Long way to develop skills solving equations starts with solving the very first and relatively simple equations. By such equations we mean equations, on the left side of which is the sum, difference, product or quotient of two numbers, one of which is unknown, and on the right side there is a number. That is, these equations contain an unknown term, minuend, subtrahend, multiplier, dividend, or divisor. The solution of such equations will be discussed in this article.

Here we will give the rules that allow us to find an unknown term, multiplier, etc. Moreover, we will immediately consider the application of these rules in practice, solving characteristic equations.

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So, we substitute the number 5 instead of x in the original equation 3 + x = 8, we get 3 + 5 = 8 - this equality is correct, therefore, we correctly found the unknown term. If during the check we received an incorrect numerical equality, then this would indicate to us that we incorrectly solved the equation. The main reasons for this may be either the application of the wrong rule, or computational errors.

How to find the unknown minuend, subtrahend?

The connection between addition and subtraction of numbers, which we already mentioned in the previous paragraph, allows us to obtain a rule for finding an unknown minuend through a known subtrahend and difference, as well as a rule for finding an unknown subtrahend through a known minuend and difference. We will formulate them in turn, and immediately give the solution of the corresponding equations.

To find the unknown minuend, you need to add the subtrahend to the difference.

For example, consider the equation x−2=5 . It contains an unknown minuend. The above rule tells us that in order to find it, we must add the known subtrahend 2 to the known difference 5, we have 5+2=7. Thus, the required minuend is equal to seven.

If you omit the explanations, then the solution is written as follows:
x−2=5 ,
x=5+2 ,
x=7 .

For self-control, we will perform a check. We substitute the found reduced into the original equation, and we obtain the numerical equality 7−2=5. It is correct, therefore, we can be sure that we have correctly determined the value of the unknown minuend.

You can move on to finding the unknown subtrahend. It is found by adding according to the following rule: to find the unknown subtrahend, it is necessary to subtract the difference from the minuend.

We solve an equation of the form 9−x=4 using the written rule. In this equation, the unknown is the subtrahend. To find it, we need to subtract the known difference 4 from the known reduced 9 , we have 9−4=5 . Thus, the required subtrahend is equal to five.

Here is a short version of the solution to this equation:
9−x=4 ,
x=9−4 ,
x=5 .

It remains only to check the correctness of the found subtrahend. Let's make a check, for which we substitute the found value 5 instead of x into the original equation, and we get the numerical equality 9−5=4. It is correct, therefore the value of the subtrahend that we found is correct.

And before moving on to the next rule, we note that in the 6th grade, a rule for solving equations is considered, which allows you to transfer any term from one part of the equation to another with the opposite sign. So, all the rules considered above for finding an unknown term, reduced and subtracted, are fully consistent with it.

To find the unknown factor, you need to...

Let's take a look at the equations x 3=12 and 2 y=6 . In them, the unknown number is the factor on the left side, and the product and the second factor are known. To find the unknown factor, you can use the following rule: to find the unknown factor, you need to divide the product by the known factor.

This rule is based on the fact that we gave the division of numbers a meaning opposite to the meaning of multiplication. That is, there is a connection between multiplication and division: from the equality a b=c , in which a≠0 and b≠0, it follows that c:a=b and c:b=c , and vice versa.

For example, let's find the unknown factor of the equation x·3=12 . According to the rule, we need to divide the known product 12 by the known factor 3. Let's do : 12:3=4 . So the unknown factor is 4 .

Briefly, the solution of the equation is written as a sequence of equalities:
x 3=12 ,
x=12:3 ,
x=4 .

It is also desirable to check the result: we substitute the found value instead of the letter in the original equation, we get 4 3 \u003d 12 - the correct numerical equality, so we correctly found the value of the unknown factor.

And one more thing: acting according to the studied rule, we actually perform the division of both parts of the equation by a non-zero known multiplier. In grade 6, it will be said that both parts of the equation can be multiplied and divided by the same non-zero number, this does not affect the roots of the equation.

How to find the unknown dividend, divisor?

As part of our topic, it remains to figure out how to find the unknown dividend with a known divisor and quotient, as well as how to find an unknown divisor with a known dividend and quotient. The relationship between multiplication and division already mentioned in the previous paragraph allows you to answer these questions.

To find the unknown dividend, you need to multiply the quotient by the divisor.

Let's consider its application with an example. Solve the equation x:5=9 . To find the unknown divisible of this equation, according to the rule, it is necessary to multiply the known quotient 9 by the known divisor 5, that is, we perform the multiplication of natural numbers: 9 5 \u003d 45. Thus, the desired dividend is 45.

Let's show a short notation of the solution:
x:5=9 ,
x=9 5 ,
x=45 .

The check confirms that the value of the unknown dividend is found correctly. Indeed, when substituting the number 45 into the original equation instead of the variable x, it turns into the correct numerical equality 45:5=9.

Note that the analyzed rule can be interpreted as the multiplication of both parts of the equation by a known divisor. Such a transformation does not affect the roots of the equation.

Let's move on to the rule for finding the unknown divisor: to find the unknown divisor, divide the dividend by the quotient.

Consider an example. Find the unknown divisor from equation 18:x=3 . To do this, we need to divide the known dividend 18 by the known quotient 3, we have 18:3=6. Thus, the required divisor is equal to six.

The solution can also be formulated as follows:
18:x=3 ,
x=18:3 ,
x=6 .

Let's check this result for reliability: 18:6=3 is the correct numerical equality, therefore, the root of the equation is found correctly.

It is clear that this rule can only be applied when the quotient is different from zero, so as not to encounter division by zero. When the quotient is zero, two cases are possible. If in this case the dividend is equal to zero, that is, the equation has the form 0:x=0 , then this equation satisfies any non-zero value of the divisor. In other words, the roots of such an equation are any numbers that are not equal to zero. If, when the quotient is equal to zero, the dividend is different from zero, then for any values ​​​​of the divisor, the original equation does not turn into a true numerical equality, that is, the equation has no roots. To illustrate, we present the equation 5:x=0 , it has no solutions.

Sharing Rules

Consistent application of the rules for finding the unknown term, minuend, subtrahend, multiplier, dividend and divisor allows solving equations with a single variable of a more complex form. Let's deal with this with an example.

Consider the equation 3 x+1=7 . First, we can find the unknown term 3 x , for this we need to subtract the known term 1 from the sum 7, we get 3 x=7−1 and then 3 x=6 . Now it remains to find the unknown factor by dividing the product of 6 by the known factor 3 , we have x=6:3 , whence x=2 . So the root of the original equation is found.

To consolidate the material, we present a brief solution of another equation (2·x−7):3−5=2 .
(2 x−7):3−5=2 ,
(2 x−7):3=2+5 ,
(2 x−7):3=7 ,
2 x−7=7 3 ,
2x−7=21 ,
2x=21+7 ,
2x=28 ,
x=28:2 ,
x=14 .

Bibliography.

  • Mathematics.. 4th grade. Proc. for general education institutions. At 2 o'clock, Part 1 / [M. I. Moro, M. A. Bantova, G. V. Beltyukova and others]. - 8th ed. - M.: Education, 2011. - 112 p.: ill. - (School of Russia). - ISBN 978-5-09-023769-7.
  • Mathematics: studies. for 5 cells. general education institutions / N. Ya. Vilenkin, V. I. Zhokhov, A. S. Chesnokov, S. I. Shvartsburd. - 21st ed., erased. - M.: Mnemosyne, 2007. - 280 p.: ill. ISBN 5-346-00699-0.

§ 1 How to find the unknown term

How to find the root of an equation if one of the terms is unknown? In this lesson, we will consider a method for solving equations based on the relationship between the terms and the value of the sum.

Let's solve this problem.

There were 6 red tulips and 3 yellow ones in the flower bed. How many tulips grew in the flower bed? Let's write down the solution. So, 6 red and 3 yellow tulips grew, therefore, we can write down the expression 6 + 3, after completing the addition, we get the result - 9 tulips grew in the flower bed.

Let's write down the solution. So, 6 red and 3 yellow tulips grew, therefore, we can write down the expression 6 + 3, after completing the addition, we get the result - 9 tulips grew in the flower bed. 6 + 3 = 9.

Let's change the condition of the problem. 9 tulips grew in the flower bed, 6 were plucked. How many tulips are left?

To find out how many tulips are left in the flower bed, you need to subtract the plucked flowers from the total number of 9 tulips, there are 6 of them.

Let's do the calculations: 9-6 we get the result 3. There are 3 tulips left in the flower bed.

Let's change this problem again. 9 tulips grew, 3 were plucked. How many tulips are left?

The solution will look like this: from the total number of tulips 9, it is necessary to subtract the plucked flowers, there are 3 of them. There are 6 tulips left.

Let's take a closer look at the equalities and try to figure out how they are related.

As you can see, these equalities contain the same numbers and reciprocal actions: addition and subtraction.

Let's return to the solution of the first problem and consider the expression 6 + 3 = 9.

Let's remember what the numbers are called when adding:

6 is the first term

3 - second term

9 - sum value

Now let's think about how we got the differences 9 - 6 = 3 and 9 - 3 = 6?

In the equation 9 - 6 = 3, the first term 6 was subtracted from the value of the sum 9, and the second term 3 was obtained.

In the equation 9 - 3 = 6, the second term3 was subtracted from the value of the sum9, and the first term6 was obtained.

Therefore, if the first term is subtracted from the value of the sum, then the second term will be obtained, and if the second term is subtracted from the value of the sum, then the first term will be obtained.

Let's formulate a general rule:

To find the unknown term, you need to subtract the known term from the value of the sum.

§ 2 Examples of solving equations with an unknown term

Let's look at equations with unknown terms and try to find roots using this rule.

Let's solve the equation X + 5 = 7.

The first term in this equation is unknown. To find it, we use the rule: to find the unknown first term X, it is necessary to subtract the second term 5 from the value of the sum 7.

So X \u003d 7 - 5,

find the difference 7 - 5 \u003d 2, X \u003d 2.

Let's check if we have found the root of the equation correctly. To carry out the check, it is necessary to substitute the number 2 in the equation instead of X:

7 = 7 - got the correct equality. We conclude: the number 2 is the root of the equation X+5=7.

Let's solve one more equation 8 + Y =17.

The second term in this equation is unknown.

To find it, it is necessary to subtract the first term 8 from the value of the sum 17.

Let's check: we substitute the number 9 instead of Y. We get:

17 = 17 - got the correct equality.

Therefore, the number 9 is the root of the equation 8 + Y = 17.

So, in the lesson we got acquainted with the method of solving equations based on the relationship between the terms and the value of the sum. To find the unknown term, you need to subtract the known term from the value of the sum.

List of used literature:

  1. I.I. Arginskaya, E.I. Ivanovskaya, S.N. Kormishin. Mathematics: Textbook for Grade 2: At 2h. - Samara: Educational Literature Publishing House: Fedorov Publishing House, 2012.
  2. Arginskaya I.I. Collection of tasks in mathematics for independent, verification and control work in elementary school. - Samara: Fedorov Corporation, Educational Literature Publishing House, 2006.

Used images:

Lesson 80-81. Topic: "Solving Equations"

Goals: learn to solve equations with an unknown term; repeat the ratio of units of length; to consolidate the skills of calculation in a column; develop the ability to reason and think logically.

Planned results: students will learn how to solve equations for finding an unknown term; perform written calculations using the techniques learned; understand the reasons for the success / failure of educational activities.

During the classes

I . Organizing time

II . Actualization of knowledge

Mathematical dictation

1. How much is 67 less than 89? (On 22.)

2. Subtract 4 tens from 7 tens. (30.)

3. Increase 23 by 32. (55.)

4. What number did I decrease by 27 to get 23? (50.)

5. By how much should 43 be increased to get 70? (On 27.)

6. Subtract 10 from the sum of numbers 9 and 6. (5.)

7. What number must be subtracted from 64 to get 37? (27.)

8. To what number did they add 0 and get 44? (44.)

9. To 21 add the difference between the numbers 14 and 6. (29.) 10. The sum of the numbers 33, 16.4 and 27. (80.)

(Verification. Self-assessment.)

III . Self-determination to activity

Make three more examples using this example. 6+4=10

(The teacher writes examples on the blackboard.) 4 + 6=10 10-4 = 6 10-6 = 4

What rule did you apply when composing the overlay example? (The sum does not change from a rearrangement of the terms.)

What rule did you apply when composing the subtraction example? (If one term is subtracted from the sum, then another term is obtained.)

- To find out the topic of the lesson, solve the crossword puzzle.

1. They are numeric and alphabetic. (Expressions.)

2. The numbers that add up are called. (Terms.)

3. The number to be subtracted from. (Minuend.)

4. Mathematical sign of subtraction. (Minus.)

5. An equality that contains an unknown number. (The equation.)

6. The sum of the lengths of the sides of the figure. (Perimeter.)

7. Expression with a plus sign. (Sum.)

8. A record containing an equals sign. (Equality.)

9. The smallest two-digit number. (Ten.) 10. Latin letter. (X.)

What happened in the highlighted line? (Solution of equations.)

Lesson topic: "Solving equations with an unknown term." What tasks will we set for ourselves?

IV . Work on the topic of the lesson

1. Work according to the textbook

Look at the dominoes on p. 7 textbook and examples written side by side. How are subtraction examples obtained? What rule did you use to make them? Finish the conclusion. ( To find the unknown term, subtract the known term from the sum.)

№ 1 (p. 7).(Oral performance.)

№2 (p. 7).(Collective execution with detailed explanation.)

2. Independent solution of equations

Option 1 Option 2

x + 45 = 92 75+x = 81

26 + x = 50 x + 22 = 70

(Two students write the solution on a flip board. Check. Self-assessment.)

Solution:

x + 45 = 92 75 + x = 81

x = 92-45 x = 81-75

x = 47 X= 6

26+x=50x+22=70

x = 50 – 26 x = 70 - 22

3. Work according to the textbook

№3(p. 7).(Oral performance.)

№4 (p. 7). (Independent implementation. For those who have difficulty, the teacher gives an assistant card with a solution program.) 1) How many glasses of raspberries did your sister collect?

2) How many glasses of raspberries did you collect together? (Check. Self-assessment.)

V . Physical education minute

I go and you go - one, two, three. (Steps in place.)

I sing and you sing - one, two, three. (Clap hands.)

We go and sing - one, two, three. (Jumping in place.)

We live very friendly - one, two, three. (Steps in place.)

VI . Consolidation of the studied material

Textbook workNo. 1 (p. 14).

What units of length do you know?

How many millimeters are in 1 cm? (Self-execution. Verification.) Solution:

5 cm 3 mm = 53 mm

3 cm 8 mm = 38 mmNo. 2 (p. 14).

(Self-execution. Verification.)

1) Solution:

AB = 3 cm 5 mm, CD= 5 cm 5 mm;

5 cm 5 mm - 3 cm 5 mm = 2 cm.

Answer: segment length CD 2 cm more than the length of the segment AB.

2) Solution: ECMO= 2 cm + 4 cm + 1 cm 5 mm = 7 cm 5 mm. No. 3 (p. 14).

(Self-execution. Verification. Self-assessment.)

Solution:

2 cm = 20 mm

4 cm 2 mm > 40 mm 30 mm = 3 cm

4 cm 5 mm < 5 cm

VII . Reflection

(“Test yourself” (textbook, p. 7). Self-fulfillment. Check.)

Solution: 15+x = 35 x = 35-15 x = 20

VIII . Summing up the lesson

What kind of equations did you remember today?

How to find the unknown term?

Who needs help?

Homework: Workbook: No. 10, 11 (p. 6).

To learn how to solve equations quickly and successfully, you need to start with the simplest rules and examples. First of all, you need to learn how to solve equations, on the left of which is the difference, sum, quotient or product of some numbers with one unknown, and on the right is another number. In other words, in these equations there is one unknown term and either the minuend with the subtrahend, or the divisible with a divisor, etc. It is about equations of this type that we will talk with you.

This article is devoted to the basic rules that allow you to find factors, unknown terms, etc. We will immediately explain all the theoretical provisions with specific examples.

Finding the unknown term

Let's say we have some number of balls in two vases, say 9 . We know that there are 4 marbles in the second vase. How to find the quantity in the second? Let's write this problem in mathematical form, denoting the number to be found as x. According to the original condition, this number together with 4 form 9, so we can write the equation 4 + x = 9. On the left, we got a sum with one unknown term, on the right, the value of this sum. How to find x? To do this, you need to use the rule:

Definition 1

To find the unknown term, subtract the known from the sum.

In this case, we give subtraction a meaning that is the opposite of addition. In other words, there is a certain connection between the operations of addition and subtraction, which can be expressed in literal form as follows: if a + b \u003d c, then c - a \u003d b and c - b \u003d a, and vice versa, from the expressions c - a \u003d b and c − b = a we can deduce that a + b = c .

Knowing this rule, we can find one unknown term using the known and the sum. Which term we know, the first or the second, is not important in this case. Let's see how to apply this rule in practice.

Example 1

Let's take the equation that we got above: 4 + x = 9. According to the rule, we need to subtract from the known sum, equal to 9, the known term, equal to 4. Subtract one natural number from another: 9 - 4 = 5 . We got the term we need, equal to 5.

Typically, solutions to such equations are written as follows:

  1. The original equation is written first.
  2. Next, we write down the equation that we got after we applied the rule for calculating the unknown term.
  3. After that, we write the equation that turned out after all the actions with numbers.

This form of writing is needed in order to illustrate the successive replacement of the original equation with equivalent ones and to display the process of finding the root. The solution to our simple equation above would be correctly written as:

4 + x = 9 , x = 9 − 4 , x = 5 .

We can check the correctness of the received answer. Let's substitute what we got into the original equation and see if the correct numerical equality comes out of it. Substitute 5 into 4 + x = 9 and get: 4 + 5 = 9 . The equality 9 = 9 is correct, which means that the unknown term was found correctly. If the equality turned out to be wrong, then we should go back to the solution and double-check it, since this is a sign of a mistake. As a rule, most often this is a computational error or the application of an incorrect rule.

Finding the unknown subtrahend or minuend

As we mentioned in the first paragraph, there is a certain relationship between the processes of addition and subtraction. With its help, you can formulate a rule that will help you find the unknown minuend when we know the difference and the subtrahend, or the unknown subtrahend through the minuend or the difference. We write these two rules in turn and show how to apply them to solve problems.

Definition 2

To find the unknown minuend, add the minuend to the difference.

Example 2

For example, we have an equation x - 6 = 10 . Reduced unknown. According to the rule, we need to add the subtracted 6 to the difference 10, we get 16. That is, the original minuend is sixteen. Let's write the solution in its entirety:

x − 6 = 10 , x = 10 + 6 , x = 16 .

Let's check the result by adding the resulting number to the original equation: 16 - 6 = 10. Equality 16 - 16 will be correct, which means that we have calculated everything correctly.

Definition 3

To find the unknown subtrahend, subtract the difference from the minuend.

Example 3

Let's use the rule to solve the equation 10 - x = 8 . We do not know what is being subtracted, so we need to subtract the difference from 10, i.e. 10 - 8 = 2. Hence, the required subtrahend is equal to two. Here is the entire solution entry:

10 - x = 8 , x = 10 - 8 , x = 2 .

Let's check for correctness by substituting a deuce in the original equation. Let's get the correct equality 10 - 2 = 8 and make sure that the value we found will be correct.

Before moving on to other rules, we note that there is a rule for transferring any terms from one part of the equation to another with the sign reversed. All of the above rules are fully consistent with it.

Finding the unknown multiplier

Let's look at two equations: x 2 = 20 and 3 x = 12. In both, we know the value of the product and one of the factors, we need to find the second. To do this, we need to use another rule.

Definition 4

To find the unknown factor, you need to divide the product by the known factor.

This rule is based on a sense that is the opposite of multiplication. There is the following relationship between multiplication and division: a b = c when a and b are not equal to 0, c: a = b, c: b = c and vice versa.

Example 4

Calculate the unknown factor in the first equation by dividing the known quotient 20 by the known factor 2 . We carry out the division of natural numbers and get 10. Let's write down the sequence of equalities:

x 2 = 20 x = 20: 2 x = 10 .

We substitute the ten in the original equality and we get that 2 10 \u003d 20. The value of the unknown multiplier was done correctly.

Let us clarify that if one of the factors is zero, this rule cannot be applied. So, we cannot solve the equation x 0 = 11 with its help. This notation doesn't make sense because the solution is to divide 11 by 0 , and division by zero is undefined. We talked about such cases in more detail in the article devoted to linear equations.

When we apply this rule, we are essentially dividing both sides of the equation by a different factor than 0 . There is a separate rule according to which such a division can be carried out, and it will not affect the roots of the equation, and what we wrote about in this paragraph is fully consistent with it.

Finding an unknown dividend or divisor

Another case we need to consider is finding the unknown dividend if we know the divisor and the quotient, and also finding the divisor when the quotient and the dividend are known. We can formulate this rule with the help of the connection between multiplication and division already mentioned here.

Definition 5

To find the unknown dividend, multiply the divisor by the quotient.

Let's see how this rule applies.

Example 5

Let's use it to solve the equation x: 3 = 5 . We multiply the known quotient and the known divisor among ourselves and get 15, which will be the divisible we need.

Here is a summary of the entire solution:

x: 3 = 5, x = 3 5, x = 15.

The check shows that we calculated everything correctly, because when dividing 15 by 3, it really turns out 5. True numerical equality is evidence of the correct decision.

This rule can be interpreted as multiplying the right and left sides of the equation by the same number other than 0. This transformation does not affect the roots of the equation in any way.

Let's move on to the next rule.

Definition 6

To find the unknown divisor, you need to divide the dividend by the quotient.

Example 6

Let's take a simple example - Equation 21: x = 3 . To solve it, we divide the known divisible 21 by the quotient 3 and get 7. This will be the desired divisor. Now we make the decision correctly:

21:x=3, x=21:3, x=7.

Let's make sure the result is correct by substituting the seven in the original equation. 21: 7 = 3, so the root of the equation was calculated correctly.

It is important to note that this rule only applies when the quotient is non-zero, otherwise we would again have to divide by 0 . If the quotient is zero, two options are possible. If the dividend is also zero and the equation looks like 0: x \u003d 0, then the value of the variable will be any, that is, this equation has an infinite number of roots. But an equation with a quotient equal to 0, with a dividend other than 0, will not have solutions, since there are no such divisor values. An example would be equation 5: x = 0, which does not have any root.

Consistent application of rules

Often in practice there are more complex problems in which the rules for finding terms, minuends, subtrahends, factors, dividends and quotients must be applied sequentially. Let's take an example.

Example 7

We have an equation like 3 x + 1 = 7 . We calculate the unknown term 3 x , subtracting one from 7. We end up with 3 · x = 7 − 1 , then 3 · x = 6 . This equation is very easy to solve: divide 6 by 3 and get the root of the original equation.

Here is a shorthand for solving yet another equation (2 x − 7): 3 − 5 = 2:

(2 x - 7) : 3 - 5 = 2 , (2 x - 7) : 3 = 2 + 5 , (2 x - 7) : 3 = 7 , 2 x - 7 = 7 3 , 2 x − 7 = 21 , 2 x = 21 + 7 , 2 x = 28 , x = 28: 2 , x = 14 .

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Learning objectives- Solve equations using the selection method and on the basis of the connection between addition and subtraction.

Lesson Objectives

All students will be able to:
find the root of an equation by fitting

Most students will be able to:
be able to write and solve simple equations to find an unknown term

Some students will be able to:
draw up and solve equations on the basis of the drawing.

Previous Knowledge: understanding of the number system within 100; ability to make comparisons and use the language of comparison.

During the classes

Creating a collaborative environment
(psychological moments)

A cheerful bell rang.
Are you ready to start the lesson?
Let's listen, discuss
And help each other!

Grouping

Target: the grouping of students into groups increases the cognitive interest in the lesson, cohesion to work in a group.
Repetition of the group work rule

Actualization of life experience

Brainstorming Strategy Using thick and thin questions.
- What is an equation? (Equality with the unknown is called an equation)
What is the unknown in an equation?
What does it mean to solve an equation? (Means to find the unknown)
- What are the components of addition?

Grading: Three Claps
Starter "View video" (educational cartoon)
Method "Freeze Frame!"

Goal setting for the lesson
- Have you guessed what we will do at the lesson today?
- What will help us achieve the goals of the lesson (learn new things, learn how to solve such mathematical records) (your experience, teacher, textbook)
Children formulate the purpose of the lesson, I summarize.
- Today in the lesson you will learn how to solve equations with unknown terms

Study. Textbook work.
Target: Explore textbook material. 46

Task 1. Game according to the textbook "Cars in the tunnel"
Group work. Strategy "Think, Discuss, Share". Interdisciplinary communication literacy training (listening and speaking)

Game "Cars in the tunnel"

How many cars are in the tunnel?
6 + x = 18 and 2 + x = 14.
Answer: 12 wagons.

Descriptor:
- makes an equation according to the drawing
- finds the value of the letter by selection.
- makes a conclusion (formulate a rule)

Feedback "Traffic light"
Here I am using an equation simulation with the goal
formation of the ability to solve equations with an unknown term.

Task 2. Work in pairs. "Help the Hero"

Game "Help the hero"

For pair work, I use collaborative learning, which transfers knowledge and skills between students.
Self-assessment by descriptor: "Thumb"

Dynamic pause. Musical exercise.

Task 3. Work in a group. "Think, find a couple, share!"

Descriptors:
- the whole group works;
- composes and solves independently equations based on the drawing;
- makes a conclusion (formulate a rule).

Feedback "Wheel"
Application (teacher - observes, helps, checks, student - solves questions, demonstrates knowledge)

Peer review on slides
Here I use group work to improve the learning process.

Task 4. Playing in a pair of "Cube" (try it)

Group work: "Think, find a couple, share!"

Descriptor:
- substitutes the dropped number
- Solve the equation.

Here I use an active method in a playful way that leads to a deeper understanding of the solution of an equation with an unknown term.
Evaluation by descriptors "Traffic light"

Task 5. Individual task
differentiated tasks.
Tasks are selected for students with different levels of knowledge.

Descriptor:

  1. finds the root of the equation by the number ray;
  2. finds the root of the equation using mathematical numbers and signs;
  3. draws up an equation from the picture.

Self-assessment "Traffic light" (checking according to the standard).
- You guys did a great job with this!
Here I use a differentiated approach for the individual learning needs of each student.

Summary of the lesson. Reflection "Method" Interview "
What did we work on today in class?
How to find the unknown term?
What is the unknown term? (Part)
- Have you achieved your goal?
- What will those guys who experienced difficulties when working with equations do? (Students' statements)

Target: the teacher will find out if the students understood the topic of the lesson and their miscalculations in order to eliminate them in the next lesson. (students' statement) (here I use more satisfactorily the needs of the students)
Mutual evaluation "2 stars, 1 wish"

Reflection "Ladder of success" (children place emoticons)
- I can solve an equation with an unknown term.
- I can teach another...
- I'm having a hard time...
- I did not get anything …

Target: self-assessment of their achievements for the lesson.

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