Presentation for elementary grades "solving equations, finding unknown factors." Video lesson “Solving equations based on the relationship between the terms and the sum

Learning Objectives- solve equations using the method of selection and based on the connection between addition and subtraction.

Lesson Objectives

All students will be able to:
find the root of an equation using the selection method

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:
Based on the drawing, compose and solve equations independently.

Previous knowledge: understanding the system of numbers within 100; ability to make comparisons and use comparative language.

During the classes

Creating a collaborative environment
(psychological minutes)

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

Grouping

Target: uniting students into groups increases cognitive interest in the lesson and cohesion in group work.
Reviewing the rules for working in groups

Updating life experience

Strategy " Brainstorm"Using thick and subtle issue.
- What is an equation? (Equality with an unknown is called an equation)
- How is the unknown indicated in the equation?
- What does it mean to solve an equation? (Means to find the unknown)
- What are the components of addition?

Rating: Three claps
Starter "Watch a video" (educational cartoon)
The "Freeze Frame" method

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

Study. Work according to the textbook.
Target: Research the textbook material p. 46

Task 1. Game based on the textbook "Cars in the tunnel"
Group work. “Think, discuss, share” strategy. Interdisciplinary connection teaching literacy (listening and speaking)

Game "Cars in the tunnel"

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

Descriptor:
- makes up an equation based on the drawing
- finds the meaning of a letter using the selection method.
- draws a conclusion (formulates a rule)

Feedback "Traffic light"
Here I am using equation modeling with the purpose
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 cooperative learning that transfers knowledge and skills between students.
Self-assessment by descriptor: "Thumb"

Dynamic pause. Musical physical exercise.

Task 3. Group work. "Think, find a pair, share!"

Descriptors:
- the whole group works;
- composes and solves equations independently based on the drawing;
- draws a conclusion (formulates 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. Game in pairs "Cube" (try it)

Group work: “Think, find a pair, share!”

Descriptor:
- substitutes the drawn number
- solves the equation independently.

Here I am using active method in game form which leads to a deeper understanding of the solution to an equation with an unknown term.
Assessment based on traffic light descriptors

Task 5. Individual task
Differentiated tasks.
The tasks are selected for students with at different levels knowledge.

Descriptor:

1. finds the root of an equation using a number line;
2. finds the root of an equation using mathematical numbers and signs;
3. makes up an equation from the picture.

Self-assessment "Traffic Light" (test against standard).
- Well done, you completed this task!
Here I use a differentiated approach to individual learning needs for each student.

Lesson summary. Reflection "Interview Method"
- What did we work on in class today?
- How to find unknown term?
- What is the unknown term? (Part)
- Have you achieved your goal?
- What will those guys who had difficulty working with equations do? (Student statements)

Target: The teacher will find out whether the students understood the topic of the lesson and their mistakes so that they can be corrected in the next lesson. (students' statement) (here I use the students' needs more satisfactorily)
Peer evaluation "2 stars, 1 wish"

Reflection “Ladder of success” (children post emoticons)
- I can solve an equation with an unknown term.
- I can teach someone else...
- I find it difficult to...
- I did not get anything …

Target: self-assessment of your achievements during the lesson.

To download material or!

Lesson 80-81. Topic: “Solving Equations”

Goals: learn to solve equations with unknown terms; repeat the ratio of units of length; consolidate calculation skills in a column; develop reasoning and logical thinking skills.

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

During the classes

I . Organizing time

II . Updating knowledge

Mathematical dictation

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

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

3. Increase 23 by 32. (55.)

4. What number did I reduce by 27 and get 23? (50.)

5. How much should you increase 43 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 you add 0 and get 44? (44.)

9. To 21 add the difference between the numbers 14 and 6. (29.) 10. Sum of numbers 33, 16,4 and 27. (80.)

(Check. Self-assessment.)

III . Self-determination for activity

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

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

What rule did you apply when creating the overlay example? (The sum does not change by rearranging the terms.)

What rule did you use when creating the subtraction example? (If you subtract one term from the sum, you get another term.)

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

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

2. The numbers that are added are called. (Additions.)

3. The number from which to subtract. (Minuend.)

4. Math sign subtraction. (Minus.)

5. 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. An entry that contains an equal sign. (Equality.)

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

What happened in the highlighted line? (Solving 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 textbooks and examples recorded side by side. How are subtraction examples obtained? What rule did you use to compile them? Finish the conclusion. ( To find the unknown term, you need to subtract the known term from the sum.)

№ 1 (p. 7).(Oral performance.)

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

2. Independent solution equations

Option 1 Option 2

x + 45 = 92 75+x = 81

26+x = 50 x + 22 = 70

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

Solution:

x + 45 = 92 75 + x = 81

x = 92-45 x = 81-75

x = 47 X= 6

26+x=50 x + 22 = 70

x = 50 – 26 x = 70 - 22

3. Work according to the textbook

№3 (p. 7).(Oral performance.)

№4 (p. 7). (Self-execution For those who have difficulties, the teacher gives a help card with a solution program.) 1) How many glasses of raspberries did the sister collect?

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

V . Physical education minute

I'm walking and you're walking - one, two, three. (Steps in place.)

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

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

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

VI . Reinforcing the material learned

Working from the textbookNo. 1 (p. 14).

What units of length do you know?

How many millimeters are in 1 cm? (Independent execution. Check.) Solution:

5 cm 3 mm = 53 mm

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

(Independent execution. Check.)

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).

(Independent implementation. Checking. 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). Independent implementation. Test.)

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

VIII . Summing up the lesson

What type of equations did you remember today?

How to find an unknown term?

Who needs help?

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

Equations are one of difficult topics easy to learn, but at the same time they are a powerful enough tool for solving most problems.

Using equations, various processes occurring in nature are described. Equations are widely used in other sciences: economics, physics, biology and chemistry.

IN this lesson We will try to understand the essence of the simplest equations, learn to express unknowns and solve several equations. As you learn new materials, the equations will become more complex, so understanding the basics is very important.

Preliminary Skills Lesson content

What is an equation?

An equation is an equality that contains a variable whose value you want to find. This value must be such that when substituted into the original equation, the correct numerical equality is obtained.

For example, the expression 2 + 2 = 4 is an equality. When calculating the left side, the correct numerical equality is obtained 4 = 4.

But the equality is 2 + x= 4 is an equation because it contains a variable x, the value of which can be found. The value must be such that when substituting this value into the original equation, the correct numerical equality is obtained.

In other words, we must find a value at which the equal sign would justify its location - the left side must be equal to the right side.

Equation 2 + x= 4 is elementary. Variable value x is equal to the number 2. For any other value, equality will not be observed

They say that the number 2 is root or solving the equation 2 + x = 4

Root or solution to the equation- this is the value of the variable at which the equation turns into a true numerical equality.

There may be several roots or none at all. Solve the equation means to find its roots or prove that there are no roots.

The variable included in the equation is otherwise called unknown. You have the right to call it what you prefer. These are synonyms.

Note. The phrase “solve an equation” speaks for itself. Solving an equation means “equalizing” the equation—making it balanced so that the left side equals the right side.

Express one thing through the other

The study of equations traditionally begins with learning to express one number included in an equality through a number of others. Let's not break this tradition and do the same.

Consider the following expression:

8 + 2

This expression is the sum of the numbers 8 and 2. The value of this expression is 10

8 + 2 = 10

We got equality. Now you can express any number from this equality through other numbers included in the same equality. For example, let's express the number 2.

To express the number 2, you need to ask the question: “what must be done with the numbers 10 and 8 to get the number 2.” It is clear that to obtain the number 2, you need to subtract the number 8 from the number 10.

That's what we do. We write down the number 2 and through the equal sign we say that to obtain this number 2 we subtracted the number 8 from the number 10:

2 = 10 − 8

We expressed the number 2 from the equality 8 + 2 = 10. As can be seen from the example, there is nothing complicated about this.

When solving equations, in particular when expressing one number in terms of others, it is convenient to replace the equal sign with the word “ There is" . This must be done mentally, and not in the expression itself.

So, expressing the number 2 from the equality 8 + 2 = 10, we got the equality 2 = 10 − 8. This equality can be read as follows:

2 There is 10 − 8

That is, a sign = replaced by the word "is". Moreover, the equality 2 = 10 − 8 can be translated from mathematical language into full-fledged human language. Then it can be read as follows:

Number 2 There is difference between number 10 and number 8

Number 2 There is difference between number 10 and number 8.

But we will limit ourselves to only replacing the equal sign with the word “is,” and we will not always do this. Elementary expressions can be understood without translating mathematical language into human language.

Let us return the resulting equality 2 = 10 − 8 to its original state:

8 + 2 = 10

Let's express the number 8 this time. What needs to be done with the remaining numbers to get the number 8? That's right, you need to subtract 2 from the number 10

8 = 10 − 2

Let us return the resulting equality 8 = 10 − 2 to its original state:

8 + 2 = 10

This time we will express the number 10. But it turns out that there is no need to express the ten, since it is already expressed. It is enough to swap the left and right parts, then we get what we need:

10 = 8 + 2

Example 2. Consider the equality 8 − 2 = 6

Let us express the number 8 from this equality. To express the number 8, the remaining two numbers must be added:

8 = 6 + 2

Let us return the resulting equality 8 = 6 + 2 to its original state:

8 − 2 = 6

Let's express the number 2 from this equality. To express the number 2, you need to subtract 6 from 8

2 = 8 − 6

Example 3. Consider the equality 3 × 2 = 6

Let's express the number 3. To express the number 3, you need 6 divided by 2

Let's return the resulting equality to its original state:

3 × 2 = 6

Let us express the number 2 from this equality. To express the number 2, you need 6 divided by 3

Example 4. Consider the equality

Let us express the number 15 from this equality. To express the number 15, you need to multiply the numbers 3 and 5

15 = 3 × 5

Let us return the resulting equality 15 = 3 × 5 to its original state:

Let us express the number 5 from this equality. To express the number 5, you need 15 divided by 3

Rules for finding unknowns

Let's consider several rules for finding unknowns. They may be familiar to you, but it doesn’t hurt to repeat them again. In the future, they can be forgotten, as we learn to solve equations without applying these rules.

Let's go back to the first example we looked at in previous topic, where in the equality 8 + 2 = 10 it was necessary to express the number 2.

In the equality 8 + 2 = 10, the numbers 8 and 2 are the terms, and the number 10 is the sum.

To express the number 2, we did the following:

2 = 10 − 8

That is, from the sum of 10 we subtracted the term 8.

Now imagine that in the equality 8 + 2 = 10, instead of the number 2, there is a variable x

8 + x = 10

In this case, the equality 8 + 2 = 10 becomes the equation 8 + x= 10 and the variable x unknown term

Our task is to find this unknown term, that is, to solve the equation 8 + x= 10 . To find an unknown term, the following rule is provided:

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

Which is basically what we did when we expressed two in the equality 8 + 2 = 10. To express term 2, we subtracted another term 8 from the sum 10

2 = 10 − 8

Now, to find the unknown term x, we must subtract the known term 8 from the sum 10:

x = 10 − 8

If you calculate the right side of the resulting equality, you can find out what the variable is equal to x

x = 2

We have solved the equation. Variable value x equals 2. To check the value of a variable x sent to the original equation 8 + x= 10 and substitute x. It is advisable to do this with any solved equation, since you cannot be absolutely sure that the equation has been solved correctly:

As a result

The same rule would apply if the unknown term was the first number 8.

x + 2 = 10

In this equation x is the unknown term, 2 is the known term, 10 is the sum. To find an unknown term x, you need to subtract the known term 2 from the sum 10

x = 10 − 2

x = 8

Let's return to the second example from the previous topic, where in the equality 8 − 2 = 6 it was necessary to express the number 8.

In the equality 8 − 2 = 6, the number 8 is the minuend, the number 2 is the subtrahend, and the number 6 is the difference

To express the number 8, we did the following:

8 = 6 + 2

That is, we added the difference of 6 and the subtracted 2.

Now imagine that in the equality 8 − 2 = 6, instead of the number 8, there is a variable x

x − 2 = 6

In this case the variable x takes on the role of the so-called unknown minuend

To find an unknown minuend, the following rule is provided:

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

This is what we did when we expressed the number 8 in the equality 8 − 2 = 6. To express the minuend of 8, we added the subtrahend of 2 to the difference of 6.

Now, to find the unknown minuend x, we must add the subtrahend 2 to the difference 6

x = 6 + 2

If you calculate the right side, you can find out what the variable is equal to x

x = 8

Now imagine that in the equality 8 − 2 = 6, instead of the number 2, there is a variable x

8 − x = 6

In this case the variable x takes on the role unknown subtrahend

To find an unknown subtrahend, the following rule is provided:

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

This is what we did when we expressed the number 2 in the equality 8 − 2 = 6. To express the number 2, we subtracted the difference 6 from the minuend 8.

Now, to find the unknown subtrahend x, you again need to subtract the difference 6 from the minuend 8

x = 8 − 6

We calculate the right side and find the value x

x = 2

Let's return to the third example from the previous topic, where in the equality 3 × 2 = 6 we tried to express the number 3.

In the equality 3 × 2 = 6, the number 3 is the multiplicand, the number 2 is the multiplier, the number 6 is the product

To express the number 3 we did the following:

That is, we divided the product of 6 by the factor of 2.

Now imagine that in the equality 3 × 2 = 6, instead of the number 3 there is a variable x

x× 2 = 6

In this case the variable x takes on the role unknown multiplicand.

To find an unknown multiplicand, the following rule is provided:

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

This is what we did when we expressed the number 3 from the equality 3 × 2 = 6. We divided the product 6 by the factor 2.

Now to find the unknown multiplicand x, you need to divide the product 6 by the factor 2.

Calculating the right side allows us to find the value of a variable x

x = 3

The same rule applies if the variable x is located instead of the multiplier, not the multiplicand. Let's imagine that in the equality 3 × 2 = 6, instead of the number 2 there is a variable x.

In this case the variable x takes on the role unknown multiplier. To find an unknown factor, the same procedure is provided as for finding an unknown multiplicand, namely, dividing the product by a known factor:

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

This is what we did when we expressed the number 2 from the equality 3 × 2 = 6. Then to get the number 2 we divided the product of 6 by its multiplicand 3.

Now to find the unknown factor x We divided the product of 6 by the multiplicand of 3.

Calculating the right side of the equality allows you to find out what x is equal to

x = 2

The multiplicand and the multiplier together are called factors. Since the rules for finding a multiplicand and a multiplier are the same, we can formulate general rule finding the unknown factor:

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

For example, let's solve the equation 9 × x= 18. Variable x is an unknown factor. To find this unknown factor, you need to divide the product 18 by the known factor 9

Let's solve the equation x× 3 = 27. Variable x is an unknown factor. To find this unknown factor, you need to divide the product 27 by the known factor 3

Let's return to the fourth example from the previous topic, where in an equality we needed to express the number 15. In this equality, the number 15 is the dividend, the number 5 is the divisor, and the number 3 is the quotient.

To express the number 15 we did the following:

15 = 3 × 5

That is, we multiplied the quotient of 3 by the divisor of 5.

Now imagine that in the equality, instead of the number 15, there is a variable x

In this case the variable x takes on the role unknown dividend.

To find an unknown dividend, the following rule is provided:

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

This is what we did when we expressed the number 15 from the equality. To express the number 15, we multiply the quotient of 3 by the divisor of 5.

Now, to find the unknown dividend x, you need to multiply the quotient 3 by the divisor 5

x= 3 × 5

x .

x = 15

Now imagine that in the equality, instead of the number 5, there is a variable x .

In this case the variable x takes on the role unknown divisor.

To find an unknown divisor, the following rule is provided:

This is what we did when we expressed the number 5 from the equality. To express the number 5, we divide the dividend 15 by the quotient 3.

Now to find the unknown divisor x, you need to divide the dividend 15 by the quotient 3

Let's calculate the right side of the resulting equality. This way we find out what the variable is equal to x .

x = 5

So, to find unknowns, we studied the following rules:

• To find the unknown term, you need to subtract the known term from the sum;
• To find the unknown minuend, you need to add the subtrahend to the difference;
• To find the unknown subtrahend, you need to subtract the difference from the minuend;
• To find an unknown multiplicand, you need to divide the product by the factor;
• To find an unknown factor, you need to divide the product by the multiplicand;
• To find an unknown dividend, you need to multiply the quotient by the divisor;
• To find an unknown divisor, you need to divide the dividend by the quotient.

Components

We will call components the numbers and variables included in the equality

So, the components of addition are terms And sum

The subtraction components are minuend, subtrahend And difference

The components of multiplication are multiplicand, factor And work

The components of division are the dividend, divisor and quotient.

Depending on which components we are dealing with, the corresponding rules for finding unknowns will apply. We studied these rules in the previous topic. When solving equations, it is advisable to know these rules by heart.

Example 1. Find the root of the equation 45 + x = 60

45 - term, x- unknown term, 60 - sum. We are dealing with the components of addition. We recall that to find an unknown term, you need to subtract the known term from the sum:

x = 60 − 45

Let's calculate the right side and get the value x equal to 15

x = 15

So the root of the equation is 45 + x= 60 is equal to 15.

Most often, an unknown term must be reduced to a form in which it can be expressed.

Example 2. Solve the equation

Here, unlike the previous example, the unknown term cannot be expressed immediately, since it contains a coefficient of 2. Our task is to bring this equation to a form in which it could be expressed x

In this example, we are dealing with the components of addition—the terms and the sum. 2 x is the first term, 4 is the second term, 8 is the sum.

In this case, term 2 x contains a variable x. After finding the value of the variable x term 2 x will take a different look. Therefore, term 2 x can be completely taken as an unknown term:

Now we apply the rule for finding the unknown term. Subtract the known term from the sum:

Let's calculate the right side of the resulting equation:

We have a new equation. Now we are dealing with the components of multiplication: the multiplicand, the multiplier, and the product. 2 - multiplicand, x- multiplier, 4 - product

In this case, the variable x is not just a multiplier, but an unknown multiplier

To find this unknown factor, you need to divide the product by the multiplicand:

Let's calculate the right side and get the value of the variable x

To check, send the found root to the original equation and substitute x

Example 3. Solve the equation 3x+ 9x+ 16x= 56

Immediately express the unknown x it is forbidden. First you need to bring this equation to a form in which it could be expressed.

We present on the left side of this equation:

We are dealing with the components of multiplication. 28 - multiplicand, x- multiplier, 56 - product. Wherein x is an unknown factor. To find an unknown factor, you need to divide the product by the multiplicand:

From here x equals 2

Equivalent equations

In the previous example, when solving the equation 3x + 9x + 16x = 56 , we have given similar terms on the left side of the equation. As a result, we obtained a new equation 28 x= 56 . Old equation 3x + 9x + 16x = 56 and the resulting new equation 28 x= 56 is called equivalent equations, since their roots coincide.

Equations are called equivalent if their roots coincide.

Let's check it out. For the equation 3x+ 9x+ 16x= 56 we found the root equal to 2. Let's first substitute this root into the equation 3x+ 9x+ 16x= 56 , and then into equation 28 x= 56, which was obtained by bringing similar terms on the left side of the previous equation. We must get the correct numerical equalities

According to the order of operations, multiplication is performed first:

Let's substitute root 2 into the second equation 28 x= 56

We see that both equations have the same roots. So the equations 3x+ 9x+ 16x= 6 and 28 x= 56 are indeed equivalent.

To solve the equation 3x+ 9x+ 16x= 56 We used one of them - reduction of similar terms. The correct identity transformation of the equation allowed us to obtain the equivalent equation 28 x= 56, which is easier to solve.

From identical transformations to this moment We only know how to reduce fractions, add similar terms, take the common factor out of brackets, and also open brackets. There are other conversions you should be aware of. But for general idea about identical transformations of equations, the topics we have studied are quite sufficient.

Let's consider some transformations that allow us to obtain the equivalent equation

If you add the same number to both sides of the equation, you get an equation equivalent to the given one.

and similarly:

If you subtract the same number from both sides of an equation, you get an equation equivalent to the given one.

In other words, the root of the equation will not change if the same number is added to (or subtracted from both sides) the same number.

Example 1. Solve the equation

Subtract 10 from both sides of the equation

We got equation 5 x= 10 . We are dealing with the components of multiplication. To find an unknown factor x, you need to divide the product 10 by the known factor 5.

and substitute x found value 2

We got the correct numerical equality. This means the equation is solved correctly.

Solving the equation we subtracted the number 10 from both sides of the equation. As a result, we obtained an equivalent equation. The root of this equation, like the equation is also equal to 2

Example 2. Solve equation 4( x+ 3) = 16

Subtract the number 12 from both sides of the equation

There will be 4 left on the left side x, and on the right side the number 4

We got equation 4 x= 4 . We are dealing with the components of multiplication. To find an unknown factor x, you need to divide the product 4 by the known factor 4

Let's return to the original equation 4( x+ 3) = 16 and substitute x found value 1

We got the correct numerical equality. This means the equation is solved correctly.

Solving equation 4( x+ 3) = 16 we subtracted the number 12 from both sides of the equation. As a result, we obtained the equivalent equation 4 x= 4 . The root of this equation, like equation 4( x+ 3) = 16 is also equal to 1

Example 3. Solve the equation

Let's expand the brackets on the left side of the equality:

Add the number 8 to both sides of the equation

Let us present similar terms on both sides of the equation:

There will be 2 left on the left side x, and on the right side the number 9

In the resulting equation 2 x= 9 we express the unknown term x

Let's return to the original equation and substitute x found value 4.5

We got the correct numerical equality. This means the equation is solved correctly.

Solving the equation we added the number 8 to both sides of the equation. As a result, we got an equivalent equation. The root of this equation, like the equation also equal to 4.5

The next rule that allows us to obtain an equivalent equation is as follows

If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one.

That is, the root of the equation will not change if we move a term from one part of the equation to another, changing its sign. This property is one of the important and one of the often used when solving equations.

Consider the following equation:

The root of this equation is equal to 2. Let us substitute x this root and check whether the numerical equality is correct

The result is a correct equality. This means that the number 2 is indeed the root of the equation.

Now let's try to experiment with the terms of this equation, moving them from one part to another, changing the signs.

For example, term 3 x is located on the left side of the equation. Let's move it to the right side, changing the sign to the opposite:

The result is an equation 12 = 9x − 3x . on the right side of this equation:

x is an unknown factor. Let's find this well-known factor:

From here x= 2 . As you can see, the root of the equation has not changed. So the equations are 12 + 3 x = 9x And 12 = 9x − 3x are equivalent.

In fact, this transformation is a simplified method of the previous transformation, where the same number was added (or subtracted) to both sides of the equation.

We said that in the equation 12 + 3 x = 9x term 3 x was moved to the right side, changing sign. In reality, the following happened: term 3 was subtracted from both sides of the equation x

Then similar terms were given on the left side and the equation was obtained 12 = 9x − 3x. Then similar terms were again given, but on the right side, and the equation 12 = 6 was obtained x.

But the so-called “translation” is more convenient for such equations, which is why it has become so widespread. When solving equations, we will often use this particular transformation.

The equations 12 + 3 are also equivalent x= 9x And 3x− 9x= −12 . This time the equation is 12 + 3 x= 9x term 12 was moved to the right side, and term 9 x to the left. We should not forget that the signs of these terms were changed during the transfer

The next rule that allows us to obtain an equivalent equation is as follows:

If both sides of the equation are multiplied or divided by the same number, not equal to zero, you get an equation equivalent to the given one.

In other words, the roots of an equation will not change if both sides are multiplied or divided by the same number. This action is often used when you need to solve an equation containing fractional expressions.

First, let's look at examples in which both sides of the equation will be multiplied by the same number.

Example 1. Solve the equation

When solving equations containing fractional expressions, it is customary to first simplify the equation.

IN in this case we are dealing with just such an equation. To simplify this equation, both sides can be multiplied by 8:

We remember that for , we need to multiply the numerator of a given fraction by this number. We have two fractions and each of them is multiplied by the number 8. Our task is to multiply the numerators of the fractions by this number 8

Now the interesting part happens. The numerators and denominators of both fractions contain a factor of 8, which can be reduced by 8. This will allow us to get rid of the fractional expression:

As a result, the simplest equation remains

Well, it’s not hard to guess that the root of this equation is 4

x found value 4

The result is a correct numerical equality. This means the equation is solved correctly.

When solving this equation, we multiplied both sides by 8. As a result, we got the equation. The root of this equation, like the equation, is 4. This means that these equations are equivalent.

The factor by which both sides of the equation are multiplied is usually written before the part of the equation, and not after it. So, solving the equation, we multiplied both sides by a factor of 8 and got the following entry:

This did not change the root of the equation, but if we had done this while at school, we would have been reprimanded, since in algebra it is customary to write a factor before the expression with which it is multiplied. Therefore, it is advisable to rewrite the multiplication of both sides of the equation by a factor of 8 as follows:

Example 2. Solve the equation

On the left side, the factors of 15 can be reduced by 15, and on the right side, the factors of 15 and 5 can be reduced by 5

Let's open the brackets on the right side of the equation:

Let's move the term x from the left side of the equation to the right side, changing the sign. And we move term 15 from the right side of the equation to the left side, again changing the sign:

We present similar terms in both sides, we get

We are dealing with the components of multiplication. Variable x

Let's return to the original equation and substitute x found value 5

The result is a correct numerical equality. This means the equation is solved correctly. When solving this equation, we multiplied both sides by 15. Further performing identical transformations, we obtained the equation 10 = 2 x. The root of this equation, like the equation equals 5. This means that these equations are equivalent.

Example 3. Solve the equation

On the left side you can reduce two threes, and the right side will be equal to 18

The simplest equation remains. We are dealing with the components of multiplication. Variable x is an unknown factor. Let's find this well-known factor:

Let's return to the original equation and substitute x found value 9

The result is a correct numerical equality. This means the equation is solved correctly.

Example 4. Solve the equation

Multiply both sides of the equation by 6

Let's open the brackets on the left side of the equation. On the right side, the factor 6 can be raised to the numerator:

Let's reduce what can be reduced on both sides of the equations:

Let's rewrite what we have left:

Let's use the transfer of terms. Terms containing the unknown x, we group on the left side of the equation, and the terms free of unknowns - on the right:

Let us present similar terms in both parts:

Now let's find the value of the variable x. To do this, divide the product 28 by the known factor 7

From here x= 4.

Let's return to the original equation and substitute x found value 4

The result is a correct numerical equation. This means the equation is solved correctly.

Example 5. Solve the equation

Let's open the parentheses on both sides of the equation where possible:

Multiply both sides of the equation by 15

Let's open the brackets on both sides of the equation:

Let's reduce what can be reduced on both sides of the equation:

Let's rewrite what we have left:

Let's expand the brackets where possible:

Let's use the transfer of terms. We group the terms containing the unknown on the left side of the equation, and the terms free of unknowns on the right. Do not forget that during the transfer, the terms change their signs to the opposite:

Let us present similar terms on both sides of the equation:

Let's find the value x

The resulting answer can be divided into a whole part:

Let's return to the original equation and substitute x found value

It turns out to be a rather cumbersome expression. Let's use variables. Let's put the left side of the equality into a variable A, and the right side of the equality into a variable B

Our task is to make sure whether the left side is equal to the right. In other words, prove the equality A = B

Let's find the value of the expression in variable A.

Variable value A equals . Now let's find the value of the variable B. That is, the value of the right side of our equality. If it is also equal, then the equation will be solved correctly

We see that the value of the variable B, as well as the value of variable A is . This means that the left side is equal to the right side. From this we conclude that the equation is solved correctly.

Now let's try not to multiply both sides of the equation by the same number, but to divide.

Consider the equation 30x+ 14x+ 14 = 70x− 40x+ 42 . Let's solve it using the usual method: we group terms containing unknowns on the left side of the equation, and terms free of unknowns - on the right. Next, performing the known identity transformations, we find the value x

Let's substitute the found value 2 instead x into the original equation:

Now let's try to separate all the terms of the equation 30x+ 14x+ 14 = 70x− 40x+ 42 by some number. We note that all terms of this equation have a common factor of 2. We divide each term by it:

Let's perform a reduction in each term:

Let's rewrite what we have left:

Let's solve this equation using the well-known identity transformations:

We got root 2. So the equations 15x+ 7x+ 7 = 35x− 20x+ 21 And 30x+ 14x+ 14 = 70x− 40x+ 42 are equivalent.

Dividing both sides of the equation by the same number allows you to remove the unknown from the coefficient. In the previous example when we got equation 7 x= 14, we needed to divide the product 14 by the known factor 7. But if we had freed the unknown from the factor 7 on the left side, the root would have been found immediately. To do this, it was enough to divide both sides by 7

We will also use this method often.

Multiplication by minus one

If both sides of the equation are multiplied by minus one, you get an equation equivalent to this one.

This rule follows from the fact that multiplying (or dividing) both sides of an equation by the same number does not change the root of the given equation. This means that the root will not change if both its parts are multiplied by −1.

This rule allows you to change the signs of all components included in the equation. What is it for? Again, to get an equivalent equation that is easier to solve.

Consider the equation. What is the root of this equation?

Add the number 5 to both sides of the equation

Let's look at similar terms:

Now let's remember about. What is the left side of the equation? This is the product of minus one and a variable x

That is, the minus sign in front of the variable x does not refer to the variable itself x, but to one, which we do not see, since coefficient 1 is not usually written down. This means that the equation actually looks like this:

We are dealing with the components of multiplication. To find X, you need to divide the product −5 by the known factor −1.

or divide both sides of the equation by −1, which is even simpler

So the root of the equation is 5. To check, let's substitute it into the original equation. Do not forget that in the original equation the minus is in front of the variable x refers to an invisible unit

The result is a correct numerical equation. This means the equation is solved correctly.

Now let's try to multiply both sides of the equation by minus one:

After opening the brackets, the expression is formed on the left side, and the right side will be equal to 10

The root of this equation, like the equation, is 5

This means that the equations are equivalent.

Example 2. Solve the equation

In this equation, all components are negative. It is more convenient to work with positive components than with negative ones, so let’s change the signs of all components included in the equation. To do this, multiply both sides of this equation by −1.

It is clear that when multiplied by −1, any number will change its sign to the opposite. Therefore, the procedure of multiplying by −1 and opening the brackets is not described in detail, but the components of the equation with opposite signs are immediately written down.

Thus, multiplying an equation by −1 can be written in detail as follows:

or you can simply change the signs of all components:

The result will be the same, but the difference will be that we will save ourselves time.

So, multiplying both sides of the equation by −1, we get the equation. Let's solve this equation. Subtract 4 from both sides and divide both sides by 3

When the root is found, the variable is usually written on the left side, and its value on the right, which is what we did.

Example 3. Solve the equation

Let's multiply both sides of the equation by −1. Then all components will change their signs to opposite ones:

Subtract 2 from both sides of the resulting equation x and give similar terms:

Let's add one to both sides of the equation and give similar terms:

Equating to zero

We recently learned that if we move a term in an equation from one part to another, changing its sign, we will get an equation equivalent to the given one.

What happens if you move from one part to another not just one term, but all the terms? That's right, in the part where all the terms were taken away there will be zero left. In other words, there will be nothing left.

As an example, consider the equation. Let's solve this equation as usual - we will group the terms containing unknowns in one part, and leave the numerical terms free of unknowns in the other. Next, performing the known identity transformations, we find the value of the variable x

Now let's try to solve the same equation by equating all its components to zero. To do this, we move all the terms from the right side to the left, changing the signs:

Let us present similar terms on the left side:

Add 77 to both sides and divide both sides by 7

An alternative to the rules for finding unknowns

Obviously, knowing about identical transformations of equations, you don’t have to memorize the rules for finding unknowns.

For example, to find the unknown in an equation, we divided the product 10 by the known factor 2

But if you divide both sides of the equation by 2, the root will be found immediately. On the left side of the equation in the numerator the factor 2 and in the denominator the factor 2 will be reduced by 2. And the right side will be equal to 5

We solved equations of the form by expressing the unknown term:

But you can use the identical transformations that we studied today. In the equation, term 4 can be moved to the right side by changing the sign:

On the left side of the equation, two twos will cancel out. The right side will be equal to 2. Hence .

Or you could subtract 4 from both sides of the equation. Then you would get the following:

In the case of equations of the form, it is more convenient to divide the product by a known factor. Let's compare both solutions:

The first solution is much shorter and neater. The second solution can be significantly shortened if you do the division in your head.

However, it is necessary to know both methods and only then use the one you prefer.

When there are several roots

An equation can have multiple roots. For example the equation x(x+ 9) = 0 has two roots: 0 and −9.

In Eq. x(x+ 9) = 0 it was necessary to find such a value x at which the left side would be equal to zero. The left side of this equation contains the expressions x And (x+9), which are factors. From the product laws we know that a product is equal to zero if at least one of the factors is equal to zero (either the first factor or the second).

That is, in Eq. x(x+ 9) = 0 equality will be achieved if x will be equal to zero or (x+9) will be equal to zero.

x= 0 or x + 9 = 0

By setting both of these expressions to zero, we can find the roots of the equation x(x+ 9) = 0 . The first root, as can be seen from the example, was found immediately. To find the second root you need to solve the elementary equation x+ 9 = 0 . It is easy to guess that the root of this equation is −9. Checking shows that the root is correct:

−9 + 9 = 0

Example 2. Solve the equation

This equation has two roots: 1 and 2. Left side equation is the product of expressions ( x− 1) and ( x− 2) . And the product is equal to zero if at least one of the factors is equal to zero (or the factor ( x− 1) or factor ( x − 2) ).

Let's find something like this x under which the expressions ( x− 1) or ( x− 2) become zero:

We substitute the found values ​​one by one into the original equation and make sure that for these values ​​the left-hand side is equal to zero:

When there are infinitely many roots

An equation can have infinitely many roots. That is, by substituting any number into such an equation, we get the correct numerical equality.

Example 1. Solve the equation

The root of this equation is any number. If you open the brackets on the left side of the equation and add similar terms, you get the equality 14 = 14. This equality will be obtained for any x

Example 2. Solve the equation

The root of this equation is any number. If you open the brackets on the left side of the equation, you get the equality 10x + 12 = 10x + 12. This equality will be obtained for any x

When there are no roots

It also happens that the equation has no solutions at all, that is, it has no roots. For example, the equation has no roots, since for any value x, the left side of the equation will not be equal to the right side. For example, let . Then the equation will take the following form

Example 2. Solve the equation

Let's expand the brackets on the left side of the equality:

Let's look at similar terms:

We see that the left side is not equal to the right side. And this will be the case for any value y. For example, let y = 3 .

Letter equations

An equation can contain not only numbers with variables, but also letters.

For example, the formula for finding speed is a literal equation:

This equation describes the speed of a body during uniformly accelerated motion.

A useful skill is the ability to express any component included in a letter equation. For example, to determine distance from an equation, you need to express the variable s .

Multiply both sides of the equation by t

Variables on the right side t let's cut it by t

In the resulting equation, we swap the left and right sides:

We have a formula for finding the distance, which we studied earlier.

Let's try to determine time from the equation. To do this you need to express the variable t .

Multiply both sides of the equation by t

Variables on the right side t let's cut it by t and rewrite what we have left:

In the resulting equation v×t = s divide both parts by v

Variables on the left v let's cut it by v and rewrite what we have left:

We have the formula for determining time, which we studied earlier.

Suppose the train speed is 50 km/h

v= 50 km/h

And the distance is 100 km

s= 100 km

Then the letter will take the following form

Time can be found from this equation. To do this you need to be able to express the variable t. You can use the rule for finding an unknown divisor by dividing the dividend by the quotient and thus determining the value of the variable t

or you can use identical transformations. First multiply both sides of the equation by t

Then divide both sides by 50

Example 2 x

Subtract from both sides of the equation a

Let's divide both sides of the equation by b

a + bx = c, then we will have a ready-made solution. It will be enough to substitute the required values ​​into it. Those values ​​that will be substituted for letters a, b, c usually called parameters. And equations of the form a + bx = c called equation with parameters. Depending on the parameters, the root will change.

Let's solve the equation 2 + 4 x= 10 . It looks like a letter equation a + bx = c. Instead of performing identical transformations, we can use a ready-made solution. Let's compare both solutions:

We see that the second solution is much simpler and shorter.

For a ready-made solution, it is necessary to make a small remark. Parameter b must not be equal to zero (b ≠ 0), since division by zero by is allowed.

Example 3. A literal equation is given. Express from this equation x

Let's open the brackets on both sides of the equation

Let's use the transfer of terms. Parameters containing a variable x, we group on the left side of the equation, and parameters free from this variable - on the right.

On the left side we take the factor out of brackets x

Let's divide both sides by the expression a − b

On the left side, the numerator and denominator can be reduced by a − b. This is how the variable is finally expressed x

Now, if we come across an equation of the form a(x − c) = b(x + d), then we will have a ready-made solution. It will be enough to substitute the required values ​​into it.

Let's say we are given the equation 4(x− 3) = 2(x+ 4) . It looks like an equation a(x − c) = b(x + d). Let's solve it in two ways: using identical transformations and using a ready-made solution:

For convenience, let's take it out of the equation 4(x− 3) = 2(x+ 4) parameter values a, b, c, d . This will allow us not to make a mistake when substituting:

As in the previous example, the denominator here should not be equal to zero ( a − b ≠ 0) . If we encounter an equation of the form a(x − c) = b(x + d) in which the parameters a And b will be the same, we can say without solving it that this equation has no roots, since the difference identical numbers equal to zero.

For example, the equation 2(x − 3) = 2(x + 4) is an equation of the form a(x − c) = b(x + d). In Eq. 2(x − 3) = 2(x + 4) options a And b the same. If we start solving it, we will come to the conclusion that the left side will not be equal to the right side:

Example 4. A literal equation is given. Express from this equation x

Let's bring the left side of the equation to a common denominator:

Multiply both sides by a

On the left side x let's put it out of brackets

Divide both sides by the expression (1 − a)

Linear equations with one unknown

The equations discussed in this lesson are called linear equations of the first degree with one unknown.

If the equation is given in the first degree, does not contain division by the unknown, and also does not contain roots from the unknown, then it can be called linear. We have not yet studied powers and roots, so in order not to complicate our lives, we will understand the word “linear” as “simple”.

Most of the equations solved in this lesson ultimately came down to a simple equation in which you had to divide the product by a known factor. For example, this is equation 2( x+ 3) = 16 . Let's solve it.

Let's open the brackets on the left side of the equation, we get 2 x+ 6 = 16. Let's move term 6 to the right side, changing the sign. Then we get 2 x= 16 − 6. Calculate the right side, we get 2 x= 10. To find x, divide the product 10 by the known factor 2. Hence x = 5.

Equation 2( x+ 3) = 16 is linear. It comes down to equation 2 x= 10, to find the root of which it was necessary to divide the product by a known factor. This simplest equation is called linear equation of the first degree with one unknown in canonical form. The word "canonical" is synonymous with the words "simple" or "normal".

A linear equation of the first degree with one unknown in canonical form is called an equation of the form ax = b.

Our resulting equation 2 x= 10 is a linear equation of the first degree with one unknown in canonical form. This equation has the first degree, one unknown, it does not contain division by the unknown and does not contain roots from the unknown, and it is presented in canonical form, that is, in the simplest form in which the value can be easily determined x. Instead of parameters a And b our equation contains the numbers 2 and 10. But such an equation can also contain other numbers: positive, negative or equal to zero.

If in a linear equation a= 0 and b= 0, then the equation has infinitely many roots. Indeed, if a equal to zero and b equals zero, then the linear equation ax= b will take the form 0 x= 0 . For any value x the left side will be equal to the right side.

If in a linear equation a= 0 and b≠ 0, then the equation has no roots. Indeed, if a equal to zero and b equal to any number, not equal to zero, say the number 5, then the equation ax = b will take the form 0 x= 5 . The left side will be zero, and the right side will be five. And zero is not equal to five.

If in a linear equation a≠ 0, and b equals any number, then the equation has one root. It is determined by dividing the parameter b per parameter a

Indeed, if a equal to some number that is not zero, say the number 3, and b equal to some number, say the number 6, then the equation will take the form .
From here.

There is another form of recording linear equation first degree with one unknown. It looks like this: ax−b= 0 . This is the same equation as ax = b

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§ 1 How to find an unknown term

How to find the root of an equation if one of the terms is unknown? In this lesson we will look at 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 growing in the flowerbed. How many tulips were there in the flowerbed? Let's write down the solution. So, there were 6 red ones and 3 yellow tulip Therefore, we can write down the expression 6+3, after performing the addition, we get the result - there were 9 tulips growing in the flowerbed.

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

Let's change the problem condition. There were 9 tulips growing in the flowerbed, 6 were picked. How many tulips are left?

To find out how many tulips are left in the flowerbed, you need to subtract the picked 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 flowerbed.

Let's transform this problem again. There were 9 tulips growing, 3 were picked. How many tulips are left?

The solution will look like this: from the total number of tulips 9, you need to subtract the picked flowers, there are 3 of them. There are 6 tulips left.

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

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

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

Let's remember what numbers are called when adding:

6 is the first term

3 - second term

9 - amount value

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

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

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

Therefore, if you subtract the first term from the value of the sum, you get the second term, and if you subtract the second term from the value of the sum, you get the first term.

Let's formulate a general rule:

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

§ 2 Examples of solving equations with an unknown term

Let's look at equations with unknown terms and try to find the 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.

This means X = 7 - 5,

let's find the difference 7 - 5 = 2, X = 2.

Let's check if we found the root of the equation correctly. To check, you need to substitute the number 2 instead of X in the equation:

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

Let's solve another equation 8 + Y = 17.

The second term in this equation is unknown.

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

Let's check: substitute the number 9 for Y. We get:

17 = 17 - we 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 connection between the terms and the value of the sum. To find the unknown term, you need to subtract the known term from the sum value.

List of used literature:

1. I.I. Arginskaya, E.I. Ivanovskaya, S.N. Kormishina. Mathematics: Textbook for 2nd grade: At 2 o'clock. - Samara: Publishing house " Educational literature»: Publishing House"Fedorov", 2012.
2. Arginskaya I.I. Collection of tasks in mathematics for independent, test and tests V primary school. - Samara: Fedorov Corporation, Educational Literature Publishing House, 2006.

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