The basic property of the 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
The basic property of the equation 2: the result of multiplying or dividing the two sides of the 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).
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, 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. [Edit this paragraph] One Yuan Linear Equation People's Education Edition will study the third chapter of the first volume of seventh grade mathematics, Hebei Education Edition will study the seventh chapter of the second volume of seventh grade mathematics, and Jiangsu Education Edition will study the first chapter of fifth grade.
Definition: An integral equation with only one unknown number 1 is called a linear equation with one variable. The usual form is kx+b=0(k, b is constant, k≠0).
Universal solution:
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.
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.
Homosolution equation: If two equations have the same solution, they 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.
7. Inspection (Jian 'an Three Tones)
⒏ wrote a reply.
Example of instructional design
Teaching objectives
1. Make students master the methods and steps of solving simple application problems with linear equations; And will enumerate the simple application problems of solving one-dimensional linear equations;
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.
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 there are X kilograms of flour, then 15% x kilograms will be shipped out.
x- 15%x=42 500，
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)
What the teacher wants to point out is: (1) The expression of these two equal relations is different from "original weight-shipped weight = remaining weight", but the essence is the same, so you can choose one of the equations 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 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. [Edit this paragraph] The second volume of seventh-grade mathematics of People's Education Edition will learn binary linear equations (groups), and the ninth chapter of seventh-grade mathematics of Hebei Education Edition will learn.
Definition of bivariate linear equation: The bivariate linear integral equation with exponent 1 is called bivariate linear equation.
Definition of binary linear equations: Two linear equations with two unknowns are called 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. [Edit this paragraph] Definition of ternary linear equation: Similar to binary linear equation, three combined linear equations contain three unknowns.
Solution of ternary linear equations: similar to binary linear equations, the elimination method is used to eliminate them 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. Within one month, user A paid more 16 yuan than user B, and user B paid more than user C for 7.5 yuan. It is known that user C is short of water 10 ton, and user B uses more than 10 ton but less than 20 ton. Q: How much water fee do users A, B and C pay each month (calculated 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.
Definition: An integral equation has an unknown number, and the highest order of the unknown number is 2. Such an equation is called an unary quadratic equation.
The transformation from a linear equation to a quadratic equation is a qualitative change. Usually, quadratic equation is much more complicated in concept and solution than linear equation.
General form: ax 2+bx+c = 0 (a ≠ 0)
There are four general solutions:
Formula method (direct Kaiping method)
4. Matching method
3. Factorization method
4. Cross multiplication
Cross multiplication can decompose some quadratic trinomials. The key of this method is to decompose the quadratic coefficient A into the product of two factors a 1 and A2 A 1. 6? 1a2, decompose the constant term c into the product of two factors, c 1 and C2? 6? 1c2, and make a 1c2+a2c 1 just be a linear term b, then the result can be written directly: when factorizing factors in this way, we should pay attention to observation, try and realize that it is actually the inverse process of binomial multiplication. When the first coefficient is not 1, it often needs to be tested many times, so be sure to pay attention to the sign of each coefficient.
example
Example 1 Factorization 2x 2-7x+3.
Analysis: first decompose the quadratic coefficient and write it in the upper left corner and lower left corner of the crosshair, then decompose the constant term and divide it into two parts.
Don't write it in the upper right corner and lower right corner of the crosshair, and then cross multiply to find the algebraic sum to make it equal to the coefficient of the first term.
Quadratic coefficient decomposition (positive factor only):
2= 1×2=2× 1;
Decomposition of constant term:
3= 1×3=3× 1=(-3)×(- 1)=(- 1)×(-3).
Draw a cross line to represent the following four situations:
1 1
╳
2 3
1×3+2× 1
=5
1 3
╳
2 1
1× 1+2×3
=7
1 - 1
╳
2 -3
1×(-3)+2×(- 1)
=-5
1 -3
╳
2 - 1
1×(- 1)+2×(-3)
=-7
After observation, the fourth case is correct, because after cross multiplication, the algebraic sum of the two terms is exactly equal to the coefficient of the first term -7.
Solution 2x 2-7x+3 = (x-3) (2x- 1).
Generally speaking, for the quadratic trinomial ax2+bx+c(a≠0), if the quadratic term coefficient A can be decomposed into the product of two factors, that is, a=a 1a2, the constant term C can be decomposed into the product of two factors, that is, c=c 1c2, and A/KLOC-.
a 1 c 1
╳
a2 c2
a 1c2+a2c 1
Cross-multiply diagonally, and then add to get a 1c2+a2c 1. If it is exactly equal to the first term coefficient b of the quadratic trinomial ax2+bx+c, that is, a 1c2+a2c 1=b, the quadratic trinomial can be decomposed into two factors a65438+.
ax2+bx+c =(a 1x+c 1)(a2x+C2)。
The way to help us decompose the quadratic trinomial by drawing cross lines like this is usually called cross multiplication.
Example 2 Factorizing 6x 2-7x-5.
Analysis: According to the method of example 1, the quadratic term coefficient 6 and the constant term -5 are decomposed and arranged respectively, and there are eight different arrangement methods, one of which is
2 1
╳
3 -5
2×(-5)+3× 1=-7
Is correct, so the original polynomial can be factorized by cross multiplication.
Solution 6x 2-7x-5 = (2x+ 1) (3x-5)
It is pointed out that through the examples of 1 and 2, it can be seen that when a quadratic trinomial factor whose quadratic coefficient is not 1 is solved by cross integration, it often needs many observations to determine whether the factor can be solved by cross integration.
For the quadratic trinomial with quadratic coefficient of 1, cross multiplication can also be used to decompose the factors. At this time, you only need to consider how to decompose the constant term. For example, if x 2+2x-15 is decomposed, the cross multiplication is
1 -3
╳
1 5
1×5+ 1×(-3)=2
So x 2+2x- 15 = (x-3) (x+5).
Example 3 Factorization 5x 2+6xy-8y 2.
Analysis: This polynomial can be regarded as a quadratic trinomial about x, and-8y 2 is regarded as a constant term. When we decompose the coefficients of quadratic terms and constant terms, we only need to decompose 5 and -8, and then use the crosshairs to decompose them. After observation, we chose a suitable group, namely
1 2
╳
5 -4
1×(-4)+5×2=6
Solution 5x 2+6xy-8y 2 = (x+2y) (5x-4y).
It is pointed out that the original formula is decomposed into two linear formulas about x and y.
Example 4 Factorization (x-y)(2x-2y-3)-2.
Analysis: This polynomial is the product of two factors and the difference of another factor. Only by multiplying the polynomials first can the deformed polynomials be factorized.
Q: What are the characteristics of the factorial of the product of two products? What is the simplest method of polynomial multiplication?
A: If the common factor 2 is proposed for the first two items in the second factor, it will become 2(x-y), which is twice that of the first factor. Then multiply (x-y) as a whole, the original polynomial can be transformed into a quadratic trinomial about (x-y), and the factor can be decomposed by cross multiplication.
Solution (x-y)(2x-2y-3)-2
=(x-y)[2(x-y)-3]-2
=2(x-y) ^2-3(x-y)-2
=[(x-y)-2][2(x-y)+ 1]
=(x-y-2)(2x-2y+ 1)。
1 -2
╳
2 1
1× 1+2×(-2)=-3
It is pointed out that decomposing (x-y) into a whole is another application of holistic thinking method in mathematics.
Example 5 x 2+2x- 15
Analysis: the constant term (-15) < 0 can be decomposed into the product of two numbers with different signs, and can be decomposed into (-1)( 15), or (1)(- 15) or (3).
(-5) or (-3)(5), in which only the sum of -3 and 5 in (-3)(5) is 2.
=(x-3)(x+5)
Summary: ① Factorization of X2+(P+Q) X+PQ formula.
The characteristics of this kind of quadratic trinomial formula are: the coefficient of quadratic term is1; Constant term is the product of two numbers; The coefficient of a linear term is the sum of two factors of a constant term. So we can directly decompose some quadratic trinomial factors with the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q).
② Factorization of KX2+MX+N formula
If it can be decomposed into k = AC, n = BD and AD+BC = M, then
kx^2+mx+n=(ax+b)(cx+d)
A b
╳
c d
1, direct Kaiping method:
The direct Kaiping method is a method to solve a quadratic equation with a direct square root. Solving (x-m)2=n (n≥0) by direct Kaiping method.
The solution is an equation of x = m.
Example 1. Solve the equation (1) (3x+1) 2 = 7 (2) 9x2-24x+16 =1.
Analysis: (1) This equation is obviously easy to do by direct flattening, (2) The left side of the equation is completely flat (3x-4)2, and the right side =11>; 0, so
This equation can also be solved by direct Kaiping method.
(1) solution: (3x+ 1)2=7×
∴(3x+ 1)2=5
∴ 3x+ 1 = (be careful not to lose the solution)
∴x=
The solution of the original equation is x 1=, x2=.
(2) Solution: 9x2-24x+16 =11.
∴(3x-4)2= 1 1
∴3x-4=
∴x=
The solution of the original equation is x 1=, x2=.
2. Matching method: use matching method to solve the equation ax2+bx+c=0 (a≠0).
First, move the constant c to the right of the equation: AX2+BX =-C.
Convert the quadratic term to 1: x2+x =-
Add half the square of the first order coefficient on both sides of the equation: x2+x+( )2=- +( )2.
The left side of the equation becomes completely flat: (x+ )2=
When b2-4ac≥0, x+=
∴x= (this is the root formula)
Example 2. Solving Equation 3x2-4x-2=0 by Matching Method
Solution: Move the constant term to the right of equation 3x2-4x=2.
Transform the quadratic term into 1: x2-x =
Add half the square of the coefficient of the first order term on both sides of the equation: x2-x+( )2= +( )2.
Formula: (x-)2=
Direct square: x-=
∴x=
The solution of the original equation is x 1=, x2=.
3. Formula method: convert the quadratic equation of one variable into a general form, and then calculate the value of the discriminant △=b2-4ac. B2-4ac≥0, release all items.
Substitute the values of coefficients A, B and C into the formula x=(b2-4ac≥0) to get the root of the equation.
Example 3. Solving Equation 2x2-8x=-5 by Formula Method
Solution: Change the equation into a general form: 2x2-8x+5=0.
∴a=2，b=-8，c=5
B2-4ac =(-8)2-4×2×5 = 64-40 = 24 & gt； 0
∴x= = =
The solution of the original equation is x 1=, x2=.
4. Factorial decomposition method: the equation is deformed into a form with one side zero, and the quadratic trinomial on the other side is decomposed into the product of two linear factors, so that,
Two linear factors are equal to zero respectively, and two linear equations are obtained. The roots obtained by solving these two linear equations are two of the original equations.
Root. This method of solving a quadratic equation with one variable is called factorization.
Example 4. Solve the following equation by factorization:
( 1)(x+3)(x-6)=-8(2)2 x2+3x = 0
(3) 6x2+5x-50=0 (optional study) (4)x2-2(+)x+4=0 (optional study)
(1) Solution: (x+3)(x-6)=-8 Simplified sorting.
X2-3x- 10=0 (the equation has a quadratic trinomial on the left and zero on the right).
(x-5)(x+2)=0 (factorization factor on the left side of the equation)
∴x-5=0 or x+2=0 (converted into two linear equations)
∴x 1=5,x2=-2 is the solution of the original equation.
(2) Solution: 2x2+3x=0
X(2x+3)=0 (factorize the left side of the equation by increasing the common factor)
∴x=0 or 2x+3=0 (converted into two linear equations)
∴x 1=0, x2=- is the solution of the original equation.
Note: Some students easily lose the solution of x=0 when doing this kind of problem. It should be remembered that there are two solutions to the quadratic equation of one variable.
(3) Solution: 6x2+5x-50=0
(2x-5)(3x+ 10)=0 (pay special attention to symbols when factorizing by cross multiplication).
2x-5 = 0 or 3x+ 10=0.
∴x 1=, x2=- is the solution of the original equation.
(4) Solution: x2-2(+ )x+4 =0 (∵4 can be decomposed into 2.2, ∴ this problem can be factorized).
(x-2)(x-2 )=0
∴x 1=2, x2=2 is the solution of the original equation.
5. Cross multiplication
The formula in the form of y = x 2+(p+q) x+pq can be factorized.
The characteristics of this kind of quadratic trinomial formula are: the coefficient of quadratic term is1; Constant term is the product of two numbers; The coefficient of a linear term is the sum of two factors of a constant term. So we can directly decompose some quadratic trinomial factors with the coefficient of1:x 2+(p+q) x+pq = (x+p) (x+q).
Binary Quadratic Equation: An integral equation with two unknowns, the highest degree of which is 2. [Edit this paragraph] Note that generally speaking, the linear equation with n unknowns is an equation with the number of unknowns 1, and the coefficient of the linear term is not equal to 0;
N-dimensional linear equations are composed of several N-dimensional linear equations (except one-dimensional linear equations);
One-dimensional A-degree equation is an equation with an unknown term whose highest degree is A (except one-dimensional linear equation);
One-dimensional A-degree equation is an equation composed of several one-dimensional A-degree equations (except one-dimensional linear equation);
N-dimensional A-degree equation is an equation with n unknowns and the highest degree of the unknowns is A (except one-dimensional linear equation);
N-dimensional A-degree equation is an equation composed of several N-dimensional A-degree equations (except one-dimensional linear equation);
Among equations (groups), equations (groups) with more unknowns than equations are called indefinite equations (groups), and such equations (groups) generally have countless solutions.
Chicken rabbit cage formula
Solution 1: (rabbit foot number × total foot number-total foot number) ÷ (rabbit foot number-chicken foot number)
= Number of chickens
Total number-number of chickens = number of rabbits
Solution 2: (total number of feet-chicken feet × total number of feet) ÷ (rabbit feet-chicken feet)
= the number of rabbits
Total-number of rabbits = number of chickens
Solution 3: total number of feet ÷2- total number of heads = number of rabbits.
Total number of rabbits = number of chickens.
Solution 4 (Equation): X= total number of feet ÷2- total number of heads (X= number of rabbits)
Total number of rabbits = number of chickens.
Solution 5 (Equation): X= (total number of feet-number of chicken feet × number of chickens) ÷ (number of rabbit feet-number of chicken feet) (X= number of rabbits)
Total number of rabbits = number of chickens.
Solution 6 (Equation): X=: (rabbit's foot number × total foot number-total foot number) ÷ (rabbit's foot number-chicken's foot number) (X= chicken's foot number)
Total number-number of chickens = number of rabbits
* In short, find out the equivalence relation, list the equations according to the equivalence relation, and solve the equations according to the properties of the equations.