How to solve equations using the substitution method. System of equations

To solve the system linear equations with two variables using the substitution method we proceed as follows:

1) express one variable through another in one of the equations of the system (x through y or y through x);

2) we substitute the resulting expression into another equation of the system and obtain a linear equation with one variable;

3) solve the resulting linear equation with one variable and find the value of this variable;

4) we substitute the found value of the variable into expression (1) for another variable and find the value of this variable.

Examples. Solve a system of linear equations using the substitution method.

Let's express X through y from the 1st equation. We get: x=7+y. Let's substitute the expression (7+y) instead X into the 2nd equation of the system.

We got the equation: 3 · (7+y)+2y=16. This is an equation with one variable at. Let's solve it. Let's open the brackets: 21+3y+2y=16. Collecting terms with a variable at on the left side, and free terms on the right. When transferring a term from one side of an equality to another, we change the sign of the term to the opposite.

We get: 3y+2y=16-21. We present similar terms in each part of the equality. 5y=-5. We divide both sides of the equality by the coefficient of the variable. y=-5:5; y=-1. Substitute this value at into the expression x=7+y and find X. We get: x=7-1; x=6. A pair of variable values ​​x=6 and y=-1 is a solution to this system.

Write down: (6; -1). Answer: (6; -1). It is convenient to write these arguments as shown below, i.e. systems of equations - on the left below each other. On the right are displays, necessary explanations, checking the solution, etc.

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. Method algebraic addition : 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 get 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.


A system of linear equations with two unknowns is two or more linear equations for which it is necessary to find all their common solutions. We will consider systems of two linear equations in two unknowns. General form a system of two linear equations with two unknowns is presented in the figure below:

( a1*x + b1*y = c1,
( a2*x + b2*y = c2

Here x and y are unknown variables, a1, a2, b1, b2, c1, c2 are some real numbers. A solution to a system of two linear equations in 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. Consider one of the ways to solve a system of linear equations, namely the substitution method.

Solution algorithm by substitution method

Algorithm for solving a system of linear equations using the substitution method:

1. Select one equation (it is better to choose the one where the numbers are smaller) and express one variable from it in terms of another, for example, x in terms of y. (you can also use y through x).

2. Substitute the resulting expression instead of the corresponding variable into another equation. Thus, we get a linear equation with one unknown.

3. Solve the resulting linear equation and obtain a solution.

4. We substitute the resulting solution into the expression obtained in the first paragraph, and obtain the second unknown from the solution.

5. Check the resulting solution.

Example

To make it more clear, let's solve a small example.

Example 1. Solve the system of equations:

(x+2*y =12
(2*x-3*y=-18

Solution:

1. From the first equation of this system we express the variable x. We have x= (12 -2*y);

2. Substitute this expression into the second equation, we get 2*x-3*y=-18; 2*(12 -2*y) - 3*y = -18; 24 - 4y - 3*y = -18;

3. Solve the resulting linear equation: 24 - 4y - 3*y = -18; 24-7*y =-18; -7*y = -42; y=6;

4. Substitute the result obtained into the expression obtained in the first paragraph. x= (12 -2*y); x=12-2*6 = 0; x=0;

5. We check the resulting solution; to do this, we substitute the found numbers into the original system.

(x+2*y =12;
(2*x-3*y=-18;

{0+2*6 =12;
{2*0-3*6=-18;

{12 =12;
{-18=-18;

We got the correct equalities, therefore, we found the solution correctly.

Solving systems of equations using the substitution method

Let us remember what a system of equations is.

A system of two equations with two variables is two equations written below each other, joined by a curly brace. Solving a system means finding a pair of numbers that will solve both the first and second equations at the same time.

In this lesson we will get acquainted with such a method of solving systems as the substitution method.

Let's look at the system of equations:

You can solve this system graphically. To do this, we will need to construct graphs of each of the equations in one coordinate system, transforming them to the form:

Then find the coordinates of the intersection point of the graphs, which will be the solution to the system. But the graphical method is not always convenient, because differs in low accuracy, or even inaccessibility. Let's try to take a closer look at our system. Now it looks like:

You can notice that the left sides of the equations are equal, which means the right sides must also be equal. Then we get the equation:

This is a familiar equation with one variable that we can solve. Let's move the unknown terms to left side, and known ones - to the right, not forgetting to change the +, - signs when transferring. We get:

Now let’s substitute the found value of x into any equation of the system and find the value of y. In our system, it is more convenient to use the second equation y = 3 - x; after substitution we get y = 2. Now let’s analyze the work done. First, in the first equation we expressed the y variable in terms of the x variable. Then the resulting expression - 2x + 4 was substituted into the second equation instead of the variable y. Then we solved the resulting equation with one variable x and found its value. And finally, we used the found value of x to find another variable y. Here the question arises: was it necessary to express the variable y from both equations at once? Of course not. We could express one variable in terms of another in only one equation of the system and use it instead of the corresponding variable in the second. Moreover, you can express any variable from any equation. Here the choice depends solely on the convenience of the account. Mathematicians called this procedure an algorithm for solving systems of two equations with two variables using the substitution method. Here's what it looks like.

1. Express one of the variables in terms of another in one of the equations of the system.

2.Substitute the resulting expression instead of the corresponding variable into another equation of the system.

3.Solve the resulting equation with one variable.

4.Substitute the found value of the variable into the expression obtained in step one and find the value of another variable.

5.Write the answer in the form of a pair of numbers that were found in the third and fourth steps.

Let's look at another example. Solve the system of equations:

Here it is more convenient to express the variable y from the first equation. We get y = 8 - 2x. The resulting expression must be substituted for y in the second equation. We get:

Let's write this equation separately and solve it. First, let's open the brackets. We get the equation 3x - 16 + 4x = 5. Let's collect the unknown terms on the left side of the equation, and the known ones on the right, and present similar terms. We get the equation 7x = 21, hence x = 3.

Now, using the found value of x, you can find:

Answer: a pair of numbers (3; 2).

Thus, in this lesson we learned to solve systems of equations with two unknowns in an analytical, accurate way, without resorting to dubious graphical methods.

List of used literature:

  1. Mordkovich A.G., Algebra 7th grade in 2 parts, Part 1, Textbook for educational institutions/ A.G. Mordkovich. – 10th ed., revised – Moscow, “Mnemosyne”, 2007.
  2. Mordkovich A.G., Algebra 7th grade in 2 parts, Part 2, Problem book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, “Mnemosyne”, 2007.
  3. HER. Tulchinskaya, Algebra 7th grade. Blitz survey: a manual for students of general education institutions, 4th edition, revised and expanded, Moscow, Mnemosyne, 2008.
  4. Alexandrova L.A., Algebra 7th grade. Thematic testing work V new form for students of general education institutions, edited by A.G. Mordkovich, Moscow, “Mnemosyne”, 2011.
  5. Alexandrova L.A. Algebra 7th grade. Independent work for students of general education institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, “Mnemosyne”, 2010.

Systems of equations have been widely used in the economic industry with 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 ​​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. IN school course Mathematics 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 optimal algorithm solutions 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 of the 7th grade program secondary school 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 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, they perform term-by-term addition and multiplication of equations by different numbers. The ultimate goal mathematical operations is an equation with one variable.

For Applications this method practice and observation are required. 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 arithmetic action one of the coefficients of the variable must 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 a standard 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 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 is less than zero, then there is 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 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 - 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 variables of 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 Gaussian 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 studying under the program in-depth study in math 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.