X y solution of a system of equations. Systems of linear equations

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

Equation systems are used not only in the field of mathematics, but also in physics, chemistry and biology, when solving problems of finding the population size.

A system of linear equations is a term for 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, the value of which must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving the equation by plotting its graph will look like a straight line, all points of which are the solution of the polynomial.

Types of systems of linear equations

The simplest are examples of 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 a system of equations - it means to find such values ​​(x, y) for which the system becomes a true equality, or to establish that there are no suitable values ​​of x and y.

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

If the systems have one common solution or there is no solution, they are called equivalent.

Homogeneous systems of linear equations are systems whose right 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 not homogeneous.

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 variables or more.

Faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not so. The number of equations in the system does not depend on the variables, there can be an arbitrarily large number of them.

Simple and complex methods for solving systems of equations

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

The main task in teaching methods of solving 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 applying a particular method.

Solving examples of systems of linear equations of the 7th class of the program secondary school quite simple and explained in great detail. In any textbook on mathematics, this section is given enough attention. The solution of examples of systems of linear equations by the method of Gauss and Cramer is studied in more detail in the first courses of higher educational institutions.

Solution of systems by the substitution method

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

Let's give an example of a system of linear equations of the 7th class by 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 get the value of Y. Last step this is a test of the received values.

It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and the expression of 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, the substitution solution is also impractical.

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

Solution using algebraic addition

When searching for a solution to systems by the addition method, term-by-term addition and multiplication of equations by various numbers. The ultimate goal of mathematical operations is an equation with one variable.

For applications this method it takes practice and observation. It is not easy to solve a system of linear equations using the addition method with the number of variables 3 or more. Algebraic addition is useful when the equations contain fractions and decimal numbers.

Solution action algorithm:

1. Multiply both sides of the equation by some number. As a result of the arithmetic operation, 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.

Solution method by introducing a new variable

A new variable can be introduced if the system needs to find 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 with respect to the entered unknown, and the resulting value is used to determine the original variable.

It can be seen from the example that by introducing a new variable t, it was possible to reduce the 1st equation of the system to a standard square 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 multipliers of the polynomial. In the given example, a=1, b=16, c=39, hence 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 only one solution: x= -b / 2*a.

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

A visual method for solving systems

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

The graphic method has a number of nuances. Consider several examples of solving systems of linear equations in a visual way.

As can be seen from the example, two points were constructed for each line, 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.

In the following example, it is required to find a graphical solution to the 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 the system has a solution or not, it is always necessary to build a graph.

Matrix and its varieties

Matrices are used for abbreviation systems 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 is equal. A matrix-vector is a single-column matrix with an infinitely possible number of rows. A matrix with units along one of the diagonals and other zero elements is called identity.

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

Rules for transforming a system of equations into a matrix

With regard to systems of equations, the coefficients and free members of the equations are written as numbers of the matrix, one equation is one row of the matrix.

A matrix row is called non-zero if at least one element of the row is not equal to 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 columns of the matrix must strictly correspond to the variables. This means that the coefficients of the variable x can only be written in one column, for example the first, the coefficient of the unknown y - only in the second.

When multiplying a matrix, all matrix elements are successively 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| - matrix determinant. |K| must not be equal to zero, then the system has a solution.

The determinant is easily calculated for a two-by-two matrix, it is only necessary to multiply the elements diagonally 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 column and row numbers of the elements do not repeat in the product.

Solution of examples of systems of linear equations by the matrix method

The matrix method of finding a solution makes it possible 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 the variables, and b n are the free terms.

Solution of systems by the Gauss method

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

The Gaussian method is very similar to substitution and algebraic addition solutions, but is more systematic. In the school course, the Gaussian solution is used for systems of 3 and 4 equations. The purpose of the method is to bring the system to the form of an inverted trapezoid. By 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, and 3 and 4 - with 3 and 4 variables, respectively.

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. The solution of 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 the advanced study program in math and physics classes.

For ease of recording calculations, it is customary to do the following:

Equation coefficients 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 side. Roman numerals denote the numbers of equations in the system.

First, they write down the matrix with which to work, then all the actions carried out with one of the rows. The resulting matrix is ​​written after the "arrow" sign and continue to perform the necessary algebraic operations until the result is achieved.

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

This notation 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 a certain amount of experience. Not all methods are applied. Some ways of finding solutions are more preferable in a particular area of ​​human activity, while others exist for the purpose of learning.

1. Substitution method: from any equation of the system we express one unknown in terms of 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 into the second equation of the system. Let's get the system equivalent to the original.

After bringing such 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 of this system is a pair of numbers .

2. Algebraic addition method: by adding two equations, 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. Adding the two equations of this system, we arrive at the system

After reducing 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 of this system is a pair of numbers .

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

Task. Solve the system of equations:

Solution. Let's write this system differently:

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

Let's solve it by the substitution method. From the first equation of the system, we express u through v and substitute 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.

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Instruction

Addition method.
You need to write two strictly under 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 of this system of equations: x=116, y=11.

Graphic way.
It consists in the practical finding of the coordinates of the point at which the lines are mathematically written in the system of equations. You should draw graphs of both lines separately in the same coordinate system. General view: - y \u003d kx + 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 \u003d 4

Y \u003d -3x + 1.
A straight line is built according to the first one, for convenience it needs to be written down: y \u003d 2x-4. Come up with (easier) values ​​for x, substituting it into the equation, solving it, find y. Two points are obtained, along which a straight line is built. (see pic.)
x 0 1

y -4 -2
A straight line is constructed according to the second equation: y \u003d -3x + 1.
Also build a line. (see pic.)

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).

Related videos

Helpful advice

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

Sources:

• Algebra Grade 8
• solve an equation with two unknowns online
• Examples of solving systems of linear equations with two

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

You will need

• -Ruler and pencil;
• -calculator.

Instruction

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

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

This method is considered to be very clear. The main disadvantage is that the calculated unknowns have approximate values. A more accurate result is given by the so-called algebraic methods.

Any solution to a system of equations is worth checking. To do this, substitute the obtained values ​​instead of the variables. You can also find its solution in several ways. 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 do a certain set of actions with these numbers.

You will need

• - paper;
• - Pen or pencil.

Instruction

Imagine that you have 8 rabbits in front of you, and you only have 5 carrots. Think you need to buy more carrots so that each rabbit gets a carrot.

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

When you substituted a number for x, you were doing the same operation as subtracting 5 from 8. Thus, to find unknown term, subtract the known term from the sum.

Let's say you have 20 rabbits and only 5 carrots. Let's compose . An equation is an equality that holds only for certain values ​​of the letters included in it. The letters whose values ​​you want to find are called. Write an equation with one unknown, call it x. When solving our problem about rabbits, the following equation is obtained: 5 + x = 20.

Let's find the difference between 20 and 5. When subtracting, the number from which it is subtracted is 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 rabbits.

Make a check: 5 + 15 = 20. The equation is correct. Of course, when we are talking about such simple ones, it is not necessary to perform a check. However, when it comes to equations with three-digit, four-digit, and so on, it is imperative to check to be absolutely sure of the result of your work.

Related videos

Helpful advice

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

To find the unknown subtrahend, it is necessary 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 a sufficient number of equations. You can try to solve it using the substitution method or using the Cramer method. Cramer's method, in addition to solving the system, allows one to evaluate whether the system is solvable before finding the values ​​of the unknowns.

Instruction

The substitution method consists in sequentially one unknown through two others and substituting the result obtained 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 get a linear expression for z through the coefficients of the equations of the system. Now go "back": plug z into the second equation and find y, then plug z and y into the first equation and find x. The process is generally shown in the figure until z is found. Further, the record in general form will be too cumbersome, in practice, substituting , you can quite easily find all three unknowns.

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

Related videos

note

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

Helpful 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 the 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.

Instruction

If two of the three systems have only two of the three unknowns, try expressing some variables in terms of the others and plugging them into the equation with three unknown. Your goal with this is to turn it into a normal the equation with the unknown. 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 by or a variable so that two unknowns are reduced at once. If there is such an opportunity, use it, most likely, the subsequent decision will not be difficult. Do not forget that when multiplying by a number, you must multiply both the left side and the right side. Similarly, when subtracting equations, remember that the right hand side must also be subtracted.

If previous methods did not help, use the general method for solving any equations with three unknown. To do this, rewrite the equations in the form a11x1 + a12x2 + a13x3 \u003d b1, a21x1 + a22x2 + a23x3 \u003d b2, a31x1 + a32x2 + a33x3 \u003d b3. Now make a matrix of coefficients at x (A), a matrix of unknowns (X) and a matrix of free ones (B). Pay attention, multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix, a matrix of free members, that is, A * X \u003d B.

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

You can also find a solution to a system of three equations using the Cramer method. To do this, find the third-order determinant ∆ corresponding to the matrix of the system. 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 of equations with three unknowns

Starting to solve a system of equations, figure out what these equations are. The methods of solving linear equations are well studied. Nonlinear equations are most often not solved. There are only one special cases, each of which is practically individual. Therefore, the study of methods of solution should begin with linear equations. Such equations can be solved even purely algorithmically.

the denominators of the found unknowns are exactly the same. Yes, and the numerators are visible some patterns of 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 know general system equations from n equations.

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

A system is a union of two or more equalities, each of which has two or more unknowns. There are two main ways to solve systems of linear equations that are used in the framework school curriculum. One of them is called the method, the other is the addition method.

Standard form of a system of two equations

At standard form the first equation is a1*x+b1*y=c1, the second equation is 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 are some numerical coefficients presented in specific equations. In turn, x and y are unknowns whose values ​​need to be determined. The desired values ​​turn both equations simultaneously into true equalities.

Solution of the system by 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 a few simple steps. The first of these is to transform any of the equations in such a way that the numerical coefficients for the variable x or y in both equations coincide in absolute value, but differ in sign.

For example, let a system consisting of two equations be 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 factor of -2, which will lead it to the form -12x-4y=-12. The correct choice of the coefficient is one of the key tasks in the process of solving the system by 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 equal in value but opposite in sign coefficients will lead it to the form -10x=-4. After that, it is necessary to solve this simple equation, from which it unambiguously follows that x=0.4.

The last step in the solution process is the substitution of the found value of one of the variables into any of the initial 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 system shown in the example.

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 an equality of the form 0.4*6+1.8*2=6 is obtained, which is true.

Related videos

We will analyze two types of solving systems of equations:

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

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

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

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

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

Example #1:

Let's solve by the substitution method

Solving the system of equations by the 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, hence it turns out that it is easiest to express the variable x from the second equation.
x=3+10y

2. After expressing, we substitute 3 + 10y in the first equation instead of the variable x.
2(3+10y)+5y=1

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

The solution of 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 paragraph where we expressed 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 by term-by-term addition (subtraction).

Solving a system of equations by the addition method

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

1. Select a variable, let's say we select 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 a total coefficient of 6.

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

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

2. From the first equation, subtract the second 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 in any of the equations, let's say in 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 point of intersection will be x=4.6; y=6.4
Answer: (4.6; 6.4)

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