With this mathematical program, you can solve the system of two linear equations with two variable substitution method and the method of addition.

The program not only gives an answer to the problem, but also gives a detailed solution with the explanations of the steps of the solution in two ways: substitution method and method of addition.

This program may be useful to students of high schools of general education schools when preparing for tests and exams, when checking knowledge before the exam, parents for monitoring the solution of many problems in mathematics and algebra. Or maybe you are too expensive to hire a tutor or buy new textbooks? Or you just want to make your homework in mathematics or algebra as possible? In this case, you can also use our programs with a detailed solution.

Thus, you can conduct your own training and / or training of your younger brothers or sisters, while the level of education in the field of solved tasks increases.

Rules for entering equations

As a variable can be any Latin Letter.
For example: \\ (x, y, z, a, b, c, o, p, q \\), etc.

When entering equations you can use brackets. At the same time, the equations are first simplified. Equations after simplification must be linear, i.e. Views AX + BY + C \u003d 0 with an accuracy of the order of the elements.
For example: 6x + 1 \u003d 5 (x + y) +2

In the equations you can use not only integer, but also fractional numbers in the form of decimal and ordinary fractions.

The rules for entering decimal fractions.
The whole and fractional part in decimal fractions can be separated as a point and the comma.
For example: 2.1N + 3.5m \u003d 55

Rules for entering ordinary fractions.
Only an integer can act as a numerator, denominator and a whole part of the fraction.
The denominator cannot be negative.
When entering a numeric fraction, the numerator separated from the denominator to the fission mark: /
The whole part is separated from the fraraty ampersand sign: &

Examples.
-1 & 2 / 3Y + 5 / 3X \u003d 55
2.1p + 55 \u003d -2/7 (3.5p - 2 & 1 / 8q)

Solve the system of equations

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A bit of theory.

Solving systems of linear equations. Method of substitution

Sequence of actions when solving a system of linear equations by the method of substitution:
1) expressed from some system equation one variable across the other;
2) substitute to another equation of the system instead of this variable the resulting expression;

$$ \\ left \\ (\\ begin (array) (L) 3x + y \u003d 7 \\\\ -5x + 2y \u003d 3 \\ End (Array) \\ Right. $$

Express from the first E equation through x: y \u003d 7-3x. Substituting into the second equation instead of y expression 7-zx, we get the system:
$$ \\ left \\ (\\ begin (array) (L) y \u003d 7-3x \\\\ -5x + 2 (7-3x) \u003d 3 \\ END (Array) \\ Right. $$

It is easy to show that the first and second system have the same solutions. In the second system, the second equation contains only one variable. Let this equation:
$$ -5x + 2 (7-3x) \u003d 3 \\ rightarrow -5x + 14-6x \u003d 3 \\ rightarrow -11x \u003d -11 \\ rightarrow x \u003d 1 $$

Substituting into the equality y \u003d 7-3x instead of x number 1, we find the appropriate value y:
$$ y \u003d 7-3 \\ CDOT 1 \\ RIGHTARROW Y \u003d 4 $$

Couple (1; 4) - Solution Solution

Systems of equations with two variables having the same solutions called equivalent. Systems that do not have solutions are also considered equivalent.

Solution of systems of linear equations by the method of addition

Consider another method of solving systems of linear equations - the method of addition. When solving systems in this method, as in solving a substitution method, we move from this system to another equivalent system in which one of the equations contains only one variable.

Sequence of actions in solving a system of linear equations by the method of addition:
1) multiplies the metering equation of the system, selecting multipliers so that the coefficients for one of the variables became opposite numbers;
2) fold the metering and right parts of the system equations;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.

Example. Resolving the system of equations:
$$ \\ left \\ (\\ begin (array) (L) 2x + 3Y \u003d -5 \\\\ X-3Y \u003d 38 \\ END (Array) \\ Right. $$

In the equations of this system, the coefficients at y are opposite numbers. Having folded the left and right parts of the equations, we obtain equation with one variable 3x \u003d 33. We replace one of the equations of the system, for example, the first, equation 3x \u003d 33. We receive the system
$$ \\ left \\ (\\ begin (array) (L) 3x \u003d 33 \\\\ X-3Y \u003d 38 \\ END (Array) \\ Right. $$

From equation 3x \u003d 33 we find that x \u003d 11. Substituting this value x into the equation \\ (x-3y \u003d 38 \\), we obtain the equation with the variable y: \\ (11-3y \u003d 38 \\). Let this equation:
\\ (- 3Y \u003d 27 \\ RIGHTARROW Y \u003d -9 \\)

Thus, we found the solution of the system of equations by the method of addition: \\ (x \u003d 11; y \u003d -9 \\) or \\ ((11; -9) \\)

Taking advantage of the fact that in the system equations, the coefficients at y are opposite numbers, we reduced its solution to solving the equivalent system (inserting both parts of each of the equations of the source symbol), in which one of the equations contains only one variable.

Books (textbooks) Abstracts EGE and OGE Tests online games, puzzles Building graphs of functions Spell dictionary of the Russian language Dictionary of youth Slang School catalog of Russia Catalog of Dzuzov Russia Catalog of universities in Russia List of tasks

More reliable than the graphic method, which was considered in the previous paragraph.

Method of substitution

We used this method in the 7th grade to solve systems of linear equations. The algorithm that was developed in the 7th grade is quite suitable for solving systems of any two equations (not necessarily linear) with two variables x and y (of course, the variables can be designated and other letters that does not matter). In fact, we used this algorithm in the previous paragraph when the problem of the two-digit number led to a mathematical model representing a system of equations. We solved this system of equations higher by substitution method (see Example 1 of § 4).

The algorithm for using the substitution method when solving a system of two equations with two variables X, y.

1. Express y through X from one system equation.
2. Substitute the expression obtained instead of in another system equation.
3. To solve the obtained equation regarding x.
4. To substitute alternately each of the roots found in the third step of the equation instead of the expression y through x, obtained in the first step.
5. Record the answer in the form of par values \u200b\u200b(x; y), which were found according to the third and fourth steps.

4) We will substitute alternately each of the found values \u200b\u200bin the formula X \u003d 5 - Zu. If
5) Couples (2; 1) and solving a given system of equations.

Answer: (2; 1);

Method of algebraic addition

This method, like the substitution method, is familiar to you from the course of the 7th class algebra, where it was used to solve systems of linear equations. The essence of the method is reminding the following example.

Example 2. Solve the system of equations

Multiply all members of the first system equation by 3, and the second equation will be left unchanged:
Submount the second equation of the system from its first equation:

As a result of the algebraic addition of two equations of the source system, an equation was obtained, simpler than the first and second equation of a given system. With this simpler equation, we have the right to replace any equation of a given system, such as the second. Then the specified system of equations will be replaced by a simpler system:

This system can be solved by substitution. From the second equation we find this expression instead of in the first equation of the system, we get

It remains to substitute the found values \u200b\u200bin the formula

If x \u003d 2, then

Thus, we found two solutions of the system:

Method of introducing new variables

With the method of introducing a new variable in solving rational equations from one variable, you learned to know the 8th grade algebras. The essence of this method in solving systems of equations is the same, but from a technical point of view there are some features that we discuss in the following examples.

Example 3. Solve the system of equations

We introduce a new variable then the first equation of the system can be rewritten in a simpler form: it will solve the equation relative to the variable T:

Both of these values \u200b\u200bsatisfy the condition, and therefore are the roots of the rational equation with the variable T. But it means, or from where we find that x \u003d 2, or
Thus, using the introduction method of the new variable, we managed to "bundle" the first system equation, quite complicated, on two simpler equations:

x \u003d 2 y; y - 2x.

What's next? And then each of the two obtained ordinary equations should be alternately considered in the system with equation x 2 - in 2 \u003d 3, which we have not yet remembered. In other words, the task is reduced to solving two systems of equations:

It is necessary to find solutions of the first system, the second system and all received vapors enable in response. Perform the first system of equations:

We use the substitution method, especially since everything is ready for him: we will substitute the expression 2y instead of x in the second equation of the system. Receive

Since x \u003d 2u, we are found respectively x 1 \u003d 2, x 2 \u003d 2. Thereby, two solutions of a given system were obtained: (2; 1) and (-2; -1). We will solve the second system of equations:

We will use the method of substitution again: we will substitute the expression 2x instead of in the second equation of the system. Receive

This equation does not have roots, it means that the equation system has no solutions. Thus, in response, only solutions of the first system must be included.

Answer: (2; 1); (-2; -1).

The method of introducing new variables when solving systems of two equations with two variables is used in two versions. First option: One new variable is introduced and is used in only one system equation. This is how the case was in the example 3.The order: two new variables are introduced and are used simultaneously in both equations of the system. So will be the case in example 4.

Example 4. Solve the system of equations

We introduce two new variables:

Take into account that then

This will reduce the specified system in a much simpler form, but relative to new variables A and B:

Since a \u003d 1, then from the equation a + 6 \u003d 2 we find: 1 + 6 \u003d 2; 6 \u003d 1. Thus, relative to variables A and B we received one solution:

Returning to the variables x and y, we get a system of equations

Apply to solve this system, the method of algebraic addition:

Since then from equation 2x + y \u003d 3 we find:
Thus, relative to the variables x and we got one solution:

Finished this paragraph with a brief, but quite a serious theoretical conversation. You have already accumulated some experience in solving different equations: linear, square, rational, irrational. You know that the basic idea of \u200b\u200bsolving the equation consists in a gradual transition from one equation to another, simpler, but equivalent specified. In the previous paragraph, we introduced the concept of equivalence for equations with two variables. Using this concept and systems of equations.

Definition.

Two systems of equations with variable x and y are called equivalent if they have the same solutions or if both systems do not have solutions.

All three methods (substitutions, algebraic addition and the introduction of new variables), which we discussed in this paragraph are absolutely correct in terms of equivalence. In other words, using these methods, we replace one system of equations of another, simpler, but equivalent initial system.

Graphic method solving systems of equations

We have already learned how to solve systems of equations with such common and reliable ways as the substitution method, algebraic addition and introducing new variables. And now let's remember with you, the method you have already studied in the previous lesson. That is, let's repeat that you know about the graphical solution.

The method of solving systems of the equation graphically is the construction of a graph for each of the specific equations that are included in this system and are in the same coordinate plane, as well as where it is required to find the intersection of points of these graphs. To solve this system of equations are the coordinates of this point (x; y).

It should be remembered that for the graphical system of the equations, it is intended to have either one single right solution or infinite multiple solutions, or not to have solutions at all.

And now on each of these solutions, we will stop more. And so, the system of equations may have a single solution if direct, which are graphs of the system equations, intersect. If these straight parallels are, such a system of equations is absolutely not solutions. In the case of coincidence of direct graphs of the system equations, then such a system allows you to find a variety of solutions.

Well, now let's consider the algorithm for solving a system of two equations with 2 unknown graphic methods:

First, at first we are building a graph of the 1st equation;
The second stage will be the construction of a graph, which refers to the second equation;
Thirdly, we need to find the intersection points of the graphs.
And as a result, we obtain the coordinates of each intersection point, which will be solved by the system of equations.

Let's consider this method in more detail on the example. We are given a system of equations that need to be solved:

Solving equations

1. At first, we will build a graph of this equation: x2 + y2 \u003d 9.

But it should be noted that this graph of the equations will be a circle that has a center at the beginning of the coordinates, and its radius will be equal to three.

2. The next step is the construction of a graph of such an equation, as: y \u003d x - 3.

In this case, we must build a straight and find points (0; -3) and (3; 0).

3. We look at us. We see that the straight line crosses the circle at two points A and B.

Now we are looking for the coordinates of these points. We see that the coordinates (3; 0) correspond to the point A, and the coordinates (0; -3), respectively, point V.

And what do we get in the end?

Received with the intersection of a line with a circle of numbers (3; 0) and (0; -3), just and are solutions of both system equations. And from this it follows that these numbers are and solutions of this system of equations.

That is, the response of this decision is numbers: (3; 0) and (0; -3).

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Instruction

Method of addition.
You need to record two strictly with each other:

549 + 45U + 4U \u003d -7, 45U + 4U \u003d 549-7, 49U \u003d 542, y \u003d 542: 49, U≈11.
In an arbitrarily selected (from the system), the equation insert instead of the already found "gamepec" number 11 and calculate the second unknown:

X \u003d 61 + 5 * 11, x \u003d 61 + 55, x \u003d 116.
Answer this system of equations: x \u003d 116, y \u003d 11.

Graphic method.
It is the practical location of the coordinate of the point in which direct, mathematically recorded in the system of equations. You should draw graphs of both direct separately in one coordinate system. General view: - y \u003d kx + b. To build a straight line, it is enough to find the coordinates of two points, and x is selected arbitrarily.
Let the system be given: 2x - y \u003d 4

Y \u003d -3x + 1.
It is built on the first to be built for convenience, it should be written: y \u003d 2x-4. Invent (easier) values \u200b\u200bfor X, substituting it into the equation, deciding it, to find heret. Two points are obtained for which the straight line is built. (see Fig.)
x 0 1.

u -4 -2.
It is built on the second equation: y \u003d -3x + 1.
Also build a straight line. (see Fig.)

in 1 -5
Find the coordinates of the intersection points of two built direct on the graph (if the straight lines do not intersect, the system of equations does not have - so).

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Helpful advice

If the same system of equations is solved by three different ways, the answer will turn out the same (if the solution is true).

Sources:

• Grade 8 algebra
• solve equation with two unknown online
• Examples of solving systems of linear equations with two

System equations It is a combination of mathematical records, each of which contains a number of variables. There are several ways to solve them.

You will need

• -Ruler and pencil;
• -calculator.

Instruction

Consider the sequence of solving the system, which consists of linear equations having a form: A1x + B1y \u003d C1 and A2x + B2Y \u003d C2. Where x and y are unknown variables, and B, C are free members. When applying this method, each system is the coordinates of the points corresponding to each equation. To begin with, in each case, express one variable through the other. Then set the variable x several any values. There are two enough. Submold to the equation and find y. Build a coordinate system, mark the received points on it and spend directly through them. Similar calculations must be carried out for other parts of the system.

The system has a single solution if the built direct intersects and one common point. It is incomplete if parallel to each other. And it has infinitely a lot of solutions when straighty merge with each other.

This method is considered very visual. The main disadvantage is that the calculated unknowns have approximate values. A more accurate result gives the so-called algebraic methods.

Any solution of the system of equations is worth checking. To do this, substitute the values \u200b\u200bobtained instead of variables. You can also find it a solution to several methods. If the solution is correct, then everyone should turn out the same.

Often there are equations in which one of the terms is unknown. To solve the equation, you need to remember and do a certain set of actions with these numbers.

You will need

• - paper;
• - Pen or pencil.

Instruction

Imagine that you are 8 rabbits in front of you, and you have only 5 carrots. Think, carrots you need to buy so that every rabbit gets in carrots.

Imagine this problem in the form of equation: 5 + x \u003d 8. We will substitute in place x number 3. Indeed, 5 + 3 \u003d 8.

When you substituted the number in place x, you have done the same operation as when subtracting 5 out of 8. So to find unknown The term, deduct from the sum of the famous term.

Suppose you have 20 rabbits and only 5 carrots. Make up. The equation is equality that is performed only with some values \u200b\u200bof the letters of it. The letters whose values \u200b\u200bneed to be found are called. Make an equation with one unknown, name it x. When solving our problem about rabbits, the following equation is obtained: 5 + x \u003d 20.

We will find the difference between 20 and 5. When subtracting, the number from which is subtracted, diminished. The number that is deducted is called, and the end result is called a difference. So, x \u003d 20 - 5; x \u003d 15. You need to buy 15 carrots for rabbits.

Take the check: 5 + 15 \u003d 20. The equation is solved correctly. Of course, when it comes to such simple, check is optional. However, when there are equations with three-digit, four-digit and the numbers, it is necessary to check in order to be absolutely confident as a result of its work.

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Helpful advice

To find an unknown diminished, it is necessary to add deductible to the difference.

To find an unknown ready, it is necessary to take a difference from the decrease.

Tip 4: How to solve a system of three equations with three unknown

The system of three equations with three unknown may not have solutions, despite the sufficient number of equations. You can try to solve it using a substitution method or using the Cramer method. The drive method in addition to solving the system allows you to evaluate whether the system is solvable before finding the unknown values.

Instruction

The substitution method consists in a sequential one unknown through two other and substitution of the result obtained in the system equation. Let a system of three equations in general form:

a1X + B1Y + C1Z \u003d D1

a2x + B2Y + C2Z \u003d D2

a3X + B3Y + C3Z \u003d D3

Express the X: X \u003d (D1 - B1Y - C1Z) / A1 from the first equation - and substitute on the second and third equations, then express Y from the second equation and substitute in the third. You will receive a linear expression for z through the coefficients of the system equations. Now go "Back": Substitute Z in the second equation and find Y, and then Z and Y substitute in the first and find x. The process in general is displayed in Figure before being Z. Next, the recording in general will be too cumbersome, in practice, substituting, you are quite easy to find all three unknown.

The Cramer Method is to compile the system matrix and calculating the determinant of this matrix, as well as three more auxiliary matrices. The system matrix is \u200b\u200bcomposed of coefficients with unknown members of equations. The column containing the numbers in the right parts of the equations, the column of the right parts. It is not used in the system, but used when solving the system.

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note

All equations in the system must supply additional information independent to other equations. Otherwise, the system will be undecid and the unequivocal solution will not be found not possible.

Helpful advice

After solving the system of equations, substitute the found values \u200b\u200bin the source system and check that they satisfy all equations.

In itself the equation With three unknown It has many solutions, so most often it is complemented by two more equations or conditions. Depending on whether the source data is largely dependent.

You will need

• - a system of three equations with three unknowns.

Instruction

If two of the three systems have only two unknowns out of three, try to express one variables through others and substitute them in the equation With three unknown. Your goal is to turn it into ordinary the equation With an unknown. If this, further solution is quite simple - substitute the value found to other equations and find all other unknown.

Some system of equations can be subtracted from one equation of the other. See if it is possible to multiply one of or a variable so that two unknown people decreased immediately. If such an opportunity is, take advantage of it, most likely, the subsequent decision will not be difficult. Do not forget that when multiplying the number must be multiplied as the left part and the right. Similarly, when subtracting the equations it is necessary to remember that the right side should also be deducted.

If the previous methods did not help, use the general way to solve any equations with three unknown. To do this, rewrite the equations in the form of A11x1 + A12x2 + A13x3 \u003d B1, A21x1 + A22x2 + A23x3 \u003d B2, A31x1 + A32x2 + A33x3 \u003d B3. Now make a matrix of coefficients at x (A), an unknown matrix (x) and a free matrix (B). Note, multiplying the coefficient matrix on the unknown matrix, you will receive a matrix, a matrix of free members, that is, a * x \u003d c.

Find the matrix A to the degree (-1) after finding, pay attention, it should not be equal to zero. After that, multiply the resulting matrix on the matrix B, as a result you will receive a desired matrix x, indicating all values.

Find a solution of a system of three equations can also be using the Cramer method. To do this, find the third order determinant δ corresponding to the system matrix. Then successively find three more determinants Δ1, Δ2 and Δ3, substituting instead of the values \u200b\u200bof the corresponding columns the value of free members. Now find x: x1 \u003d δ1 / δ, x2 \u003d δ2 / δ, x3 \u003d δ3 / δ.

Sources:

• solutions of equations with three unknown

Getting Started by solving the system of equations, deal with what equations are. Solving linear equations are quite well studied. Nonlinear equations are most often not solved. There are only some particular cases, each of which is almost individual. Therefore, the study of decision techniques should be started with equations of exactly linear. Such equations can be solved even purely algorithmically.

dannels for those found unknowns are made the same. Yes, and the numbers are viewed by some patterns of their construction. If the dimension of the system of equations would be greater than two, then the exclusion method would lead to very bulky calculations. To avoid them, purely algorithmic solutions solutions have been developed. The easiest of them is the Cramer Algorithm (Cramer Formula). To learn, the general system of equations from n equations should be found.

The system N of linear algebraic equations with n unknown has the form (see Fig. 1a). In it, AIJ - system coefficients,
xj - unknown, Bi are free members (i \u003d 1, 2, ..., n; j \u003d 1, 2, ..., p). The compact system can be recorded in the matrix form ah \u003d b. Here a - matrix of the coefficients of the system, x - the column-column of unknown, B is a matrix-column-column of free members (see Fig. 1b). According to the Kramer method, each unknown xi \u003d Δi / δ (i \u003d 1.2 ..., n). The determinant δ matrix coefficients is called the main, and Δi auxiliary. For each unknown, the auxiliary determinant is found by replacing the i-th column of the main determinant on the column of free members. Details of the Cramer method for the case of second and third order systems is presented in Fig. 2.

The system is a combination of two or more equalities, each of which has two or more unknown. There are two basic ways to solve systems of linear equations that are used as part of the school program. One of them wears the name of the method, the other - the method of addition.

Standard view of the system of two equations

With standard form, the first equation has the form A1 * X + B1 * Y \u003d C1, the second equation has the form A2 * X + B2 * Y \u003d C2 and so on. For example, in the case of two parts of the system in both of the above A1, A2, B1, B2, C1, C2 are some numerical coefficients presented in specific equations. In turn, x and y are unknown, whose values \u200b\u200bmust be determined. The desired values \u200b\u200bare drawn by both equations simultaneously into faithful equalities.

Solution of the system by the method of addition

In order to solve the system, that is, finding those X and Y values \u200b\u200bthat will turn them into loyal equality, several simple steps must be taken. The first of these is to transform any of the equations in such a way that the numerical coefficients for the variable x or y in both equations coincide in the module, but differed by the sign.

For example, let the system consisting of two equations specify. The first of them has the form 2x + 4y \u003d 8, the second has the form 6x + 2y \u003d 6. One embodiment of the task is to multiply the second equation on the coefficient -2, which will lead it to the form -12x-4y \u003d -12. The correct choice of the coefficient is one of the key tasks in the process of solving the system by the method of addition, since it determines the entire further course of the procedure for finding unknowns.

Now it is necessary to make the addition of two system equations. Obviously, the mutual destruction of the variables with equal in value, but the coefficients opposite by the sign will result in its type -10x \u003d -4. After that, it is necessary to solve this simple equation, from which it unambiguously follows that x \u003d 0.4.

The last step in the solution process is the substitution of the found value of one of the variables to any of the initial equalities in the system. For example, substituting x \u003d 0.4 in the first equation, it is possible to obtain an expression 2 * 0.4 + 4Y \u003d 8, from where y \u003d 1.8. Thus, x \u003d 0.4 and y \u003d 1.8 are the roots of the system given in the example.

In order to make sure that the roots were found correctly, it is useful to check, substituting the found values \u200b\u200binto the second equation of the system. For example, in this case, the equality of the form 0.4 * 6 + 1.8 * 2 \u003d 6 is obtained, which is correct.

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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 by 2, and the second to 3 and we obtain the total 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)

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