Axiom (Theorem) for Judging Parallel Lines (1) Two straight lines are cut by a third straight line. If the congruence angle is equal, the two lines are parallel (referred to as "congruence angle is equal, and the two lines are parallel"). (2) Two straight lines are cut by the third line. If the internal dislocation angles are equal, the two straight lines are parallel (referred to as "the internal dislocation angles are equal and the two straight lines are parallel"). Then these two lines are parallel (hereinafter referred to as "the internal angles on the same side are complementary, and the two lines are parallel"). 2. Nature axiom (theorem) of parallel lines If two parallel lines are cut by the third line, then (1) is equal to the complementary angle (hereinafter referred to as "two parallel lines are equal to the complementary angle"). 2. The basic property of the equation with equal internal dislocation angles (hereinafter referred to as "two parallel lines with equal internal angles") is 1: both sides of the equation add (or subtract) the same number or the same algebraic expression at the same time, and the result is still the equation. Represented by letters: if a = b, c is a number or an algebraic expression. Then: [1] A+C = B+C [2] Basic properties of the equation A-C = B-C 2: The result obtained by multiplying or dividing 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). 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. 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. Edit this paragraph. The unary equation will be studied in the fourth chapter of the first volume of seventh grade mathematics in People's Education Edition, and the seventh chapter of the second volume of seventh grade mathematics in Hebei Education Edition. 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). General solution: 1. Both sides of denominator equation are multiplied by the least common multiple of denominator at the same time. 4. Generally, the brackets are removed first, then the brackets are removed, and finally the braces are removed. 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). 5] coefficient 1 equation, both sides are divided by unknown coefficient at the same time, and the solution of the equation is obtained. Homosolution equation: If two equations have the same solution, they are called homosolution equations. The same solution principle of the equation: 1. 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 way to do the application problem of linear equation with one variable: 1. Examine the problem carefully; 2. Analyze known quantity and unknown quantity; 3. Find an equal relationship; 3. Solve the equation; 3. test; Write an answer example: 1, 3 times of a certain number MINUS 2 equals the sum of a certain number and 4, and find a certain number. (solve it by arithmetic first, and the students will answer it. ) teacher blackboard) solution 1: (4+2) ÷ (3- 1) = 3. A: A certain number is 3. Secondly, it is solved by algebraic method, which is guided by the teacher and completed by the students. Solution 2: Let a certain number be x, then there is 3x-2 = X+4. Listing equations and finding solutions to application problems by solving equations have a feeling of making it difficult to make it easy, which is also one of the purposes of learning to solve application problems by using linear equations with one variable. 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. We will explain how to find an equality relationship and the methods and steps to transform this equality relationship into an equation through examples. Secondly, teachers and students will analyze and study the methods and steps of solving simple application problems with linear equations of one variable. Example 2 After the flour stored in the flour warehouse was shipped out 15%, there was still 42,500 kg left. How much flour is there in this warehouse? Teacher-student analysis: 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 raw flour has x kilograms, how many kilograms can the flour be expressed? 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 will be shipped out. From the meaning of the question, X- 15% X = 42500, so X = 50,000. A: There are 50,000 kilograms of flour. Now let the students discuss: the equal relationship in this question is except if it is, what is it? (Also, original weight = shipping weight+remaining weight; Original weight-residual weight = residual weight) The teacher should point out that (1) these two equal relations and the expression of "original weight-residual weight = residual weight" are different in form, but the essence is the same, so you can choose one of the equations at will; (2) The process of solving the equation in Example 2 is relatively simple, and students should pay attention to imitation. According to the analysis and solution process in Example 2, first, students should think about the methods and steps to solve application problems by making a linear equation. Then, give feedback by asking questions; Finally, according to the students' summary, the teacher summed it up 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. Editing this binary linear equation (group) can be learned in the second book of seventh grade mathematics of People's Education Edition and the ninth chapter of the second book of seventh grade mathematics of Hebei Education Edition. 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 methods of elimination: substitution elimination method: solving the equations X+Y = 5 16x+ 13Y = 89② solution: getting x = 5-y3 from ①, bringing ③ into ②, getting 6(5-y)+ 13y=89, and getting Examples of addition, subtraction and elimination: Solve the equation system x+y=5① x-y=9② Solution: ①+②, and get 2x= 14, that is, x=7 brings x=7 into ①, and gets 7+y=5, and the solution is y=-2 ∴x=7, and y=-2. Binary linear equations have three solutions: 1. There is a set of solutions such as the equation X+Y = 5 16x+ 13Y = 89②, and the solutions are x=-24/7 and y=59/7. 2. There are countless groups of solutions, such as the system of equations X+Y = 6 12x+2Y = 12②. Because these two equations are actually one equation (also called "the equation has two equal real roots"), this equation group has countless solutions. 3. There is no solution, such as the equation set X+Y = 4 12x+2Y = 10②, because the simplified equation ② is x+y=5, which contradicts the equation ①, so this kind of equation set has no solution. Edit the definition of the ternary linear equation in this paragraph: similar to the binary linear equation, it is composed of three ternary linear equations. Solution of ternary linear equations: similar to binary linear equations, the elimination method is used to eliminate them step by step. Analysis of typical problems: 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: Suppose A uses X tons of water, B uses Y tons of water and C uses Z tons of water. Obviously, the water consumption of user A exceeds 20 tons, so the payment of user A is 0.9 *10+1.6 *10+2.4 * (x-20) = 2.4x-23; The payment of user B is 0.9 * 10+ 1.6 *. = 1.6y-7 Payment by Party C: 0.9z2.4x-23 =1.6y-7+061.6y-7 = 0.9z+7.5 Simplified 3x-2y = 40-(65438k < 7 when k=4, y= 13, x=22, substitute (2) to get z=7 when k=5, y= 16, substitute (2), z has no integer solution when k=6, and substitute (y= 19. /CA & gt； Editing this one-dimensional quadratic equation will be studied in the first volume of ninth-grade mathematics of People's Education Edition and in Chapter 29 of the first volume of ninth-grade mathematics of Hebei Education Edition. 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: 1. Formula method (direct Kaiping method) 1. Matching method 1. Cross multiplication 1. Factorization method (due to limited energy, I hope someone can help me) 1, directly Kaiping. The equation with the shape of (x-m)2=n (n≥0) is solved by direct Kaiping method, and the solution is x = m+0. Solve the equation (1) (3x+1) 2 = 7 (2) 9x2-24x+6544. 0, so this equation can also be solved by direct Kaiping method. (1) solution: (3x+1) 2 = 7x ∴ (3x+1) 2 = 5 ∴ 3x+1= (be careful not to lose the solution) ∴ x = ∴. Kloc-0/ 1∴3x-4 =∴x = x2 = 2 ... Matching method: solve the equation ax2+bx+c=0 (a≠0) by matching method. First, move the constant c to the right of the equation: ax2+bx=-c, convert the quadratic term coefficient into 1:X2+X =- and add the square of half the linear term coefficient to both sides of the equation: X2+X+ (). X+= ∴ X = (This is the formula for finding the root) Example 2. Solve equation 3x2-4x-2=0 by matching method: move the constant term to the right of equation 3x2-4x=2, and convert the quadratic term into 1: x2-x = the square of half the coefficient of the primary term on both sides of the equation: x2-x+ (). 2= direct square root: 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. When b2-4ac≥0, take all the coefficients A. Example 3. Solve the equation 2x2-8x=-5 by formula: transform the equation into a general form: 2x2-8x+5 = 0 ∴ A = 2, b =-8, c = 5b2-4ac = (-8) 2-4× 2× 5 = 64-40 = 24. 0 ∴x= = = ∴ The solution of the original equation is X 1 =, X2 = .4. Factorial decomposition method: the equation is deformed to zero on one side, and the quadratic trinomial on the other side is decomposed into the product of two linear factors, so that the two linear factors are equal to zero respectively, and two unary linear equations are obtained and solved. This method of solving a quadratic equation with one variable is called factorization. Example 4. Solve the following equations by factorization: (1) (x+3) (x-6) =-8 (2) 2x2+3x = 0 (3) 6x2+5x-50 = 0 (optional research) (4) x2-2 (+) x+4. =-8 Simplify the arrangement to get x2-3x- 10=0 (quadratic trinomial on the left and 0 on the right) (x-5) (x+2) = 0 (factorization on the left side of the equation) ∴x-5=0 or x+2=0 (converted into two linear equations) \ 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 the symbols when factorizing cross multiplication) ∴2x-5=0 or 3x+/kloc-0 = 0 ∴ x656;. (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 original equation. Binary Quadratic Equation: An integral equation with two unknowns, the highest degree of which is 2. Edit the comments in this paragraph. Generally speaking, a linear equation with n variables is an equation with n unknowns, the number of unknowns is 1, and the coefficient of 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. Complementarity (meaning "two lines are parallel and complementary to each other's internal angles")