Shabalina Natalya Alekseevna. MKOU Tuturskaya Secondary School
Mathematics 3rd grade.
Topic: Property - dividing a sum by a number.
Goal: getting to know something new arithmetic property, developing the ability to use it when solving expressions.
Planned results.
Subject:
Know the name of the new property;
Know algorithms for solving expressions using this property;
Be able to compare different calculation methods and choose the most convenient one.
Personal:
Realize the importance of studying properties for ease of calculation;
The need to come to the aid of a classmate in case of difficulties arises,
Self-assessment of one’s own actions and achievements.
Metasubject:
Independent setting of lesson goals;
Independent construction of speech utterances regarding ways of solving expressions;
Independent determination of solution methods and formulation of action algorithms;
Determining the meaning of a schematic representation of a property;
Collective discussion of methods of action.
1 Oral counting with the goal of the lesson.
I hand out cards with the first learning task (hereinafter referred to as HL)
UZ No. 1 (communicative)
Notes:
I note to myself who was the first to solve this or that expression. They will not be able to solve the last one, so please comment on the first three. I especially rely on the guys who were the first to find the correct values. They say the most rational ways. If they are not found, please find them frontally. No. 1 - applied the combining property (grouped): (27 + 3) + (16 + 4) No. 2 - rounded the minuend: 50-7 No. 3 - applied the property of multiplying a sum by a number (15 + 5).3
Based on this assignment,state the purpose of the lesson.
They may say: “Learn to solve new examples. Find out how to solve such examples.” If they don’t tell you about the method, I remind you that the three examples were not solved in the same way, but different ones were used… what? (methods) Please establish a logical sequence of these goals. 2 targets (personification of goals) with corresponding signatures appear on the board (1 - find out new way, 2-learn to solve with it) I remind you: “Whoever understands that he has already achieved the goal, approach the board as usual and aim your arrow at the bull’s-eye.”
2 Setting the topic of the lesson.
Let's start looking for ways to solve a difficult example, and a new property of arithmetic operations will help, which you will try to name yourself. But let's look at it using a simpler example.
On the board there is a model and expressions:
(6+4).2 6-4 (6+4):2
Having selected an expression for the model, we determine the name of the property.
Let's discuss the model. On it we divide both red and blue into 2 parts at the same time, therefore, the last expression is suitable. Please read the expression (the sum of 6 and 4 is divided by 2)
What should we call the property?
(They try it themselves. If it doesn’t work, please name it by analogy with the studied property of multiplication.)
Dividing a sum by a number.
Let's formulate goal No. 1 more precisely. (If they can’t, then I focus on a new property. The goal is to find a way or ways to divide a sum by a number.)
4 Search for solutions.
I divide the class into pairs or triplets. I am distributing 6 red and 4 blue circles, cards with LS No. 2 (cognitive)
I give no more than 5 minutes. The method is presented using demonstration figures on a typesetting canvas.
1 way:
Without paying attention to the color, they “mixed” it into a sum, and it was divided in half (6+4): 2=5
Let's clarify the algorithm.
First, they found the amount, and then divided it by the number.
Method 2:
We divided the red ones separately, then divided the blue ones, and then added them up in each part (6:2)+(4:2)=5
Let's clarify the algorithm.
We divided each term of the sum separately, and then added the results of the division.
If suddenly no one finds the first method, I ask you to find it, not paying attention to the color of the figures. If they don’t find the second one, I remind you that for some reason the mugs are given in two colors.
Perhaps some of the children will already see the achievement of the first goal. If everyone remains silent, I’ll ask: “Why did you perform this task?” (We went to the first goal and achieved it, but we have not yet achieved the second, because we do not yet know whether the methods found will be useful for solving more complex examples.)
How can I check this? (If they don’t tell you themselves, please remember what difficulty they encountered in UZ No. 1. So we need to try to solve the example (70+8):6
I propose to solve it yourself in notebooks in two ways, using algorithms on the screen. I check and ask who achieved the second goal (these children draw their arrow into the “bull’s-eye” on the board)
What if someone hasn't hit that target yet? (The “experts” will teach - the law of the class.) Any of those who solved the example comes to the board and shows their method with a clear pronunciation of the algorithm.
Why study both methods? We conclude that you need to choose a convenient solution.
5 Primary consolidation
I offer two KZs to choose from and say that one is very difficult. I advise those who have not achieved the second goal on their own to take KZ No. 3 (a) - reflective. Those who are more confident in themselves, let them take UZ No. 3 (b)
UK No. 3 (a)-reflexive
That's better. The ability to use the most convenient method is a real skill. Look Pay attention to the expressions and terms in the sums. Look to solution algorithms. Choose for each example a convenient way and write it down it after the = sign (13+17):3= (24+27):3= Take a sample solution from your teacher and test yourself. Evaluate your work according to the following criteria: Correctly applied both methods and did not make any computational errors - “I accurately hit 2 targets” Correctly applied both methods, but made computational errors - “I hit the target, but almost missed” Correctly applied one method or none - “We still need to practice by learning the algorithms” |
UZ No. 3(b)-reflexive
6 Reflection
If you wish, I ask you to discuss the self-assessment of the work in the lesson from the point of view of achieving the goals of one of the children who completed CL No. 3 (a) and one of those who completed CL No. 3 (b)
7 D.Z. optionally.
Solve the number from the textbook to reinforce the solution methods.
Increased difficulty task (distributing cards)
What numbers can be inserted into the expression (___ + ___): ___ so that each of them is divisible by 2, and their sum is divisible by 2. Write down as many options as possible. Think about the pattern in the selection of these numbers.
On this lesson Students are given the opportunity to repeat tabular cases of multiplication and division, become familiar with the rule for dividing a sum by a number, and also practice performing various tasks on the topic of the lesson.
Read and compare the expressions written on the board.
(6 + 4) + 2
(6 + 4) - 2
(6 + 4) * 2
(6 + 4) : 2
You noticed that in each expression the sum of the numbers is 6 + 4.
Let's read the expressions.
(6 + 4) + 2
The sum of numbers 6 + 4 is increased by 2.
(6 + 4) - 2
The sum of numbers 6 + 4 is reduced by 2.
(6 + 4) * 2
The sum of the numbers 6 + 4 is doubled.
(6 + 4) : 2
The sum of numbers 6 + 4 is halved
Do you think the values of these amounts will be the same?
Let's check. Let's calculate the values of the expressions. Remember that we perform the first action in parentheses.
(6 + 4) + 2 = 12
(6 + 4) - 2 = 8
(6 + 4) * 2 = 20
(6 + 4) : 2 = 5
We got different values.
Let's look at how a sum can be divided by a number.
Rice. 1. Dividing a sum by a number
Method 1.
First we added up the blue and red squares, and then divided their number into two equal parts.
(6 + 4) : 2 = 10: 2 = 5
Method 2.
First, we can divide the blue squares into two equal parts, then divide the red squares into two equal parts, and then add the results.
(6 + 4) : 2 = 6: 2 + 4: 2 = 3 + 2 = 5
When performing actions different ways the result is the same. Therefore we can draw a conclusion.
To divide a sum by a number, you can divide each term by that number,
and add up the resulting quotients.
(6 + 4) : 2 = 6: 2 + 4: 2
Let's apply the acquired knowledge in practice. Let's calculate the values of the expressions.
(64 + 72) : 8
(36 + 81) : 9
(80 + 16) : 4
To divide the sum by a number, divide each term by this number, and add the resulting values of the quotients.
(64 + 72) : 8 = 64: 8 + 72: 8 = 8 + 9 = 17
(36 + 81) : 9 = 36: 9 + 81: 9 = 4 + 9 = 13
(80 + 16) : 4 = 80: 4 + 16: 4 = 20 + 4 = 24
Consider the expressions. What do they have in common?
(36 + 6) : 6
(10 + 32) : 6
(34 + 8) : 6
(24 + 18) : 6
Right. In each expression, you must divide the sum by the number 6.
Let's divide the expressions into two groups.
In the first we write down those expressions where we can apply the property of dividing a sum by a number. In these expressions, each term of the sum is divided by 6.
(36 + 6) : 6
(24 + 18) : 6
In the second group we will write expressions where the summands of the sum are not divisible by 6, this means that the property of dividing a sum by a number cannot be applied to them.
(10 + 32) : 6
(34 + 8) : 6
Let's complete the task.
Which of these numbers can be written as a sum of two terms, in which each of the terms is divisible by 7?
35, 43, 28, 14, 7, 47, 56, 49, 63, 26, 70
First, we write down the numbers that are divisible by the number 7 without a remainder.
35, 28, 14, 7, 56, 49, 63, 70
Let's make up expressions and find their meanings.
(35 + 28) : 7 = 35: 7 + 28: 7 = 5 + 4 = 9
(70 + 14) : 7 = 70: 7 + 14: 7 = 10 + 2 = 12
(56 + 49) : 7 = 56: 7 + 49: 7 = 8 + 7 = 15
Let's complete the following task.
Fill in the missing numbers using the rule of dividing the sum by the number.
(… + …) : 8 = 8 + 6
(… + …) : 9 = 9 + 5
(… + …) : 3 = 8 + 5
Let's think like this.
(… + …) : 8 = 8 + 6
The first term was divided by 8 and we got the number 8. So it was the number 64. The second term was divided by 8 and we got the number 6. So it was the number 48. Let's write down the solution.
(64 + 48) : 8 = 8 + 6
(… + …) : 9 = 9 + 5
The first term was divided by 9 and we got the number 9. So it was the number 81. The second term was divided by 9 and we got the number 5. So it was the number 45. Let's write down the solution.
(81 + 45) : 9 = 9 + 5
(… + …) : 3 = 8 + 5
The first term was divided by 3 and we got the number 8. So it was the number 24. The second term was divided by 3 and we got the number 5. So it was the number 15. Let's write down the solution.
(24 + 15) : 3 = 8 + 5
Today in class we learned about the rule of dividing a sum by a number and practiced solving examples on the topic of the lesson.
Bibliography
- M.I. Moreau, M.A. Bantova and others. Mathematics: Textbook. 3rd grade: in 2 parts, part 1. - M.: “Enlightenment”, 2012.
- M.I. Moreau, M.A. Bantova and others. Mathematics: Textbook. 3rd grade: in 2 parts, part 2. - M.: “Enlightenment”, 2012.
- M.I. Moro. Math lessons: Guidelines for the teacher. 3rd grade. - M.: Education, 2012.
- Regulatory document. Monitoring and evaluation of learning outcomes. - M.: “Enlightenment”, 2011.
- “School of Russia”: Programs for primary school. - M.: “Enlightenment”, 2011.
- S.I. Volkova. Mathematics: Test work. 3rd grade. - M.: Education, 2012.
- V.N. Rudnitskaya. Tests. - M.: “Exam”, 2012.
1. Property of dividing two equals natural numbers:
If a natural number is divided by its equal number, the result is one.
It remains to give a couple of examples. The quotient of the natural number 405 divided by its equal number 405 is 1; The result of dividing 73 by 73 is also 1.
2. Property of dividing a natural number by one:
The result of dividing a given natural number by one is that natural number.
Let us write down the formulated property of division in literal form: a: 1 = a.
Let's give examples. The quotient of the natural number 23 divided by 1 is the number 23, and the result of dividing the natural number 10,388 by one is the number 10,388.
3. Division of natural numbers does not have the commutative property.
If the dividend and the divisor are equal natural numbers, then due to the property of dividing equal natural numbers, discussed in the first paragraph of this article, we can swap them. In this case, the result of division will be the same natural number 1.
In other words, if the dividend and the divisor are equal natural numbers, then in this case division has the commutative property. 5: 5 = 1 and 5: 5 = 1
In other cases, when the dividend and the divisor are not equal natural numbers, the commutative property of division does not apply.
So, in general, division of natural numbers does NOT have the commutative property.
Using letters, the last statement is written as a: b ≠ b: a, where a and b are some natural numbers, and a ≠ b.
4. The property of dividing the sum of two natural numbers by a natural number:
dividing the sum of two natural numbers by a given natural number is the same as adding the quotients of dividing each term by a given natural number.
Let's write this property of division using letters. Let a, b and c be natural numbers such that a can be divided by c and b can be divided by c, then (a + b) : c = a: c + b: c. On the right side of the written equality, division is performed first, followed by addition.
Let us give an example that confirms the validity of the property of dividing the sum of two natural numbers by a given natural number. Let us show that the equality (18 + 36) : 6 = 18: 6 + 36: 6 is correct. First, let's calculate the value of the expression from the left side of the equality. Since 18 + 36 = 54, then (18 + 36) : 6 = 54: 6. From the multiplication table of natural numbers we find 54: 6 = 9. We proceed to calculating the value of the expression 18:6+36:6. From the multiplication table we have 18: 6 = 3 and 36: 6 = 6, therefore 18: 6 + 36: 6 = 3 + 6 = 9. Therefore, the equality (18 + 36) : 6 = 18: 6 + 36: 6 is correct .
5. The property of dividing the difference of two natural numbers by a natural number:
divide the difference of two numbers by given number- this is the same as subtracting from the quotient of the minuend and the given number the quotient of the subtrahend and the given number.
Using letters, this property of division can be written as follows: (a - b) : c = a: c - b: c, where a, b and c are natural numbers such that a is greater than or equal to b, and also both a and b can be divided by c.
As an example confirming the property of division under consideration, we will show the validity of the equality (45 - 25) : 5 = 45: 5 - 25: 5. Since 45 - 25 = 20 (if necessary, study the material in the article subtracting natural numbers), then (45 - 25) : 5 = 20: 5. Using the multiplication table, we find that the resulting quotient is equal to 4. Now let’s calculate the value of the expression 45: 5 - 25: 5, which is on the right side of the equality. From the multiplication table we have 45: 5 = 9 and 25: 5 = 5, then 45: 5 - 25: 5 = 9 - 5 = 4. Therefore, the equality (45 - 25) : 5 = 45: 5 - 25: 5 is true .
6. The property of dividing the product of two natural numbers by a natural number:
the result of dividing the product of two natural numbers by a given natural number that is equal to one of the factors is equal to the other factor.
Here is the literal form of this division property: (a · b) : a = b or (a · b) : b = a, where a and b are some natural numbers.
“Dividing multi-digit numbers into single-digit numbers” - The dividend is found like this: b) The number being divided by is called the divisor; a) The number that is divided is called a divisor; A) add a divisor to the quotient; If the digit of the incomplete dividend less than divisor, then in the quotient 0. Algorithm of actions. Which statement is true? c) The number that results from division is called a divisor.
“Minimum Subtractable Difference” - The tests are just beginning... Assignment: put them in ascending order. + = Difference - =. Sum. Let's ask the cunning fox to help Ivan Tsarevich find the chest. Who's ready to open the chest? Minuend. Difference. Who became Ivan's true friend? Addend addend sum difference minuend subtrahend. Presentation for a 1st grade math lesson.
“Division problems” - Create a problem and solve it. Decipher the puzzles: 10: 5 = 2 (z.). What figures does it consist of? 9: 3 = 3 (t.). Tribune. Gun. Arrange the signs of arithmetic operations: 12: 4 = 3 (w). Seven hundred. The specific meaning of the division action. Solve the problem. Fill in the empty cell. Catch some fish. Again. Mathematics class Moro M.I.
“Sum and difference of cubes” - Perform squaring. (2x – 1)2 (9 – n)2 (–3a + 5)2. Factor into: Present as a cube: 8x3 64c6 b12. Present in the form of a cube: 125у3 x3 а9b6 8n6y15. Factorization of the sum and difference of cubes.
“Multiplication and division of numbers” - 3. Indicate the number that will be obtained if 709 is increased by 61 times. Preparation for testing in mathematics. 1. Indicate the value of the product if the first factor is 6248, and the second is 9. 6. Indicate the number that must be inserted into the “window” so that the equality: 24 = 2003 becomes true. 9. Provide a correctly solved example. 5. Indicate the value of the product of the numbers 4379 and 8.
“Division by a two-digit number” - We’ll immediately get into a fairy tale, If we find the key. Geometric material. Consolidation of what has been learned. Division. Physical education minute. Continue working on developing the ability to perform written division by two-digit numbers. Problem solving. Target. 24x5. 149376:64. 38232:72. Hooray. To double digits. 36x4. Frontal work.
IN initial course Mathematics theorems on the divisibility of a sum are “represented” in the form of the book “Dividing a sum by a number.” This property is used in division double digit number to a single digit.
In the M2M textbook, the method of introducing children to this property is similar to the method of studying the property of multiplying a sum by a number. Namely: first, students analyze two ways to solve a problem, using a drawing for this purpose, then, using a specific example, two methods of action when dividing a sum by a number are explained, i.e., the case is considered when each term is divided by a given number.
Consider two ways to solve the example: (6+9):3 ;
Calculate the sum and divide the result by the number: (6+9):3=15:3=5;
Divide each term by a number, and then add the results: (6+9):3=6:3+9:3=2+3=5. Compare the results.
The new method of action is reinforced during the exercises: Clear the meaning of each expression in two ways: (10+4):2, (8+12):4, (12+15):3.
In the M2I textbook, a different methodological approach is used to introduce students to the property of dividing a sum by a number.
Students are given the following task: Guess! What is the rule for writing the expressions in each column? Calculate their values: 54:9 (36+18):9 36:9+18:9; 63:7 (49+14):7 49:7+14:7.
In the process of completing this task, students become aware of a new way of doing things. Namely: the dividend is represented as the sum of two terms, each of which is divided by a given number, then each term is divided by this number and the resulting results are added. To learn a new way of acting, various tasks are performed. Moreover, the expressions used in the tasks include only tabular cases of division, so students do not experience difficulties in applying the new method of action.
24. Methodology for introducing the concept of “equation”.
Numeric expression;
Expression with variable;
Equality and inequality;
The equation.
2) Reveal their content.
The concept of an equation is one of the basic algebraic concepts studied in a mathematics course in primary school. In elementary school, only 1st degree equations with one unknown are considered, and most methods recommend introducing children exclusively to the simplest equations.
The simplest equations are those in which to find the root it is enough to perform a single step. But according to some other methods, in addition to the indicated equations, it is recommended to introduce students to more complex equations such as:
The basis for solving an equation in elementary school is the connection between the components of arithmetic operations and their result.
Tasks facing the teacher:
Introduce students to the concept of an equation and its solution;
Develop a conscious skill in solving equations.
Preparatory work:
Offer elementary school students to solve the equation in implicit form, i.e. offer a record like:
Insert the missing number into the box to get the correct equation.
This task can be offered at various stages of education in primary school. Depending on the stage of learning at which these tasks are offered, students can act in 2 ways:
1. If children do not yet know the connections between the components of actions and their results, then they perform the specified tasks using the selection method. Those. Substitute different numbers into the window and check whether the equality is true.
2. If the specified tasks are offered when children are already familiar with the connections between the components of actions and their results, then they find them using this connection.
From the above, we can conclude that at the stage of preparing students to become familiar with the concept of an equation, they become familiar with the equation in implicit form and the method of solving equations by the selection method => 2nd method of solving equations - the selection method.
Ditto preparatory stage should include familiarizing primary school students with the components of various arithmetic operations, their results and the relationship between them. If students are not familiarized with these concepts at the proper level and the children do not consciously learn the rules for finding unknown terms, subtrahends, minuends, etc., then familiarization with solving an equation will not be performed at the proper level. Throughout the entire process of studying mathematics at the elementary level, until familiarization with the equation, it is necessary to carry out work aimed at developing in students solid skills in finding unknown components of arithmetic operations.
Introduction to the concept of equation.
Children are invited to record:
Then it is reported that in mathematics an unknown number is usually denoted by special letters, the main one of which is “ X».
and it is reported that the equality presented is called an equation. In order for children to form the concept of an equation, you need to offer a number of expressions:
Children must identify from the indicated objects those that are equations, explaining their choice. At the same time, they must indicate the essential properties of the equations (equality, there is X).
Along with the concept of “equation,” children develop an idea of what it means to solve an equation. They must fully understand the fact that solving an equation means finding a number that, when substituted into the equation for the unknown, turns the latter into a true numerical equality. The concept of “equation root” is not introduced, although certain techniques allow the introduction of this term (according to Elkonin-Davydov).
Already at the stage of studying the equation at the beginning, it is a good idea to engage in propaedeutics of the concept of “domain of definition of the equation.” This work is carried out especially effectively...
X-10=2 (9 is not possible, because...)
15:x=5 (you can’t use 5, because...)
When considering this kind of equations, it is concluded that not every number can be a solution to these equations.
In order for the work on studying equations to be effective, children need to be offered equations with a variety of tasks:
Solve the equation and test;
Check the equations being solved and find the error;
Make equations with numbers: x, 10, 12
12's=10, etc.
From the given equations, solve only those that can be solved using the subtraction action:
10's=8, etc.
Of the given equations, solve only those that can be solved by addition;
Children are given an equation in which the action sign is missing
and a solution was given
Special attention When considering a concept, the equation should be given verification. It is very important that when checking the solution of equations, students approach this work not formally, but consciously. To do this, they should be offered problem situations in which they need to perform specific actions to check solved equations, namely, offer an already solved equation and ask, without solving it, to establish whether an error has been made or not. To control the actions of students in this process, it is necessary to invite them to talk about their actions out loud.
25. Methodology for introducing the concept of “expression” (numeric expressions and expressions with a variable).
In elementary school mathematics courses, children are introduced to the following algebraic concepts:
Numeric expression;
Expression with variable;
Equality and inequality;
The equation.
Tasks facing the teacher:
1) To form an idea among students about these concepts.
2) Reveal their content.
NUMERICAL EXPRESSION.
Tasks:
2) Introduce the rules for the order of performing actions in expressions. Learn to use them in calculations.
3) Teach children to perform some identical transformations of expressions.
Students are introduced to the concept of a numerical expression from the first days of school with the introduction of one or another arithmetic operation.
Introducing elementary school children to the concept of addition: children are shown a numerical expression called a sum. The teacher must remember that the action sign placed between the numbers has a double meaning. On the one hand, it shows the actions that should be performed on numbers, and on the other hand, it shows the designation of a given numerical expression. Hence, the concept of “numerical expressions” is inextricably linked with the concept of “arithmetic operations” and in the formation of these concepts, one contributes to the formation of the other.
Familiarization with numerical expressions occurs gradually, and first students become familiar with the simplest expressions (with one action sign), and then with more complex expressions (2 or more actions). A very important stage is the stage of comparing expressions. By comparing expressions, children become familiar with concepts such as equality and inequality.
As expressions become more complex, in order to find their meanings, it becomes necessary to familiarize elementary school students with the rules for performing actions in expressions.
Getting to know these rules also happens gradually:
1) First, children become familiar with the rule for performing actions in an expression that includes actions of one level, and there are no parentheses.
2) Then students become familiar with the rules for performing actions in expressions with actions of the same step and parentheses.
3) Then - expressions with actions of different levels, but without parentheses.
4) Then - expressions with actions of two steps and brackets.
Familiarization with all the rules is as follows: The teacher informs that the children must remember.
In order for children to learn the rules introduced, they should be offered a variety of tasks:
1) Calculate the value of this expression, having previously indicated the procedure.
2) Arrange the brackets to get the correct equalities.
3) From the given pairs of examples, write down only those in which the calculations were performed according to the rules of the order of actions.
After explaining the errors, you can give the task: using brackets, change the expression so that it has the specified value.
4) Children are asked to indicate the order of actions in the following entries:
Particular attention when forming concepts numerical expressions should be addressed to children identity transformations(a transformation is identical if one expression produces another expression that is identically equal to it).
Identical transformations performed by elementary school students:
1) Replacing +, -, :, x with their values.
2) Rearrangement of terms.
3) Opening the brackets.
All identical transformations that elementary school students perform are based on the rules for performing operations on numbers and the properties of certain arithmetic operations (commutative, associative, distributive, the rule for multiplying a sum by a number, the rule for subtracting a sum from a number, operations with 0 and 1, etc. .d.)
When studying each property, students make sure that in the expressions certain type you can perform actions in different ways, but the meanings of the expressions will not change.
In the future, students use certain properties for identical transformations of expressions.
1) the student reads the expression;
2) remembers the corresponding property;
3) based on this property, it transforms the expression.
In order to ensure that the transformations are correct, students are advised to find the meaning of the same expression in another way.
If the resulting value matches the first one, then the conversion was performed correctly.
To develop mathematical speech and consciously carry out transformations, it is necessary to invite children to give an explanation of the actions performed.
EXPRESSION WITH VARIABLE.
Tasks:
1) Give an idea of expressions containing a variable.
2) Learn to find the value of an expression for different values of a variable.
When studying mathematics in primary school, students are exposed to expressions with variables at various stages. Introducing and working with these mathematical concepts allows students to generalize the concept of expression.
A good preparation is a task where the variable is presented in an implicit form (empty window, dots)
For example: 3+
Insert each of the following numbers 1, 2, 3 into the box and find the sum.
Gradually, children are led to the idea that in mathematics, instead of a missing number, you can write a letter, and, giving the letter certain meanings, get different meanings expressions.
Also, values with variables are used when becoming familiar with formulas for finding perimeter and area.
It should be noted that the amount of knowledge students acquire on this topic differs depending on the mathematics textbook.
For example:
Peterson, Istomina, Aleksandrova – the scope and content of expressions with a variable are significantly expanded and are actively used (formation of the properties of arithmetic operations in students)
