★ Emphasis★ Solution of one-dimensional linear equation, one-dimensional quadratic equation and two-dimensional linear equations; Related application problems of the equation (especially travel and engineering problems)

☆ Summary ☆

I. Basic concepts

1. equation, its solution (root), its solution, its solution (group)

2. Classification:

Second, the basis of solving equation-the nature of equation

1.a=b←→a+c=b+c

2.a=b←→ac=bc (c≠0)

Third, the solution

1. Solution of linear equation with one variable: remove denominator → remove brackets → move terms → merge similar terms →

The coefficient becomes 1→ solution.

2. Solution of linear equations: ① Basic idea: "elimination method" ② Method: ① Replacement method.

② addition and subtraction

Fourth, a quadratic equation

1. Definition and general form:

2. Solution: (1) direct leveling method (pay attention to characteristics)

(2) Matching method (pay attention to the step-inferring the root formula)

(3) Formula method:

(4) factorization method (feature: left =0)

3. The discriminant of the root:

4. The relationship between root and coefficient top:

Inverse theorem: If, then the quadratic equation with one root is:.

5. Common equation:

5. Equations that can be transformed into quadratic equations

1. Fractional equation

(1) definition

(2) Basic ideas:

⑶ Basic solution: ① Denominator removal ② Substitution method (such as).

(4) Root test and method

2. Unreasonable equation

(1) definition

(2) Basic ideas:

(3) Basic solution: ① Multiplication method (pay attention to skills! ! (2) substitution method (example), (4) root test and method.

3. Simple binary quadratic equation

A binary quadratic equation consisting of a binary linear equation and a binary quadratic equation can be solved by method of substitution.

Six, column equation (group) to solve application problems

summary

Solving practical problems by using equations (groups) is an important aspect of integrating mathematics with practice in middle schools. The specific steps are as follows:

(1) review the questions. Understand the meaning of the question. Find out what is a known quantity, what is an unknown quantity, and what is the equivalent relationship between problems and problems.

⑵ Set an element (unknown). ① Direct unknowns ② Indirect unknowns (often both). Generally speaking, the more unknowns, the easier it is to list the equations, but the more difficult it is to solve them.

⑶ Use algebraic expressions containing unknowns to express related quantities.

(4) Find the equation (some are given by the topic, some are related to this topic) and make the equation. Generally speaking, the number of unknowns is the same as the number of equations.

5] Solving equations and testing.

[6] answer.

To sum up, the essence of solving application problems by column equations (groups) is to first transform practical problems into mathematical problems (setting elements and column equations), and then the solutions of practical problems (column equations and writing answers) are caused by the solutions of mathematical problems. In this process, the column equation plays a role of connecting the past with the future. Therefore, the column equation is the key to solve the application problem.

Two commonly used equality relations

1. Travel problem (uniform motion)

Basic relationship: s=vt

(1) Meeting problem (at the same time):

+ = ;

(2) Follow-up questions (start at the same time):

If A starts in t hours, B starts, and then catches up with A at B, then

(3) sailing in the water:

2. batching problem: solute = solution × concentration

Solution = solute+solvent

3. Growth rate:

4. Engineering problems: Basic relationship: workload = working efficiency × working time (workload is often considered as "1").

5. Geometric problems: Pythagorean theorem, area and volume formulas of geometric bodies, similar shapes and related proportional properties.

Third, pay attention to the relationship between language and analytical formula.

Such as more, less, increase, increase to (to), at the same time, expand to (to), ...

Another example is a three-digit number, where A has 100 digits, B has 10 digits and C has one digit. Then this three-digit number is: 100a+ 10b+c, not abc.

Fourth, pay attention to writing equal relations from the language narrative.

For example, if X is greater than Y by 3, then x-y=3 or x=y+3 or X-3 = Y, and if the difference between X and Y is 3, then x-y=3. Pay attention to unit conversion

Such as the conversion of "hours" and "minutes"; Consistency of s, v and t units, etc.

Seven, application examples (omitted)

Chapter VI One-dimensional Linear Inequalities (Groups)

★ Essentials ★ Properties and Solutions of One-variable Linear Inequality

☆ Summary ☆

1. Definition: A > B, A < B, a≥b, a≤b, A ≠ B.

2. Unary linear inequality: ax > b, ax < b, ax≥b, ax≤b, ax≠b(a≠0).

3. One-dimensional linear inequalities;

4. the essence of inequality: (1) a > b ←→ a+c > b+c

⑵a & gt； b←→AC & gt； BC (c>0)

⑶a & gt； b←→AC & lt； BC (c<0)

(4) (transitivity) a>b, b & gtc→a & gt；; c

⑸a & gt； b，c & gtd→a+c & gt； b+d。

5. Solution of one-dimensional linear inequality, solution of one-dimensional linear inequality

6. Solution of one-dimensional linear inequality group, solution of one-dimensional linear inequality group (representing the solution set on the number axis)

The teaching of equation application can be said to run through the whole senior grade and junior high school, which plays a very important role in students' mathematics learning activities (the teaching hours of equation and its application problems in the whole junior high school are 4 1 hour, accounting for about 1 1.5% of the whole junior high school mathematics hours), and the teaching of unary equation application problems is the most basic starting part of all equation application problems teaching. However, because junior one students have strong mechanical memory, but their analytical ability is still weak, in order to improve the teaching effect of junior one math application problems, in addition to gradually improving students' mathematical analytical ability and giving students timely guidance on problem-solving methodology, it is also a problem that every math teacher must consider and seriously explore.

Obviously, the key to solving application problems with tabular equations is to list the corresponding equations from the implicit equivalence relationship in the problems. Through years of teaching practice, the author believes that there are basically the following methods for the teaching of mathematical application problems in junior high schools:

First, the series method. In other words, the relevant equations are directly listed by the words "and", "less" and "times" in the title.

Example 1 There are 27 people working in A, and there are 19 people working in B. Now, 20 people are transferred to support, so that the number of A is twice that of B. How many people should A and B be transferred respectively?

Analysis: Obviously, after the personnel transfer is completed, the number of people in A = 2×b B.

Solution: If X people are transferred to A, then (20-x) people are transferred to B.

27+x=2( 19+20-x)，

The solution is X = 17.

20-x = 20- 17 = 3 (person)

Answer: 17 people will be transferred to a, and 3 people will be transferred to b. ..

Second, the formula method. Formulas familiar to students, such as "distance = speed× time", "total work = working efficiency× working time", "profit = selling price-purchase price" and "profit rate = profit/purchase price", are all tools to solve the application problems of related equations.

Example 2 The purchase price of a commodity is 1800 yuan, and the original price is 2250 yuan. What is the minimum discount for this product if it is required to be sold at a discount rate of not less than 5%?

Analysis: According to the profit rate formula, just list the equation.

Solution: let's assume that the lowest discount is x, according to the meaning of the question:

5% =(2250 x- 1800)/ 1800，

The result is x = 0.84.

The lowest discount is 8.4%.

Third, the total score method. That is to say, according to the fact that the total amount is equal to the sum of the components, the equations are listed. In this way, we should be careful not to omit the ingredients.

Example 3 "Passers-by! Diophantine is buried here. Please calculate the following questions, and you will know how many winters and summers he has experienced in his life. One sixth of his life was a happy childhood, and one twelfth was a carefree teenager. In the past one seventh of the time, he has established a happy family. Five years later, my son was born, only to find that he died four years earlier than his father and lived only half his age. The poor old man lost his son in his later years, and he spent his later years in grief. Please calculate, how old can Diophantine live to see death? "

Analysis: This question is the "epitaph" of the famous Diophantine. In the problem, the total age of Diophantine is cleverly divided into several parts. When solving a problem, you only need to use its total age = the sum of the ages of all parts to get the answer.

Solution: Diophantine lived for X years. According to the meaning of the question:

x=x/6+x/ 12+x/7+5+x/2+4

The result is x = 84.

A: Diophantu lived to be 84 years old.

From the answer to this question, we can also know that Diophantu, a great mathematician in ancient Greece, got married at the age of 33, got a son at the age of 38, died at the age of 80, and his son lived for 42 years.

Fourth, the same method. The principle of solving this kind of problem is that if the same quantity can be expressed by two different algebraic expressions, then the two algebraic expressions must be equal.

A group of students set out from school to receive military training in the army. The speed is 5 km/h. When they walked 4.5 km, a correspondent returned to the school to report the news, and then he immediately caught up with the team. The correspondent's speed is 14 km/h, and he catches up with the team at a distance of 6 km from the army and asks what is the distance between the school and the army. (Ignore notification time)

Analysis: The key to the answer to this question is that the time it takes for the correspondent to go back to school and catch up with the team is the same as the time it takes for the team to walk 4.5 kilometers to 6 kilometers from the army.

Solution: The distance from the school to the army is X kilometers. According to the meaning of the question:

(x-4.5-6)/5=(x+4.5-6)/ 14，

Solution: x = 15.5

A: The distance from the school to the army is 15.5 kilometers.

Of course, the above four methods are not used in isolation. For example, the answer of Example 4 must use the formula: "distance = speed × time". Moreover, the solution to a topic is often not unique. For example, the solution of 1 can also be solved by the total score method:

Solution: Let's assume that the number of people in location B is X and the number of people in location A is 2x. After distribution, the total number of people is 27+ 19+20 = 66.

x+2x=27+ 19+20，

X = 22，

∴ 2x = 44, so 44-27 = 17 (person) and 22- 19 = 39 (person).

Answer: 17 people will be transferred to a, and 3 people will be transferred to b. ..

It can be seen that the training of the methodology of equation application problems not only enables most students to "follow the map" when answering related questions, but also has great significance in cultivating the divergence and diversity of students' thinking, making it possible to solve multiple problems.

Classification and collection of application problems of linear equations with one variable;

(1) Travel issues:

1. From place A to place B, it takes someone 3.6 hours longer to walk than to take a bus. It is known that the walking speed is 8 kilometers per hour and the bus speed is 40 kilometers per hour. If the distance between A and B is X kilometers, the equation is _ _ _ _ _ _ _ _ _.

2. Party A and Party B set out at the same time at a distance of 18km, face to face, and meet at 1 hour 48. If Party A starts 40 minutes earlier than Party B, they will meet 30 minutes away from Party B 1 hour, so that the speed of Party A and Party B can be calculated. ..

Someone goes to school by bike from home. If you drive at a speed of 15 km/h, you can arrive 15 minutes earlier than the scheduled time; If you drive 9 kilometers per hour, you can arrive 15 minutes later than the scheduled time; How many kilometers is the distance from home to school?

On the 4.800-meter runway, two people are practicing middle and long distance running. A runs 320 meters per minute, and B runs 280 meters per minute. They both started from the same place and direction at the same time, and met for the first time after t minutes, and t is equal to minutes.

5. The passenger train is 200 meters long and the freight train is 280 meters long. They are driving in opposite directions on parallel tracks. It takes 16 seconds from the time when two cars meet to the time when two cars leave each other. As we all know, the speed ratio of passenger cars to trucks is 3: 2. How many meters do two cars travel per second?

Clock problem:

/kloc-When does the minute hand and the hour hand coincide between 0/0.6 and 7 o'clock? (textbook review questions)

Navigation problems:

12. A ship is sailing between two docks. The current speed is 3 kilometers per hour. It takes 2 hours to sail downstream and 3 hours to sail upstream. What's the distance between the two docks?

13. An airplane flies between two cities with the wind speed of 24 kilometers per hour. It takes 2 hours and 50 minutes to fly with the wind and 3 hours to fly against the wind. Find out the distance between the two cities.

(2) Engineering problems:

1. For a project, it takes 10 days for Party A to do it alone, and 15 days for Party B to do it alone. Four days of cooperation, how many days does the remaining Party B need?

2. A project is completed by two teams, Team A and Team B. It takes 16 days for Team A to complete it alone, and 12 days for Team B to complete it alone. If Team A works for four days first, and then two teams work together, how many days will it take to complete 5/6 of the project?

As we all know, the pool has a water inlet pipe and a water outlet pipe. The water inlet pipe can fill the empty pool 15 hours, and the water outlet pipe can fill the pool for 24 hours.

(1) How much water can be injected per hour if the water inlet pipe is opened separately?

(2) If the outlet pipe is opened separately, how much water can be released per hour?

(3) If two pipes are opened at the same time, what is the effect per hour? How to form?

(4) For an empty pool, if the water inlet pipe is opened for 2 hours first, and then the two pipes are opened at the same time, how long will it take to fill the pool?

(3) the difference with the times (production, work and other issues):

1. It takes 40 hours for one person to sort out a batch of books. Now it is planned that some people will work for four hours first, and then two people will work with them for eight hours to finish the work. Assuming that these people are equally efficient, how many people should be arranged to work first.

2. The price of water and electricity in a residential area in Yuechi County is: per ton of water 1.55 yuan, 0.67 yuan per kilowatt hour, and per cubic meter of natural gas 1.47 yuan. A resident paid 67.54 yuan +0 1.54 yuan in June 2006, including 5 tons of water and 30,000 yuan.

3. It is known that the taxi charging standard in our city is as follows: 2 yuan is charged if the mileage does not exceed 2 kilometers; If the mileage exceeds 2 kilometers, the excess part will be charged at 1.4 yuan per kilometer except 2 yuan.

(1) If someone drives a taxi for x kilometers (x >: 2), how much should he pay? (column algebra, no simplification) (8 points)

(2) A tourist takes a taxi from the passenger center to Sanxingdui and pays the fare 10.4 yuan. Try to estimate how many kilometers it is from the passenger center to Sanxingdui.

Competition integral problem:

10. An enterprise conducted an English test for candidates, and the test questions consisted of 50 multiple-choice questions. According to the grading standard, if the answer to each question is correct, 3 points will be scored; if it is not selected, 0 points will be scored; if it is wrong, 1 point will be deducted. It is known that someone didn't do five questions and scored 103, then this person chose the wrong question.

1 1. Eight classes in Grade 7 of a school held a football friendly match, which adopted a scoring system: 3 points for winning a game, 1 point for drawing a game, and 0 point for losing a game. After a class played 1 with seven other teams, it scored 17 points with an unbeaten record. How many games did this class win?

Age problem:

12.A is older than B. 15 years old. Five years ago, A was twice as big as B. Now B is _ _ _ _ _.

13. Xiaohua's father is 25 years older than Xiaohua now. Eight years later, Xiaohua's father is three times older than Xiaohua and five years older. Ask Xiaohua's age now.

Proportion problem:

14. The length of a part on the drawing is 32 cm, and the actual length is 4 cm. Then measure the length of another part of the diagram as 12cm, and find the actual length of this part.

15. In the first phase, the exchange rate between Japanese yen and RMB was 25.2: 1, so how much RMB can I exchange for 500,000 yen?

16. Teacher Wei went to the market to buy vegetables, and found that if 10 Jin of vegetables were put on the scale, the pointer on the dial turned to 180. As the picture shows, the next day, Teacher Wei gave the students two questions:

(1) If you put 0.5 kg of vegetables on the scale, how many angles will the pointer turn?

(2) If the pointer turns to 540, how many kilograms are these dishes?