With this video I begin a series of lessons dedicated to systems of equations. Today we will talk about solving systems linear equations addition method- this is one of the most simple ways, but at the same time one of the most effective.
The addition method consists of three simple steps:
- Look at the system and choose a variable that has identical (or opposite) coefficients in each equation;
- Perform algebraic subtraction (for opposite numbers - addition) of equations from each other, and then bring similar terms;
- Solve the new equation obtained after the second step.
If everything is done correctly, then at the output we will get a single equation with one variable— it won’t be difficult to solve it. Then all that remains is to substitute the found root into the original system and get the final answer.
However, in practice everything is not so simple. There are several reasons for this:
- Solving equations using the addition method implies that all lines must contain variables with equal/opposite coefficients. What to do if this requirement is not met?
- Not always, after adding/subtracting equations in the indicated way, we get a beautiful construction that can be easily solved. Is it possible to somehow simplify the calculations and speed up the calculations?
To get the answer to these questions, and at the same time understand a few additional subtleties that many students fail at, watch my video lesson:
With this lesson we begin a series of lectures devoted to systems of equations. And we will start from the simplest of them, namely those that contain two equations and two variables. Each of them will be linear.
Systems is 7th grade material, but this lesson will also be useful for high school students who want to brush up on their knowledge of this topic.
In general, there are two methods for solving such systems:
- Addition method;
- A method of expressing one variable in terms of another.
Today we will deal with the first method - we will use the method of subtraction and addition. But to do this, you need to understand the following fact: once you have two or more equations, you can take any two of them and add them to each other. They are added member by member, i.e. “X’s” are added to “X’s” and similar ones are given, “Y’s” with “Y’s” are similar again, and what is to the right of the equal sign is also added to each other, and similar ones are also given there.
The results of such machinations will be a new equation, which, if it has roots, they will certainly be among the roots of the original equation. Therefore, our task is to do the subtraction or addition in such a way that either $x$ or $y$ disappears.
How to achieve this and what tool to use for this - we’ll talk about this now.
Solving easy problems using the addition method
So, we learn to use the addition method using the example of two simple expressions.
Task No. 1
\[\left\( \begin(align)& 5x-4y=22 \\& 7x+4y=2 \\\end(align) \right.\]
Note that $y$ has a coefficient of $-4$ in the first equation, and $+4$ in the second. They are mutually opposite, so it is logical to assume that if we add them up, then in the resulting sum the “games” will be mutually destroyed. Add it up and get:
Let's solve the simplest construction:
Great, we found the "x". What should we do with it now? We have the right to substitute it into any of the equations. Let's substitute in the first:
\[-4y=12\left| :\left(-4 \right) \right.\]
Answer: $\left(2;-3 \right)$.
Problem No. 2
\[\left\( \begin(align)& -6x+y=21 \\& 6x-11y=-51 \\\end(align) \right.\]
The situation here is completely similar, only with “X’s”. Let's add them up:
We have the simplest linear equation, let's solve it:
Now let's find $x$:
Answer: $\left(-3;3 \right)$.
Important points
So, we have just solved two simple systems of linear equations using the addition method. Key points again:
- If there are opposite coefficients for one of the variables, then it is necessary to add all the variables in the equation. In this case, one of them will be destroyed.
- We substitute the found variable into any of the system equations to find the second one.
- The final response record can be presented in different ways. For example, like this - $x=...,y=...$, or in the form of coordinates of points - $\left(...;... \right)$. The second option is preferable. The main thing to remember is that the first coordinate is $x$, and the second is $y$.
- The rule of writing the answer in the form of point coordinates is not always applicable. For example, it cannot be used when the variables are not $x$ and $y$, but, for example, $a$ and $b$.
In the following problems we will consider the technique of subtraction when the coefficients are not opposite.
Solving easy problems using the subtraction method
Task No. 1
\[\left\( \begin(align)& 10x-3y=5 \\& -6x-3y=-27 \\\end(align) \right.\]
Note that there are no opposite coefficients here, but there are identical ones. Therefore, we subtract the second from the first equation:
Now we substitute the value $x$ into any of the system equations. Let's go first:
Answer: $\left(2;5\right)$.
Problem No. 2
\[\left\( \begin(align)& 5x+4y=-22 \\& 5x-2y=-4 \\\end(align) \right.\]
We again see the same coefficient of $5$ for $x$ in the first and second equations. Therefore, it is logical to assume that you need to subtract the second from the first equation:
We have calculated one variable. Now let's find the second one, for example, by substituting the value $y$ into the second construction:
Answer: $\left(-3;-2 \right)$.
Nuances of the solution
So what do we see? Essentially, the scheme is no different from the solution of previous systems. The only difference is that we do not add equations, but subtract them. We are doing algebraic subtraction.
In other words, as soon as you see a system consisting of two equations in two unknowns, the first thing you need to look at is the coefficients. If they are the same anywhere, the equations are subtracted, and if they are opposite, the addition method is used. This is always done so that one of them disappears, and in the final equation, which remains after subtraction, only one variable remains.
Of course, that's not all. Now we will consider systems in which the equations are generally inconsistent. Those. There are no variables in them that are either the same or opposite. In this case, to solve such systems, an additional technique is used, namely, multiplying each of the equations by a special coefficient. How to find it and how to solve such systems in general, we’ll talk about this now.
Solving problems by multiplying by a coefficient
Example #1
\[\left\( \begin(align)& 5x-9y=38 \\& 3x+2y=8 \\\end(align) \right.\]
We see that neither for $x$ nor for $y$ the coefficients are not only mutually opposite, but also in no way correlated with the other equation. These coefficients will not disappear in any way, even if we add or subtract the equations from each other. Therefore, it is necessary to apply multiplication. Let's try to get rid of the $y$ variable. To do this, we multiply the first equation by the coefficient of $y$ from the second equation, and the second equation by the coefficient of $y$ from the first equation, without touching the sign. We multiply and get a new system:
\[\left\( \begin(align)& 10x-18y=76 \\& 27x+18y=72 \\\end(align) \right.\]
Let's look at it: at $y$ the coefficients are opposite. In such a situation, it is necessary to use the addition method. Let's add:
Now we need to find $y$. To do this, substitute $x$ into the first expression:
\[-9y=18\left| :\left(-9 \right) \right.\]
Answer: $\left(4;-2 \right)$.
Example No. 2
\[\left\( \begin(align)& 11x+4y=-18 \\& 13x-6y=-32 \\\end(align) \right.\]
Again, the coefficients for none of the variables are consistent. Let's multiply by the coefficients of $y$:
\[\left\( \begin(align)& 11x+4y=-18\left| 6 \right. \\& 13x-6y=-32\left| 4 \right. \\\end(align) \right .\]
\[\left\( \begin(align)& 66x+24y=-108 \\& 52x-24y=-128 \\\end(align) \right.\]
Our new system is equivalent to the previous one, however, the coefficients of $y$ are mutually opposite, and therefore it is easy to apply the addition method here:
Now let's find $y$ by substituting $x$ into the first equation:
Answer: $\left(-2;1 \right)$.
Nuances of the solution
The key rule here is the following: we always multiply only by positive numbers - this will save you from stupid and offensive mistakes associated with changing signs. In general, the solution scheme is quite simple:
- We look at the system and analyze each equation.
- If we see that neither $y$ nor $x$ the coefficients are consistent, i.e. they are neither equal nor opposite, then we do the following: we select the variable that we need to get rid of, and then we look at the coefficients of these equations. If we multiply the first equation by the coefficient from the second, and the second, correspondingly, multiply by the coefficient from the first, then in the end we will get a system that is completely equivalent to the previous one, and the coefficients of $y$ will be consistent. All our actions or transformations are aimed only at getting one variable in one equation.
- We find one variable.
- We substitute the found variable into one of the two equations of the system and find the second.
- We write the answer in the form of coordinates of points if we have variables $x$ and $y$.
But even such a simple algorithm has its own subtleties, for example, the coefficients of $x$ or $y$ can be fractions and other “ugly” numbers. We will now consider these cases separately, because in them you can act somewhat differently than according to the standard algorithm.
Solving problems with fractions
Example #1
\[\left\( \begin(align)& 4m-3n=32 \\& 0.8m+2.5n=-6 \\\end(align) \right.\]
First, notice that the second equation contains fractions. But note that you can divide $4$ by $0.8$. We will receive $5$. Let's multiply the second equation by $5$:
\[\left\( \begin(align)& 4m-3n=32 \\& 4m+12.5m=-30 \\\end(align) \right.\]
We subtract the equations from each other:
We found $n$, now let's count $m$:
Answer: $n=-4;m=5$
Example No. 2
\[\left\( \begin(align)& 2.5p+1.5k=-13\left| 4 \right. \\& 2p-5k=2\left| 5 \right. \\\end(align )\right.\]
Here, as in the previous system, there are fractional coefficients, but for none of the variables the coefficients do not fit into each other an integer number of times. Therefore, we use the standard algorithm. Get rid of $p$:
\[\left\( \begin(align)& 5p+3k=-26 \\& 5p-12.5k=5 \\\end(align) \right.\]
We use the subtraction method:
Let's find $p$ by substituting $k$ into the second construction:
Answer: $p=-4;k=-2$.
Nuances of the solution
That's all optimization. In the first equation, we did not multiply by anything at all, but multiplied the second equation by $5$. As a result, we received a consistent and even identical equation for the first variable. In the second system we followed a standard algorithm.
But how do you find the numbers by which to multiply equations? After all, if we multiply by fractions, we get new fractions. Therefore, the fractions must be multiplied by a number that would give a new integer, and after that the variables must be multiplied by coefficients, following the standard algorithm.
In conclusion, I would like to draw your attention to the format for recording the response. As I already said, since here we have not $x$ and $y$, but other values, we use a non-standard notation of the form:
Solving complex systems of equations
As a final note to today's video tutorial, let's look at a couple of really complex systems. Their complexity will consist in the fact that they will have variables on both the left and right. Therefore, to solve them we will have to apply preprocessing.
System No. 1
\[\left\( \begin(align)& 3\left(2x-y \right)+5=-2\left(x+3y \right)+4 \\& 6\left(y+1 \right )-1=5\left(2x-1 \right)+8 \\\end(align) \right.\]
Each equation carries a certain complexity. Therefore, let's treat each expression as with a regular linear construction.
In total, we get the final system, which is equivalent to the original one:
\[\left\( \begin(align)& 8x+3y=-1 \\& -10x+6y=-2 \\\end(align) \right.\]
Let's look at the coefficients of $y$: $3$ fits into $6$ twice, so let's multiply the first equation by $2$:
\[\left\( \begin(align)& 16x+6y=-2 \\& -10+6y=-2 \\\end(align) \right.\]
The coefficients of $y$ are now equal, so we subtract the second from the first equation: $$
Now let's find $y$:
Answer: $\left(0;-\frac(1)(3) \right)$
System No. 2
\[\left\( \begin(align)& 4\left(a-3b \right)-2a=3\left(b+4 \right)-11 \\& -3\left(b-2a \right )-12=2\left(a-5 \right)+b \\\end(align) \right.\]
Let's transform the first expression:
Let's deal with the second one:
\[-3\left(b-2a \right)-12=2\left(a-5 \right)+b\]
\[-3b+6a-12=2a-10+b\]
\[-3b+6a-2a-b=-10+12\]
In total, our initial system will take the following form:
\[\left\( \begin(align)& 2a-15b=1 \\& 4a-4b=2 \\\end(align) \right.\]
Looking at the coefficients of $a$, we see that the first equation needs to be multiplied by $2$:
\[\left\( \begin(align)& 4a-30b=2 \\& 4a-4b=2 \\\end(align) \right.\]
Subtract the second from the first construction:
Now let's find $a$:
Answer: $\left(a=\frac(1)(2);b=0 \right)$.
That's it. I hope this video tutorial will help you understand this difficult topic, namely solving systems of simple linear equations. There will be many more lessons on this topic: we will look at more complex examples, where there will be more variables, and the equations themselves will already be nonlinear. See you again!
In this lesson we will continue to study the method of solving systems of equations, namely the method of algebraic addition. First, let's look at the application of this method using the example of linear equations and its essence. Let's also remember how to equalize coefficients in equations. And we will solve a number of problems using this method.
Topic: Systems of equations
Lesson: Algebraic addition method
1. Method of algebraic addition using linear systems as an example
Let's consider algebraic addition method using the example of linear systems.
Example 1. Solve the system
If we add these two equations, then y cancels out, leaving an equation for x.
If we subtract the second from the first equation, the x's cancel each other out, and we get an equation for y. This is the meaning of the algebraic addition method.
We solved the system and remembered the method of algebraic addition. Let's repeat its essence: we can add and subtract equations, but we must ensure that we get an equation with only one unknown.
2. Method of algebraic addition with preliminary equalization of coefficients
Example 2. Solve the system
The term is present in both equations, so the algebraic addition method is convenient. Let's subtract the second from the first equation.
Answer: (2; -1).
Thus, after analyzing the system of equations, you can see that it is convenient for the method of algebraic addition, and apply it.
Let's consider another linear system.
3. Solution of nonlinear systems
Example 3. Solve the system
We want to get rid of y, but the coefficients of y are different in the two equations. Let's equalize them; to do this, multiply the first equation by 3, the second by 4.
Example 4. Solve the system 
Let's equalize the coefficients for x
You can do it differently - equalize the coefficients for y.
We solved the system by applying the algebraic addition method twice.
The algebraic addition method is also applicable when solving nonlinear systems.
Example 5. Solve the system
Let's add these equations together and we'll get rid of y.
The same system can be solved by applying the algebraic addition method twice. Let's add and subtract from one equation another.
Example 6. Solve the system 
Answer: 
Example 7. Solve the system 
Using the method of algebraic addition we will get rid of the xy term. Let's multiply the first equation by .
The first equation remains unchanged, instead of the second we write the algebraic sum.
Answer: 
Example 8. Solve the system
Multiply the second equation by 2 to isolate a perfect square.
Our task was reduced to solving four simple systems.
4. Conclusion
We examined the method of algebraic addition using the example of solving linear and nonlinear systems. On next lesson Let's consider the method of introducing new variables.
1. Mordkovich A.G. et al. Algebra 9th grade: Textbook. For general education Institutions.- 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.
2. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill.
3. Makarychev Yu. N. Algebra. 9th grade: educational. for general education students. institutions / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, I. E. Feoktistov. — 7th ed., rev. and additional - M.: Mnemosyne, 2008.
4. Alimov Sh. A., Kolyagin Yu. M., Sidorov Yu. V. Algebra. 9th grade. 16th ed. - M., 2011. - 287 p.
5. Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. — 12th ed., erased. - M.: 2010. - 224 p.: ill.
6. Algebra. 9th grade. In 2 parts. Part 2. Problem book for students of general education institutions / A. G. Mordkovich, L. A. Aleksandrova, T. N. Mishustina and others; Ed. A. G. Mordkovich. — 12th ed., rev. - M.: 2010.-223 p.: ill.
1. College section. ru in mathematics.
2. Internet project “Tasks”.
3. Educational portal“I WILL SOLVE THE USE.”
1. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill. No. 125 - 127.
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Very often, students find it difficult to choose a way to solve systems of equations.
In this article we will look at one of the ways to solve systems - the substitution method.
If a common solution to two equations is found, then these equations are said to form a system. In a system of equations, each unknown represents the same number in all equations. To show that given equations form a system, they are usually written one below the other and joined by a curly brace, for example
We note that for x = 15 and y = 5, both equations of the system are correct. This pair of numbers is the solution to the system of equations. Each pair of unknown values that simultaneously satisfies both equations of the system is called a solution to the system.
A system can have one solution (as in our example), infinitely many solutions, or no solutions.
How to solve systems using the substitution method? If the coefficients for some unknown in both equations are equal in absolute value (if they are not equal, then we equate), then by adding both equations (or subtracting one from the other), you can obtain an equation with one unknown. Then we solve this equation. We determine one unknown. We substitute the resulting value of the unknown into one of the system equations (the first or the second). We find another unknown. Let's look at examples of the application of this method.
Example 1. Solve the system of equations
Here the coefficients for y are equal in absolute value, but opposite in sign. Let's try to add the equations of the system term by term.
We substitute the resulting value x = 4 into some equation of the system (for example, into the first one) and find the value y:
2 *4 +y = 11, y = 11 – 8, y = 3.
Our system has a solution x = 4, y = 3. Or the answer can be written in parentheses as the coordinates of a point, x in the first place, y in the second.
Answer: (4; 3)
Example 2. Solve system of equations
Let's equalize the coefficients of the variable x, to do this we multiply the first equation by 3, and the second by (-2), we get
Be careful when adding equations
Then y = - 2. Substitute the number (-2) instead of y into the first equation, and we get
4x + 3(-2) = - 4. Solve this equation 4x = - 4 + 6, 4x = 2, x = ½.
Answer: (1/2; - 2)
Example 3. Solve the system of equations
Multiply the first equation by (-2)
Solving the system
we get 0 = - 13.
The system has no solutions, since 0 is not equal to (-13).
Answer: there are no solutions.
Example 4. Solve the system of equations
We notice that all coefficients of the second equation are divisible by 3,
let's divide the second equation by three and we get a system that consists of two identical equations.
This system has infinitely many solutions, since the first and second equations are the same (we got only one equation with two variables). How can we imagine the solution to this system? Let's express the variable y from the equation x + y = 5. We get y = 5 – x.
Then answer will be written like this: (x; 5-x), x – any number.
We looked at solving systems of equations using the addition method. If you have any questions or something is not clear, sign up for a lesson and we will solve all the problems with you.
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Algebraic addition method
You can solve a system of equations with two unknowns in various ways- graphical method or variable replacement method.
In this lesson we will get acquainted with another method of solving systems that you will probably like - this is the method of algebraic addition.
Where did the idea of putting something in systems come from? When solving systems main problem is the presence of two variables, because we do not know how to solve equations with two variables. This means that one of them must be excluded in some legal way. And such legitimate ways are mathematical rules and properties.
One of these properties is: the sum of opposite numbers is zero. This means that if one of the variables has opposite coefficients, then their sum will be equal to zero and we will be able to exclude this variable from the equation. It is clear that we do not have the right to add only terms with the variable we need. You need to add the entire equations, i.e. separately add similar terms on the left side, then on the right. As a result, we get a new equation containing only one variable. Let's look at what has been said with specific examples.
We see that in the first equation there is a variable y, and in the second opposite number-y. This means that this equation can be solved by addition.
One of the equations is left as it is. Any one you like best.
But the second equation will be obtained by adding these two equations term by term. Those. We add 3x with 2x, we add y with -y, we add 8 with 7.
We obtain a system of equations
The second equation of this system is a simple equation with one variable. From it we find x = 3. Substituting the found value into the first equation, we find y = -1.
Answer: (3; - 1).
Sample design:
Solve a system of equations using the algebraic addition method
There are no variables with opposite coefficients in this system. But we know that both sides of the equation can be multiplied by the same number. Let's multiply the first equation of the system by 2.
Then the first equation will take the form:
Now we see that the variable x has opposite coefficients. This means that we will do the same as in the first example: we will leave one of the equations unchanged. For example, 2y + 2x = 10. And we get the second by addition.
Now we have a system of equations:
We easily find from the second equation y = 1, and then from the first equation x = 4.
Sample design:
Let's summarize:
We learned how to solve systems of two linear equations with two unknowns using the algebraic addition method. Thus, we now know three main methods for solving such systems: graphical, variable replacement method and addition method. Almost any system can be solved using these methods. In more complex cases, a combination of these techniques is used.
List of used literature:
- Mordkovich A.G., Algebra 7th grade in 2 parts, Part 1, Textbook for general education institutions / A.G. Mordkovich. – 10th ed., revised – Moscow, “Mnemosyne”, 2007.
- Mordkovich A.G., Algebra 7th grade in 2 parts, Part 2, Problem book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, “Mnemosyne”, 2007.
- HER. Tulchinskaya, Algebra 7th grade. Blitz survey: a manual for students of general education institutions, 4th edition, revised and expanded, Moscow, Mnemosyne, 2008.
- Alexandrova L.A., Algebra 7th grade. Thematic testing work V new form for students of general education institutions, edited by A.G. Mordkovich, Moscow, “Mnemosyne”, 2011.
- Alexandrova L.A. Algebra 7th grade. Independent work for students of general education institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, “Mnemosyne”, 2010.
OGBOU "Education Center for Children with Special Educational Needs in Smolensk"
Center for Distance Education
Algebra lesson in 7th grade
Lesson topic: Method of algebraic addition.
- Lesson type: Lesson of initial presentation of new knowledge.
Purpose of the lesson: control the level of acquisition of knowledge and skills in solving systems of equations using the method of substitution; developing skills and abilities to solve systems of equations using addition.
Lesson objectives:
Subject: learn to solve systems of equations with two variable method addition.
Metasubject: Cognitive UUD: analyze (highlight the main thing), define concepts, generalize, draw conclusions. Regulatory UUD: determine the goal, problem in educational activities. Communicative UUD: express your opinion, giving reasons for it. Personal UUD: f to form a positive motivation for learning, to create a positive emotional attitude of the student towards the lesson and the subject.
Form of work: individual
Lesson steps:
1) Organizational stage.
organize the student’s work on the topic through creating an attitude towards integrity of thinking and understanding of this topic.
2. Questioning the student on the material assigned for homework, updating knowledge.
Purpose: to test the student’s knowledge acquired during the implementation homework, identify errors, do work on errors. Review the material from the previous lesson.
3. Studying new material.
1). develop the ability to solve systems of linear equations using addition;
2). develop and improve existing knowledge in new situations;
3). cultivate control and self-control skills, develop independence.
http://zhakulina20090612.blogspot.ru/2011/06/blog-post_25.html
Goal: preserve vision, relieve eye fatigue while working in class.
5. Consolidation of the studied material
Purpose: to test the knowledge, skills and abilities acquired in the lesson
6. Lesson summary, information about homework, reflection.
Progress of the lesson (work in electronic document Google):
1. Today I wanted to start the lesson with Walter’s philosophical riddle.
What is the fastest, but also the slowest, the largest, but also the smallest, the longest and shortest, the most expensive, but also cheaply valued by us?
Time
Let's remember the basic concepts on the topic:
Before us is a system of two equations.
Let's remember how we solved systems of equations in the last lesson.
Substitution method
Once again, pay attention to the solved system and tell me why we cannot solve each equation of the system without resorting to the substitution method?
Because these are equations of a system with two variables. We can solve equations with only one variable.
Only by obtaining an equation with one variable were we able to solve the system of equations.
3. We proceed to solve the following system:
Let's choose an equation in which it is convenient to express one variable through another.
There is no such equation.
Those. In this situation, the previously studied method is not suitable for us. What is the way out of this situation?
Find a new method.
Let's try to formulate the purpose of the lesson.
Learn to solve systems using a new method.
What do we need to do to learn how to solve systems using a new method?
know the rules (algorithm) for solving a system of equations, complete practical tasks
Let's start developing a new method.
Pay attention to the conclusion we made after solving the first system. It was possible to solve the system only after we obtained a linear equation with one variable.
Look at the system of equations and think about how to get one equation with one variable from two given equations.
Add up the equations.
What does it mean to add equations?
Separately compose the sum of the left sides, the sum of the right sides of the equations and equate the resulting sums.
Let's try. We work together with me.
13x+14x+17y-17y=43+11
We have obtained a linear equation with one variable.
Have you solved the system of equations?
The solution to the system is a pair of numbers.
How to find y?
Substitute the found value of x into the system equation.
Does it matter which equation we substitute the value of x into?
This means that the found value of x can be substituted into...
any equation of the system.
We got acquainted with a new method - the method of algebraic addition.
While solving the system, we discussed the algorithm for solving the system using this method.
We have reviewed the algorithm. Now let's apply it to problem solving.
The ability to solve systems of equations can be useful in practice.
Let's consider the problem:
The farm has chickens and sheep. How many of both are there if they together have 19 heads and 46 legs?
Knowing that there are 19 chickens and sheep in total, let’s create the first equation: x + y = 19
4x - the number of legs of sheep
2у - number of legs in chickens
Knowing that there are only 46 legs, let’s create the second equation: 4x + 2y = 46
Let's create a system of equations:
Let's solve the system of equations using the solution algorithm using the addition method.
Problem! The coefficients in front of x and y are not equal and not opposite! What to do?
Let's look at another example!
Let's add one more step to our algorithm and put it in first place: If the coefficients in front of the variables are not the same and not opposite, then we need to equalize the modules for some variable! And then we will act according to the algorithm.
4. Electronic physical education for the eyes: http://zhakulina20090612.blogspot.ru/2011/06/blog-post_25.html
5. We complete the problem using the algebraic addition method, fixing new material and find out how many chickens and sheep there were on the farm.
Additional tasks:
6. 
Reflection.
I give a grade for my work in class -...
6. Internet resources used:
Google services for education
Mathematics teacher Sokolova N. N.