1. If x = 2 is the solution of equation 2x-a = 7, then a = _ _ _ _ _ _
5. The equation 2x-4 = 3m and x+2 = m about x have the same root, so m = _ _ _ _ _ _ _ _
7. If m-n = 1, then the value of 4-2m+2n is _ _ _ _ _ _ _
8. A school teacher went out for four days on holiday. It is known that the sum of the dates of these four days is 42, so the dates of these four days are _ _ _ _ _ _ _ _ _.
2. Multiple choice questions (5 points for each question, 30 points for * * *)
2. If x = 2 is the solution of equation k (2x- 1) = kx+7, then the value of k is ().
A. 1 B. - 1 C. 7 D. -7
A classroom has five lights, 40 watts and 60 watts. The total wattage is 260 watts, so the number of 40-watt and 60-watt bulbs is () respectively.
A. 1，4b 2，3c 3，2d 4， 1
4. A store's turnover last month was 10,000 yuan, and this month is higher than last month by 15%, so this month's turnover is ().
A.(m+ 1)？ 15% ten thousand yuan
C.( 1+ 15%) m yuan D. (1+ 15%) 2m yuan.
Aunt Li deposited 2000 yuan in the bank for one year. After deducting 20% interest tax, the sum of principal and interest is 2 120 yuan. If the annual interest rate of this deposit is x, then equation () can be obtained.
A.2000( 1+x)=2 120
C.2000( 1+x？ 80%)= 2 120d . 2000( 1+x？ 20%)=2 120
Xiaoming's father bought two carpets. He told Xiao Ming that the area of the rug is exactly 1/3 of that of the rug, and the two rugs add up to 20 square meters. Xiao Ming quickly came to the conclusion that the area of two carpets is (unit: square meters) ().
Three. Problem solving (each question 10, ***30)
1. Solve equation:
The x value of.
Xiao Ming found that the unit price of his favorite walkman is the same in shop A and shop B, and so is the unit price of his schoolbag. The sum of the unit price of the walkman and the schoolbag is 452 yuan, and the unit price of the walkman is four times less than that of the schoolbag. 8 yuan.
(1) What is the unit price of Xiaoming's favorite walkman and schoolbag?
(2) If all the goods in Supermarket A get a 20% discount, Supermarket B will return the shopping voucher to 30 yuan for every 100 yuan (if it is less than 100 yuan, the shopping voucher is universal), and Xiaoming will only bring 400 yuan money. If he only buys these two things in one supermarket, can you explain which supermarket he can buy? If you can choose both, which supermarket is more economical to buy?
Test answer
Fill in the blanks
1.-3 2.9 3.x = 0 4.2x-1= 0 and so on.
5.-8
Hint: Equation 2x-4 = 3m
From the equation x=m-2 = m, x+2=m-2 is obtained.
The solution is m =-8.
6.-3
So m =-3.
7.2
Prompt: 4-2m+2n = 4-2 (m-n) = 4-2×1= 2.
8.9, 10, 1 1, 12.
2. Multiple choice problem
1.D 2。 C 3。 B 4。 C 5。 C 6。 C
Three. solve problems
1. solution: without brackets:
Simplify and get:
To move an item, you must:
Simplify and get:
2. Solution:
3. Solution: (1) If the unit price of the schoolbag is X yuan, then the unit price of the walkman is (4x-8) yuan.
According to the meaning of the question, you must:
A: The unit price of Walkman is 360 yuan, and the unit price of schoolbag is 92 yuan.
(2) If purchased in Supermarket A: 452× 80% = 36 1.6 (RMB)
At this time, Xiao Ming still has: 400-36 1.6 = 38.4 yuan.
If he bought it in supermarket B, he first spent 360 yuan to buy a walkman and returned the shopping voucher:
At this time, buy a schoolbag, and there is still a surplus: (90+40)-92 = 38 yuan.
Because 38.4 >; 38
Therefore, Xiao Ming can choose between two supermarkets, but it is more economical for him to buy in supermarket A. ...& gt& gt
Question 2: How to learn the equation well? Hehe, I'm just the opposite of you. My equation is better. First of all, the unknown must be clear, and it will not be difficult in the future. According to the conditions, the equations with their own unknowns are listed, and some topics need to use the unknowns many times. This is an empirical problem. Come on! I believe you can learn it well! ! These methods only play a transitional role, and it is not necessary to really learn the equation well. One more thing: when looking at a topic, look at the problem first, then carefully look at what conditions are available, and see what is known and what is unknown. Then think about what conditions are needed to find the answer, then use the known conditions to obtain those conditions (some simple questions will be given directly), and finally find the answer. Solving an application problem with a linear equation only changes the answer or the conditions required for finding the answer into X, so as to analyze the problem better. If you are good at math, it is not too difficult to get a linear equation. The following is the format of a general linear equation with one variable: solution: (copy the problem, just change "what" into X or set it according to the meaning of the problem) According to the meaning of the problem (generally speaking, many words can be omitted to explain it, which is very popular among middle school teachers and students): formulation (that is, you have to substitute X into the formula, just like you look up arithmetic, and find the known conditions with X as the answer) Solve the equation (that is, you have to find the known conditions with X as the answer).
Question 3: How to learn to solve the equation is actually quite simple, that is, to solve it according to the relationship between the parts.
For example, addend = and-another addend.
Subtraction = minuend-difference
Question 4: How can we learn to solve equations, read every step in the textbook clearly, and then do the problem? .
Question 5: How to learn mathematical equations well? 20-point equation (English: EQUATION) is an equation that represents the equal relationship between two mathematical equations (such as two numbers, functions, quantities and operations), and there is usually an equal sign "=" between them. Equations do not need reverse thinking, but can be listed directly, including unknowns. It has many forms, such as one-dimensional linear equation, two-dimensional linear equation and so on. Widely used in mathematics, physics and other scientific applications.
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.
Binary linear equation (group)
Example of substitution elimination method: solving equations x+y = 5 16x+ 13y = 89 ②.
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. Example of addition, subtraction and elimination: solve the equation 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 to define an integral equation with unknown number, and the highest order of unknown number is 2. This equation is called 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 3. Factorization method 4. Cross multiplication and cross multiplication can factorize 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. A2, decompose the constant term c into two factors, the product of c 1 and C2? C2, and make a 1c2+a2c 1 just a linear term b, then it can be directly written as a result: when decomposing factors in this way, we should pay attention to observation and try to realize that its essence is 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 1 Factorization 2x 2-7x+3. Analysis: First, the quadratic term coefficient is decomposed and written in the upper left corner and the lower left corner of the intersection line, then the constant term is decomposed and written in the upper right corner and the lower right corner of the intersection line, and then the algebraic sum is obtained by cross multiplication to make it equal to the linear term coefficient. Decomposition of quadratic coefficient (positive factor only): 2 =/kloc-0 /× 2 = 2×1; Decomposition constant term: 3 =1× 3 = 3×1= (-3) × (-1)× (-3) The following four cases are represented by drawing a cross line:11w 23/kl. Kloc-0/× 654385438+0× (-3)+2× (-1) =-51-3w2-11× (-1)+2. Solution 2x 2-7x+3 = (x-3) (2x- 1). Generally speaking, for the quadratic trinomial ax2+bx+c(a≠0), if the coefficient of the second term A can be decomposed into the product of two factors, that is, A =>
Question 6: How to learn the method of solving equations In elementary school, solving equations is based on the relationship between known numbers and obtained numbers in four operations. We can solve the equation in the following three ways.
First, directly according to the relationship between the known number and the obtained number in the four operations, find the value of the unknown.
For example: 3.6÷x=0.9. This is the division formula, and X is the divisor, which means that the quotient of X divided by 3.6 is 0.9. According to the relationship that divisor equals dividend divided by quotient in division, find the value of x.
Solving equation: 3.6÷x=0.9
Solution: x=3.6÷0.9
x=4
Second, regard the term containing the unknown x as a number and gradually find the value of the unknown.
For example: 2x-6= 14. The term (2x) containing an unknown number is regarded as a number. So 6 is subtraction, 2x is minuend, and 14 is difference. Find out how much 2x equals first, and then find out the value of x further.
Solving equation: 2x-6= 14
Solution: 2x= 14+6
2x=20
x = 20 \2
x= 10
Thirdly, through calculation, the original equation is simplified first, and then the solution of the equation is solved step by step.
For example: 3x-2.5x4 = 5; Calculate 2.5×4 first, and then find the unknown value according to the previous method.
Solving equation: 3x-2.5×4=5
Solution: 3x- 10=5
3x=5+ 10
3x= 15
x= 15÷3
x=5
Another example is: 4.5x+5.5x+3 = 30; Calculate 4.5x+5.5x first, and then find the unknown value according to the previous method.
Solving equation: 4.5x+5.5x+3=30.
Solution: (4.5+5.5)x+3=30.
10x+3=30
10x=30-3
10x=27
x = 27 \ 10
x=2.7
Exercise:
Solve the following equation.
1.2-x = 0.4 2.5x = 63x+5 = 20 6x- 14 = 10
7x-2x = 5(8+x)×8 = 120 5.4-3x = 2×2. 1 5x-2x-7 = 14
Question 7: How to learn mathematical equations well? 1, the key to learning mathematical equations well: first, understand the physical meaning after mathematical equations. The second is to write more and practice more.
2. Mathematical physical equations refer to some partial differential equations (sometimes including integral equations and some ordinary differential equations) which are obtained after some simplification in physics, mechanics, engineering technology and other problems and reflect the relationship between physical quantities in the objective world. Specifically, there are three common mathematical equations:
① Wave equation reflecting wave phenomenon
② Transport equation reflecting the transport process.
③ Equation reflecting stable field