Goals and objectives:
Educational:
- Consider a method for solving "complex" equations of the form: (x + 3): 8 \u003d 5 and derive an action algorithm for solving them.
- Improve computing skills.
Developing:
- To develop the ability to analyze, reason, explain the way of action of equations of the form: (x + 3): 8 \u003d 5.
Educational:
- Form the ability to work in pairs (listen to the opinion of a friend, discuss a problem, come to a consensus).
Health-saving:
- Learn to take care of your health.
Equipment:
- Multimedia projector and screen;
- A computer;
- Presentation;
- Support memo;
- Tasks on cards.
During the classes:
I. Organizational moment.
- The bell rang. Check your readiness for the math lesson. Everyone is ready.
Let's make sure of this!
- BLITZ: How to find the unknown term? (subtracted, decremented, dividend, divisor, factor).
- Well done! Sit down. We can safely start working. Open your notebooks. Write down the number, cool work.
II. Updating basic knowledge.
1) - I suggest you do a warm-up. Attention to the screen!
(Appendix 1. Presentation -Slide 1).
| 100 ∙ 29 32 ∙ 20 4800: 2 |
a ∙ 15 9000 - in from: 317 |
x ∙ 80 \u003d 640 k: 50 \u003d 500 c + 90 \u003d 34 + 56 |
- Divide the recording data into groups. Who divided by 2? For 3 groups?
Discussion!!! By what principle did he divide…. , and …..?
- Name the numerical expressions. Name the letters. Rest? (Equations.)
(Slide 2)
- Find the values \u200b\u200bof numerical expressions.
- Find the meanings of literal expressions if
a \u003d 0, b \u003d 1, c \u003d 317
- Find “extra” among the equations. Prove it!
- Find the root of 1 equation, 2 equations. (Simple.)
- What needs to be done first to solve a complex equation of this kind? (Simplify.) - How? (Perform action.) What?
- Simplify the equation. Find the root.
III. Topic, tasks.
- Who wants to learn how to solve complex equations of a new kind? Raise a hand! Well done! This means that you are not afraid of difficulties and are ready for new discoveries!
- The topic of our lesson is "Solving" complex "equations of a new kind."
(Since the term "complex" equation is conditional, I enclosed it in quotation marks.)
- Let's define educational tasks:
1. Learn to solve complex equations of a new kind.
2. Make up an algorithm for solving. (Algorithm - order, sequence of actions.)
3. Learn to comment on the solution of equations.
4. Improve computing skills.
Physical education 1.
IV. Work on the topic. Formulation of the problem. Opening a new one.
1) From number 488. Textbook.
- I want to suggest that you visit researchers again now.
□ + 30 \u003d 50 This post is on the board!
- Read the expression. 1 slug. 2 slug. Amount value.
- Is this an equation? Why?
- Insert the expression into the "window"
□ + 30 \u003d 50 - What is the name of the entry? (Difficult ur.) - Does it look like the one that we already know how to solve? - Why?
- Try to find a way to solve this equation. PLEASE NOTE, I did not accidentally sign the components of the action! Checkout without verification!
2) Explanation: - What (what component) is the literal expression 4 ∙ х (this is 1 term) in this sum.
This means that 1 term is a literal expression 4 ∙ х and it is unknown!
The rule does not change! How to find unknown 1 slug?
|
\u003d 50 - 30 - Know how to solve? |
3) - Open the tutorial with. 149 № 488. Read how Misha reasoned.
V. Derivation of the algorithm. Securing a new one.
1) Solve the equation: (x + 3): 8 \u003d 5 1 to the board.
The task! - Try to determine the sequence!
2) Derivation of the algorithm.
- How do you understand that the components will be called: dividend, divisor, quotient value.
- Which division is the first or the last? \u003d Where to start?
3). Algorithm (Slide 3).
- I'll define the last action and name the components.
- I will define the unknown component and remember the rule for finding it.
- I'll write a new equation and simplify it.
- I'll solve a simple equation.
4) Reading the memo for commenting.
five). No. 489. Textbook. Commenting.
Physical education 2 (for eyes).
6). Teamwork. Work in pairs.
1) (y– 5) ∙ 4 \u003d 28
2) 3 ∙ a - 7 \u003d 14
3) (24 + d): 8 \u003d 7
4) 63: (14 - x) \u003d 7
Fill out the self-check table!
| The equation. | 1 | 2 | 3 | 4 |
| Decision. | ||||
Linear equations. Solution, examples.
Attention!
There are additional
materials in Special section 555.
For those who are "not very ..."
And for those who "very much ...")
Linear equations.
Linear equations are not the most difficult topic in school mathematics. But there are tricks there that can puzzle even a trained student. Shall we figure it out?)
Usually a linear equation is defined as an equation of the form:
ax + b = 0 Where a and b - any numbers.
2x + 7 \u003d 0. Here a \u003d 2, b \u003d 7
0.1x - 2.3 \u003d 0 Here a \u003d 0.1, b \u003d -2.3
12x + 1/2 \u003d 0 Here a \u003d 12, b \u003d 1/2
Nothing complicated, right? Especially if you don't notice the words: "where a and b are any numbers"... And if you notice, but carelessly think?) After all, if a \u003d 0, b \u003d 0 (any numbers are possible?), then you get a funny expression:
But that's not all! If, say, a \u003d 0, and b \u003d 5, it turns out quite something out of the ordinary:
Which strains and undermines the credibility of mathematics, yes ...) Especially in exams. But from these strange expressions it is also necessary to find the X! Which is not there at all. And, surprisingly, this X is very easy to find. We will learn how to do this. In this tutorial.
How do you know a linear equation by its appearance? It depends on the appearance.) The trick is that linear equations are not only equations of the form ax + b = 0 , but also any equations that are reduced to this form by transformations and simplifications. And who knows whether it can be reduced or not?)
A linear equation can be clearly recognized in some cases. Say, if we have an equation in which there are only unknowns in the first degree, and numbers. And in the equation there is no fractions divided by unknown , it is important! And division by number, or a numeric fraction - please! For instance:
This is a linear equation. There are fractions, but there are no x's in the square, in the cube, etc., and there are no x's in the denominators, i.e. not division by x... And here is the equation
cannot be called linear. Here the x's are all in the first degree, but there is division by expression with x... After simplifications and transformations, you can get a linear equation, and a quadratic, and anything you like.
It turns out that you cannot find out a linear equation in some tricky example until you almost solve it. This is upsetting. But assignments usually don't ask about the type of equation, right? The tasks are given equations solve. This makes me happy.)
Solving linear equations. Examples.
The entire solution to linear equations consists of identical transformations of equations. By the way, these transformations (as many as two!) Underlie the solutions all equations of mathematics. In other words, the solution any the equation begins with these very transformations. In the case of linear equations, it (the solution) is based on these transformations and ends with a full-fledged answer. It makes sense to go to the link, right?) Moreover, there are also examples of solving linear equations.
Let's start with the simplest example. Without any pitfalls. Suppose we need to solve this equation.
x - 3 \u003d 2 - 4x
This is a linear equation. X is all in the first degree, there is no division by X. But, in fact, we do not care what equation it is. We need to solve it. The scheme is simple. Collect everything with x on the left side of the equation, everything without x (number) on the right.
To do this, you need to transfer - 4x to the left, with a sign change, of course, but - 3 - to the right. By the way, this is first identical transformation of equations. Are you surprised? So, we didn't follow the link, but in vain ...) We get:
x + 4x \u003d 2 + 3
We give similar ones, we believe:
What do we lack for complete happiness? Yes, so that there was a clean X on the left! The five is in the way. Getting rid of the top five with second identical transformation of equations. Namely, we divide both sides of the equation by 5. We get a ready answer:
An elementary example, of course. This is for warm-up.) It is not very clear why I recalled identical transformations here? Okay. We take the bull by the horns.) Let's decide something more impressive.
For example, here's the equation:
Where do we start? With x - to the left, without x - to the right? Could be so. Little steps along the long road. Or you can immediately, in a universal and powerful way. If, of course, you have in your arsenal identical transformations of equations.
I ask you a key question: what do you dislike most about this equation?
95 people out of 100 will answer: fractions ! The answer is correct. So let's get rid of them. Therefore, we start right away with second identity transformation... What do you need to multiply the fraction on the left so that the denominator can be reduced completely? Right, 3. And on the right? By 4. But mathematics allows us to multiply both sides by the same number... How do we get out? And let's multiply both sides by 12! Those. by a common denominator. Then both the three and the four will decrease. Do not forget that you need to multiply each part wholly... This is what the first step looks like:
Expand the brackets:
Note! Numerator (x + 2) I bracketed! This is because when you multiply fractions, the numerator is multiplied entirely, entirely! And now the fractions can be reduced:
Expand the remaining brackets:
Not an example, but sheer pleasure!) Now we recall the spell from the elementary grades: with an x \u200b\u200b- to the left, without an x \u200b\u200b- to the right! And apply this transformation:
Here are similar ones:
And we divide both parts by 25, i.e. apply the second transformation again:
That's all. Answer: x=0,16
Take a note: to bring the original confused equation to a pleasant form, we used two (only two!) identical transformations - transfer left-right with a change of sign and multiplication-division of the equation by the same number. This is a universal way! We will work this way with any equations! Absolutely any. That is why I am repeating these identical transformations all the time.)
As you can see, the principle of solving linear equations is simple. We take the equation and simplify it using identical transformations until we get the answer. The main problems here are in computation, not in principle of solution.
But ... There are such surprises in the process of solving the most elementary linear equations that they can drive you into a strong stupor ...) Fortunately, there can be only two such surprises. Let's call them special cases.
Special cases when solving linear equations.
First surprise.
Suppose you come across an elementary equation, something like:
2x + 3 \u003d 5x + 5 - 3x - 2
Slightly bored, we transfer it with an X to the left, without an X - to the right ... With a change of sign, everything is chin-chinar ... We get:
2x-5x + 3x \u003d 5-2-3
We consider, and ... oh shit !!! We get:
This equality in itself is not objectionable. Zero is indeed zero. But X is gone! And we must write in the answer what is x. Otherwise, the decision does not count, yes ...) Dead end?
Calm! In such doubtful cases, the most general rules save. How to solve equations? What does it mean to solve an equation? It means, find all the x values \u200b\u200bthat, when substituted into the original equation, will give us the correct equality.
But we have true equality already happened! 0 \u003d 0, how much more accurate ?! It remains to figure out at what xx it turns out. What x values \u200b\u200bcan be substituted in initial equation if these x's will shrink to zero anyway? Come on?)
Yes!!! Xs can be substituted any! What you want. At least 5, at least 0.05, at least -220. They will shrink anyway. If you don't believe, you can check.) Substitute any x values \u200b\u200bin initial equation and count. All the time, the pure truth will be obtained: 0 \u003d 0, 2 \u003d 2, -7.1 \u003d -7.1 and so on.
Here's the answer: x - any number.
The answer can be written in different mathematical symbols, the essence does not change. This is an absolutely correct and complete answer.
Second surprise.
Let's take the same elementary linear equation and change only one number in it. This is what we will solve:
2x + 1 \u003d 5x + 5 - 3x - 2
After the same identical transformations, we get something intriguing:
Like this. Solved a linear equation, got a strange equality. In mathematical terms, we got incorrect equality. And in simple terms, this is not true. Rave. But nevertheless, this nonsense is a very good reason for solving the equation correctly.)
Again, we think based on the general rules. What x, when substituted in the original equation, will give us true equality? Yes, none! There are no such x's. Whatever you substitute, everything will be reduced, delirium will remain.)
Here's the answer: no solutions.
This is also a complete answer. In mathematics, such answers are common.
Like this. Now, I hope, the loss of x in the process of solving any (not only linear) equation will not confuse you at all. The matter is already familiar.)
Now that we have figured out all the pitfalls in linear equations, it makes sense to solve them.
If you like this site ...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
In this video, we will analyze a whole set of linear equations that are solved using the same algorithm - that is why they are called the simplest ones.
To begin with, let's define: what is a linear equation and what is the simplest of them?
A linear equation is one in which there is only one variable, and only in the first degree.
The simplest equation means the construction:
All other linear equations are reduced to the simplest ones using the algorithm:
- Expand parentheses, if any;
- Move the terms containing the variable to one side of the equal sign, and the terms without the variable to the other;
- Bring similar terms to the left and right of the equal sign;
- Divide the resulting equation by the coefficient of the variable $ x $.
Of course, this algorithm does not always help. The fact is that sometimes, after all these machinations, the coefficient at the variable $ x $ turns out to be zero. In this case, two options are possible:
- The equation has no solutions at all. For example, when you get something like $ 0 \\ cdot x \u003d 8 $, i.e. there is zero on the left and a nonzero number on the right. In the video below, we will look at several reasons at once why this situation is possible.
- The solution is all numbers. The only case when this is possible is the equation has been reduced to the construction $ 0 \\ cdot x \u003d 0 $. It is quite logical that no matter what $ x $ we substitute, we will still get "zero equal to zero", i.e. correct numeric equality.
Now let's see how it all works in real-life problems.
Examples of solving equations
Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality containing exactly one variable, and it goes only to the first degree.
Such constructions are solved in about the same way:
- First of all, you need to expand the parentheses, if any (as in our last example);
- Then bring similar
- Finally, seize the variable, i.e. everything that is associated with a variable - the terms in which it is contained - should be transferred in one direction, and everything that is left without it should be transferred to the other side.
Then, as a rule, you need to bring similar ones on each side of the resulting equality, and after that it remains only to divide by the coefficient at the "x", and we will get the final answer.
In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Usually mistakes are made either when expanding parentheses, or when calculating "pluses" and "minuses".
In addition, it happens that a linear equation has no solutions at all, or so that the solution is the entire number line, i.e. any number. We will analyze these subtleties in today's lesson. But we will start, as you already understood, with the simplest tasks.
Scheme for solving the simplest linear equations
First, let me once again write the whole scheme for solving the simplest linear equations:
- Expand the brackets, if any.
- We secrete the variables, i.e. everything that contains "x" is transferred to one side, and without "x" - to the other.
- We present similar terms.
- We divide everything into the coefficient at the "x".
Of course, this scheme does not always work, there are certain subtleties and tricks in it, and now we will get to know them.
Solving real-life examples of simple linear equations
Problem number 1
The first step requires us to expand the brackets. But they are not in this example, so we skip this stage. In the second step, we need to seize the variables. Please note: we are talking only about individual terms. Let's write:
We present similar terms on the left and right, but this has already been done. Therefore, we move on to the fourth step: divide by coefficient:
\\ [\\ frac (6x) (6) \u003d - \\ frac (72) (6) \\]
So we got the answer.
Problem number 2
In this problem, we can observe the brackets, so let's expand them:
Both on the left and on the right we see approximately the same construction, but let's proceed according to the algorithm, i.e. we secrete the variables:
Here are similar ones:
At what roots it is performed. Answer: for any. Therefore, we can write that $ x $ is any number.
Problem number 3
The third linear equation is more interesting:
\\ [\\ left (6-x \\ right) + \\ left (12 + x \\ right) - \\ left (3-2x \\ right) \u003d 15 \\]
There are a few parentheses here, but they are not multiplied by anything, they just have different signs in front of them. Let's open them up:
We carry out the second step already known to us:
\\ [- x + x + 2x \u003d 15-6-12 + 3 \\]
Let's count:
We carry out the last step - we divide everything by the coefficient at "x":
\\ [\\ frac (2x) (x) \u003d \\ frac (0) (2) \\]
Things to remember when solving linear equations
Aside from too simple tasks, I would like to say the following:
- As I said above, not every linear equation has a solution - sometimes there are simply no roots;
- Even if there are roots, there may be zero among them - there is nothing wrong with that.
Zero is the same number as the rest, you should not discriminate against it in any way or assume that if you get zero, then you did something wrong.
Another feature is related to parenthesis expansion. Please note: when there is a "minus" in front of them, then we remove it, but in brackets we change the signs to opposite... And then we can open it according to standard algorithms: we will get what we saw in the calculations above.
Understanding this simple fact will allow you to avoid stupid and hurtful mistakes in high school, when such actions are taken for granted.
Solving complex linear equations
Let's move on to more complex equations. Now the constructions will become more complex and a quadratic function will arise when performing various transformations. However, you should not be afraid of this, because if, according to the author's intention, we are solving a linear equation, then in the process of transformation all monomials containing a quadratic function will necessarily cancel.
Example # 1
Obviously, the first step is to expand the parentheses. Let's do it very carefully:
Now for privacy:
\\ [- x + 6 ((x) ^ (2)) - 6 ((x) ^ (2)) + x \u003d -12 \\]
Here are similar ones:
Obviously, this equation has no solutions, so we will write in the answer:
\\ [\\ varnothing \\]
or no roots.
Example No. 2
We follow the same steps. First step:
Move everything with the variable to the left, and without it to the right:
Here are similar ones:
Obviously, this linear equation does not have a solution, so we write it this way:
\\ [\\ varnothing \\],
or there are no roots.
Solution nuances
Both equations are completely solved. Using these two expressions as an example, we once again made sure that even in the simplest linear equations, everything may not be so simple: there can be either one root, or none, or infinitely many. In our case, we considered two equations, in both there are simply no roots.
But I would like to draw your attention to another fact: how to work with parentheses and how to open them if there is a minus sign in front of them. Consider this expression:
Before disclosing, you need to multiply everything by "X". Note: multiplies each individual term... Inside there are two terms - respectively, two terms and multiplied.
And only after these seemingly elementary, but very important and dangerous transformations are performed, you can expand the parenthesis from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the parentheses, which means that everything that goes down just changes signs. In this case, the brackets themselves disappear and, most importantly, the leading minus also disappears.
We do the same with the second equation:
It is not by chance that I draw attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and again learn to solve such simple equations.
Of course, the day will come and you will hone these skills to automatism. You no longer have to perform so many transformations every time, you will write everything in one line. But while you are just learning, you need to write each action separately.
Solving even more complex linear equations
What we are going to solve now is difficult to call the simplest task, but the meaning remains the same.
Problem number 1
\\ [\\ left (7x + 1 \\ right) \\ left (3x-1 \\ right) -21 ((x) ^ (2)) \u003d 3 \\]
Let's multiply all the elements in the first part:
Let's do the seclusion:
Here are similar ones:
We carry out the last step:
\\ [\\ frac (-4x) (4) \u003d \\ frac (4) (- 4) \\]
Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, they mutually annihilated, which makes the equation exactly linear and not square.
Problem number 2
\\ [\\ left (1-4x \\ right) \\ left (1-3x \\ right) \u003d 6x \\ left (2x-1 \\ right) \\]
Let's take the first step neatly: multiply every element in the first bracket by every element in the second. In total, there should be four new terms after the transformations:
Now let's carefully perform the multiplication on each term:
Let's move the terms with "x" to the left, and without - to the right:
\\ [- 3x-4x + 12 ((x) ^ (2)) - 12 ((x) ^ (2)) + 6x \u003d -1 \\]
Here are similar terms:
Once again, we received the final answer.
Solution nuances
The most important note about these two equations is as follows: as soon as we start multiplying the brackets in which there is more than it is a term, then this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we get four terms.
Algebraic sum
With the last example, I would like to remind the students what an algebraic sum is. In classical mathematics, by $ 1-7 $ we mean a simple construction: subtract seven from one. In algebra, we mean by this the following: to the number "one" we add another number, namely "minus seven". This is how the algebraic sum differs from the usual arithmetic one.
Once, when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.
In conclusion, let's look at a couple more examples that will be even more complex than the ones we just looked at, and to solve them we will have to slightly expand our standard algorithm.
Solving equations with a fraction
To solve such problems, we will have to add one more step to our algorithm. But first, I will remind our algorithm:
- Expand brackets.
- Secreate variables.
- Bring similar ones.
- Divide by factor.
Alas, this wonderful algorithm, for all its effectiveness, is not entirely appropriate when we are faced with fractions. And in what we will see below, we have a fraction on the left and on the right in both equations.
How to work in this case? Everything is very simple! To do this, you need to add one more step to the algorithm, which can be done both before and after the first action, namely, get rid of fractions. Thus, the algorithm will be as follows:
- Get rid of fractions.
- Expand brackets.
- Secreate variables.
- Bring similar ones.
- Divide by factor.
What does “get rid of fractions” mean? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numeric according to the denominator, i.e. everywhere in the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, then we get rid of fractions.
Example # 1
\\ [\\ frac (\\ left (2x + 1 \\ right) \\ left (2x-3 \\ right)) (4) \u003d ((x) ^ (2)) - 1 \\]
Let's get rid of the fractions in this equation:
\\ [\\ frac (\\ left (2x + 1 \\ right) \\ left (2x-3 \\ right) \\ cdot 4) (4) \u003d \\ left (((x) ^ (2)) - 1 \\ right) \\ cdot four\\]
Pay attention: everything is multiplied by "four" once, ie. just because you have two parentheses doesn't mean you need to multiply each of them by four. Let's write:
\\ [\\ left (2x + 1 \\ right) \\ left (2x-3 \\ right) \u003d \\ left (((x) ^ (2)) - 1 \\ right) \\ cdot 4 \\]
Now let's open:
Solve the variable:
We carry out the reduction of similar terms:
\\ [- 4x \u003d -1 \\ left | : \\ left (-4 \\ right) \\ right. \\]
\\ [\\ frac (-4x) (- 4) \u003d \\ frac (-1) (- 4) \\]
We have got the final solution, we go to the second equation.
Example No. 2
\\ [\\ frac (\\ left (1-x \\ right) \\ left (1 + 5x \\ right)) (5) + ((x) ^ (2)) \u003d 1 \\]
Here we perform all the same actions:
\\ [\\ frac (\\ left (1-x \\ right) \\ left (1 + 5x \\ right) \\ cdot 5) (5) + ((x) ^ (2)) \\ cdot 5 \u003d 5 \\]
\\ [\\ frac (4x) (4) \u003d \\ frac (4) (4) \\]
The problem has been solved.
That, in fact, is all that I wanted to tell today.
Key points
The key findings are as follows:
- Know the algorithm for solving linear equations.
- Ability to open brackets.
- Do not worry if you have quadratic functions somewhere, most likely they will shrink in the process of further transformations.
- Roots in linear equations, even the simplest ones, are of three types: one single root, the whole number line is a root, there are no roots at all.
I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site, solve the examples presented there. Stay tuned for a lot more interesting things!
How to learn to solve simple and complex equations
Dear Parents!
Without basic mathematical training, it is impossible to formulate the education of a modern person. At school, mathematics serves as a reference subject for many related disciplines. In post-school life, lifelong education becomes a real necessity, which requires basic general school training, including mathematics.
In elementary school, not only knowledge on the main topics is laid, but also logical thinking, imagination and spatial representations develop, as well as an interest in this subject is formed.
Observing the principle of continuity, we will focus on the most important topic, namely "The relationship between the components of actions in solving compound equations."
With this lesson, you can easily learn how to solve complicated equations. In the lesson, you will learn in detail the step-by-step instructions for solving complicated equations.
Many parents are perplexed by the question - how to get children to learn how to solve simple and complex equations. If the equations are simple, that's half the trouble, but there are also complex ones - for example, integral ones. By the way, for information, there are also such equations, over the solution of which the best minds of our planet are fighting and for the solution of which very significant monetary awards are issued. For example, if you rememberPerelmanand an unclaimed cash bonus of several million.
However, let's go back to simple mathematical equations and repeat the types of equations and names of components. A little warm-up:
_________________________________________________________________________
WARM UP
Find the extra number in each column:
2) What word is missing in each column?
3) Connect the words from the first column with the words from the 2nd column.
"Equation" "Equality"
4) How do you explain what equality is?
5) What about the "equation"? Is this equality? What is special about it?
term sum
decreasing difference
subtracted product
factorequality
dividend
the equation
Conclusion: An equation is equality with a variable whose value must be found.
_______________________________________________________________________
I invite each group to write on a piece of paper with a felt-tip pen the equations: (on the board)
| Group 1 - with an unknown term; Group 2 - with an unknown diminished; Group 3 - with an unknown deducted; Group 4 - with an unknown divisor; Group 5 - with an unknown dividend; Group 6 - with an unknown multiplier. | 1 group x + 8 \u003d 15 Group 2 x - 8 \u003d 7 3 group 48 - x \u003d 36 4 group 540: x \u003d 9 5 group x: 15 \u003d 9 6 group x * 10 \u003d 360 |
One of the group must read his equation in mathematical language and comment on their solution, that is, speak the operation being performed with known components of the actions (algorithm).
Conclusion: We are able to solve simple equations of all kinds according to the algorithm, read and write literal expressions.
I propose to solve a problem in which a new type of equations appears.
Conclusion: We got acquainted with the solution of equations, one of the parts of which contains a numerical expression, the value of which must be found and a simple equation obtained.
________________________________________________________________________
Consider another version of the equation, the solution of which is reduced to solving a chain of simple equations. Here is one of the introduction of compound equations.
| a + b * c (x - y): 3 2 * d + (m - n) Are write equations? Why? What are these actions called? Read them, calling the last action: | Not. These are not equations, because the equation must have an "\u003d" sign. Expressions a + b * c - the sum of the number a and the product of the numbers b and c; (x - y): 3 - the quotient of the difference between the numbers x and y; 2 * d + (m - n) - the sum of the doubled number d and the difference between the numbers m and n. |
I invite everyone to write down a sentence in mathematical language:
The product of the difference between the numbers x and 4 and the number 3 is 15.
CONCLUSION: The problem situation that has arisen motivates the setting of the lesson goal: to learn to solve equations in which the unknown component is an expression. Such equations are compound equations.
__________________________________________________________________________
Or maybe the types of equations already studied will help us? (algorithms)
Which known equation does our equation look like? X * a \u003d b
VERY IMPORTANT QUESTION: What is the expression on the left side - sum, difference, product or quotient?
(x - 4) * 3 \u003d 15 (Product)
Why? (since the last action is multiplication)
Output:Such equations have not yet been considered. But you can decide if the expressionx - 4 put a card (y is a game), and you get an equation that can be easily solved using a simple algorithm for finding the unknown component.
When solving compound equations, it is necessary to select an action at an automated level at each step, commenting, naming the components of the action.
| Simplify part |
| Not ↓ Yes |
(y - 5) *
4
=
28
y - 5 = 28: 4
y - 5 \u003d 7
y \u003d 5 +7
y \u003d 12
(12 - 5) * 4 = 28
28 \u003d 28 (and)
Output:In classes with different backgrounds, this work may be organized in different ways. In more prepared classes, even for the initial consolidation, expressions can be used in which not two, but three or more actions, but their solution requires more steps with each step simplifying the equation, until a simple equation is obtained. And each time one can observe how the unknown component of actions changes.
_____________________________________________________________________________
CONCLUSION:
When it comes to something very simple, understandable, we often say: "The matter is clear, like two and two is four!"
But before thinking that two times two is four, people had to study for many, many thousand years.
Many of the rules from school textbooks of arithmetic and geometry were known to the ancient Greeks more than two thousand years ago.
Everywhere, where it is necessary to count something, measure, compare, one cannot do without mathematics.
It is difficult to imagine how people would live if they did not know how to count, measure, compare. This is what mathematics teaches.
Today you plunged into school life, played the role of students and I invite you, dear parents, to evaluate your skills on a scale.
| My skills | Date and estimate |
| Action components. | |
| Equation with an unknown component. | |
| Reading and writing expressions. | |
| Find the root of an equation in a simple equation. | |
| Find the root of an equation that contains a numeric expression. | |
| Find the root of an equation in which the unknown action component is an expression. |