The process from beginning to end is a meeting process, which is equivalent to meeting people from beginning to end.
In the process of catching up, if the catching-up time is x seconds and the speed of the leading team is 90 m/min = 1.5 m/s, then the driving distance of the leading team is1.5 m; If the pursuer's speed is 3m/s, the distance traveled by the pursuer is 3x meters. From the reciprocal relationship in the pursuit problem, "the distance between the pursuer and the pursued = the original distance apart", there are:
3x- 1.5x=450 ∴x=300
In the process of meeting, let the meeting time be y seconds, and the speed of the team and the returnees has not changed, then the distance traveled by the followers is 1.5y meters, and the distance traveled by the returnees is 3y meters. According to the equality relation in the encounter problem, "the distance traveled by A+the distance traveled by B = the total distance" is 3Y+ 1.5Y = 450 ∴.
Therefore, the round-trip time is x+y=300+ 100=400 (seconds).
Example 2 If a car travels from A to B at a speed of 40km per hour, it will be half an hour late; If you drive at a speed of 45 kilometers per hour, it will be half an hour ahead of schedule. Find the distance between a and b.
Comments: We usually call it a "priority issue", such as starting first and then arriving, then starting first and arriving first, the fast one arriving early, and the slow one arriving late. This kind of problem mainly considers the amount of time, examines the time relationship between the two, and finds the equal relationship from the time interval. In this problem, if the distance between A and B is X kilometers, the speed is 40 kilometers per hour, and the time is hours; When the speed is 45 km/ h, the time is hours, and the interval from morning till night is 1 hour, so there are
- = 1 ∴ x = 360
Example 3 A ship travels between A and B, and it takes 6 hours to sail downstream and 8 hours to sail upstream. The current known speed is 2 kilometers per hour. Find the distance between a and b.
Note: If the distance between A and B is x kilometers, the downstream speed is km/h and the upstream speed is km/h.. Important equivalence relations in navigation problems are as follows:
-2= +2 ∴ x = 96
2. Engineering problems
The basic quantities of engineering problems are: workload, work efficiency and working time. The relationship is: ① workload = work efficiency × work time. ② Working hours =, ③ Working efficiency =.
In engineering problems, the total workload is generally regarded as the whole 1. If the time to complete all the work is t, the work efficiency is 0. There are two common equivalence relations: ① If the workload is regarded as equivalence relation, the sum of partial workload = total workload. (2) If the time is equal, the time difference for completing the same work = the time spent.
In engineering problems, it should also be noted that the workload in some problems is given a clear quantity, which cannot be regarded as all 1, and the work efficiency is also the work speed.
Example 4. It takes 20 days for Party A to process a workpiece alone, and only 10 is needed for Party B to complete the task. They are now required to complete the task within 12 days. How many days does B need to work before A can finish the task on schedule?
Comments: Taking the workload of all tasks as a whole 1, we can know from the time when Party A and Party B completed it separately that Party A's work efficiency is, assuming that Party B needs to work for X days, Party A will continue to handle it for (12-x) days, and Party B's workload is, according to the meaning of the question, += 1.
Example 5. Harvest a wheat field, cut 4 acres per hour, and it is expected to be cut in a few hours. After harvesting, new farm tools are used to harvest, and the work efficiency is improved to 1.5 times. Therefore, it is completed in advance 1 hour. How many acres is this wheat field?
Remarks: Assuming that the wheat field has X mu, that is, the total workload is X mu, the operation efficiency before switching to new tools is 4 mu/hour, the estimated time for harvesting X mu is hours, and the operation time for harvesting X mu is /4= hour; After using the new tool, the working efficiency is 1.5×4=6 mu/hour, and the time for cutting the remaining mu is /6= hour, so the actual time is (+) hours. According to the meaning of the question, "finish ahead of schedule 1 hour".
-(+)= 1 ∴ x =36
Example 6. A pool is equipped with three water pipes, A, B and C, which are water inlet pipe and drainage pipe respectively. It takes 10 hour for A to open alone, 6 hours for B to open alone, and 15 hour for C to discharge alone. Now that all three pipes are open, how long will it take to fill the pool?
Comments: According to the topic design, the working efficiency of A, B and C are respectively,,-(the working efficiency of the water inlet pipe is regarded as positive, and the efficiency of the drainage pipe is recorded as negative). If the pool can be filled within x hours, the workload of A, B and C are respectively, and the total workload of the three water pipes is 1, of which +-= 1 ∴.
3. Economic problems
In recent years, economic application problems related to life and production practice are prominent types of mathematical innovation problems in senior high school entrance examination. Economic problems are mainly reflected in three categories: ① sales profit, ② preferential treatment (promotion) and ③ deposit and loan. The basic quantities of these three types of problems are different. When looking for the equation relationship, we must think in connection with the real life situation, so as to better understand the essence of the problem and list the equations correctly.
(1) Sales profit. There are four basic quantities of profit: cost (purchase price), sales price (income), profit and profit rate. The basic relationships are as follows: ① Profit = sales price (income)-cost (purchase price)-cost (purchase price) = sales price (income)-profit; (2) Profit rate = profit = cost (purchase price) × profit rate. In the sales problem with discount, the actual sales price = list price × discount rate. On the issue of discount, the purchase price is always equal.
(2) preferential treatment (promotion). There are many promotional activities in daily life, and different shopping (consumption) methods can get different discounts. In this kind of problem, generally from the "under what circumstances the effect is the same" to analyze. Based on the obtained value, the number greater than it and the number less than it are tested to predict its changing trend.
(3) the problem of deposit and loan. The problem of deposit and loan is closely related to daily life, and it is also one of the best problem scenarios to choose when the senior high school entrance examination is put forward. There are three basic quantities in the deposit and loan problem: principal, interest and interest tax, and related interest rates, the sum of principal and interest, and tax rates. The relationship is as follows: ① Interest = principal × interest rate × number of periods; ② Interest tax = interest × tax rate; ③ Sum of principal and interest (principal and interest) = principal+interest-interest tax.
Example 7. A store first bought 10 pieces of a commodity at the price of 15 yuan, and then went to Shenzhen to buy 40 pieces of the same commodity at the price of 12.5 yuan. If the store wants to make a profit 12% when selling this commodity, what should the price of this commodity be?
Remarks: If the sales price is X yuan per piece, the sales revenue is (10+40)x yuan, while the cost (purchase price) is (5×/kloc-0+40×12.5), the profit rate is 12%. By relation (1)
( 10+40)x-(5× 10+40× 12.5)=(5× 10+40× 12.5)× 12% ∴x= 14.56
Example 8. A commodity is going to be sold at a discount because of the change of seasons. If you sell at a 15% discount, you will lose 25 yuan; if you sell at a 10% discount, you will earn 20 yuan. What's the price of this commodity?
Comments: Set price of X yuan, 25% discount price of 75% X, profit -25 yuan, purchase price of 75% X-(-25) = 75% X+25; 10% off sales price 90% x, profit 20 yuan, purchase price 90% x-20. Through the purchase price, there are
75%x+25=90%x-20 ∴ x = 300
Example 9. Li Yong earned his salary by working during the holiday, and he immediately deposited it in the bank for half a year. The annual interest rate is 2. 16%. Deduct 20% interest tax when withdrawing money. Li Yong's classmate * * * got the principal and interest of 504.32 yuan. How much did Li Yong save six months ago?
Comment: The unknown required by this question is the principal. Assuming that the deposit principal is X yuan, the annual interest rate is 2. 16%, and the number of installments is 0.5 years, the interest is 0.5× 2. 16% X, and the interest tax is 20% × 0.5× 2.16% X. From the perspective of deposit and loan problems, the relationship ③ is as follows.
Example 10. A clothing store sells discount shopping cards. If you buy this card with 200 yuan, you can use this card to buy a 20% discount in this store. Under what circumstances is it cost-effective to buy a card for shopping?
Comments: Shopping discounts should first consider "all the same under any circumstances". Assuming that shopping for X yuan to buy a card has the same effect as not buying a card, the amount spent on buying a card is (200+80% x) yuan, and the amount spent on not buying a card is X yuan, so there are
200+80%x = x ∴ x = 1000
When X > 1000, for example, x=2000, the credit card consumption is: 200+80% × 2000 = 1800 (yuan).
The cost of not buying a card is: 2000 yuan (RMB). It is cost-effective to buy a card at this time.
When x < 1000, if x=800, the card purchase expenditure is: 200+80% × 800 = 840 yuan.
The cost of not buying a card is: 800 yuan. It is not cost-effective to buy a card at this time.
4. Solution (mixture) problem
There are four basic quantities in solution (mixture) problem: solute (pure substance), solvent (impurity), solution (mixture) and concentration (content). The relationship is as follows: ① Solution = solute+solvent (mixture = pure substance+impurity); ② Concentration = x100% = x100% purity (content) = x100% = x100%; ③ From ① ②, we can get: solute = concentration× solution = concentration× (solute+solvent). In the solution problem, the key quantity is "solute": "solute is constant", and the total amount of solute before mixing is equal to the dissolved mass after mixing, which is the main equivalence relationship in many equation application problems.
Example 1 1. Mix1000g of 80% alcohol into 60% alcohol, and a classmate doesn't consider adding 300g of water. (1) Try to explain whether the student added too much water through calculation. (2) How many grams of 20% alcohol should be added if the water is not too much? If you add too much water, how many grams should you add to 95% alcohol?
Comments: There are two kinds of concentration changes of solution problems: dilution (reducing the concentration of high-concentration solution by adding solvent or low-concentration solution) and thickening (increasing the concentration of low-concentration solution by evaporating solvent, adding solute and adding high-concentration solution). In the process of concentration change, it is not difficult to find the equation relationship and list the equations by grasping the two key quantities of solute and solvent and analyzing them with relevant formulas.
In this question, before (1) adding water, the original solution is 1000g, the concentration is 80%, and the solute (pure alcohol) is1000× 80% g; Assuming that the concentration is 60% after adding X grams of water, the solution becomes (1000+x) grams, and the solute (pure alcohol) is (1000+X) × 60% grams. The solute did not change before and after adding water, which was (1000+x) × 60% =1000× 80%.
X = > 300 The student didn't add too much water.
(2) Suppose you want to add y grams of 20% alcohol. At this time, the total amount of solution is (1000+300+y) g, the concentration is 60%, and the solute (pure alcohol) is (1000+300+y) × 60%; The concentrations of the original two solutions were 1000× 80% and 20% Y, respectively, and the dissolution quality remained unchanged before and after mixing, (1000+300+y )× 60% =1000× 80%+20% ∴ y. .
5. Numbers.
The number problem is a common mathematical problem. The numerical problems in the application of linear equations with one variable are mostly integers, so we should pay attention to the relationship between the number of digits, the number of digits on the number of digits and the value: any number = ∑ (number of digits × bit weight), such as two digits =10a+b; Three digits =100a+10b+C. When solving numerical problems, we should pay attention to the application of the idea of overall setting elements.
Example 12. For three digits, the sum of the three digits is 17, the number in the hundredth digit is 7 larger than that in the tenth digit, and the number in the single digit is 3 times that in the tenth digit. Find this number.
Note: Let the number in the ten digits be x, then the number in the one digit is 3x, the number in the hundredth digit is (x+7), and the number in the three digits is 100(x+7)+ 10x+3x. (x+7)+x+3x= 17 ∴x=2.
∴ 100(x+7)+ 10x+3x=900+20+6=926
Example 13. The highest digit of six digits is 1. If you move this number to the right of the single digit, the number you get is equal to three times the original number, so find the original number.
Comments: After the number on the highest digit of six digits is moved to single digits, the last five digits move forward by 1 digit, that is, the number on each digit is enlarged by 10 times, and the last five digits can be regarded as a complete unknown. Let the five digits after removing the highest digit 1 be x, then the original number is 10+x, and the shifted number is 10x+ 1, which is10x+1according to the meaning of the question.
∴x = 42857, the original number was 142857.
6. Distribution and proportion
Distribution and proportion problems are very common in daily life, such as reasonable arrangement of workers' production, proportional selection of engineering materials, adjustment of the number of people or goods, etc. The key to the distribution problem is to understand the partial quantity, the total quantity and the relationship between them. On the issue of distribution, the main consideration is "the total amount remains unchanged"; On the question of proportion, we mainly consider the relationship between total quantity and partial quantity, or the proportional relationship between quantity and quantity.
Example 14. There are several books on each shelf. If you take 100 books from shelf B and put them on shelf A, there are five times more books on shelf A than those left on shelf B. If you take 100 books from shelf A and put them on shelf B, all the books are equal. How many books are there on each shelf?
Comments: The difficulty of this problem is to set the unknown correctly, and use an algebraic expression containing the unknown to represent the number of books on another shelf. In the distribution problem, the quantity allocated is equal, that is, the extra quantity of the original party is divided equally. According to the topic "Take 100 books from shelf A to shelf B, two books are equal", it can be known that shelf A has 200 more books than shelf B. Therefore, if shelf B has x books, shelf A has (x+200) books ... Take 100 books from shelf B and put them on shelf A, leaving the rest on shelf B. There are five times as many books on shelf A as on shelf B, that is, six times as many books on shelf B, where (x+200)+100 = 6 (x-100) ∴ x =180x+200 = 380.
Example 15. There are 13 lights and ceiling fans in the classroom. It is known that each cable tube has 3 lamps or 2 ceiling fans, and there are 5 such cables. How many indoor lights are there?
Comment: This is a question about the distribution of switch cables. If there are X tubes, there are (13-x) ceiling fans, and there are tubes and ceiling fans. According to the meaning of the question, "* * * has five lines", with += 5 ∴ x = 9.
Example 16. Twenty-two workers in a workshop participated in the production of a kind of nut and screw. Everyone produces 120 screws or 200 nuts on average every day. One screw needs two nuts. How many workers should be assigned to produce screws and nuts to make the products produced every day just match?
Comments: For the problem of product matching (worker distribution), we should correctly find the quantitative relationship between them according to the matching relationship (proportional relationship) of products, and list the equations according to the equal relationship. There are x workers producing nuts, and the number of nuts produced is 200x, so there are (22-x) individuals producing screws, and the number of screws produced is 120 (22-x). From "one screw needs two nuts", that is, "the number of nuts is twice that of screws", there is 200x = 2× 120 (22-x).
∴x= 12 22-x= 10
Example 17. The blank of floor brick factory is made of clay, sand, gypsum and water in the ratio of 25: 2: 1: 6. Now the first three materials have been weighed, which is 5600 kg. How many kilograms of water should I add to stir? What are the weights of the first three materials?
Comments: The general method to solve the proportion problem is to set the unknown number according to the proportion and list the equations according to the equation relationship in the problem setting. The ratio of four kinds of blanks in this topic is 25∶2∶ 1∶6, and the four kinds of blanks are 25x, 2x, x and 6x kg respectively. The first three kinds of blanks are * * * 5600kg, and there are 25x+2x+x=5600.
∴x = 200 25x = 5000 2x = 400 x = 200 6x = 1200
Example 18. Give some apples to the children, each person has m apples and 14 apples, each person has 9 apples, so the last person gets 6 apples. How many people are there?
Comment: This is a distribution problem. If there are x children, each person is divided into m apples, and there are 14 apples. The total number of apples is mx+ 14, each person has 9 apples, and the last person has 6 apples, so the total number of apples is 9 (X- 1)+6. The total number of apples remains the same, with
MX+14 = 9 (x-1)+6 ∴ x = ∫ x, and m is an integer ∴ 9-M = 1 x = 17.
Example 19. Export 1 ton pork can be exchanged for 5 tons of steel, and the price of 7 tons of pork is equal to the price of 4 tons of sugar. There are 288 tons of sugar now. How many tons of steel can be exchanged for exporting these sugars?
Comment: This question can be transformed into a question of proportion. Pork: steel = 1: 5, pork: sugar = 7: 4, pork: steel: sugar = 7: 35: 4, assuming that X tons of steel can be exchanged, X: 288 = 35: 4 ∴ X = 2620.
7. Problems that need to be solved with intermediate (indirect) unknowns
In some application problems, it is difficult to list equations by setting direct unknowns, but it is easier to list equations by setting indirect unknowns according to the conditions in the problem, and then get the results through intermediate unknowns.
Example 20. The sum of four numbers, A, B, C and D, is 43, A multiplied by 2 multiplied by 8, B multiplied by 3, C multiplied by 4, and D multiplied by 5 to get four equal numbers. Find the numbers a, b, c and d.
Comments: This problem needs four quantities, which can be solved by equations later. It is very troublesome to use a linear equation of one variable to solve it, and set a certain number as an unknown number and the rest as an unknown number. Here, the numbers obtained by the changes of A, B, C and D are equal, so let this equal number be X, then A, B, C and D, and the sum of the four numbers is 43, where ++= 43 ∴ X = 36.
∴ = 14 = 12 =9 =8
Example 2 1. A county middle school football league * * has 10 rounds (that is, each team needs 10 games), in which/kloc-0 wins 3 points, draws/kloc-0 wins/kloc-0 points, and loses/kloc-0 points. Xiangming middle school football team lost three games less than the draw in this league, and the score was 19. How many games did Xiangming Middle School win in this league?
Comments: If the number of wins is directly set as unknown in this question, the number of negative and peaceful games cannot be expressed by the unknown formula, but the number of wins can be expressed by setting flat or negative games. Therefore, if the X field is flat, it is a negative X-3 field and a win 10-(X+X-3) field, that is, 3 [10-(x+x-3)]+x =19 ∴ x = .
8. The problem of setting not seeking (setting intermediate parameters)
In some application problems, the given known conditions are not enough to meet the needs of basic quantitative relations, and some do not need to be solved. At this time, we can set this quantity as a known condition and then eliminate it in the calculation. This will help us understand the nature of the problem.
Ex. 22: It takes five days and nights for a ship to travel from Chongqing to Shanghai, and seven days and nights for a ship to sail from Shanghai to Chongqing. How many days and nights does it take to put bamboo cards from Chongqing to Shanghai? (The speed of a bamboo raft is the speed of water)
Analysis: Navigation problems should grasp three basic quantities: distance, speed and time. Generally, there are two known quantities to find the third unknown quantity. In this problem, the amount of time is known, and the amount of time is also calculated, so it is necessary to set an intermediate parameter in the distance and speed to list the equation. Considering the constant distance, let the distance between the two places be one kilometer, then the downstream speed is 0, the upstream speed is 0, and the water speed is X, so there is -X =+X ∴ X =, and the time from Chongqing to Shanghai is Y day and night, so there is X = A ∴ X = 35.
Example 23. Two teachers from a school took several students to travel and contacted two travel agencies with the same price tag. After negotiation, the preferential terms of Travel Agency A are: 1 All teachers charge, and the rest are charged at 7.5% discount; B The preferential terms of travel agency are: 20% discount for all teachers and students.
(1) When the number of students is equal, are the fees charged by Travel Agency A and Travel Agency B the same?
(2) If the accounting results show that the preferential price of travel agency A is cheaper than that of travel agency B, what is the number of students?
Comments: In this question, the bid price and the number of students of the two travel agencies are unknown, which are indispensable basic quantities when doing the equation, but the bid price does not need to be solved. The bid price in (1) is a yuan, the number of students is x, the fee for travel agency A is a+0.75a(x+ 1) yuan, the fee for travel agency B is 0.8a(x+2) yuan, and a+0.75a (x+1) = 0.8a (.
(2) If the number of students is Y, the charge of travel agency A is a+0.75a(x+ 1) and that of travel agency B is 0.8a(x+2), that is, 0.8A (X+2)-[A+0.75A (X+1)] =