Instructions

Addition method.
You need to write two strictly below each other:

549+45y+4y=-7, 45y+4y=549-7, 49y=542, y=542:49, y≈11.
In an arbitrarily chosen (from the system) equation, insert the number 11 instead of the already found “game” and calculate the second unknown:

X=61+5*11, x=61+55, x=116.
The answer to this system of equations is x=116, y=11.

Graphic method.
It consists of practically finding the coordinates of the point at which the lines are mathematically written in a system of equations. The graphs of both lines should be drawn separately in the same coordinate system. General view: – y=khx+b. To construct a straight line, it is enough to find the coordinates of two points, and x is chosen arbitrarily.
Let the system be given: 2x – y=4

Y=-3x+1.
A straight line is constructed using the first one, for convenience it should be written down: y=2x-4. Come up with (easier) values for x, substituting it into the equation, solving it, and finding y. We get two points along which a straight line is constructed. (see picture)
x 0 1

y -4 -2
A straight line is constructed using the second equation: y=-3x+1.
Also construct a straight line. (see picture)

y 1 -5
Find the coordinates of the intersection point of two constructed lines on the graph (if the lines do not intersect, then the system of equations does not have - so).

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Useful advice

If the same system of equations is solved by three in different ways, the answer will be the same (if the solution is correct).

Sources:

8th grade algebra
solve an equation with two unknowns online
Examples of system solutions linear equations with two

System equations is a collection 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.

Instructions

Let's consider the sequence of solving the system, which consists of linear equations having the form: a1x + b1y = c1 and a2x + b2y = c2. Where x and y are unknown variables, and b,c are free terms. When applying this method, each system represents the coordinates of points corresponding to each equation. To begin, in each case, express one variable in terms of another. Then set the variable x to any number of values. Two is enough. Substitute into the equation and find y. Construct a coordinate system, mark the resulting points on it and draw a line through them. Similar calculations must be carried out for other parts of the system.

The system has the only solution, if the constructed lines intersect and have one common point. It is incompatible if parallel to each other. And it has infinitely many solutions when the lines merge with each other.

This method is considered very visual. The main disadvantage is that the calculated unknowns have approximate values. More accurate results are provided by the so-called algebraic methods.

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

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

You will need

- a sheet of paper;
- pen or pencil.

Instructions

Imagine that there are 8 rabbits in front of you, and you only have 5 carrots. Think about it, you still need to buy more carrots so that each rabbit gets one.

Let's present this problem in the form of an equation: 5 + x = 8. Let's substitute the number 3 in place of x. Indeed, 5 + 3 = 8.

When you substituted a number for x, you did the same thing as when you subtracted 5 from 8. So, to find unknown term, subtract the known term from the sum.

Let's say you have 20 rabbits and only 5 carrots. Let's make it up. An equation is an equality that holds only for certain values of the letters included in it. The letters whose meanings need to be found are called . Write an equation with one unknown, call it x. When solving our rabbit problem, we get the following equation: 5 + x = 20.

Let's find the difference between 20 and 5. When subtracting, the number from which it is subtracted is the one being reduced. The number that is subtracted is called , and the final result is called the difference. So, x = 20 – 5; x = 15. You need to buy 15 carrots for the rabbits.

Check: 5 + 15 = 20. The equation is solved correctly. Of course, when it comes to such simple ones, checking is not necessary. However, when you have equations with three-digit, four-digit, etc. numbers, you definitely need to check to be absolutely sure of the result of your work.

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Useful advice

To find the unknown minuend, you need to add the subtrahend to the difference.

To find the unknown subtrahend, you need to subtract the difference from the minuend.

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

A system of three equations with three unknowns may not have solutions, despite sufficient quantity equations. You can try to solve it using the substitution method or using Cramer's method. Cramer's method, in addition to solving the system, allows you to evaluate whether the system is solvable before finding the values of the unknowns.

Instructions

The substitution method consists of sequentially sequentially one unknown through two others and substituting the resulting result into the equations of the system. Let a system of three equations be given in general view:

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

Express x from the first equation: x = (d1 - b1y - c1z)/a1 - and substitute into the second and third equations, then express y from the second equation and substitute into the third. You will obtain a linear expression for z through the coefficients of the system equations. Now go “backward”: substitute z into the second equation and find y, and then substitute z and y into the first and solve for x. The process is generally shown in the figure before finding z. Further writing in general form will be too cumbersome; in practice, by substituting , you can quite easily find all three unknowns.

Cramer's method consists of constructing a system matrix and calculating the determinant of this matrix, as well as three more auxiliary matrices. The system matrix is composed of coefficients for the unknown terms of the equations. A column containing the numbers on the right-hand sides of equations, a column of right-hand sides. It is not used in the system, but is used when solving the system.

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Please note

All equations in the system must provide additional information independent of other equations. Otherwise, the system will be underdetermined and it will not be possible to find an unambiguous solution.

Useful advice

After solving the system of equations, substitute the found values into the original system and check that they satisfy all the equations.

By itself equation with three unknown has many solutions, so most often it is supplemented by two more equations or conditions. Depending on what the initial data are, the course of the decision will largely depend.

You will need

- a system of three equations with three unknowns.

Instructions

If two of the three systems have only two of the three unknowns, try to express some variables in terms of the others and substitute them into equation with three unknown. Your goal in this case is to turn it into normal equation with an unknown person. If this is , the further solution is quite simple - substitute the found value into other equations and find all the other unknowns.

Some systems of equations can be subtracted from one equation by another. See if it is possible to multiply one of or a variable so that two unknowns are canceled at once. If there is such an opportunity, take advantage of it; most likely, the subsequent solution will not be difficult. Remember that when multiplying by a number, you must multiply both the left side and the right side. Likewise, when subtracting equations, you must remember that the right-hand side must also be subtracted.

If the previous methods did not help, use in a general way solutions to any equations with three unknown. To do this, rewrite the equations in the form a11x1+a12x2+a13x3=b1, a21x1+a22x2+a23x3=b2, a31x1+a32x2+a33x3=b3. Now create a matrix of coefficients for x (A), a matrix of unknowns (X) and a matrix of free variables (B). Please note that by multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix of free terms, that is, A*X=B.

Find matrix A to the power (-1) by first finding , note that it should not be equal to zero. After this, multiply the resulting matrix by matrix B, as a result you will receive the desired matrix X, indicating all the values.

You can also find a solution to a system of three equations using Cramer's method. To do this, find the third-order determinant ∆ corresponding to the system matrix. Then successively find three more determinants ∆1, ∆2 and ∆3, substituting the values of the free terms instead of the values of the corresponding columns. Now find x: x1=∆1/∆, x2=∆2/∆, x3=∆3/∆.

Sources:

solutions to equations with three unknowns

When starting to solve a system of equations, figure out what kind of equations they are. Methods for solving linear equations have been studied quite well. Nonlinear equations are most often not solved. There are only one special cases, each of which is practically individual. Therefore, the study of solution techniques should begin with linear equations. Such equations can even be solved purely algorithmically.

Instructions

Begin your learning process by learning how to solve a system of two linear equations with two unknowns X and Y by elimination. a11*X+a12*Y=b1 (1); a21*X+a22*Y=b2 (2). The coefficients of the equations are indicated by indices indicating their locations. Thus, the coefficient a21 emphasizes the fact that it is written in the first place in the second equation. In generally accepted notation, the system is written by equations located one below the other and jointly denoted by a curly bracket on the right or left (for more details, see Fig. 1a).

The numbering of equations is arbitrary. Choose the simplest one, such as one in which one of the variables is preceded by a coefficient of 1 or at least an integer. If this is equation (1), then further express, say, the unknown Y in terms of X (the case of excluding Y). To do this, transform (1) to the form a12*Y=b1-a11*X (or a11*X=b1-a12*Y when excluding X)), and then Y=(b1-a11*X)/a12. Substituting the latter into equation (2) write a21*X+a22*(b1-a11*X)/a12=b2. Solve this equation for X.
a21*X+a22*b1/a12-a11*a22*X/a12=b2; (a21-a11*a22/a12)*X=b2-a22*b1/a12;
X=(a12* b2-a22*b1)/(a12*a21-a11*a22) or X=(a22* b1-a12*b2)/(a11*a22-a12*a21).
Using the found connection between Y and X, you will finally obtain the second unknown Y=(a11* b2-a21*b1)/(a11*a22-a12*a21).

If the system had been specified with specific numerical coefficients, then the calculations would have been less cumbersome. But general solution makes it possible to consider the fact that the unknowns found are exactly the same. Yes, and the numerators show some patterns in their construction. If the dimension of the system of equations were greater than two, then the elimination method would lead to very cumbersome calculations. To avoid them, purely algorithmic solutions have been developed. The simplest of them is Cramer's algorithm (Cramer's formulas). For you should find out general system equations from n equations.

System n linear algebraic equations with n unknowns has the form (see Fig. 1a). In it, aij are the coefficients of the system,
xj – unknowns, bi – free terms (i=1, 2, ... , n; j=1, 2, ... , n). Such a system can be written compactly in matrix form AX=B. Here A is the matrix of system coefficients, X is the column matrix of unknowns, B is the column matrix of free terms (see Figure 1b). According to Cramer's method, each unknown xi =∆i/∆ (i=1,2…,n). The determinant ∆ of the coefficient matrix is called the main one, and ∆i the auxiliary one. For each unknown, the auxiliary determinant is found by replacing the i-th column of the main determinant with a column of free terms. The Cramer method for the case of second and third order systems is presented in detail in Fig. 2.

The system is a combination of two or more equalities, each of which contains two or more unknowns. There are two main ways to solve systems of linear equations that are used within school curriculum. One of them is called the method, the other - the addition method.

Standard form of a system of two equations

At standard form the first equation has the form a1*x+b1*y=c1, the second equation has the form a2*x+b2*y=c2 and so on. For example, in the case of two parts of the system in both given a1, a2, b1, b2, c1, c2 - some numerical odds, presented in specific equations. In turn, x and y represent unknowns whose values need to be determined. The required values turn both equations simultaneously into true equalities.

Solving the system using the addition method

In order to solve the system, that is, to find those values of x and y that will turn them into true equalities, you need to take several simple steps. The first of them is to transform either equation so that the numerical coefficients for the variable x or y in both equations are the same in magnitude, but different in sign.

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

Now it is necessary to add the two equations of the system. Obviously, the mutual destruction of variables with coefficients equal in value but opposite in sign will lead to the form -10x=-4. After this, it is necessary to solve this simple equation, from which it clearly follows that x = 0.4.

The last step in the solution process is to substitute the found value of one of the variables into any of the original equalities available in the system. For example, substituting x=0.4 into the first equation, you can get the expression 2*0.4+4y=8, from which y=1.8. Thus, x=0.4 and y=1.8 are the roots of the example system.

In order to make sure that the roots were found correctly, it is useful to check by substituting the found values into the second equation of the system. For example, in this case we get an equality of the form 0.4*6+1.8*2=6, which is true.

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Systems of equations are widely used in the economic sector for mathematical modeling of various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) or equipment placement.

Systems of equations are used not only in mathematics, but also in physics, chemistry and biology, when solving problems of finding population size.

A system of linear equations is two or more equations with several variables for which it is necessary to find a common solution. Such a sequence of numbers for which all equations become true equalities or prove that the sequence does not exist.

Linear equation

Equations of the form ax+by=c are called linear. The designations x, y are the unknowns whose value must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving an equation by plotting it will look like a straight line, all points of which are solutions to the polynomial.

Types of systems of linear equations

The simplest examples are considered to be systems of linear equations with two variables X and Y.

F1(x, y) = 0 and F2(x, y) = 0, where F1,2 are functions and (x, y) are function variables.

Solve system of equations - this means finding values (x, y) at which the system turns into a true equality or establishing that suitable values of x and y do not exist.

A pair of values (x, y), written as the coordinates of a point, is called a solution to a system of linear equations.

If systems have one common solution or no solution exists, they are called equivalent.

Homogeneous systems of linear equations are systems whose right-hand side is equal to zero. If the right part after the equal sign has a value or is expressed by a function, such a system is heterogeneous.

The number of variables can be much more than two, then we should talk about an example of a system of linear equations with three or more variables.

When faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not the case. The number of equations in the system does not depend on the variables; there can be as many of them as desired.

Simple and complex methods for solving systems of equations

There is no general analytical method for solving such systems; all methods are based on numerical solutions. The school mathematics course describes in detail such methods as permutation, algebraic addition, substitution, as well as graphical and matrix methods, solution by the Gaussian method.

The main task when teaching solution methods is to teach how to correctly analyze the system and find the optimal solution algorithm for each example. The main thing is not to memorize a system of rules and actions for each method, but to understand the principles of using a particular method

Solving examples of systems of linear equations in the 7th grade general education curriculum is quite simple and explained in great detail. In any mathematics textbook, this section is given enough attention. Solving examples of systems of linear equations using the Gauss and Cramer method is studied in more detail in the first years of higher education.

Solving systems using the substitution method

The actions of the substitution method are aimed at expressing the value of one variable in terms of the second. The expression is substituted into the remaining equation, then it is reduced to a form with one variable. The action is repeated depending on the number of unknowns in the system

Let us give a solution to an example of a system of linear equations of class 7 using the substitution method:

As can be seen from the example, the variable x was expressed through F(X) = 7 + Y. The resulting expression, substituted into the 2nd equation of the system in place of X, helped to obtain one variable Y in the 2nd equation. Solving this example is easy and allows you to get the Y value. The last step is to check the obtained values.

It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and expressing the variable in terms of the second unknown will be too cumbersome for further calculations. When there are more than 3 unknowns in the system, solving by substitution is also impractical.

Solution of an example of a system of linear inhomogeneous equations:

Solution using algebraic addition

When searching for solutions to systems using the addition method, they perform term-by-term addition and multiplication of equations by different numbers. The ultimate goal of mathematical operations is an equation in one variable.

Application of this method requires practice and observation. Solving a system of linear equations using the addition method when there are 3 or more variables is not easy. Algebraic addition is convenient to use when equations contain fractions and decimals.

Solution algorithm:

Multiply both sides of the equation by a certain number. As a result of the arithmetic operation, one of the coefficients of the variable should become equal to 1.
Add the resulting expression term by term and find one of the unknowns.
Substitute the resulting value into the 2nd equation of the system to find the remaining variable.

Method of solution by introducing a new variable

A new variable can be introduced if the system requires finding a solution for no more than two equations; the number of unknowns should also be no more than two.

The method is used to simplify one of the equations by introducing a new variable. The new equation is solved for the introduced unknown, and the resulting value is used to determine the original variable.

The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to a standard quadratic trinomial. You can solve a polynomial by finding the discriminant.

It is necessary to find the discriminant value by well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the factors of the polynomial. In the given example, a=1, b=16, c=39, therefore D=100. If the discriminant is greater than zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant less than zero, then there is only one solution: x= -b / 2*a.

The solution for the resulting systems is found by the addition method.

Visual method for solving systems

Suitable for 3 equation systems. The method consists in constructing graphs of each equation included in the system on the coordinate axis. The coordinates of the intersection points of the curves will be the general solution of the system.

The graphical method has a number of nuances. Let's look at several examples of solving systems of linear equations in a visual way.

As can be seen from the example, for each line two points were constructed, the values of the variable x were chosen arbitrarily: 0 and 3. Based on the values of x, the values for y were found: 3 and 0. Points with coordinates (0, 3) and (3, 0) were marked on the graph and connected by a line.

The steps must be repeated for the second equation. The point of intersection of the lines is the solution of the system.

The following example requires finding a graphical solution to a system of linear equations: 0.5x-y+2=0 and 0.5x-y-1=0.

As can be seen from the example, the system has no solution, because the graphs are parallel and do not intersect along their entire length.

The systems from examples 2 and 3 are similar, but when constructed it becomes obvious that their solutions are different. It should be remembered that it is not always possible to say whether a system has a solution or not; it is always necessary to construct a graph.

The matrix and its varieties

Matrices are used to concisely write a system of linear equations. A matrix is a special type of table filled with numbers. n*m has n - rows and m - columns.

A matrix is square when the number of columns and rows are equal. A matrix-vector is a matrix of one column with an infinitely possible number of rows. A matrix with ones along one of the diagonals and other zero elements is called identity.

An inverse matrix is a matrix when multiplied by which the original one turns into a unit matrix; such a matrix exists only for the original square one.

Rules for converting a system of equations into a matrix

In relation to systems of equations, the coefficients and free terms of the equations are written as matrix numbers; one equation is one row of the matrix.

A matrix row is said to be nonzero if at least one element of the row is not zero. Therefore, if in any of the equations the number of variables differs, then it is necessary to enter zero in place of the missing unknown.

The matrix columns must strictly correspond to the variables. This means that the coefficients of the variable x can be written only in one column, for example the first, the coefficient of the unknown y - only in the second.

When multiplying a matrix, all elements of the matrix are sequentially multiplied by a number.

Options for finding the inverse matrix

The formula for finding the inverse matrix is quite simple: K -1 = 1 / |K|, where K -1 - inverse matrix, and |K| is the determinant of the matrix. |K| must not be equal to zero, then the system has a solution.

The determinant is easily calculated for a two-by-two matrix; you just need to multiply the diagonal elements by each other. For the “three by three” option there is a formula |K|=a 1 b 2 c 3 + a 1 b 3 c 2 + a 3 b 1 c 2 + a 2 b 3 c 1 + a 2 b 1 c 3 + a 3 b 2 c 1 . You can use the formula, or you can remember that you need to take one element from each row and each column so that the numbers of columns and rows of elements are not repeated in the work.

Solving examples of systems of linear equations using the matrix method

The matrix method of finding a solution allows you to reduce cumbersome entries when solving systems with a large number variables and equations.

In the example, a nm are the coefficients of the equations, the matrix is a vector x n are variables, and b n are free terms.

Solving systems using the Gaussian method

In higher mathematics, the Gaussian method is studied together with the Cramer method, and the process of finding solutions to systems is called the Gauss-Cramer solution method. These methods are used to find variables of systems with a large number of linear equations.

Gauss's method is very similar to solutions using substitutions and algebraic addition, but more systematic. In the school course, the solution by the Gaussian method is used for systems of 3 and 4 equations. The purpose of the method is to reduce the system to the form of an inverted trapezoid. By means of algebraic transformations and substitutions, the value of one variable is found in one of the equations of the system. The second equation is an expression with 2 unknowns, while 3 and 4 are, respectively, with 3 and 4 variables.

After bringing the system to the described form, the further solution is reduced to the sequential substitution of known variables into the equations of the system.

In school textbooks for grade 7, an example of a solution by the Gauss method is described as follows:

As can be seen from the example, at step (3) two equations were obtained: 3x 3 -2x 4 =11 and 3x 3 +2x 4 =7. Solving any of the equations will allow you to find out one of the variables x n.

Theorem 5, which is mentioned in the text, states that if one of the equations of the system is replaced by an equivalent one, then the resulting system will also be equivalent to the original one.

The Gauss method is difficult for students to understand high school, but is one of the most interesting ways to develop the ingenuity of children enrolled in advanced study programs in mathematics and physics classes.

For ease of recording, calculations are usually done as follows:

The coefficients of the equations and free terms are written in the form of a matrix, where each row of the matrix corresponds to one of the equations of the system. separates the left side of the equation from the right. Roman numerals indicate the numbers of equations in the system.

First, write down the matrix to be worked with, then all the actions carried out with one of the rows. The resulting matrix is written after the "arrow" sign and the necessary algebraic operations are continued until the result is achieved.

The result should be a matrix in which one of the diagonals is equal to 1, and all other coefficients are equal to zero, that is, the matrix is reduced to a unit form. We must not forget to perform calculations with numbers on both sides of the equation.

This recording method is less cumbersome and allows you not to be distracted by listing numerous unknowns.

The free application of any method of solution will require care and certain experience. Not all methods are of an applied nature. Some methods of finding solutions are more preferable in a particular area of human activity, while others exist for educational purposes.

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

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

This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.

This way you can conduct your own training and/or training of yours. younger brothers or sisters, while the level of education in the field of problems being solved increases.

Rules for entering equations

Any Latin letter can act as a variable.
For example: $x, y, z, a, b, c, o, p, q$, etc.

When entering equations you can use parentheses. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of elements.
For example: 6x+1 = 5(x+y)+2

In equations, you can use not only whole numbers, but also fractions in the form of decimals and ordinary fractions.

Rules for entering decimal fractions.
Integer and fractional parts in decimals can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &

Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)

Solve system of equations

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A little theory.

Solving systems of linear equations. Substitution method

The sequence of actions when solving a system of linear equations using the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression into another equation of the system instead of this variable;

$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$

Let's express y in terms of x from the first equation: y = 7-3x. Substituting the expression 7-3x into the second equation instead of y, we obtain the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$

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

Substituting 1 instead of x into the equality y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$

Pair (1;4) - solution of the system

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

Solving systems of linear equations by addition

Let's consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by substitution, we move from this system to another, equivalent system, in which one of the equations contains only one variable.

The sequence of actions when solving a system of linear equations using the addition method:
1) multiply the equations of the system term by term, selecting factors so that the coefficients of one of the variables become opposite numbers;
2) add the left and right sides of the system equations term by term;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.

Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$

In the equations of this system, the coefficients of y are opposite numbers. Adding the left and right sides of the equations term by term, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$

From the equation 3x=33 we find that x=11. Substituting this x value into the equation $x-3y=38$ we get an equation with the variable y: $11-3y=38$. Let's solve this equation:
$-3y=27 \Rightarrow y=-9 $

Thus, we found the solution to the system of equations by addition: $x=11; y=-9$ or $(11;-9)$

Taking advantage of the fact that in the equations of the system the coefficients of y are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both sides of each of the equations of the original system), in which one of the equations contains only one variable.

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We are already familiar with the concept of a linear equation in two unknowns. Equations can be present in one problem either individually or several equations at once. In such cases, the equations are combined into a system of equations.

What is a system of linear equations

System of equations- these are two or more equations for which it is necessary to find all their common solutions. Usually, to write a system of equations, they are written in a column and one common curly bracket is drawn. A recording of the system of linear equations is presented below.

( 4x + 3y = 6
( 2x + y = 4

This entry means that a system of two equations with two variables is given. If there were three equations in the system, then we would be talking about a system of three equations. And so on for any number of equations.

If all the equations present in a system are linear, then we say that a system of linear equations is given. In the example above, a system of two linear equations is presented. As noted above, the system may have general solutions. We will talk about the term “general solution” below.

What is the solution?

A solution to a system of two equations with two unknowns is a pair of numbers (x,y) such that if we substitute these numbers into the equations of the system, then each of the equations of the system turns into a true equality.

For example, we have a system of two linear equations. The solution to the first equation will be all pairs of numbers that satisfy this equation.

For the second equation, the solution will be pairs of numbers that satisfy this equation. If there is a pair of numbers that satisfies both the first and second equations, then this pair of numbers will be the solution to a system of two linear equations in two unknowns.

Graphic solution

Graphically, the solution to a linear equation is all the points of a certain line on the plane.

For a system of linear equations, we will have several straight lines (according to the number of equations). And the solution to the system of equations will be the point at which ALL lines intersect. If there is no such point, then the system will have no solutions. The point at which all lines intersect belongs to each of these lines, therefore the solution is called general.

By the way, plotting graphs of the equations of a system and finding their common point is one of the ways to solve a system of equations. This method is called graphical.

Other ways to solve linear equations

There are other ways to solve systems of linear equations in two variables. Basic methods for solving systems of linear equations with two unknowns.