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Application of linear equation of one variable in junior high school mathematics (with answers)
Example 1: In order to save electricity in summer, air conditioners often take two measures: raising the set temperature and cleaning the equipment. At first, a hotel raised the set temperature of A and B air conditioners by 1℃. Results A air conditioner saves 27 degrees of electricity every day than B air conditioner. Then clean the equipment of air conditioner B, so that the total electricity saving of air conditioner B is 1. 1 times of air conditioner A only after the temperature rises, while the regulated electricity of air conditioner A remains unchanged, so that the two air conditioners can save 405 degrees of electricity every day. How many degrees can they save every day after the temperature rises 1℃?

Analysis: There are four unknowns in this question: A, B, A and B after air heating. The equal relationship includes the regulated electric quantity of air A after heating-air B after heating = 27, air B after cleaning equipment = 1. 1 × air B after heating, air A after heating = air A after cleaning equipment+air B after cleaning equipment = 405. According to the first three equality relations, an unknown number represents four unknowns, and then the equation is listed according to the last equality relation.

Suppose that only after the temperature increases by 1℃, the second air conditioner saves X degrees of electricity every day, and the first air conditioner saves X degrees of electricity every day.

1. 1x+(x+27)=405

Solution: x= 180

Answer: Only increase the temperature 1℃, and the A-type air conditioner saves 207 degrees of electricity every day, and the B-type air conditioner saves 180 degrees of electricity every day.

Second, segmented; The application of piecewise linear equations refers to a kind of application problems with the same unknowns and different constraints in different ranges. When solving this kind of problem, we must first determine the segmentation of the given data, and then solve it reasonably according to its segmentation.

Example 2: The prices of bananas in the fruit wholesale market are as follows:

The quantity (kg) of bananas purchased shall not exceed 20kg, but shall not exceed 40kg or more.

Price per kilogram 6 yuan 5 yuan 4 yuan

The quantity (kg) of bananas purchased shall not exceed 20kg but not exceed 40kg. The price per catty is more than 40 kilograms. 6 yuan 5 yuan Top 4 bought 50kg bananas twice (the second time was more than the first time), and * * paid 264 yuan. How many Jin of bananas did Zhang Qiang buy for the first time and the second time respectively?

Analysis: Because Zhang Qiang bought 50 kilograms of bananas twice (the second time was more than the first time), the second time he bought more than 25 kilograms, and the first time he bought less than 25 kilograms. Because 50 kilograms of bananas * * * paid 264 yuan, with an average price of 5.28 yuan, the price of bananas for the first time must be 6 yuan/kg, that is, less than 20 kilograms, and the price of bananas for the second time may be.

1) When the first banana purchase quantity is less than 20kg and the second banana purchase quantity is more than 20kg but not more than 40kg, it is assumed that the first banana purchase quantity is x kg and the second banana purchase quantity is (50-x) kg. According to the meaning, we get 6x+5 (50-x) = 264, and we get x = 65438+.

2) When the first banana purchase quantity is less than 20kg and the second banana purchase quantity is more than 40kg, it is assumed that the first banana purchase quantity is x kg and the second banana purchase quantity is (50-x) kg.

According to the meaning of the question, we get: 6x+4 (50-x) = 264, and we get: x = 32.

Answer: I bought 14kg bananas for the first time and 36kg bananas for the second time.

3 to participate in the medical insurance of insurance companies, hospitalized patients have the right to reimbursement by stages. The reimbursement rules formulated by insurance companies are as follows.

Proportion of reimbursement of hospitalization medical expenses (yuan) (%)

No more than the part of 500 yuan 0

Excess 500 ~ 1000 yuan 60

Excess1000 ~ 3,000 yuan 80

After someone is hospitalized, the amount reimbursed by the insurance company is 1 100 yuan, so the medical expenses for this person's hospitalization are ().

A, 1000 yuan b, 1250 yuan c, 1500 yuan d, 2000 yuan.

Let's assume that this person's hospitalization expense is X yuan, which is 500× 60%+(x-1000) 80% =1100.

Solution: x = 2000

So the answer to this question is D.

Third, the scheme-based one-dimensional linear equation often gives two schemes to calculate the same unknown, and then connects the algebraic expressions representing the two schemes with an equal sign to form a one-dimensional linear equation.

Example 4: Junior three students in a school participate in social practice activities. It was originally planned to rent a number of 30-seat buses, but there were still 15 people without seats. (65,438+0) Assuming that the original plan was to rent X buses with 30 seats, try to use an algebraic expression containing X to represent the total number of third-grade students in the school; (2) It is decided to rent 40 buses, one less than the original planned 30 buses, and one of the 40 buses is not full, only 35 people. Please check the total number of third-grade students in this school.

Analysis: There are two schemes to indicate the total number of students in Grade Three. The number of 30-seat buses is 30x+ 15, and the number of 40-seat buses is 40 (x-2)+35.

(1) The total number of junior three students in our school is 30x+ 15.

(2) From the meaning of the question: 30x+ 15 = 40 (x-2)+35.

Solution: x = 6 30x+15 = 30× 6+15 =195 (person) Answer: There are *** 195 students in Grade Three.