catalogue
basic concept
Historical story of equation
Mathematical term
One-dimensional linear equation
Example of instructional design
Binary linear equation (group)
Ternary linear equation
N-element linear equation
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basic concept
Unknown: usually x.y.z is unknown, and other letters can be set. The unknowns of two equations in a problem cannot be the same!
The concept of "yuan"
During the Song and Yuan Dynasties, Chinese mathematicians founded Tianyuan, which was used to express unknowns and then set up equations. Later, people set up geographical elements, cultural elements and Thai elements to represent unknowns, and some elements are called multivariate equations. The representative work of this method is the Round Sea Mirror written by mathematician Ye Li (1248), in which "setting a Tianyuan" is equivalent to "setting an unknown x", so the unknown is called "yuan" when the equation is abbreviated now. For example, an unknown equation is called "unary equation". In ancient times, more than two unknowns were also called "Tianyuan", "Geo-element" and "Human element".
"Quadratic": The concept of quadratic in the equation is similar to algebraic expression. Refers to the sum of all unknown indicators in a project with unknown numbers. The term with the highest degree is the degree of the equation.
"Solution": the solution of the equation, also called the root of the equation. The value of an unknown quantity in an equation. Generally expressed as "x=a", where x represents an unknown number and a is a constant.
Solving an equation: The process of finding the solution of an equation or the value of an unknown quantity in an equation is called solving an equation.
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Historical story of equation
1. About 3600 years ago, the mathematical problems written by ancient Egyptians on papyrus involved equations with unknowns.
2. Around 825 AD, Al-Hua Lazimi, a mathematician in Central Asia, wrote a book called Elimination and Reduction, focusing on solving equations.
2. One of the nine chapters of arithmetic.
"Biography of Ma Yan" in the later Han Dynasty is good at nine chapters of arithmetic. Tang Note: "Liu Hui has nine chapters of arithmetic, the first, the second, the third, the fourth, the fifth, and the sixth. Bai Shangshu noted in Nine Chapters of Arithmetic Equation that "Fang" is a square and "Cheng" is an expression or an expression. In a problem, if there are several related data, arrange these related data side by side into a square, which is called' equation'. The so-called' equation' is today's augmented matrix. "
3. The concept of "yuan":
During the Song and Yuan Dynasties, Chinese mathematicians founded Tianyuan, which was used to express unknowns and then set up equations. The representative work of this method is the Round Sea Mirror written by mathematician Ye Li (1248), in which "setting a Tianyuan" is equivalent to "setting an unknown x", so the unknown is called "yuan" when the equation is abbreviated now. For example, an unknown equation is called "unary equation". In ancient times, more than two unknowns were also called "Tianyuan", "Geo-element" and "Human element".
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Mathematical term
An equation with unknowns is called an equation, which is the logical definition of middle school. The definition of equation includes function definition method and relationship definition, while the unknown equation is not necessarily an equation, such as 0x=0, it is not an equation, and it should be defined as follows:
Equation in the form of f (x 1, x2, x3) ... xn) = g (x 1, x2, x3...xn), where f (x 1, x2, x3...xn) and g (x65438+).
Basic properties of equation 1
Adding (or subtracting) the same number or the same algebraic expression on both sides of the equation at the same time, the result is still an equation. Represented by letters: if a=b, c is a number or an algebraic expression. Then: (1) A+C = B+C (2) A-C = B-C.
Basic properties of equation 2
Multiplying or dividing two sides of an equation by the same number that is not 0 is still an equation.
(3) If a=b, then b=a (symmetry of the equation).
(4) If a = b and b = c, then a=c (transitivity of the equation).
Represented by letters: if a=b, c is a number or an algebraic expression (not 0). Then:
a×c=b×c a÷c=b÷c
Some concepts of equation
Solution of the equation: the value of the unknown that makes the left and right sides of the equation equal is called the solution of the equation.
Solving equations: The process of solving equations is called solving equations.
The basis of solving the equation: 1. Shift terminology; 2. Basic properties of the equation; 3. Merge similar projects; 4. Add, subtract, multiply and divide the relationship between the parts.
Steps to solve the equation: 1. What can be counted first; 2. Transform-Calculation-Result
For example:
3x=5×6
3x=30
x = 30 \3
x= 10
Moving term: after changing the sign of some terms in the equation, they move from one side of the equation to the other. Based on the basic properties of the equation 1, this deformation is called the shift term.
There are integral equations and fractional equations.
Integral equation: An algebraic expression equation with unknowns on both sides is called an integral equation.
Fractional equation: The equation with unknown number in denominator is called fractional equation.
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One-dimensional linear equation
People's Education Press will learn the fourth chapter of the first volume of fifth grade mathematics, Hebei Education Press will learn the third chapter of the second volume of fifth grade mathematics, and Beijing Normal University Press will learn the fifth chapter of the first volume of seventh grade mathematics.
The first chapter of the fifth grade of Su Jiaoban.
definition
An integral equation with only one unknown number and the unknown number is 1 is called a linear equation with one variable. The usual form is kx+b=0(k, b is constant, k≠0).
General solution steps
1. Both sides of the denominator equation are multiplied by the least common multiple of each denominator.
4. Generally, the brackets are removed first, then the brackets are removed, and finally the braces are removed. But sometimes the order can be determined according to the situation, which makes the calculation simple. According to the law of multiplicative distribution.
3. Move the unknown term to the other side of the equation, and don't forget to change the sign when moving other terms to the other side of the equation. (Generally, it is like this: for example, 5x-4x = 8； is obtained from 5x=4x+8; Bring the unknown together! ~
4. Merge similar terms to transform the original equation into ax=b(a≠0).
⒌ coefficient: the coefficient that both sides of the equation are divided by the unknown at the same time.
Find the solution of the equation.
Homotopy equation
If the solutions of two equations are the same, then these two equations are called homosolution equations.
The same solution principle of the equation;
Adding or subtracting the same number or the same equation on both sides of the equation is the same solution equation as the original equation.
2. The equation obtained by multiplying or dividing the same number whose two sides are not zero is the same as the original equation.
An important method to solve the application problem of linear equation with one variable;
1. Examine the questions carefully.
Analysis of known and unknown quantities.
[13] Find the equivalence relation.
4. Set an unknown number.
⒌ sequence equation
Solve equations.
⒎ test
⒏ wrote a reply.
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Example of instructional design
Teaching objectives
1. Make students master the methods and steps of solving simple application problems with linear equations; And will list simple application problems solved by linear equations with one variable.
2. Cultivate students' observation ability and improve their ability to analyze and solve problems.
3. Make students form the good habit of thinking correctly.
Teaching emphases and difficulties
Methods and steps of solving simple application problems with linear equations of one variable.
Classroom teaching process design
First, ask questions from students' original cognitive structure
In elementary school arithmetic, we learned the knowledge of solving practical problems with arithmetic. So, can a linear equation solve a practical problem? If it can be solved, how? What are the advantages of solving application problems with one-dimensional linear equations compared with solving application problems with arithmetic methods?
To answer these questions, let's look at the following examples.
Example 1 3 times of a certain number minus 2 equals the sum of a certain number and 4, so find a certain number.
(First, solve it by arithmetic, the students answer, and the teacher writes it on the blackboard.)
Solution 1: (4+2) ÷ (3- 1) = 3.
A: A certain number is 3.
(Secondly, solve the problem by algebraic method, with the guidance of the teacher and oral completion by the students. )
Solution 2: Let a certain number be x, then there is 3x-2 = x+4.
3x-2=x+4
(3- 1)x=2+4
2x=2+4
2x=6
x = 6 \2
x=3
X = 3 is obtained by solving.
A: A certain number is 3.
Looking at the two solutions of the example 1, it is obvious that the arithmetic method is not easy to think about, but the method of setting unknowns, listing equations and solving equations to solve application problems has a feeling of making it difficult, which is also one of the purposes of learning to solve application problems with linear equations.
We know that the equation is an equation with unknowns, and the equation represents an equal relationship. Therefore, for any condition provided in an application problem, we must first find an equal relationship from it, and then express this equal relationship as an equation.
In this lesson, we will explain how to find an equality relationship and the methods and steps to transform this equality relationship into an equation through examples.
Second, teachers and students analyze and study the methods and steps of solving simple application problems with one-dimensional linear equations.
Example 2 After 65,438+05% of the flour stored in the flour warehouse was shipped out, there were still 42,500 kilograms left. How much flour is there in this warehouse?
* * * Analysis of teachers and students:
1. What are the known and unknown quantities given in this question?
2. What is the equal relationship between known quantity and unknown quantity? (Original weight-shipping weight = remaining weight)
3. If the original flour has X kilograms, how many kilograms can the flour represent? Using the above equation relationship, how to formulate the equation?
The above analysis process can be listed as follows:
Solution: Assuming that there are x kilograms of flour, then 15%x kilograms are shipped out. From the meaning of the question, x- 15%x=42 500.
x- 15%x=42 500
( 1- 15%)x=42 500
85%x=42 500
x=42 500÷85%
x=50 000
So x = 50,000.
A: There used to be 50,000 kilograms of flour.
At this point, let the students discuss: are there any other expressions in this question besides the above expression of equal relationship? If so, what is it?
(Also, original weight = shipping weight+remaining weight; Original weight-remaining weight = shipping weight)
The teacher should point out that (1) the expressions of these two equal relations are different from "original weight-shipment weight = residual weight", but the essence is the same, so you can choose one of them to form the equation at will.
(2) The equation solving process of Example 2 is relatively simple, so students should pay attention to imitation.
According to the analysis and solution process of Example 2, please first think about the methods and steps to solve application problems by making linear equations with one variable. Then, give feedback by asking questions; Finally, according to the students' summary, the teacher summarized as follows:
(1) Carefully examine the question and thoroughly understand the meaning of the question, that is, make clear the known quantity, the unknown quantity and their relationship, and use letters (such as X) to represent a reasonable unknown quantity in the question.
(2) according to the meaning of the question, find an equivalent relationship that can express all the meanings of the application question (this is a key step)
(3) According to the relationship of equations, the equations are listed correctly, that is, the listed equations should satisfy that the quantities on both sides should be equal; The units of algebraic expressions on both sides of the equation should be the same; The conditions in the problem should be fully utilized, and none of them can be omitted or reused.
(4) Find the solutions of the listed equations.
(5) Write the answers clearly and completely after the exam. The test required here should be that the solution obtained from the test can not only make the equation valid, but also make the application problem meaningful.
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Binary linear equation (group)
definition
People's Education Edition will learn the fourth chapter of the seventh grade math book, and Hebei Education Edition will learn the ninth chapter of the seventh grade math book.
Definition of binary linear equation: A binary linear integral equation whose exponents are all 1 is called a binary linear equation.
Definition of binary linear equations: Binary linear equations are composed of two binary linear equations.
Solution of binary linear equation: the values of two unknowns that make the values of both sides of binary linear equation equal are called the solutions of binary linear equation.
Solutions of binary linear equations: Two common solutions of binary linear equations are called solutions of binary linear equations.
General solution and elimination: solve the unknowns in the equations one by one from more to less.
There are two ways to eliminate elements:
elimination by substitution
Example: Solve the system of equations x+y = 5 16x+ 13y = 89②.
Solution: Take ③ from ① with x=5-y③ to ② to get 6(5-y)+ 13y=89 and y=59/7.
Bring y=59/7 into ③ to get x=5-59/7, that is, x=-24/7.
∴x=-24/7,y=59/7
This solution is the substitution elimination method.
Addition, subtraction and elimination method
Example: Solve the system of equations x+y=9① x-y=5②.
Solution: ①+②, 2x= 14, that is, x=7.
Bring x=7 into ① to get 7+y=9 and y=2.
∴x=7,y=2
This solution is addition, subtraction and elimination.
There are three solutions to binary linear equations:
1. There is a solution.
For example, the solution of the equation set X+Y = 5 16x+ 13Y = 89 ② is x=-24/7 and y=59/7.
There are countless solutions.
For example, the equation group X+Y = 6 12x+2Y = 12②, because these two equations are actually an equation (also called "the equation has two equal real roots"), so this equation group has countless solutions.
3. No solution
For example, the equation set X+Y = 4 12x+2Y = 10②, because the simplified equation ② is x+y=5, which contradicts equation ①, so this kind of equation set has no solution.
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Ternary linear equation
definition
Similar to binary linear equations, three combinatorial linear equations contain three unknowns.
Solution of ternary linear equations
Similar to the binary linear equation, the elimination method is used to eliminate it step by step.
Typical problem analysis
In order to encourage a certain area to save water, the charging standard of tap water is as follows: if the monthly water consumption of each household does not exceed 10 ton, it will be charged according to 0.9 yuan/ton; If it exceeds 10 ton and does not exceed 20 tons, it will be charged at 1.6 yuan/ton; The part exceeding 20 tons is charged according to 2.4 yuan/ton. In a certain month, user A paid 16 yuan more than user B, and user B paid 7.5 yuan more than user C. It is known that user C is short of water 10 ton, and user B uses more water than 10 ton but less than 20 tons. Q: What is the monthly water fee for A, B, C and C users (charged by the whole ton)?
Solution: Assume that Party A uses X tons of water, Party B uses Y tons of water and Party C uses Z tons of water.
Obviously, User A used more than 20 tons of water.
Therefore, Party A's payment: 0.9 *10+1.6 *10+2.4 * (x-20) = 2.4x-23.
Payment: 0.9 *10+1.6 * (y-10) =1.6y-7.
Payment by C: 0.9z
2.4x-23= 1.6y-7+ 16
1.6y-7=0.9z+7.5
simplify
3x-2y=40……( 1)
16y-9z= 145……(2)
X=(2y+40)/3 from ( 1)
So let y =1+3k,3.
When k = 4, y = 13 and x = 22, substitute (2) to get z=7.
When k=5, y= 16, substituting (2), z has no integer solution.
When k=6, y= 19, substituting (2), z has no integer solution.
Therefore, A uses 22 tons of water, B uses 13 tons, and C uses 7 tons.
The water consumption of Party A is 29.8 yuan, that of Party B is 13.8 yuan, and that of Party C is 6.3 yuan.