Multiple choice
1. If (x+y): (x-y) = 3: 1, then x: y = ().
a、3∶ 1 B、2∶ 1 C、 1∶ 1 D、 1∶2
2. If the solution of equation -2x+ m=-3 is 3, then the value of m is ().
a、6 B、-6 C、D、- 18
3. In the equation 6x+ 1= 1, 2x=, 7x- 1=x- 1, 5x=2-x, the number of solving equations is ().
a， 1 b，2 c，3 d，4
4. According to the quantitative relationship of "the difference between the absolute value of 3 times a and -4 is equal to 9", equation () can be obtained.
a 、|3a-(-4)|=9 B 、|3a-4|=9
c、3|a|-|-4|=9 D、3a-|-4|=9
5. If the solution of the equation =4(x- 1) about x is x=3, then the value of a is ().
a、2 B、22 C、 10 D 、-2
Answer and analysis
Answer: 1, B 2, A 3, B 4, D 5, c.
Analysis:
1. Analysis: This question examines the equal deformation of the equation.
From (x+y)∩(x-y)= 3∶ 1, we know that x+y=3(x-y), which is simplified as: x+y=3x-3y.
2x-4y=0, that is, x=2y and x∶y=2∶ 1.
2. Analysis: ∫3 is the solution of equation -2x+ m=-3,
∴ -2×3+ m=-3，
That is -6+m =-3,
∴ m=-3+6,-According to the basic properties of the equation, 1
∴ m=6, according to the basic properties of Equation 2.
Choose a.
3. Analysis: the solution of 6x+ 1= 1 is 0, that of 2x= 0, that of 7x- 1=x- 1 is 0, and that of 5x=2-x is 0.
4. Omit.
5. Analysis: Because x=3 is the solution of equation =4(x- 1), substituting x=3 into the equation will satisfy the equation.
I. Multivariable types
The application problem of multivariate linear equation solution refers to the application problem with multiple unknowns and multiple equality relationships in the problem. As long as one of these unknowns is X, the other unknowns can be represented by an algebraic expression containing X according to the equality relation in the topic, and then a linear equation can be listed according to another equality relation.
Example 1: (Beijing People's Education Edition, 2005) In order to save electricity in summer, two measures are often taken: increasing the set temperature of air conditioner and cleaning equipment. At first, a hotel raised the set temperature of A and B air conditioners by 1℃. Results A air conditioner saves 27 degrees more electricity every day than B air conditioner. Then clean the equipment of air conditioner B, so that the total daily electricity saving of air conditioner B is 1. 1 times that of air conditioner A only after the temperature rises, while the specified electricity consumption of air conditioner A remains unchanged, so that the two air conditioners can save 405 degrees of electricity every day. After the temperature is increased by 1℃, how many kWh can each air conditioner save every day?
Analysis: There are four unknowns in this question: the regulated electric quantity of air A after heating, air B after heating, air A after cleaning equipment and air B after cleaning equipment. The equality relationship is as follows: a-a-b-a-b-a = 27, b-a-b =1.1× b-a-b = a-a-b = 405. According to the first three equations, one unknown number represents four unknowns, and then the equations are listed according to the last equation.
Solution: 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. According to the meaning of the question, you must:
Solution:
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 class of application problems with the same unknown quantity and different restrictions 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 price of bananas in a fruit wholesale market in Dongying in 2005 is as follows:
The number of bananas purchased
(kg) not exceeding
20 kg or more
But not more than 40 kilograms, more than 40 kilograms.
Price per kilogram 6 yuan 5 yuan 4 yuan
Zhang Qiang bought 50 kilograms of 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 kg bananas twice (the second time was more than the first time), he bought more than 25 kg for the second time and less than 25 kg for the first time. Because 50 kilograms of bananas cost 264 yuan, with an average price of 5.28 yuan, the price of bananas purchased for the first time is inevitably 6 yuan/kg, that is, less than 20 kilograms, and the price of bananas purchased for the second time may be 5 yuan or 4 yuan. We can discuss it in two situations.
Solution:
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 of the question, it is concluded that:
6x+5(50-x)=264
Solution: x = 14
50- 14 = 36 (kg)
2) When the first banana purchase is less than 20kg and the second banana purchase is more than 40kg, we assume that the first banana purchase is x kg and the second banana purchase is (50-x) kg.
6x+4(50-x)=264
Solution: x = 32 (not in line with the meaning of the question)
Answer: I bought 14kg bananas for the first time and 36kg bananas for the second time.
Example 3: (Jingmen City, Hubei Province, 2005) participated in the medical insurance of an insurance company, and hospitalized patients have the right to reimbursement by stages. The reimbursement rules formulated by insurance companies are as follows. When someone is reimbursed by the insurance company after hospitalization, the amount is 1 100 yuan, then the medical expenses of this person are ().
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
……
A, 1000 yuan b, 1250 yuan c, 1500 yuan d, 2000 yuan.
Solution: Let this person's hospitalization expenses be X yuan. According to the meaning of the question:
500×60%+(x- 1000)80% = 1 100
Solution: x = 2000
So the answer to this question is D.
Third, the scheme type.
One-dimensional linear equations based on schemes often give two schemes to calculate the same unknown number, and then connect the algebraic expressions representing the two schemes with equal signs to form one-dimensional linear equations.
Example 4: (Quanzhou City, 2005) 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.
(1) Assume that 30 buses X were originally planned to be rented, and try to express the total number of students in the third grade of the school with an algebraic expression containing X;
(2) Now it is decided to rent a 40-seat bus, one less than the original planned 30-seat bus, and one of the rented 40-seat buses is not full and only takes 35 people. Please find out 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.
The total number of people is represented by the number of 40 buses: 40 (x-2)+35.
Solution: (1) The total number of junior three students in our school is 30x+ 15.
(2) From the meaning of the problem:
30x+ 15=40(x-2)+35
Solution: x = 6
30x+15 = 30x6+15 =195 (person)
A: There are *** 195 students in Grade Three.
Fourth, the type of data processing
When dealing with linear equations with data to solve application problems, we often don't directly tell us some conditions, so we need to analyze the given data and get the data we need.
Example 5: (Haidian District, Beijing, 2004) Application problem solution: In April, 2004, China's railways increased speed for the fifth time. Assuming that the average speed of K 120 air-conditioned express train is 44 km/h higher than that before the speed increase, the train timetable before the speed increase is shown in the following table:
The start time and arrival time of the train in the driving interval last for the whole mileage.
A-B K 120 2: 00 6: 00 4 hours 264 kilometers
Please fill in the accelerated train timetable according to the information provided by the topic and write out the calculation process.
The start time and arrival time of the train in the driving interval last for the whole mileage.
A-B K 120 2: 00 264 km
Solution:
The start time and arrival time of the train in the driving interval last for the whole mileage.
A-B KK120 2: 00 4: 24 2.4: 24 2.4 hours 264 kilometers
Analysis: According to the table 1, the train speed before the speed increase is 264 ÷ 4 = 66 km/h, so as to get the speed after the speed increase, and then calculate the required value according to the data given in Table 2.
Solution: Assume that the running time of the train after speed increase is X hours.
After examination, x=2.4 meets the meaning of the question.
A: The arrival time is 4:24, which lasts 2.4 hours.
Example 6: (Zhejiang Province in 2005) It is understood that the train fare is determined by the method of "". It is known that the total mileage from Station A to Station H is 1, 500km, and the reference price for the whole journey is 1, 80 yuan. The following table shows the mileage from stations along the way to H station:
name of a station
The mileage (km) from each station to H station is15001kloc-0/30910 622 402 219 720.
For example, determine the train fare from Mile Mile to E station, and its fare is (yuan).
(1) Find the train fare from station A to station F (the result is accurate to 1 yuan);
(2) Passenger Aunt Wang goes to her daughter's house by train. After getting on the bus for two stops, she took the train ticket and asked the flight attendant, am I near the station? When the stewardess saw that Aunt Wang's ticket price was 66 yuan, she immediately said that the next stop was here. At which station does Aunt Wang get off? Write the solution process.
Solution: (1) solution 1: known.
The actual mileage from Station A to Station F is1500-219 =1281.
So the train fare from Station A to Station F is 0.121281=153.72154 (yuan).
Option 2: The train fare from Station A to Station F is (yuan).
(2) Let Aunt Wang's actual mileage be X kilometers.
The solution is x= (km).
According to the comparison table, the distance between Station D and Station G is 550 kilometers, so Aunt Wang gets off at Station D or Station G. 。
Algebra chapter 6 ability self-test questions
One-dimensional linear inequality and one-dimensional linear inequality system
Junior high school mathematics website
fractional equation
(1) Fill in the blanks
The equation about y is _ _ _ _.
(2) Choose
a . x =-3； b . x≦-3；
C. all real numbers; D. no solution.
C. no solution; D. all real numbers
a . x = 0； B.x=0，x = 1；
C.x=0，x =- 1； D. the value of algebraic expression cannot be zero.
a . a = 5； b . a = 10；
c . a = 10； D.a= 15。
a . a =-2； b . a = 2；
c . a = 1； D.a=- 1。
A. all real numbers; All real numbers of b.x ≠ 7;
C. no solution; All real numbers of d. x≦- 1, 7.
a . a = 2； B.a only has 4;
C.a = 4 or 0; D. None of the above answers are correct.
a . a > 0； B.a > 0 and a ≠1;
C.a > 0 and a ≠ 0; D.a0。
(3) Solving equations
5 1. Party A and Party B set out from place A at the same time and walked 30 kilometers to place B. Party A walked more than Party B 1 km per hour, and as a result, Party A arrived earlier than Party B 1 hour. How many kilometers did they walk per hour?
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