The method of addition, the equation of the system is replenished, with 1 - but either both (several) equations can be multiplied by any number. As a result, they come to equivalent serving, where in one of the equations there is only one variable.

To solve the system mETHOD OF MILLACE INSTRUCTION (subtraction) Follow the following steps:

1. Select the variable in which the same coefficients will be made.

2. Now you need to add or subtract equations and obtain equation with one variable.

Solution System - These are the points of intersection of the graphs of the function.

Consider at the examples.

Example 1.

Dana System:

After analyzing this system, it can be noted that the coefficients with a variable are equal to the module and different ones (-1 and 1). In this case, the equation is easy to fold soil:

Actions that are circled in red, perform in the mind.

The result of the soil addition was the disappearance of the variable y.. It is precisely in this that, in fact, the meaning of the method is to get rid of the 1st of the variables.

-4 - y. + 5 = 0 → y. = 1,

In the form of a system, the solution looks somewhere like this:

Answer: x. = -4 , y. = 1.

Example 2.

Dana System:

In this example, you can use the "school" method, but it has a rather big minus - when you express any variable from any equation, you will receive a solution in ordinary fractions. And the solution of fractions occupies enough time and the likelihood of error assumption increases.

Therefore, it is better to use the killed addiction (subtraction) of equations. We analyze the coefficients of the corresponding variables:

Need to pick up the number that can be divided and on 3 and on 4 It is necessary that this number would be the minimum possible. it the smallest common pain . If you are hard to choose a suitable number, you can multiply the coefficients :.

The next step:

1st equation multiply on,

3rd equation multiply on

This video, I begin the cycle of lessons dedicated to the systems of equations. Today we will talk about solving systems linear equations by addition - This is one of the most simple waysBut at the same time one of the most efficient.

The way of addition consists of three simple Steps:

1. Look at the system and choose a variable in which each equation is the same (or opposite) coefficients;
2. Perform an algebraic subtraction (for opposite numbers - addition) of equations from each other, after which the lifestyles are given;
3. Solve a new equation obtained after the second step.

If you do everything right, then at the exit we will get a single equation with one variable - It is not difficult to decide. Then it will only remain substituting the root found in the source system and get the final response.

However, in practice, everything is not so simple. There are several reasons for this:

• The solution of the equation by the method of addition implies that variables with the same / opposite coefficients should be present in all lines. And what if this requirement is not performed?
• Not always after the addition / subtraction of the equations in the specified way, we will get a beautiful design that is easily solved. Is it possible to somehow simplify the calculations and speed up the calculations?

To get an answer to these questions, and at the same time deal with several additional subtleties on which many students are "falling", see my video tutorial:

With this lesson, we begin the cycle of lectures on the systems of equations. And let's start from the simplest of them, namely, those that contain two equations and two variables. Each of them will be linear.

Systems are the 7th grade material, but this lesson will also be useful to high school students who want to refresh their knowledge in this topic.

In general, there are two methods for solving such systems:

1. The method of addition;
2. Method of expressing one variable across the other.

Today we will deal with the first method - we will apply the method of subtraction and addition. But for this you need to understand the following fact: as soon as you have two or more equations, you have the right to take any two of them and fold with each other. They are so far, i.e. The "Xers" add up with "Iksami" and are given similar, "igraki" with "Games" - again given similar, and what is worth the right of the sign of equality, also develops with each other, and there are also given similar.

The results of such fractions will be a new equation, which, if and has roots, they will definitely be among the roots of the original equation. Therefore, our task is to make subtraction or addition so that or $ x $, or $ y $ disappeared.

How to achieve this and what tool for this to use - we will talk about this now.

Solution of light challenges using the method of addition

So, learning to apply the method of addition on the example of two simple expressions.

Task number 1.

\\ [\\ left \\ (\\ begin (align) & 5x-4y \u003d 22 \\\\ & 7x + 4y \u003d 2 \\\\ End (Align) \\ Right. \\]

Note that in the $ y $ coefficient in the first equation of $ -4 $, and in the second - $ + $ 4. They are mutually opposed, so it is logical to assume that if we fold them, then in the resulting amount "igrek" mutually destroyed. We fold and get:

We solve the simplest design:

Fine, we found "X". What now to do with it? We have the right to substitute it to any of the equations. Substitute in the first:

\\ [- 4Y \u003d 12 \\ left | : \\ Left (-4 \\ Right) \\ Right. \\]

Answer: $ \\ left (2; -3 \\ Right) $.

Task number 2.

\\ [\\ left \\ (\\ begin (align) & -6x + y \u003d 21 \\\\ & 6x-11y \u003d -51 \\\\\\ End (Align) \\ Right. \\]

There is a completely similar situation here, only already with "Iksami". Mix them:

We got the simplest linear equation, let's decide it:

Now let's find $ x $:

Answer: $ \\ left (-3; 3 \\ Right) $.

Important moments

So, we have just solved the two simplest system of linear equations by the method of addition. Once again key points:

1. If there are opposite coefficients with one of the variables, it is necessary to add all the variables in the equation. In this case, one of them will be destroyed.
2. The found variable is substituted into any of the system equations to find the second.
3. The final response entry can be represented in different ways. For example, so - $ x \u003d ..., y \u003d ... $, or in the form of the coordinates of the points - $ \\ left (...; ... \\ right) $. The second option is preferable. The main thing is to remember that the first coordinate is $ x $, and the second is $ y $.
4. Rule write the answer in the form of the point coordinates is not always applicable. For example, it cannot be used when the role of variables are not $ x $ and $ y $, but, for example, $ A $ and $ b $.

In the following tasks, we will consider receiving subtraction when the coefficients are not opposite.

Solution of light tasks using the subtraction method

Task number 1.

\\ [\\ left \\ (\\ begin (align) & 10x-3y \u003d 5 \\\\ & -6x-3y \u003d -27 \\\\\\ End (Align) \\ Right. \\]

Note that there are no opposite coefficients here, however there are the same. Therefore, we subtract second from the first equation:

Now we substitute the value of $ x $ into any of the system equations. Let's first:

Answer: $ \\ left (2; 5 \\ Right) $.

Task number 2.

\\ [\\ left \\ (\\ begin (align) & 5x + 4y \u003d -22 \\\\ & 5x-2y \u003d -4 \\\\\\ End (Align) \\ Right. \\]

We again see the same $ 5 $ coefficient at $ x $ in the first and in the second equation. Therefore, it is logical to assume that it is necessary to subtract the second equation:

We calculated one variable. Now let's find the second, for example, substituting the value of $ y $ to the second design:

Answer: $ \\ left (-3; -2 \\ right) $.

Nuances Solutions

So what do we see? Essentially, the scheme does not differ from the solution of previous systems. The only difference is that we do not fold the equations, but deduct. We are conducting algebraic subtraction.

In other words, as soon as you see a system consisting of two equations with two unknown, the first thing you need to see is the coefficients. If they are somewhere the same, the equations are deducted, and if they are opposite - the method of addition is applied. It is always done so that one of them disappears, and in the total equation, which remained after subtraction, only one variable would remain.

Of course, this is not all. Now we will look at the systems in which the equations are generally inconsistent. Those. There are no such variables in them that would be either identical or opposite. In this case, an additional reception is used to solve such systems, namely, the multiplication of each of the equations for a special coefficient. How to find it and how to solve such systems in general, now we will talk about it.

Solving tasks by multiplying the coefficient

Example number 1.

\\ [\\ left \\ (\\ begin (align) & 5x-9y \u003d 38 \\\\ & 3x + 2y \u003d 8 \\\\\\ End (Align) \\ Right. \\]

We see that no at $ x $, nor with $ y $, the coefficients are not only opposite, but in general, they do not correlate with another equation. These coefficients will not disappear, even if we fold or subtract the equation from each other. Therefore, it is necessary to apply multiplication. Let's try to get rid of the $ y $ variable. To do this, we are dominant the first equation for the coefficient with $ y $ from the second equation, and the second equation is at $ y $ from the first equation, while not a touch mark. Multiply and get a new system:

\\ [\\ left \\ (\\ begin (align) & 10x-18y \u003d 76 \\\\ & 27x + 18y \u003d 72 \\\\\\ End (Align) \\ Right. \\]

We look at it: with $ y $ opposite coefficients. In such a situation, it is necessary to apply the method of addition. Mix:

Now it is necessary to find $ y $. To do this, we will substitute $ x $ in the first expression:

\\ [- 9Y \u003d 18 \\ left | : \\ Left (-9 \\ Right) \\ Right. \\]

Answer: $ \\ left (4; -2 \\ Right) $.

Example number 2.

\\ [\\ left \\ (\\ begin (align) & 11x + 4y \u003d -18 \\\\ & 13x-6y \u003d -32 \\\\ End (Align) \\ Right. \\]

Again the coefficients neither with one of the variables are not agreed. Doming on coefficients at $ y $:

\\ [\\ left \\ (\\ begin (align) & 11x + 4y \u003d -18 \\ left | 6 \\ Right. \\\\ & 13x-6y \u003d -32 \\ Left | 4 \\ Right. \\\\\\ End (Align) \\ Right . \\]

\\ [\\ left \\ (\\ begin (align) & 66x + 24y \u003d -108 \\\\ & 52x-24y \u003d -128 \\\\\\ End (Align) \\ Right. \\]

Our new system is equivalent to the previous one, however coefficients for $ y $ are mutually opposite, and therefore it is easy to apply the addition method:

Now we find $ Y $, substituting $ x $ in the first equation:

Answer: $ \\ left (-2; 1 \\ Right) $.

Nuances Solutions

The key rule here is the following: always multiply only on positive numbers - it will save you from stupid and offensive mistakes associated with changing signs. In general, the solution scheme is quite simple:

1. We look at the system and analyze each equation.
2. If we see that neither $ y $, none at $ x $ coefficients agreed, i.e. They are neither equal, neither opposite, then we do the following: Select the variable from which you need to get rid of, and then we look at the coefficients at these equations. If the first equation is dominated to the coefficient of the second, and the second, appropriate, doubt on the coefficient of the first, then as a result we will receive a system that is completely equivalent to the previous one, and the coefficients for $ y $ will be agreed. All of our actions or transformations are directed only to obtain one variable in one equation.
3. We find one variable.
4. We substitute the variable found into one of the two system equations and we find the second.
5. Record the answer in the form of the coordinates of the points, if we have a variable $ x $ and $ y $.

But even in such a simple algorithm there are its subtleties, for example, coefficients at $ x $ or $ y $ can be fractions and other "ugly" numbers. We now consider these cases separately, because they can act somewhat differently than according to the standard algorithm.

Solving problems with fractional numbers

Example number 1.

\\ [\\ left \\ (\\ begin (align) & 4m-3n \u003d 32 \\\\ & 0.8m + 2.5n \u003d -6 \\\\\\ End (Align) \\ Right. \\]

To begin with, we note that there are fractions in the second equation. But we note that it is possible to divide $ 4 $ by $ 0.8 $. We get $ 5 $. Let's the second equation of $ 5 perk. $ 5.

\\ [\\ left \\ (\\ begin (align) & 4m-3n \u003d 32 \\\\ & 4m + 12,5m \u003d -30 \\\\\\ End (Align) \\ Right. \\]

We subtract the equation from each other:

$ n $ We found, now consider $ M $:

Answer: $ n \u003d -4; m \u003d $ 5

Example number 2.

\\ [\\ left \\ (\\ begin (align) & 2,5p + 1,5k \u003d -13 \\ left | 4 \\ right. \\\\ & 2p-5k \u003d 2 \\ left | 5 \\ Right. \\\\\\ End (Align ) \\ RIGHT. \\]

Here, as in the previous system, fractional coefficients are present, but neither one of the variable coefficients is not fitted with an integer in each other. Therefore, we use the standard algorithm. Get rid of $ per $:

\\ [\\ left \\ (\\ begin (align) & 5p + 3k \u003d -26 \\\\ & 5p-12,5k \u003d 5 \\\\\\ End (Align) \\ Right. \\]

Use the subtraction method:

Let's find $ p $, substituting $ k $ to the second design:

Answer: $ p \u003d -4; k \u003d -2 $.

Nuances Solutions

That's all optimization. In the first equation, we did not go to multiply anything, and the second equation is domed $ 5 $. As a result, we received a consistent and even equal equation at the first variable. In the second system, we acted according to the standard algorithm.

But how to find numbers for which the equation needs? After all, if you draw on fractional numbersWe will get new fractions. Therefore, the fraction must be drawn by a number that would give a new integer, and after that, to multiply variables on the coefficients, following the standard algorithm.

In conclusion, I would like to draw your attention to the response recording format. As I already said, because here we have here not $ x $ and $ y $, but other meanings, we use a non-standard view of the form:

Solution of complex systems of equations

As a final chord to today's video, let's look at a couple of really complex systems. Their complexity will be that in them on the left, and the variables will stand on the right. Therefore, for their solution, we will have to use preliminary processing.

System No. 1.

\\ [\\ left \\ (\\ begin (align) & 3 \\ left (2x-y \\ right) + 5 \u003d -2 \\ left (x + 3y \\ Right) +4 \\\\ & 6 \\ Left (Y + 1 \\ Right ) -1 \u003d 5 \\ Left (2x-1 \\ Right) +8 \\\\\\ End (Align) \\ Right. \\]

Each equation carries a certain difficulty. Therefore, with each expression, let's do it with a conventional linear design.

Total we get the final system, which is equivalent to the original:

\\ [\\ left \\ (\\ begin (align) & 8x + 3y \u003d -1 \\\\ & -10x + 6y \u003d -2 \\\\\\ End (Align) \\ Right. \\]

Let's look at the coefficients at $ y $: $ 3 $ stacked at $ 6 $ twice, so the dominarity is the first equation of $ 2 $:

\\ [\\ left \\ (\\ begin (align) & 16x + 6y \u003d -2 \\\\ & -10 + 6Y \u003d -2 \\\\\\ End (Align) \\ Right. \\]

The coefficients at $ y $ are now equal, so we subtract second from the first equation: $$

Now we find $ y $:

Answer: $ \\ left (0; - \\ FRAC (1) (3) \\ Right) $

System No. 2.

\\ [\\ left \\ (\\ begin (align) & 4 \\ left (a-3b \\ right) -2a \u003d 3 \\ left (b + 4 \\ right) -11 \\\\ & -3 \\ left (B-2a \\ Right ) -12 \u003d 2 \\ left (A-5 \\ RIGHT) + B \\\\\\ End (Align) \\ Right. \\]

We transform the first expression:

We understand the second:

\\ [- 3 \\ left (b-2a \\ right) -12 \u003d 2 \\ left (A-5 \\ RIGHT) + B \\]

\\ [- 3b + 6a-12 \u003d 2a-10 + b \\]

\\ [- 3b + 6a-2a-B \u003d -10 + 12 \\]

Total, our initial system will take this kind:

\\ [\\ left \\ (\\ begin (align) & 2a-15b \u003d 1 \\\\ & 4a-4b \u003d 2 \\\\\\ End (Align) \\ Right. \\]

Looking at the coefficients at $ A $, we see that the first equation must be multiplied by $ 2 $:

\\ [\\ left \\ (\\ begin (align) & 4a-30b \u003d 2 \\\\ & 4a-4b \u003d 2 \\\\ End (Align) \\ Right. \\]

We subtract second from the first design:

Now we find $ A $:

Answer: $ \\ left (a \u003d \\ FRAC (1) (2); B \u003d 0 \\ Right) $.

That's all. I hope this video tutorial helps you understand this difficult topic, namely, in solving systems of simple linear equations. Then there will be many lessons dedicated to this topic: we will analyze more complex exampleswhere variables will be larger, and the equations themselves will already be non-linear. To new meetings!

OKBOU "Education Center for Children with Special Educational Needs of Smolensk"

Center for remote education

Lesson algebra in grade 7

Theme of the lesson: the method of algebraic addition.

1. 1. Type of lesson: The lesson of the primary presentation of new knowledge.

The purpose of the lesson: control the level of learning the knowledge and skills of solving systems of equations by the substitution method; Formation of skills and skills solving systems of equations by the method of addition.

Tasks lesson:

Subjects: learn to perform solutions of systems of equations with two variables by adding.

MetaPered: Cognitive Uud.: analyze (allocate the main thing), to determine the concepts, summarize, draw conclusions. Regulatory Wood : To determine the goal, the problem in training activities. Communicative Woods: Put your opinion, argueing it. Personal Wood: Forders a positive learning motivation, create a positive emotional attitude of the student to the lesson and the subject.

Form of work: individual

Stages of the lesson:

1) Organizational stage.

organize the work of the study on the topic through the creation of an integrity of thinking and understanding this topic.

2. A survey of the student on the material specified for the house, the actualization of knowledge.

Purpose: Check the knowledge of the student obtained during the execution homework, identify errors, make work on errors. Repeat the material of the last lesson.

3. Studying a new material.

one). form the ability to solve the system of linear equations by the method of addition;

2). develop and improve existing knowledge in new situations;

3). Rail the skills of control and self-control, develop independence.

http://zhakulina20090612.blogspot.ru/2011/06/blog-post_25.html

Purpose: preservation of vision, removal of fatigue with glaze hours of work at the lesson.

5. Fastening the material studied

Purpose: Check knowledge, skills and skills obtained in class

6. Outcome lesson, information about homework, reflection.

Location lesson (work in electronic document Google):

1. Today I wanted to start a lesson from the philosophical puzzle of Walter.

What is the fastest, but the most slow, the biggest, but also the smallest, the longest and short, most expensive, but cheaply appreciated by us?

Time

Recall the basic concepts on the topic:

Before us is the system of two equations.

Recall how we solved the system of equations on the past lesson.

For a substitution method

Once again, pay attention to the solid system and tell me why we cannot solve each system equation without resorting to the substitution method?

Because it is the system equations with two variables. We are able to solve the equation with only one variable.

Only by receiving the equation with one variable we managed to solve the system of equations.

3. We proceed to solving the following system:

We choose the equation in which it is convenient to express one variable through another.

There is no such equation.

Those. In this situation, we are not suitable for the previously studied method. What is the output from this situation?

Find a new method.

We will try to formulate the purpose of the lesson.

Learn to solve the systems with a new method.

What do we need to do to learn to solve the system with a new method?

know the rules (algorithm) solutions to the system of the equation, perform practical tasks

We will proceed to bring the new method.

Pay attention to the conclusion that we did 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 two data equations to obtain one equation with one variable.

Fold equations.

What does it mean to fold the equations?

Separately make up the amount of left parts, the sum of the right parts of the equations and the obtained amounts to equate.

Let's try. We work with me.

13X + 14X + 17Y-17Y \u003d 43 + 11

Received a linear equation with one variable.

Solved the system of equations?

Solution system - a pair of numbers.

How to find y?

The found value x substitute to the system equation.

It matters into what equation to substitute the value of x?

So the found value can be substituted in ...

any system equation.

We got acquainted with the new method - by algebraic addition.

Solving the system, we talked the algorithm for solving the system by this method.

Algorithm we looked at. Now we apply it to solving problems.

The ability to solve the system of equations can be useful in practice.

Consider the task:

The farm has chickens and sheep. How many others, if they have 19 heads and 46 legs together?

Knowing that all chiches and sheep 19, we will form the first equation: x + y \u003d 19

4x - the number of legs in sheep

2U - number of legs from chickens

Knowing that only 46 legs, we will form a second equation: 4x + 2y \u003d 46

Make a system of equations:

We decide the system of equations using the algorithm for solving the method of addition.

Problem! The coefficients before x and y are not equal and not opposite! What to do?

Consider another example!

Add to our algorithm one more step and put it in first place: if the coefficients before variables are not the same and not opposite, then you need to equalize the modules with some variable! And then we will act on the algorithm.

4. Eye Physical Cultiminate: http://zhakulina20090612.blogspot.ru/2011/06/blog-post_25.html

5. Especially the task by algebraic addition, consolidating new Material And we find out how many chickens and sheep were in the farm.

Additional tasks:

6.

Reflection.

I put an assessment for my job in the lesson - ...

6. Used resources-Internet:

google Services for Education

Teacher Mathematics Sokolova N. N.

We will analyze two types of solutions of the systems of the equation:

1. Solution of the system by substitution.
2. Solution of the system by the method of kindy addition (subtraction) of the system equations.

In order to solve the system of equations for a substitution method You need to follow a simple algorithm:
1. Express. From any equation, we express one variable.
2. Substitute. We substitute to another equation instead of a pronounced variable obtained.
3. We solve the obtained equation with one variable. We find the system solution.

To solve introduction system system (subtraction) need to:
1. To give a variable to which we will do the same coefficients.
2. Wide or subtract the equations, as a result, we obtain an equation with one variable.
3. We solve the resulting linear equation. We find the system solution.

The system solution is the point of intersection of the function graphs.

Consider in detail the examples of systems.

Example number 1:

By deciding by substitution

Solution of the system of equations by substitution

2x + 5y \u003d 1 (1 equation)
x-10y \u003d 3 (2 equation)

1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, it turns out that it turns out that the variable x from the second equation is easier to express.
x \u003d 3 + 10y

2. After how expressed substituted in the first equation 3 + 10y instead of a variable x.
2 (3 + 10y) + 5y \u003d 1

3. Over the resulting equation with one variable.
2 (3 + 10y) + 5y \u003d 1 (reveal brackets)
6 + 20Y + 5Y \u003d 1
25Y \u003d 1-6.
25Y \u003d -5 |: (25)
Y \u003d -5: 25
Y \u003d -0.2

The solution of the system of the equation is the points of intersection of graphs, therefore we need to find x and y, because the intersection point consists of their X and Y.Nad X, in the first paragraph where we expressed there we substitute Y.
x \u003d 3 + 10y
x \u003d 3 + 10 * (- 0.2) \u003d 1

Points are taken to record in the first place we write the variable x, and on the second variable y.
Answer: (1; -0.2)

Example number 2:

By deciding by the method of kindy addition (subtraction).

Solution of the system of equations by addition

3x-2y \u003d 1 (1 equation)
2x-3y \u003d -10 (2 equation)

1. Select the variable, let's say, select x. In the first equation in the variable x coefficient 3, in the second 2. It is necessary to make the coefficients are the same, for this we have the right to multiply equations or divide on any number. The first equation is permiterable to 2, and the second to 3 and get general coefficient 6.

3x-2y \u003d 1 | * 2
6x-4y \u003d 2

2x-3y \u003d -10 | * 3
6x-9y \u003d -30

2. The first equation will subtract the second to get rid of the variable X. The linear equation.
__6x-4y \u003d 2

5Y \u003d 32 | :five
Y \u003d 6,4.

3. Land x. We substitute in any of the equations found y, let's say in the first equation.
3x-2y \u003d 1
3x-2 * 6,4 \u003d 1
3x-12.8 \u003d 1
3x \u003d 1 + 12.8
3x \u003d 13.8 |: 3
x \u003d 4.6

Point of intersection will be x \u003d 4.6; Y \u003d 6,4.
Answer: (4,6; 6.4)

Want to prepare for exams for free? Tutor online. is free. No kidding.

Very often, students are hampered with the choice of a method for solving systems of equations.

In this article, we will consider one of the ways to solve systems - the method of substitution.

If you find common decision Two equations, they say that these equations form the system. In the system of equations, each unknown indicates the same number in all equations. To show that these equations form a system, they are usually recorded one below the other and combine the figure bracket, for example

We notice that at x \u003d 15, and y \u003d 5 both equations of the system are correct. This pair of numbers is the solution of the system of equations. Each pair of unknown values, which simultaneously satisfies both system equations is called a solution solving.

The system can have one solution (as in our example), infinitely many solutions and not have solutions.

How to solve the system of the substitution method? If the coefficients with some unknown in both equations are equal in an absolute value (but not equal, then equalize), then folding both equations (or subtracting one of the other), you can get an equation with one unknown. Then we solve this equation. We define one unknown. We substitute the obtained value of the unknown to one of the system equations (first or second). Find another unknown. Let's consider the use of this method on the examples.

Example 1. Solve the system of equations

Here the coefficients at the absolute value are equal to each other, but are opposed to the sign. Let's try reassessing the system equations.

The resulting value x \u003d 4, we substitute in some equation of the system (for example, first) and we find the value of:

2 * 4 + y \u003d 11, y \u003d 11 - 8, y \u003d 3.

Our system has a solution x \u003d 4, y \u003d 3. or the answer can be written in parentheses, as the coordinates of the point, in the first place x, on the second.

Answer: (4; 3)

Example 2.. Solve the system of equations

Equalize the coefficients with the variable x, for this you will multiply the first equation for 3, and the second on (-2), we obtain

Be careful when adding equations

Then y \u003d - 2. Substitute in the first equation instead of the number (-2), we get

4 + 3 (-2) \u003d - 4. We solve this equation 4x \u003d - 4 + 6, 4x \u003d 2, x \u003d ½.

Answer: (1/2; - 2)

Example 3. Solve the system of equations

Multiply the first equation on (-2)

We solve the system

we get 0 \u003d - 13.

The solutions system does not have, so 0 is not equal to (-13).

Answer: no solutions.

Example 4. Solve the system of equations

We notice that all coefficients of the second equation are divided by 3,

let's divide the second equation for three and we get a system that consists of two identical equations.

This system has infinitely many solutions, since the first and second equation are the same (we received only one equation with two variables). How to imagine the solution of this system? Let's express the variable from the equation x + y \u003d 5. We obtain y \u003d 5x.

Then answer Wrongs like this: (x; 5-x), x - any number.

We considered the solution of systems of equations by the method of addition. If you have any questions or something is unclear to sign up for a lesson and we will eliminate all problems.

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