The equation has an unknown number term. Solving an equation with an unknown term. What is the equation
Asks if you like mathematics.
What adjectives characterize this science?
What else do you think this science is?
Whose portrait is on the board?
Do you know why the portrait of M.V. Lomonosov in our lesson?
He said: “Mathematics must be taught later because it puts the mind in order.”
So what else is this science?
Relying on the words of M.V. Lomonosov, shall we study mathematics?
Offers a title for the entry.
Offers to solve equations, find the “extra” and prove.
Asks how to find an unknown term.
Invites the student to complete the task on the card on the board independently.
And the rest of the students are offered
Game "Yes and No". (Game presentation)
Suggests a title.
He asks what unites them.
Suggests dividing the equations into 2 groups.
Offers to explain what is the difference between the equations that were not solved, i.e. complex.
Offers to name the topic of the lesson and formulate the task.
He asks what will help him learn to solve complex equations.
He asks if we can make a simple equation out of the new equation that we know how to solve and what needs to be done for this.
Can we find the amount? How?
Explains that in mathematics this is called simplifying an equation.
He asks whether the sum can be expressed as a quotient of numbers, a difference of numbers, or a sum of numbers.
Organizes work in pairs. Offers to organize the algorithm for solving the equation and determine whether it is an algorithm for solving a simple or complex equation.
Offers to justify the answer.
Offers to check on the board.
Offers to determine what these equations are and explain the solution of the equations using an algorithm.
Offers to compare equations, distribute them according to degree of complexity, and solve more complex ones using an algorithm on the board.
Offers to solve a problem by composing an equation using an algorithm.
He suggests building a scale of knowledge, assessing your knowledge and skills, and marking their level with a pencil:
1.I know what an equation is.
2.I know how to solve a simple equation to find an unknown term.
3.I can simplify it.
4.I can solve a complicated equation to find an unknown term.
Sets a learning task: choose from three equations on a card the equation that you think you can handle and solve it yourself.
Offers checking on the board.
Offers to show on the knowledge scale with a green pen what level you are at.
Asks about difficulties encountered in solving.
Offers to take a square that matches the color of the square of your equation on the card, if the equation is solved correctly. If you decide incorrectly, take a brown square and let's build a diagram on the board.
Offers to evaluate work in class. Do you think we achieved the goal of our lesson? Have you learned how to solve complex equations?
He asks what helped him solve the equations.
Organizes a discussion about doing homework on p. 62 “Choose the task yourself.”
§ 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, 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.
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:
- 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.
- Arginskaya I.I. A collection of mathematics assignments for independent, test and assessment work in elementary school. - Samara: Fedorov Corporation, Educational Literature Publishing House, 2006.
Images used:
To learn how to solve equations quickly and successfully, you need to start with the simplest rules and examples. First of all, you need to learn how to solve equations that have a difference, sum, quotient, or product of some numbers with one unknown on the left, and another number on the right. In other words, in these equations there is one unknown term and either a minuend with a subtrahend, or a dividend with a divisor, etc. It is about equations of this type that we will talk to you.
This article is devoted to the basic rules that allow you to find factors, unknown terms, etc. We will immediately explain all theoretical principles using specific examples.
Finding the unknown term
Let's say we have a certain number of balls in two vases, for example, 9. We know that there are 4 balls in the second vase. How to find the quantity in the second? Let's write this problem in mathematical form, denoting the number that needs to be found as x. According to the original condition, this number together with 4 form 9, which means we can write the equation 4 + x = 9. On the left we have a sum with one unknown term, on the right we have the value of this sum. How to find x? To do this you need to use the rule:
Definition 1
To find the unknown term, you need to subtract the known term from the sum.
In this case, we give subtraction a meaning that is the opposite of addition. In other words, there is a certain connection between the actions of addition and subtraction, which can be expressed literally as follows: if a + b = c, then c − a = b and c − b = a, and vice versa, from the expressions c − a = b and c − b = a, we can deduce that a + b = c.
Knowing this rule, we can find one unknown term using the known term and the sum. Which exact term we know, the first or the second, in this case does not matter. Let's see how to apply this rule in practice.
Example 1
Let's take the equation that we got above: 4 + x = 9. According to the rule, we need to subtract from a known sum equal to 9 a known term equal to 4. Let's subtract one natural number from another: 9 - 4 = 5. We got the term we needed, equal to 5.
Typically, solutions to such equations are written as follows:
- The original equation is written first.
- Next, we write down the equation that resulted after we applied the rule for calculating the unknown term.
- After this, we write the equation that was obtained after all the manipulations with numbers.
This form of notation is needed to illustrate the sequential replacement of the original equation with equivalent ones and to display the process of finding the root. The solution to our simple equation above would be correctly written as:
4 + x = 9, x = 9 − 4, x = 5.
We can check the correctness of the received answer. Let's substitute what we got into the original equation and see if the correct numerical equality comes out of it. Substitute 5 into 4 + x = 9 and get: 4 + 5 = 9. The equality 9 = 9 is correct, which means the unknown term was found correctly. If the equality turned out to be incorrect, then we should go back to the solution and recheck it, since this is a sign of an error. As a rule, most often this is a computational error or the application of an incorrect rule.
Finding an unknown subtrahend or minuend
As we already mentioned in the first paragraph, there is a certain connection between the processes of addition and subtraction. With its help, we can formulate a rule that will help us find an unknown minuend when we know the difference and the subtrahend, or an unknown subtrahend through the minuend or the difference. Let's write these two rules in turn and show how to apply them to solve problems.
Definition 2
To find the unknown minuend, you need to add the subtrahend to the difference.
Example 2
For example, we have the equation x - 6 = 10. Unknown minuend. According to the rule, we need to add the subtracted 6 to the difference of 10, we get 16. That is, the original minuend is equal to sixteen. Let's write down the entire solution:
x − 6 = 10, x = 10 + 6, x = 16.
Let's check the result by adding the resulting number to the original equation: 16 - 6 = 10. The equality 16 - 16 will be correct, which means we have calculated everything correctly.
Definition 3
To find the unknown subtrahend, you need to subtract the difference from the minuend.
Example 3
Let's use the rule to solve the equation 10 - x = 8. We don’t know the subtrahend, so we need to subtract the difference from 10, i.e. 10 - 8 = 2. This means that the required subtrahend is equal to two. Here's the entire solution:
10 - x = 8, x = 10 - 8, x = 2.
Let's check for correctness by substituting the two into the original equation. Let's get the correct equality 10 - 2 = 8 and make sure that the value we found will be correct.
Before moving on to other rules, we note that there is a rule for transferring any terms from one part of the equation to another, replacing the sign with the opposite one. All the above rules fully comply with it.
Finding an unknown factor
Let's look at two equations: x · 2 = 20 and 3 · x = 12. In both, we know the value of the product and one of the factors; we need to find the second. To do this, we need to use another rule.
Definition 4
To find an unknown factor, you need to divide the product by the known factor.
This rule is based on a meaning that is the opposite of the meaning of multiplication. There is the following connection between multiplication and division: a · b = c when a and b are not equal to 0, c: a = b, c: b = c and vice versa.
Example 4
Let's calculate the unknown factor in the first equation by dividing the known quotient 20 by the known factor 2. We divide natural numbers and get 10. Let us write down the sequence of equalities:
x · 2 = 20 x = 20: 2 x = 10.
We substitute the ten into the original equality and get that 2 · 10 = 20. The value of the unknown multiplier was performed correctly.
Let us clarify that if one of the multipliers is zero, this rule cannot be applied. Thus, we cannot solve the equation x · 0 = 11 with its help. This notation makes no sense, since to solve it you need to divide 11 by 0, and division by zero is not defined. We talked about such cases in more detail in the article devoted to linear equations.
When we apply this rule, we are essentially dividing both sides of the equation by a factor other than 0. There is a separate rule according to which such a division can be carried out, and it will not affect the roots of the equation, and what we wrote about in this paragraph is completely consistent with it.
Finding an unknown dividend or divisor
Another case that we need to consider is finding the unknown dividend if we know the divisor and the quotient, as well as finding the divisor when the quotient and the dividend are known. We can formulate this rule using the connection between multiplication and division already mentioned here.
Definition 5
To find the unknown dividend, you need to multiply the divisor by the quotient.
Let's see how this rule is applied.
Example 5
Let's use it to solve the equation x: 3 = 5. We multiply the known quotient and the known divisor together and get 15, which will be the dividend we need.
Here's a summary of the entire solution:
x: 3 = 5, x = 3 5, x = 15.
Checking shows that we calculated everything correctly, because when dividing 15 by 3, it actually turns out to be 5. Correct numerical equality is evidence of a correct solution.
This rule can be interpreted as multiplying the right and left sides of the equation by the same number other than 0. This transformation does not affect the roots of the equation in any way.
Let's move on to the next rule.
Definition 6
To find an unknown divisor, you need to divide the dividend by the quotient.
Example 6
Let's take a simple example - equation 21: x = 3. To solve it, divide the known dividend 21 by the quotient 3 and get 7. This will be the required divisor. Now let’s formalize the solution correctly:
21: x = 3, x = 21: 3, x = 7.
Let's make sure the result is correct by substituting seven into the original equation. 21: 7 = 3, so the root of the equation was calculated correctly.
It is important to note that this rule only applies to cases where the quotient is not equal to zero, because otherwise we will again have to divide by 0. If zero is private, two options are possible. If the dividend is also equal to zero and the equation looks like 0: x = 0, then the value of the variable will be any, that is, this equation has an infinite number of roots. But an equation with a quotient equal to 0 and a dividend different from 0 will not have solutions, since such values of the divisor do not exist. An example would be equation 5: x = 0, which does not have any roots.
Consistent application of rules
Often in practice there are more complex problems in which the rules for finding addends, minuends, subtrahends, factors, dividends and quotients must be applied sequentially. Let's give an example.
Example 7
We have an equation of the form 3 x + 1 = 7. We calculate the unknown term 3 x by subtracting one from 7. We end up with 3 x = 7 − 1, then 3 x = 6. This equation is very simple to solve: divide 6 by 3 and get the root of the original equation.
Here is a short summary of the solution to another equation (2 x − 7) : 3 − 5 = 2:
(2 x − 7) : 3 − 5 = 2 , (2 x − 7) : 3 = 2 + 5 , (2 x − 7) : 3 = 7 , 2 x − 7 = 7 3 , 2 x − 7 = 21, 2 x = 21 + 7, 2 x = 28, x = 28: 2, x = 14.
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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 the success/failure of 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. Mathematical sign for 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. Smallest 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 of 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). (Independent completion. 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).
Math lesson notes 2nd grade
The purpose of the lesson: create the necessary conditions for students to derive the rule for finding an unknown term.Lesson objectives:
form the concepts of “equation”, “root of equation”;
create an algorithm for solving the equation;
consolidate the ability to write equations, find the root of the equation and check the accuracy of the calculation;
improve computational skills, mathematical speech, develop logical thinking;
develop self-control skills and the ability to work in pairs;
develop the ability to work according to a plan and algorithm.
Planned results:
Subject:
know and apply the rule for finding the unknown term when solving simple equations;
be able to write and solve simple equations to find an unknown term.
use mathematical terms correctly in speech.
Metasubject:
educational : search and selection of necessary information; conscious and voluntary construction of a speech utterance; establishing cause-and-effect relationships.
regulatory : identification and awareness by students of what has already been learned and what still needs to be learned, comparison of the method of action and its result with a given standard.
communicative : an emotionally positive attitude towards the collaboration process, the ability to listen to the interlocutor, taking into account different opinions and the ability to justify one’s own, respect for a different point of view.
personal : formation of adequate positive conscious self-esteem, development of cognitive interests, educational motives.
Methods:
partially search; verbal;
Technological lesson map
I .Class organization. Motivation for learning activities.
Today we have an open lesson. Guests have come to our lesson, turn to them and greet them.Sit down quietly.
I am glad that I see your lovely faces again at our next math lesson. Today's lesson is exciting, you are worried. Let's try to lift our spirits, turn to each other, smile, support each other:
Don't be sad today
Together we will be on the road!
Well done! Has your mood changed? What has it become?
Look at the board and choose your lesson setting:
I will:
Attentive
Diligent
Hardworking
For the curious
At the end of the lesson you will say whether you completed it or failed. Let's get to work.
Recording a number. Classwork.
Let's imagine the number 16 as the sum of two numbers, the difference of two numbers, as the product of two numbers, as the difference and product of numbers.
Yes. Calm, joyful, fear and excitement disappeared.
II .
Updating of reference knowledge
Goal: improving computational skills, repeating the composition of numbers
1. Place “+” or “–” signs
2. Fill out the table:
Conclusion:3. Problem
From a piece of fabric 24 m long, first 6 m were cut, and then another 4 m. How many meters of fabric remained 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 an equation makes the equation true...equality
Numerical equalities, numerical inequalities, equations, roots of equations
The equation.
An equality containing an unknown is called an equation.
The root of an equation is a number that, when substituted into the equation instead of x, produces a true numerical equality.
III .
Identifying the location and cause of the problem
Goal: Creating conditions for isolating an equation with an unknown subtrahend;
Identify the location of the problem;
Record the cause of the difficulty in external speech
IV. Formulating the topic and purpose of the lesson
Each of you must remember how equations are solved.
– Look at the diagrams on the board.
What kind of pattern do you think the lesson will be devoted to?
Open the textbook (p. 77), bookmark the textbook page and read the topic of the lesson.
Determine the purpose of the lesson.
We are still poorly able to explain how to find the unknown term
Learn to solve equations with unknown terms.
Solving equations with an unknown term
V . Discovery of new knowledge.
Goal: identifying the rule for finding an unknown subtrahend.
Group work
Find an equation in which you need to find the unknown first term, come up with an algorithm for solving it.Algorithm on slide .
Name the components of addition.
Which component is unknown? (- How to find it using “Whole” and “Part”.
Replace “Whole” and “Part” with the names of the components of the addition actions.
How to find an unknown term?
Where can we find confirmation of our assumptions?
Compare your conclusions with what the authors of the textbook suggest p. 79
Formulate a rule for finding an unknown term.
To find the unknown part, you need to subtract the known part from the whole.
VI .Physical training minute
VII . Primary consolidation with pronunciation in external speech.
Goal: applying the rule when solving equations
Work at the board
Page 79 No. 6,7
They complete the task and pronounce a new concept.
VIII . Independent work in pairs with self-test in class.
Goal: developing the ability to work in pairs, to show responsibility for one’s own choices and the results of one’s activities.
Page 79. No. 8
Ability to work in pairs using an algorithm
The rule for finding an unknown term.
IX . Systematization and repetition.
Goal: organize repetition of skills to find all ways to solve problems
Where can we apply the equation in math lessons?
In solving problems.
Solving the problem with explanation.
There were 32 books on one shelf, 8 on the other, how many books are there 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 assumptions based on working with the textbook material
X.Reflection
Goal: to develop the ability to reflect on one’s activities
What new did you learn in class today?
What was your goal? Have you reached your goal?
What was the topic of the lesson?
Evaluate the correctness of the action at the level of adequate assessment
The ability to self-assess based on the criterion of success in educational activities
Application
Self-control sheet _____________________________________________
At each stage, evaluate your work by selecting the sign in the required line «+».
StageEducational activities
Executed without error
Executed with errors
Experienced great difficulties
Start of the lesson
Lesson mood
1 step
Repetition of covered material. Verbal counting
Step 2
Setting a learning task, lesson goals
Step 3
Group work
Step 4
Primary consolidation
Work from the textbook p. 79 No. 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
X + 120 = 220y – 19= 78