What is the solution of the system at. How is a system of equations solved? Methods for solving systems of equations

A system of equations is called linear if all equations included in the system are linear. It is customary to write a system of equations using curly braces, for example:

Definition:A pair of variable values ​​that makes each equation with two variables included in the system a true equality is called solving a system of equations.

Solve the system- means finding all its solutions or proving that there are no solutions.

When solving a system of linear equations, the following three cases are possible:

the system has no solutions;

the system has exactly one solution;

the system has infinitely many solutions.
I . Solving a system of linear equations using the substitution method.

This method can also be called the "substitution method" or the method of eliminating unknowns.

Here we are given a system of two equations with two unknowns. Note that the free terms (numbers -5 and -7) are located on the left side of the equation. Let us write the system in its usual form.

Don’t forget that when moving a term from part to part, it needs to change its sign.

What does it mean to solve a system of linear equations? Solving a system of equations means finding such values ​​of variables that turn each equation of the system into a correct equality. This statement is true for any system of equations with any number of unknowns.

Let's decide.

We substitute the resulting expression into the second equation of the system instead of the variable

Let's decide given equation relative to one variable.
Open the brackets, add similar terms and find the value :

4) Next we return to substitution to calculate the value .We already know the value, all that remains is to find:

5) Couple
is the only solution to a given system.

Answer: (2.4; 2.2).

After solving any system of equations in any way, I strongly recommend checking it on a draft. This is done easily and quickly.

1) Substitute the found answer to the first equation:

– the correct equality is obtained.

2) Substitute the found answer into the second equation:

– the correct equality is obtained.

The considered solution method is not the only one; from the first equation it was possible to express , and not .

or

However, in some cases you still can’t do without fractions. You should strive to complete any task in the most rational way. This saves time and also reduces the likelihood of making a mistake.
Example 2

Solve a system of linear equations

II. Solving the system using the method algebraic addition(subtraction) equations of the system

When solving systems of linear equations, you can use not the substitution method, but the method of algebraic addition (subtraction) of the system’s equations. This method saves time and simplifies calculations, however, now everything will become clearer.

Solve a system of linear equations:

Let's take the same system as in the first example.

2) Let's solve this equation for one variable.

As you can see, as a result of term-by-term addition, we lost the variable. This, in fact, is the essence of the method - to get rid of one of the variables.

3) Now everything is simple:
– substitute into the first equation of the system (you can also into the second):

The final solution should look something like this:

Answer: (2.4; 2.2).

Solve a system of linear equations:

In this example, we can use the substitution method, but the big disadvantage is that when we express any variable from any equation, we will get a solution in ordinary fractions. Few people like working with fractions, which means it’s a waste of time, and there’s a high chance of making a mistake.

Therefore, it is advisable to use term-by-term addition (subtraction) of equations. Let's analyze the coefficients for the corresponding variables:

As we see, the numbers in pairs (14 and 7), (-9 and –2) are different, therefore, if we add (subtract) the equations right now, we will not get rid of the variable. Thus, I would like to see in one of the pairs numbers that are identical in absolute value, for example, 14 and -14 or 18 and –18.

We will consider the coefficients of the variable.

14x – 9y = 24;

7x – 2y = 17.
We select a number that is divisible by both 14 and 7, and it should be as small as possible. In mathematics, this number is called the least common multiple. If you find it difficult to select, you can simply multiply the coefficients.

As a result:

Now let’s subtract the second from the first equation term by term.

It should be noted that one could do the opposite - subtract the first from the second equation, this does not change anything.

Now we substitute the found value into one of the system equations, for example, into the first one:

Answer: (3:2)

14x – 9y = 24;

7x – 2y = 17.

Obviously, instead of a pair of coefficients (-9 and –3), we need to get 18 and –18.

We add the equations term by term and find the values ​​of the variables:

Now we substitute the found value of x into one of the equations of the system, for example, into the first one:

Answer: (3:2)

Solve a system of linear equations:

This is an example for you to solve on your own (answer at the end of the lecture).
Example 6.

Solve system of equations

Solution. The system has no solutions, since two equations of the system cannot be satisfied simultaneously (from the first equation
and from the second

Answer: There are no solutions.
Example 7.

solve system of equations

Solution. The system has infinitely many solutions, since the second equation is obtained from the first by multiplying by 2 (i.e., in fact, there is only one equation with two unknowns).

Answer: There are infinitely many solutions.
III. Solving the system using matrices.

The determinant of this system is a determinant composed of coefficients for the unknowns. This determinant

1. Solving the system using the substitution method.
2. Solving the system by term-by-term addition (subtraction) of the system equations.

In order to solve the system of equations by substitution method you need to follow a simple algorithm:
1. Express. From any equation we express one variable.
2. Substitute. We substitute the resulting value into another equation instead of the expressed variable.
3. Solve the resulting equation with one variable. We find a solution to the system.

To solve system by term-by-term addition (subtraction) method need to:
1. Select a variable for which we will make identical coefficients.
2. We add or subtract equations, resulting in an equation with one variable.
3. Solve the resulting linear equation. We find a solution to the system.

The solution to the system is the intersection points of the function graphs.

Let us consider in detail the solution of systems using examples.

Example #1:

Let's solve by substitution method

2x+5y=1 (1 equation)
x-10y=3 (2nd equation)

1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, which means that it is easiest to express the variable x from the second equation.
x=3+10y

2.After we have expressed it, we substitute 3+10y into the first equation instead of the variable x.
2(3+10y)+5y=1

3. Solve the resulting equation with one variable.
2(3+10y)+5y=1 (open the brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2

The solution to the equation system is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first point where we expressed it, we substitute y there.
x=3+10y
x=3+10*(-0.2)=1

It is customary to write points in the first place we write the variable x, and in the second place the variable y.
Answer: (1; -0.2)

Example #2:

Let's solve using the term-by-term addition (subtraction) method.

3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)

1. We choose a variable, let’s say we choose x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get overall coefficient 6.

3x-2y=1 |*2
6x-4y=2

2x-3y=-10 |*3
6x-9y=-30

2. Subtract the second from the first equation to get rid of the variable x. Solve the linear equation.
__6x-4y=2

5y=32 | :5
y=6.4

3. Find x. We substitute the found y into any of the equations, let’s say into the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6

The intersection point will be x=4.6; y=6.4
Answer: (4.6; 6.4)

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Systems of equations are widely used in the economic sector in mathematical modeling 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 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. Solution this example does not cause difficulties and allows you to obtain 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 inappropriate.

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, equations are added term by term and multiplied by various 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:

1. 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.
2. Add the resulting expression term by term and find one of the unknowns.
3. 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 the standard one quadratic trinomial. You can solve a polynomial by finding the discriminant.

It is necessary to find the value of the discriminant using the 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 points of intersection of the curves and will be general decision systems.

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 for short note systems of linear equations. A matrix is ​​a table special type 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 infinite possible number lines. 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 is the 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 of 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 variable systems with a large number of linear equations.

The Gauss method is very similar to solutions by substitution and algebraic addition, but is 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 Gaussian 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 use of any solution method will require care and some 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.

In this lesson we will look at methods for solving a system of linear equations. In a course of higher mathematics, systems of linear equations are required to be solved both in the form of separate tasks, for example, “Solve the system using Cramer’s formulas,” and in the course of solving other problems. Systems of linear equations have to be dealt with in almost all branches of higher mathematics.

First, a little theory. What in in this case stands for the mathematical word "linear"? This means that the equations of the system All variables included in the first degree: without any fancy stuff like etc., which only participants in mathematical Olympiads are delighted with.

In higher mathematics, not only letters familiar from childhood are used to denote variables.
A fairly popular option is variables with indexes: .
Or initial letters Latin alphabet, small and large:
It is not so rare to find Greek letters: – known to many as “alpha, beta, gamma”. And also a set with indices, say, with the letter “mu”:

The use of one or another set of letters depends on the section of higher mathematics in which we are faced with a system of linear equations. So, for example, in systems of linear equations encountered when solving integrals and differential equations, it is traditional to use the notation

But no matter how the variables are designated, the principles, methods and methods for solving a system of linear equations do not change. Thus, if you come across something scary like , do not rush to close the problem book in fear, after all, you can draw the sun instead, a bird instead, and a face (the teacher) instead. And, funny as it may seem, a system of linear equations with these notations can also be solved.

I have a feeling that the article will turn out to be quite long, so a small table of contents. So, the sequential “debriefing” will be like this:

– Solving a system of linear equations using the substitution method (“school method”);
– Solving the system by term-by-term addition (subtraction) of the system equations;
– Solution of the system using Cramer’s formulas;
– Solving the system using an inverse matrix;
– Solving the system using the Gaussian method.

Everyone is familiar with systems of linear equations from school mathematics courses. Essentially, we start with repetition.

Solving a system of linear equations using the substitution method

This method can also be called the “school method” or the method of eliminating unknowns. Figuratively speaking, it can also be called “an unfinished Gaussian method.”

Example 1

Here we are given a system of two equations with two unknowns. Note that the free terms (numbers 5 and 7) are located on the left side of the equation. Generally speaking, it doesn’t matter where they are, on the left or on the right, it’s just that in problems in higher mathematics they are often located that way. And such a record should not lead to confusion; if necessary, the system can always be written “as usual”: . Don’t forget that when moving a term from part to part, it needs to change its sign.

What does it mean to solve a system of linear equations? Solving a system of equations means finding many of its solutions. The solution of a system is a set of values ​​of all variables included in it, which turns EVERY equation of the system into a true equality. In addition, the system can be non-joint (have no solutions).Don't worry, it's general definition=) We will have only one value “x” and one value “y”, which satisfy each equation s-we.

There is a graphical method for solving the system, which you can familiarize yourself with in class. The simplest problems with a line. There I talked about geometrically systems of two linear equations with two unknowns. But now this is the era of algebra, and numbers-numbers, actions-actions.

Let's decide: from the first equation we express:
We substitute the resulting expression into the second equation:

We open the brackets, add similar terms and find the value:

Next, we remember what we danced for:
We already know the value, all that remains is to find:

Answer:

After ANY system of equations has been solved in ANY way, I strongly recommend checking (orally, on a draft or on a calculator). Fortunately, this is done easily and quickly.

1) Substitute the found answer into the first equation:

– the correct equality is obtained.

2) Substitute the found answer into the second equation:

– the correct equality is obtained.

Or, to put it more simply, “everything came together”

The considered solution method is not the only one; from the first equation it was possible to express , and not .
You can do the opposite - express something from the second equation and substitute it into the first equation. By the way, note that the most disadvantageous of the four methods is to express from the second equation:

The result is fractions, but why? There is a more rational solution.

However, in some cases you still can’t do without fractions. In this regard, I would like to draw your attention to HOW I wrote down the expression. Not like this: and in no case like this: .

If in higher mathematics you are dealing with fractional numbers, then try to carry out all calculations in ordinary improper fractions.

Exactly, and not or!

A comma can be used only sometimes, in particular if it is the final answer to some problem, and no further actions need to be performed with this number.

Many readers probably thought “why such a detailed explanation as for a correction class, everything is clear.” Nothing of the kind, it seems like such a simple school example, but there are so many VERY important conclusions! Here's another one:

You should strive to complete any task in the most rational way. If only because it saves time and nerves, and also reduces the likelihood of making a mistake.

If in a higher mathematics problem you encounter a system of two linear equations with two unknowns, then you can always use the substitution method (unless it is indicated that the system needs to be solved by another method). No teacher will think that you are a sucker and will reduce your grade for using the “school method” "
Moreover, in some cases it is advisable to use the substitution method when more variables.

Example 2

Solve a system of linear equations with three unknowns

A similar system of equations often arises when using the so-called method of indefinite coefficients, when we find the integral of a fractional rational function. The system in question was taken from there by me.

When finding the integral, the goal is fast find the values ​​of the coefficients, rather than using Cramer’s formulas, the inverse matrix method, etc. Therefore, in this case, the substitution method is appropriate.

When any system of equations is given, first of all it is desirable to find out whether it is possible to somehow simplify it IMMEDIATELY? Analyzing the equations of the system, we notice that the second equation of the system can be divided by 2, which is what we do:

Reference: the mathematical sign means “from this follows that” and is often used in problem solving.

Now let's analyze the equations; we need to express some variable in terms of the others. Which equation should I choose? You probably already guessed that the easiest way for this purpose is to take the first equation of the system:

Here, no matter what variable to express, one could just as easily express or .

Next, we substitute the expression for into the second and third equations of the system:

We open the brackets and present similar terms:

Divide the third equation by 2:

From the second equation we express and substitute into the third equation:

Almost everything is ready, from the third equation we find:
From the second equation:
From the first equation:

Check: Substitute the found values ​​of the variables into the left side of each equation of the system:

1)
2)
3)

The corresponding right-hand sides of the equations are obtained, thus the solution is found correctly.

Example 3

Solve a system of linear equations with 4 unknowns

This is an example for you to solve on your own (answer at the end of the lesson).

Solving the system by term-by-term addition (subtraction) of the system equations

When solving systems of linear equations, you should try to use not the “school method”, but the method of term-by-term addition (subtraction) of the equations of the system. Why? This saves time and simplifies calculations, however, now everything will become clearer.

Example 4

Solve a system of linear equations:

I took the same system as in the first example.
Analyzing the system of equations, we notice that the coefficients of the variable are identical in magnitude and opposite in sign (–1 and 1). In such a situation, the equations can be added term by term:

Actions circled in red are performed MENTALLY.
As you can see, as a result of term-by-term addition, we lost the variable. This, in fact, is what the essence of the method is to get rid of one of the variables.

1. Substitution method: from any equation of the system we express one unknown through another and substitute it into the second equation of the system.

Task. Solve the system of equations:

Solution. From the first equation of the system we express at through X and substitute it into the second equation of the system. Let's get the system equivalent to the original one.

After bringing similar terms, the system will take the form:

From the second equation we find: . Substituting this value into the equation at = 2 - 2X, we get at= 3. Therefore, the solution to this system is a pair of numbers.

2. Algebraic addition method: By adding two equations, you get an equation with one variable.

Task. Solve the system equation:

Solution. Multiplying both sides of the second equation by 2, we obtain the system equivalent to the original one. Adding the two equations of this system, we arrive at the system

After bringing similar terms, this system will take the form: From the second equation we find . Substituting this value into equation 3 X + 4at= 5, we get , where . Therefore, the solution to this system is a pair of numbers.

3. Method for introducing new variables: we are looking for some repeating expressions in the system, which we will denote by new variables, thereby simplifying the appearance of the system.

Task. Solve the system of equations:

Solution. Let's write it down this system otherwise:

Let x + y = u, xy = v. Then we get the system

Let's solve it using the substitution method. From the first equation of the system we express u through v and substitute it into the second equation of the system. Let's get the system those.

From the second equation of the system we find v 1 = 2, v 2 = 3.

Substituting these values ​​into the equation u = 5 - v, we get u 1 = 3,
u 2 = 2. Then we have two systems

Solving the first system, we get two pairs of numbers (1; 2), (2; 1). The second system has no solutions.

Exercises for independent work

1. Solve systems of equations using the substitution method.