Three equations find the unknown term. Solution of equations with an unknown term. Consistent application of rules
Abstract of a lesson in mathematics, grade 2
The purpose of the lesson: to create the necessary conditions for students to deduce the rule for finding the unknown term.Lesson Objectives:
to form the concepts of "equation", "root of the equation";
compose an algorithm for solving the equation;
reinforce the ability to compose equations, find the root of the equation and check the correctness of the calculation;
improve computational skills, mathematical speech, develop logical thinking;
develop self-control skills, the ability to work in pairs;
to form the ability to work according to a plan, an algorithm.
Planned results:
Subject:
know and apply the rule for finding the unknown term when solving simple equations;
be able to write down and solve simple equations for finding the unknown term.
use mathematical terms correctly in speech.
Metasubject:
cognitive : search and highlight the necessary information; conscious and arbitrary construction of a speech utterance; establishment of causal relationships.
regulatory : selection and awareness by students of what has already been mastered and what is still subject to assimilation, comparison of the method of action and its result with a given standard.
communicative : emotionally positive attitude to the process of cooperation, the ability to listen to the interlocutor, consideration of different opinions and the ability to substantiate their own, respect for a different point of view.
personal : the formation of an adequate positive conscious self-esteem, the development of cognitive interests, educational motives.
Methods:
partial search; verbal;
Technological lesson map
I .Organization of the class. Motivation for learning activities.
Today we have public lesson... Guests have come to our lesson, turn to them, we will greet them.Sit down quietly.
I am glad that I see your lovely faces again in our next math lesson. Today's lesson is exciting, you are alarmed. Let's try to cheer up, turn around, smile, support each other:
Don't be sad today
Together we will be on the way!
Well done! Has your mood changed? What has it become?
Look at the board and choose your setup for the lesson:
I will:
Attentive
Diligent
Hardworking
Curious
At the end of the lesson, say whether you completed it or it failed. Let's get to work.
Recording a number. Classwork.
Let's represent the number 16 as a sum of two numbers, a difference of two numbers, as a product of two numbers, as a difference and a product of numbers.
Yes. Calm, joyful, fear and excitement disappeared.
II .
Updating basic knowledge
Purpose: improving computational skills, repeating the composition of numbers
1. Put the signs "+" or "-"
2. Fill in the table:
Output:3. Task
First 6 m was cut from a piece of fabric 24 m long, and then another 4 m. How many meters of fabric were left in the piece?
4 . Solve the puzzle.
What groups can these mathematical notations be divided into?
Add ...An equation is an equality containing ...unknown number
The unknown number in the equation is called ...root of the equation
The root of the equation makes the equation true ...equality
Numeric equalities, numeric inequalities, equations, roots of equations
The equation.
Equality containing the unknown is called an equation.
The root of an equation is a number that, when substituted into the equation instead of x, results in the correct numerical equality.
III .
Identifying the place and cause of the difficulty
Purpose: Creation of conditions for the selection of an equation with an unknown subtraction;
Identify the place of difficulty;
Record the cause of the difficulty in external speech
IV. Formulation of the topic and purpose of the lesson
Each of you should remember how the equations are solved.
– Review the diagrams on the board.
What do you think, the discovery, what pattern will the lesson be devoted to?
Open the tutorial (page 77), bookmark the tutorial page and read the lesson topic.
Define the purpose of the lesson.
We, while poorly can explain how to find the unknown term
Learn to solve equations with an unknown term.
Solving Equations with an Unknown Summand
V ... Discovery of new knowledge.
Purpose: highlighting the rule for finding the unknown subtracted.
Working in groups
Find the equation in which you need to find the unknown first term, come up with an algorithm for solving it.Algorithm on the slide .
Name the components when adding.
Which component is unknown? (- How to find it using "Whole" and "Part".
Replace "Whole" and "Part" with the names of the add action components.
How to find the unknown term?
Where can we find confirmation of our assumptions?
Compare your findings with what the authors of the textbook suggest p.79
Formulate a rule for finding an unknown term.
To find the unknown part, subtract the known part from the whole.
VI .Physical culture
Vii ... Primary reinforcement with pronunciation in external speech.
Purpose: applying the rule to solving equations
Working at the blackboard
Page 79 No. 6,7
They carry out the task, pronounce a new concept.
VIII . Independent work in pairs with a self-test in the classroom.
Purpose: the formation of the ability to work in pairs, to show responsibility for their own choices and the results of their activities.
Page 79. No. 8
Ability to work in pairs using an algorithm
The rule for finding the unknown term.
IX ... Systematization and repetition.
Purpose: to organize the repetition of the skills to find all the ways to solve problems
Where can we apply the equation in math lessons?
In solving problems.
Solution to the problem with an explanation.
On one shelf there were 32 books, on the other - 8, how many books are on the third shelf if there are 100 books on three shelves.
Reserve. Work on individual cards.
Working with information
Be able to express your guess based on the work with the textbook material
X. Reflection
Purpose: to form the ability to reflect on their activities
What new things did you learn in the lesson today?
What was your goal? Have you reached your goal?
What was the topic of the lesson?
Assess the correctness of the action at the level of adequate assessment
The ability for self-assessment based on the criterion of the success of educational activities
Application
Self-check sheet ______________________________________
At each stage, evaluate your work by selecting the sign in the required line «+».
StageEducational activities
Performed without error
Completed with errors
Experienced great difficulty
Lesson start
Inspiration for the lesson
Step 1
Repetition of the passed material. Verbal counting
Step 2
Statement of the educational problem, the purpose of the lesson
Step 3
Group work
Step 4
Primary anchoring
Work according to the textbook p.79 №6.7
Step 5
Independent work
p.79 No. 6.7
Step 6
The solution of the problem.
Step 7
Application of new material in the knowledge system
NS + 120 = 220y - 19 = 78
Short term lesson planning
Subject: Mathematics
Class: 2 "D"
Date: 5.12.14
Teacher: Agitaeva G.K.
Resources: Interactive whiteboard, presentation, diagram cards, posters, colored markers,
Theme:
Solution of an equation with unknown terms.
Learning Objectives
to form the ability to solve equations with unknown terms by subtracting the same number from both parts of it;
analyze and explain the meaning of the concept of an equation;
develop attention and logical thinking;
foster positive motivation for the subject, a sense of friendship and mutual assistance.
Expected Result
They solve equations with unknown terms: analyze and explain the meaning of the concept of an equation, compose and solve compound problems.
Key ideas
An equation is an equality containing an unknown number.
Lesson steps
Organizing time... Psychological attitude.
Close your eyes, smile and mentally wish each other good luck in the lesson.
Guys, our friend came to us again today. What's his name?(Know)
He invited a guest to our lesson
(Video Dunno)
Dunno and wants to help him and you to study new topic but keeps it a secret and will name it after we complete his assignments.
There is a secret door to the land of new knowledge, and in order to open it, Dunno needs to complete Znayka's tasks and collect the key.
Verbal counting.
9+3 8+7 6+7
15-8 12-3 14-7
8+6 9+5 12-5
16-7 8+4 13-7
7+4 11-4 7+7
11-3 6+7
Logic puzzles.
There were 2 birches, 4 apple trees, 5 cherries in the garden. How many fruit trees were there in the garden? (9 fruit trees)
Sister is 9 years old, brother is 3 years old. How much older will your sister be in five years? (for 6 years)
3. Making a notebook. "A minute" of calligraphy.
Znayka asks:
What's the date today?(5)
What's the month?
How can you replace the number 12 with the sum of terms?
What can you say about him?(Two-digit. It contains 1 dec. And 2 units.
What's the next number? Previous?
And what number do you get if you swap tens and ones?
Let's write the number 12.
But do not forget that Znayka loves cleanliness and accuracy.
4 ... Mathematical dictation.
1st group
42- 22=20
38-25=13
(84-4)+10=90
1st group
50+ (10-2)=58
14-6=8
5+9=14
3rd group
58-43= 15
(25-20)+ 10=15
6+6=12
Arrange the letters in the order given in the table. We will receive both the key and the code to open the door.
58- and
20th
8 - at
14 - in
13- a
15 - n
8
12
13
14
15
20
15
58
20
at
R
a
v
n
e
n
and
e
5. Introduction to the topic
Are you familiar with this entry: □ + 4 = 12?
(Yes, this is an example with a "window")
What needs to be done to make the entry correct?(Pick up the number.)
Who will pick the right number?
Let's check?
b) Introduction of the concept.
Guys, look at this entry: x + 4 = 12.(A note appears on the board)
How does it differ from the previous one?
(Latin letter x is inserted instead of the window)
Does anyone of you know the name of such a recording?
This expression is called an equation.
6. Brainstorming. Compilation of a definition from a cluster.
Children, how would you end the phrase? Let's work in pairs. Let's make a definition
7 ... PHIZMINUTKA with Dunno and his friends.
8. Formative survey.
Find equations among the following entries:
All equations are written using what sign of action?
This means Addition.
Let's remember the components of addition.
What should be done to find the unknown term?
- What does it mean to solve an equation? (Find an unknown number to make the equality true)
Find the root of the equation. (Slide)
1 group - a + 10 = 18
Group 2 - y + 30 = 38
Group 3 - 8 + x = 38
9. Solution of the problem.
Before completing the next task, you must solve the rebus and find out what task you have preparedKnow you.
task
Open the tutorials on p.
Problem number 4.
Drawing up a task using a picture
1) 40 + 20 = 60 (tg.) Pencils
2) 40 + 60 = 100 (tg.)
B: 40+ (40 + 20) = 100 (tg.)
Answer: only 100 tenge costs paints and pencils
10. Independent work. (group)
Make an equation and find the root.
1 group? +? = 15
2 group? +? = 16
3 group? +? = 14
If the lesson was fruitful, glue it to the tree - the fruit
Interesting - flowers
Boring - leaves
P. 102 No. 3
Teacher actions
Student actions
Comments (1)
Call phase
Reflection phase
Reflection phase
The teacher greets the students.
Teacher showing presentation
The teacher reads logic puzzles.
The teacher asks questions and reminds you that each number is written in a separate cell.
The teacher distributes the tasks on the cards to the groups.
The teacher gives the key to unravel the encrypted word
The teacher asks the students to compare notes.
The teacher invites the children to do exercises with Dunno's animated friends.
The teacher asks leading questions.
The teacher gives out cards.
The teacher is distributing posters.
Children greet the teacher.
Students look at the slide and find out who they invited to the Znayka lesson
Students orally solve examples
Pupils decide and answer orally.
Children answer questions and write down the number beautifully in a notebook.
Pupils read and write down the dictation. Finds the values of the recorded expressions. Each group speaks and the other groups evaluate their work.
Pupils place numbers and letters in a table and name the cipher word.
Children in pairs on desks make up definitions.
Children do physical exercises.
Children find equations.
Children answer the questions posed.
The children collectively constitute the condition of the problem.
1 student decides at the blackboard.
Children in the group discuss and fill in posters.
Children stick stickers on the tree.
Formative grading technique
"Traffic light" (oral feedback). The teacher uses the technique to see how the students themselves
cope with the task well and, if possible, to help them.
Thumb technique.
"Verbal assessment"
(oral feedback).
The teacher praises
pupils for correct
actions performed.
so teacher
conducted an oral feedback
communication and learners
realized they were right
well done
tasks.
§ 1 How to find the unknown term
How to find the root of the 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 tulips in the flowerbed. How many tulips were there in the flowerbed? Let's write down the solution. So, there were 6 red and 3 yellow tulips, therefore, we can write down the expression 6 + 3, after completing the addition, we get the result - 9 tulips grew on the flowerbed.
Let's write down the solution. So, 6 red and 3 yellow tulips grew, therefore, we can write down the expression 6 + 3, performing addition, we get the result - 9 tulips grew on the flowerbed. 6 + 3 = 9.
Let's change the condition of the problem. 9 tulips grew on the flowerbed, 6 were plucked. How many tulips are left?
To find out how many tulips are left in the flowerbed, you need to subtract the plucked flowers from the total number of 9 tulips, there are 6 of them.
Let's make the calculations: 9-6 we get the result 3. There are 3 tulips left on the flowerbed.
Let's transform this task 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, you need to subtract the plucked flowers, there are 3. There are 6 tulips left.
Let's take a close 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 go back 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 - value of the sum
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 term 6 was subtracted from the value of the sum 9 to obtain the second term 3.
In the equality 9 - 3 = 6 from the value of the sum9, the second term3 was subtracted, and the first term6 was obtained.
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 value of the sum.
§ 2 Examples of solving equations with an unknown summand
Let's consider equations with unknown terms and try to find the roots using this rule.
Solve the equation X + 5 = 7.
The first term is unknown in this equation. To find it, we will 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.
Hence, X = 7 - 5,
find the difference 7 - 5 = 2, X = 2.
Let's check if we found the root of the equation correctly. To 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 another equation 8 + Y = 17.
The second term is unknown in this equation.
To find it, you need to subtract the first term 8 from the value of the sum 17.
Let's check: substitute 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:
- I.I. Arginskaya, E.I. Ivanovskaya, S.N. Kormishina. Mathematics: Textbook for grade 2: In 2h. - Samara: Publishing House "Educational Literature": Publishing House "Fedorov", 2012.
- Arginskaya I.I. Collection of assignments in mathematics for independent, test and control works v primary school... - Samara: Corporation "Fedorov", Publishing house "Educational Literature", 2006.
Images used:
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, subtracted, subtracted, factor, dividend, or divisor. The solution of such equations will be discussed in this article.
Here we give the rules for finding an unknown term, multiplier, etc. Moreover, we will immediately consider the application of these rules in practice, solving typical equations.
Page navigation.
So, substituting the number 5 into the original equation 3 + x = 8 instead of x, we get 3 + 5 = 8 - this equality is true, therefore, we correctly found the unknown summand. If, when checking, we received an incorrect numerical equality, then this would indicate to us that we have solved the equation incorrectly. The main reasons for this can be either the use of the wrong rule, or computational errors.
How to find the unknown diminishing, subtracted?
The relationship between addition and subtraction of numbers, which we have already mentioned in the previous paragraph, allows us to obtain the rule for finding the unknown diminished through the known subtracted and the difference, as well as the rule for finding the unknown subtracted through the known diminished and the difference. We will formulate them in turn, and immediately give the solution of the corresponding equations.
To find the unknown diminished, it is necessary to add the subtracted to the difference.
For example, consider the equation x − 2 = 5. It contains an unknown redundant. The above rule indicates to us that to find it, we must add the known subtracted 2 to the known difference 5, we have 5 + 2 = 7. Thus, the desired diminished is seven.
If we 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, in this case we obtain the numerical equality 7−2 = 5. It is correct, therefore, you can be sure that we have correctly identified the value of the unknown diminished.
You can move on to finding the unknown subtracted. It is found using addition according to the following rule: to find the unknown subtracted, it is necessary to subtract the difference from the reduced.
Using this rule, solve an equation of the form 9 − x = 4. In this equation, the unknown is the subtracted. To find it, we need to subtract the known difference 4 from the known decreasing 9, we have 9−4 = 5. Thus, the desired subtraction is 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 subtracted found. Let's check, for which we substitute the found value 5 into the original equation instead of x, and we obtain the numerical equality 9−5 = 4. It is correct, therefore the value of the subtracted found by us is correct.
And before moving on to the next rule, we note that in the 6th grade, the rule for solving equations is considered, which allows you to carry out the transfer of any term from one part of the equation to another with the opposite sign. So all the above rules for finding the unknown term, reduced and subtracted with it are fully consistent.
To find an unknown factor, you need ...
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 an unknown factor, the product must be divided by a 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, find the unknown factor of the equation x · 3 = 12. According to the rule, we need to divide famous work 12 by a known factor of 3. Let's spend: 12: 3 = 4. So the unknown factor is 4.
Briefly, the solution to the equation is written in the form of a sequence of equalities:
x 3 = 12,
x = 12: 3,
x = 4.
It is also advisable to check the result: we substitute the found value into the original equation instead of the letter, we get 4 · 3 = 12 - the correct numerical equality, so we correctly found the value of the unknown factor.
And one more thing: acting according to the learned rule, we actually divide both sides of the equation by a known factor other than zero. In grade 6, it will be said that both sides 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?
Within the framework of our topic, it remains to figure out how to find the unknown divisor with a known divisor and quotient, as well as how to find an unknown divisor with a known divisor 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 dividend of this equation, according to the rule, multiply the known quotient 9 by the known divisor 5, that is, we perform the multiplication natural numbers: 9 5 = 45. Thus, the required dividend is 45.
Let's show a short record of the solution:
x: 5 = 9,
x = 9 5,
x = 45.
The check confirms that the value of the unknown dividend was found correctly. Indeed, when the number 45 is substituted 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 sides of the equation by a known divisor. This 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, the dividend must be divided by the quotient.
Let's look at an example. Find the unknown factor 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 desired divisor is six.
The decision can be made like this:
18: x = 3,
x = 18: 3,
x = 6.
Let's check this result for reliability: 18: 6 = 3 - correct numerical equality, therefore, the root of the equation is found correctly.
It is clear that this rule can be used only when the quotient is different from zero, so as not to collide with division by zero. When the quotient is zero, then 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 any nonzero value of the divisor satisfies this equation. In other words, the roots of such an equation are any numbers that are not equal to zero. If, for a quotient equal to zero, the dividend is nonzero, then at no value of the divisor the original equation does not turn into a true numerical equality, that is, the equation has no roots. To illustrate, we give equation 5: x = 0, it has no solutions.
Sharing rules
The consistent application of the rules for finding the unknown summand, reduced, subtracted, factor, dividend and divisor makes it possible to solve equations with a single variable more than complex kind... Let's look at this with an example.
Consider the equation 3 x + 1 = 7. First, we can find the unknown term 3 x, for this it is necessary to subtract the known term 1 from the sum 7, we obtain 3 x = 7−1 and then 3 x = 6. Now it remains to find the unknown factor, dividing the product 6 by the known factor 3, we have x = 6: 3, whence x = 2. This is how the root of the original equation was found.
To consolidate the material, we present a short solution to one more 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.
Bibliography.
- Maths.... 4th grade. Textbook. for general education. institutions. At 2 pm Part 1 / [M. I. Moro, MA Bantova, GV Beltyukova and others] .- 8th ed. - M .: Education, 2011 .-- 112 p .: ill. - (School of Russia). - ISBN 978-5-09-023769-7.
- Maths: textbook. for 5 cl. general education. institutions / N. Ya. Vilenkin, V. I. Zhokhov, A. S. Chesnokov, S. I. Shvartsburd. - 21st ed., Erased. - M .: Mnemozina, 2007 .-- 280 p .: ill. ISBN 5-346-00699-0.
Equations are one of the hardest topics to learn, but they are powerful enough for most tasks.
Various processes occurring in nature are described with the help of equations. 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 how 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 contentWhat 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 it is substituted into the original equation, the correct numerical equality is obtained.
For example, the expression 3 + 2 = 5 is equal. When calculating the left side, you get the correct numerical equality 5 = 5.
But equality 3 + x= 5 is an equation because it contains the variable x, the value of which can be found. The value must be such that when you substitute this value into the original equation, you get the correct numeric equality.
In other words, we must find such a value that the equal sign would justify its location - the left side should be equal to the right side.
Equation 3+ x= 5 is elementary. Variable value x is equal to 2. For any other value, equality will not be met
The number 2 is said to be root or by solving the equation 3 + x = 5
Root or equation solution Is the value of the variable at which the equation becomes a valid numeric equality.
There may be few or no roots at all. Solve the equation means to find its roots or to prove that there are no roots.
The variable included in the equation is also called unknown... You have the right to call whatever is more convenient for you. These are synonyms.
Note... Collocation "Solve the equation" speaks for itself. Solving an equation means “equalizing” the equality — making it balanced so that the left side is equal to the right side.
Express one through the other
Traditionally, the study of equations begins with learning how to express one number in 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 in terms of 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 should be done with the numbers 10 and 8 to get the number 2”. It is clear that to get the number 2, you need to subtract the number 8 from the number 10.
So we do it. 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 you can see from the example, there is nothing complicated about it.
When solving equations, in particular when expressing one number in terms of others, it is convenient to replace the equality sign with the word there is" ... This should 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 the sign = replaced by the word "is". Moreover, the equality 2 = 10 - 8 can be translated from mathematical language into a full-fledged human language. Then it can be read as follows:
Number 2 there is difference of number 10 and number 8
Number 2 there is the difference between the number 10 and the number 8.
But we will restrict ourselves only to replacing the equal sign with the word "is", and then we will not always do this. Elementary expressions can be understood without translating mathematical language into human language.
Let's return the resulting equality 2 = 10 - 8 to its original state:
8 + 2 = 10
Let's express the number 8 this time. What do you need to do with the rest of the numbers to get the number 8? That's right, you need to subtract the number 2 from the number 10
8 = 10 − 2
Let's 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 sides, 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's return the resulting equality 8 = 6 + 2 to its original state:
8 − 2 = 6
Let us 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
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 from this equality the number 2. To express the number 2, 6 must be 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's return the resulting equality 15 = 3 × 5 to its original state:
Let us express from this equality the number 5. To express the number 5, you need 15 divided by 3
Rules for finding unknowns
Let's consider several rules for finding unknowns. You may be familiar with them, but it doesn't hurt to repeat them again. In the future, they can be forgotten, as we will learn how to solve equations without applying these rules.
Let's go back to the first example, which we considered in the previous topic, where in the equality 8 + 2 = 10 it was required to express the number 2.
In the equality 8 + 2 = 10, the numbers 8 and 2 are terms, and the number 10 is the sum.
To express the number 2, we did the following:
2 = 10 − 8
That is, the term 8 was subtracted from the sum 10.
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 turns into 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 the unknown term, the following rule is provided:
To find the unknown term, you need to subtract the known term from the sum.
What we basically 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 well-known term 8 from the sum 10:
x = 10 − 8
If you calculate the right side of the resulting equality, then you can find out what the variable is equal to x
x = 2
We've solved the equation. Variable value x is equal to 2. To check the value of the variable x send to original equation 8 + x= 10 and substitute for x. It is desirable to do this with any solved equation, since you cannot be sure that the equation is solved correctly:
As a result
The same rule would apply if the unknown term were the first number 8.
x + 2 = 10
In this equation x Is an unknown term, 2 is a known term, 10 is a sum. To find the unknown term x, it is necessary to subtract from the sum 10 the known term 2
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 required to express the number 8.
In the equality 8 - 2 = 6, the number 8 is the subtracted, the number 2 is the subtracted, the number 6 is the difference
To express the number 8, we did the following:
8 = 6 + 2
That is, add the difference 6 and the subtracted 2.
Now imagine that in the equality 8 - 2 = 6 instead of 8 there is a variable x
x − 2 = 6
In this case, the variable x takes on the role of the so-called unknown diminished
To find the unknown reduced, the following rule is provided:
To find the unknown diminished, it is necessary to add the subtracted to the difference.
This is exactly what we did when we expressed the number 8 in the equality 8 - 2 = 6. To express the subtracted 8, we added the subtracted 2 to the difference 6.
Now, to find the unknown diminutive x, we must add the subtracted 2 to the difference 6
x = 6 + 2
If you calculate the right side, then 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 deductible
To find the unknown deductible, the following rule is provided:
To find the unknown subtracted, you need to subtract the difference from the subtracted.
This is exactly 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 reduced 8.
Now, to find the unknown deductable x, again, from the reduced 8, subtract the difference 6
x = 8 − 6
Calculate the right side and find the value x
x = 2
Let's go back 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 multiplier, the number 2 is the factor, the number 6 is the product
To express the number 3, we did the following:
That is, we divided the product of 6 by a 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 multiplier.
To find the unknown multiplier, the following rule is provided:
To find the unknown multiplier, you need to divide the product by a factor.
This is exactly what we did when we expressed the number 3 from the equality 3 × 2 = 6. We divided the product of 6 by a factor of 2.
Now, to find the unknown multiplier x, you need to divide the product 6 by a factor of 2.
Calculating the right side allows us to find the value of the variable x
x = 3
The same rule applies if the variable x is located instead of a multiplier, not a multiplier. 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 factor... To find an unknown factor, the same is provided as for finding an unknown multiplier, namely, dividing the product by a known factor:
To find the unknown factor, you need to divide the product by the multiplier.
This is exactly 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 6 by the multiplier 3.
Now, to find the unknown factor x we have divided the product of 6 by the multiplier 3.
Calculation of the right-hand side of the equality allows you to find out what is x
x = 2
The multiplier and the multiplier are collectively called factors. Since the rules for finding the multiplier and the factor are the same, we can formulate a general rule for finding the unknown factor:
To find an unknown factor, you need to divide the product by a 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 it was required 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 3 by the divisor 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 exactly what we did when we expressed the number 15 from equality. To express the number 15, we multiplied the quotient 3 by the divisor 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 the unknown divisor, the following rule is provided:
This is exactly what we did when we expressed the number 5 from equality. To express the number 5, we divided 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 is how 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 diminished, it is necessary to add the subtracted to the difference;
- To find the unknown subtracted, you need to subtract the difference from the subtracted;
- To find the unknown multiplier, you need to divide the product by a factor;
- To find an unknown factor, you need to divide the product by the multiplier;
- To find the unknown dividend, you need to multiply the quotient by the divisor;
- To find the 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
Subtraction components are minuend, subtrahend and difference
The components of the multiplication are multiplicand, factor and work
The division components 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 this rule by heart.
Example 1... Find the root of equation 45 + x = 60
45 - term, x- unknown term, 60 - sum. We are dealing with components of addition. We recall that to find the unknown term, you need to subtract the known term from the sum:
x = 60 − 45
We calculate the right side, we get the value x equal to 15
x = 15
So the root of the equation 45 + x= 60 is 15.
Most often, the unknown term must be reduced to a form in which it could be expressed.
Example 2... Solve the equation
Here, unlike the previous example, the unknown term cannot be expressed immediately, since it contains a coefficient 2. Our task is to reduce this equation to a form in which it would be possible to express x
V 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.
Moreover, the term 2 x contains a variable x... After finding the value of the variable x term 2 x will take on a different form. Therefore, the 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 got a new equation. Now we are dealing with the components of multiplication: multiplication, multiplier and product. 2 - multiplicable, x- multiplier, 4 - product
In this case, the variable x is not just a factor, but an unknown factor
To find this unknown factor, you need to divide the product by the multiplier:
We calculate the right side, we get the value of the variable x
To check, we send the found root to the original equation and substitute instead x
Example 3... Solve the equation 3x+ 9x+ 16x= 56
Express the unknown immediately 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 - multiplicable, x- multiplier, 56 - product. Wherein x is an unknown factor. To find the unknown factor, you need to divide the product by the multiplier:
From here x is equal to 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, a new equation was obtained 28 x= 56. Old Equation 3x + 9x + 16x = 56 and the resulting new equation 28 x= 56 are called equivalent equations since their roots are the same.
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. First, we substitute this root into the equation 3x+ 9x+ 16x= 56 and then into equation 28 x= 56, which was obtained as a result of bringing similar terms on the left side of the previous equation. We must get correct numeric equalities
According to the order of actions, the multiplication is performed first:
Substitute the root 2 into the second equation 28 x= 56
We see that the roots of both equations coincide. Hence the equations 3x+ 9x+ 16x= 56 and 28 x= 56 are indeed equivalent.
To solve the equation 3x+ 9x+ 16x= 56 we used one of them - the reduction of similar terms. The correct identical transformation of the equation allowed us to obtain an equivalent equation 28 x= 56, which is easier to solve.
From identical transformations on this moment we can only reduce fractions, bring such terms, take the common factor out of parentheses, and also open parentheses. There are other transformations to be aware of. But for a general idea of identical transformations of equations, the topics we have studied are quite enough.
Consider some transformations that make it possible to obtain an 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 the equation, you get an equation that is equivalent to the given one.
In other words, the root of the equation does not change if the same number is added (or subtracted from both sides) to both sides of the equation.
Example 1... Solve the equation 
Subtract 10 from both sides of the equation
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 instead x found value 2
We got the correct numerical equality. So the equation is solved correctly.
Solving the equation we subtract 10 from both sides of the equation. As a result, an equivalent equation was obtained. The root of this equation, like the equations also equal to 2
Example 2... Solve equation 4 ( x+ 3) = 16
Subtract 12 from both sides of the equation
On the left side there will be 4 x, and on the right side the number 4
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 us return to the original equation 4 ( x+ 3) = 16 and substitute instead x found value 1
We got the correct numerical equality. So the equation is solved correctly.
Solving equation 4 ( x+ 3) = 16 we subtract 12 from both sides of the equation. As a result, we got 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 to both sides of the equation the number 8
We present similar terms in both sides of the equation:
On the left side there will be 2 x, and on the right side the number 9
In the resulting equation 2 x= 9 we express the unknown term x
Let's go back to the original equation and substitute instead x found value 4.5
We got the correct numerical equality. So 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 equations also equal to 4.5
The next rule, which allows you to get an equivalent equation, is as follows
If you transfer the term from one part to another in the equation, changing its sign, you get an equation equivalent to the given one.
That is, the root of the equation will not change if we transfer the term from one side of the equation to another, changing its sign. This property is one of the most important and one of the most frequently used in solving equations.
Consider the following equation:
The root of this equation is 2. Substitute instead x this root and check if the correct numerical equality is obtained
It turns out the correct equality. So the number 2 is really the root of the equation.
Now let's try to experiment with the terms of this equation, transferring them from one part to another, changing the signs.
For example, the term 3 x is located on the left side of the equality. Let's move it to the right side, changing the sign to the opposite:
The equation turned out 12 = 9x − 3x ... on the right side of this equation:
x is an unknown factor. Let's find this known factor:
From here x= 2. As you can see, the root of the equation has not changed. Hence the equations 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 the sign. In reality, the following happened: the term 3 was subtracted from both sides of the equation x
Then, on the left-hand side, similar terms were given and the equation was obtained 12 = 9x − 3x. Then again similar terms were given, but on the right side, and the equation was obtained 12 = 6 x.
But the so-called "transfer" is more convenient for such equations, which is why it has become so widespread. When solving equations, we will often use this transformation.
Equations 12 + 3 are also equivalent x= 9x and 3x - 9x= −12 ... This time in the equation 12 + 3 x= 9x term 12 was moved to the right-hand side, and term 9 x to the left. It should not be forgotten that the signs of these terms were changed during the transfer
The next rule, which allows you to get an equivalent equation, is as follows:
If both sides of the equation are multiplied or divided by the same number, which is not equal to zero, then an equation is obtained that is equivalent to the given one.
In other words, the roots of the equation will not change if both sides of it are multiplied or divided by the same number. This action is often used when you need to solve an equation containing fractional expressions.
Let's first 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 at first to simplify this equation.
In this case, we are dealing with just such an equation. In order to simplify this equation, both sides of it can be multiplied by 8:
We remember that for, you need to multiply the numerator of the 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 fun is happening. The numerators and denominators of both fractions contain a factor of 8, which can be canceled 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
It turns out the correct numerical equality. So 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. So these equations are equivalent.
It is customary to write the factor by which both sides of the equation are multiplied 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:
From this, the root of the equation did not change, but if we did it while at school, we would be made a comment, 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 15 can be reduced by 15, and on the right side, the factors 15 and 5 can be reduced by 5
Let's expand the brackets on the right side of the equation:
We carry over the term x from the left side of the equation to the right side by changing the sign. And the term 15 from the right side of the equation is transferred to the left side, again changing the sign:
Given similar terms in both parts, we obtain
We are dealing with the components of multiplication. Variable x
Let's go back to the original equation and substitute instead x found value 5
It turns out the correct numerical equality. So the equation is solved correctly. When solving this equation, we multiplied both sides by 15. Further, performing identical transformations, we got the equation 10 = 2 x... The root of this equation, like the equations is equal to 5. So these equations are equivalent.
Example 3... Solve the equation
On the left side, you can cut two triplets, 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 known factor:
Let's return to the original equation and substitute instead x found value 9
It turns out the correct numerical equality. So the equation is solved correctly.
Example 4... Solve the equation 
Let us multiply both sides of the equation by 6
Expand the brackets on the left side of the equation. On the right side, the multiplier 6 can be raised to the numerator:
Reduce in both sides of the equations what can be canceled:
Let's rewrite what we have left:
Let's use the transfer of terms. Unknown terms x, we group on the left side of the equation, and the terms free of unknowns - on the right:
Here are 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 go back to the original equation and substitute instead x found value 4
The result is correct numerical equality. So the equation is solved correctly.
Example 5... Solve the equation 
We expand the brackets in both sides of the equation where possible:
Let us multiply both sides of the equation by 15
Let's expand the brackets in both sides of the equation:
Reduce in both sides of the equation, what can be canceled:
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:
We present similar terms in both sides of the equation:
Find the value x
In the resulting answer, you can highlight the whole part:
Let's return to the original equation and substitute instead x found value
It turns out to be a rather cumbersome expression. Let's use variables. We put the left side of the equality into the variable A, and the right-hand side of the equality into the variable B
Our task is to make sure whether the left side is equal to the right one. In other words, prove the equality A = B
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 like the value of the variable A equals . This means that the left side is equal to the right side. Hence, 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 us solve it by the usual method: we group the terms containing unknowns on the left side of the equation, and the terms free of unknowns - on the right. Further, performing the well-known identity transformations, we find the value x
Substitute the found value 2 instead of x to the original equation:
Now let's try to separate all the terms of the equation 30x+ 14x+ 14 = 70x− 40x+ 42 by some number. Note that all the terms of this equation have a common factor 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 identical transformations:
Got root 2. Hence 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 removes 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 on the left side freed the unknown from the factor 7, the root would be found immediately. To do this, it was enough to divide both parts by 7
We will also use this method often.
Multiplication by minus one
If both sides of the equation are multiplied by minus one, then you get an equation equivalent to this one.
This rule follows from the fact that multiplying (or dividing) both sides of the equation by the same number does not change the root of this equation. This means that the root will not change if both parts of it 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 to both sides of the equation the number 5
Here are similar terms:
Now let's remember about. What is the left side of the equation. This is the product of minus one and the variable x
That is, the minus before the variable x, does not refer to the variable itself x, but to one, which we do not see, since it is customary not to write the coefficient 1. This means that the equation actually looks like this:
We are dealing with the components of multiplication. To find NS, you need to divide the product −5 by the known factor −1.
or divide both sides of the equation by −1, which is even easier
So, the root of the equation is 5. To check, we substitute it into the original equation. Do not forget that in the original equation the minus in front of the variable x refers to an invisible unit
The result is correct numerical equality. So the equation is solved correctly.
Now let's try to multiply both sides of the equation by minus one:
After expanding the parentheses, an 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
Hence the equations and 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 we change the signs of all the components included in the equation. To do this, multiply both sides of this equation by −1.
It is clear that from multiplication by −1, any number will change its sign to the opposite. Therefore, the procedure of multiplication by −1 and the opening of the brackets are not described in detail, but the components of the equation with opposite signs are immediately written down.
So, multiplying an equation by −1 can be written in detail as follows:
or you can simply change the signs of all components:
It will turn out the same, but the difference will be that we will save ourselves time.
So, multiplying both sides of the equation by −1, we got the equation. Let's solve this equation. Subtract 4 from both parts and divide both parts by 3
When the root is found, the variable is usually written on the left side, and its value on the right, which we did.
Example 3... Solve the equation
We multiply both sides of the equation by −1. Then all the components will change their signs to the opposite:
From both sides of the resulting equation, subtract 2 x and give similar terms:
We add unity to both sides of the equation and give similar terms: 
Equalization to zero
Recently we learned that if in an equation we transfer a term from one part to another, changing its sign, we get an equation equivalent to the given one.
And what will happen if you transfer from one part to another not one term, but all the terms? It is true that in the part from which all the terms were taken will remain zero. In other words, there will be nothing left.
As an example, consider the equation. We solve this equation, as usual - we group the terms containing unknowns in one part, and leave the numerical terms free of unknowns in the other. Further, performing the well-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 transfer all terms from the right side to the left, changing the signs:
Here are similar terms on the left:
Add 77 to both parts, and divide both parts by 7
An alternative to the rules for finding unknowns
Obviously, knowing about the identical transformations of equations, one does not need to memorize the rules for finding unknowns.
For example, to find the unknown in the equation, we divided the product 10 by the known factor 2
But if in the equation both parts are divided by 2, the root is found at once. On the left side of the equation, the factor 2 in the numerator and the factor 2 in the denominator 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 take advantage of 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. The right side will be 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 greatly shortened by doing the division in your head.
However, you need to know both methods, and only then use the one you like best.
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 the equation 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) that are factors. From the laws of multiplication, we know that the product is zero if at least one of the factors is zero (or the first factor or the second).
That is, in the equation x(x + 9) = 0 equality will be achieved if x will be zero or (x + 9) will be equal to zero.
x= 0 or x + 9 = 0
Equating both of these expressions to zero, we can find the roots of the equation x(x + 9) = 0. The first root, as you can see from the example, was found right away. 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. The check shows that the root is correct:
−9 + 9 = 0
Example 2... Solve the equation
This equation has two roots: 1 and 2. The left side of the equation is the product of expressions ( x- 1) and ( x- 2). And the product is zero if at least one of the factors is zero (or the factor ( x- 1) or factor ( x − 2) ).
Let's find this x in which the expressions ( x- 1) or ( x- 2) vanish:
We substitute in turn the found values into the original equation and make sure that for these values the left side is equal to zero:
When there are infinitely many roots
An equation can have infinitely many roots. That is, by substituting any number in such an equation, we get the correct numerical equality.
Example 1... Solve the equation 
Any number is the root of this equation. If you open the brackets on the left side of the equation and give similar terms, you get the equality 14 = 14. This equality will be obtained for any x
Example 2... Solve the equation
Any number is the root of this equation. If you expand 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, because for any value x, the left side of the equation will not equal the right side. For example, let. Then the equation takes the following form
Example 2... Solve the equation
Let's expand the brackets on the left side of the equality:
Here are similar terms:
We see that the left side is not equal to the right side. And so it will be 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 the body during uniformly accelerated motion.
A useful skill is the ability to express any component in a letter equation. For example, to determine the distance from the equation, you need to express the variable s .
Let us multiply both sides of the equation by t
On the right side, the variables t reduce by t
In the resulting equation, we swap the left and right sides:
We have obtained the formula for finding the distance, which we studied earlier.
Let's try to determine the time from the equation. To do this, you need to express the variable t .
Let us multiply both sides of the equation by t
On the right side, the variables t reduce by t and rewrite what we have left:
In the resulting equation v × t = s we divide both parts into v
On the left, the variables v reduce by v and rewrite what we have left:
We have obtained the formula for determining the 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 literal equation takes 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 the unknown divisor by dividing the dividend by the quotient and thus determine the value of the variable t
or you can use identical transformations. First multiply both sides of the equation by t
Then divide both parts by 50
Example 2 x
Subtract from both sides of the equation a
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 it is customary to call parameters... Equations of the form a + bx = c are called equation with parameters... Depending on the parameters, the root will change.
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, you need to make a small remark. Parameter b should not be zero (b ≠ 0) since division by zero by is allowed.
Example 3... A letter equation is given. Express from the given equation x
Let's expand the brackets in both sides of the equation
Let's use the transfer of terms. Parameters containing a variable x, we will group on the left side of the equation, and the parameters free of this variable - on the right.
On the left side, we take the factor out of the brackets x
Let's split both parts into expression a - b
On the left, 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.
Suppose we are given the equation 4(x - 3) = 2(x+ 4) ... It looks like an equation a (x - c) = b (x + d)... We will solve it in two ways: using identical transformations and using a ready-made solution:
For convenience, we take out from the equation 4(x - 3) = 2(x+ 4) parameter values a, b, c, d ... This will allow us not to make mistakes when substituting:
As in the previous example, the denominator here should not be zero ( a - b ≠ 0). If we come across 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 of the same numbers is 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 the equation 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 letter equation is given. Express from the given equation x
Let's bring the left side of the equation to a common denominator:
Smarter both parts by a
On the left side x put out of brackets
We divide both parts into the expression (1 - a)
Linear Equations in 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 the degrees and roots, so in order not to complicate our life, the word "linear" will be understood as "simple".
Most of the equations solved in this lesson ultimately boiled down to the simplest equation in which you had to divide the product by a known factor. Such, for example, is equation 2 ( x+ 3) = 16. Let's solve it.
Opening the brackets on the left side of the equation, we get 2 x+ 6 = 16. Move the term 6 to the right side, changing the sign. Then we get 2 x= 16 - 6. Calculate the right-hand side, we get 2 x= 10. To find x, we divide the product 10 by the known factor 2. Hence x = 5.
Equation 2 ( x+ 3) = 16 is linear. It reduced to equation 2 x= 10, to find the root of which it was required to divide the product by a known factor. This simplest equation is called linear equation of the first degree with one unknown in the canonical form... Canonical is synonymous with 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 equation 2 x= 10 is a linear equation of the first degree with one unknown in the 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 you can easily determine the value x... Instead of parameters a and b our equation contains numbers 2 and 10. But a similar equation can contain other numbers: positive, negative, or zero.
If in the linear equation a= 0 and b= 0, then the equation has infinitely many roots. Indeed, if a is equal to zero and b is equal to 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 the linear equation a= 0 and b≠ 0, then the equation has no roots. Indeed, if a is equal to zero and b is equal to some number that is 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 the linear equation a≠ 0, and b is equal to any number, then the equation has one root. It is determined by dividing the parameter b per parameter a
Indeed, if a is equal to some nonzero number, say 3, and b is equal to some number, say the number 6, then the equation will take the form.
From here.
There is another form of notation 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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