An equation contains only one unknown, and the degree of the highest term containing this unknown is 1. This equation is called a linear equation with one variable. The usual form is ax+b=0(a, b are constants, a≠0). The linear equation of one variable belongs to the whole equation, that is, both sides of the equation are algebraic expressions. Unary refers to an equation containing only one unknown, and the number of times the first term refers to the unknown is 1. A and b are known numbers, and a≠0) is called the standard form of linear equation. Where a is the coefficient of the unknown, b is a constant, and the degree of x must be 1. The linear equation in English is (linear.
equation
exist
I) Edit the properties of this paragraph.
Properties of the equation 1: When both sides of the equation add a number or subtract the same number, the equation still holds.
Property 2 of the equation: When both sides of the equation are multiplied by a number or divided by the same number that is not 0, the equation still holds.
Property 3 of the equation: When both sides of the equation are multiplied (or squared) at the same time, the equation still holds.
Solving the equation is based on these three properties of the equation. Edit the solution of the linear equation in this paragraph.
The value of the unknown that makes the left and right sides of the equation equal is called the solution of the equation.
ax=b
When a ≠ 0 and b = 0,
ax=0
x = 0；
When a≠0, x = b/a.
When a=0,
When b=0, the equation has countless solutions (note: this situation is not a linear equation, but an identity).
When a=0,
When b≠0, the equation has no solution.
Example:
(3x+ 1)/2-2 =(3x-2)/ 10-(2x+3)/5
Get the denominator (denominator multiplied by the least common multiple of both sides of the equation).
↓
5(3x+ 1)- 10×2 =(3x-2)-2(2x+3)
Without brackets,
↓
15x+5-20=3x-2-4x-6
Transferred projects,
↓
15x-3x+4x=-2-6-5+20
Merge similar projects,
↓
16x=7
The coefficient is 1,
↓
X=7/ 16。 Edit the linear equations and practical problems in this paragraph.
One-dimensional linear equations involve many practical problems, such as
Engineering problem, planting area problem, competition score problem, distance problem, encounter problem, countercurrent problem, opposition problem. From formula to equation.
When listing equations, first set letters to represent unknowns, and then write an equation containing unknowns according to the equation relationship in the problem-equation.
1.4x=24
2. 1700+ 150x=2450
3.0.52x-( 1-0.52)x=80
The above equations all contain only one unknown (element), and the number of unknowns is 1. Such equations are called linear equations (linear equations).
equation
along with
one
Unknown).
It is a way to solve practical problems by mathematics to analyze the quantitative relations in practical problems and list equations by using their equal relations. Learning practice of editing this unary linear equation group
Elementary school will learn simple linear equations, and junior high school will begin to understand the solution of linear equations and solve difficult application problems with linear equations.
One-dimensional linear equation inclusion
Engineering problems
Planting problem
Meeting problem (distance problem)
Grazing problem
Wait, edit this equation.
Multiply by the same number on both sides of the equation, or divide by the same number that is not 0, and the result is still equal.
5x-4x=-25-20
As mentioned above, moving the sign of a term on one side of an equation to the other side is called moving the term. Edit the steps to solve the linear equation in this paragraph.
Universal solution:
1. Denominator: both sides of the equation are multiplied by the least common multiple of each denominator;
2. Remove the brackets: first remove the brackets, then remove the brackets, and finally remove the braces;
3. Move the term: move all terms containing unknowns to one side of the equation, and all other terms to the other side of the equation; Move the item to change the symbol.
4. Merge similar terms: transform the equation into ax=b(a≠0);
5. Coefficient division 1: divide the unknown coefficient a on both sides of the equation to get the solution of equation x = b/a. 。
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 Write the answers and edit the teaching objectives of this teaching design example.
1. Make students master the methods and steps of solving simple application problems with linear equations of one variable, and list the simple application problems of linear equations of 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 have learned the relevant knowledge of solving practical problems by arithmetic methods. 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.
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
The flour stored in the flour warehouse was shipped out.
After 15%, there is still 42.
500 kilograms. 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:
Suppose there are x kilograms of flour, then 15% x kilograms are shipped out. According to the meaning of the question, x- 15% x = 42.
500, so
x=50
000.
A: Yes.
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)
Teachers should point out that:
(1) The expression of these two equal relations is different from "original weight-shipment weight = residual weight", but the essence is the same. You can choose any one of the constituent equations.
(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, ask questions and give feedback.
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 the 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) solving 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.