Solve the system with two unknowns - this means finding all pairs of variable values that satisfy each of the given equations. Each such pair is called system solution.
Example:
The pair of values \(x=3\);\(y=-1\) is a solution to the first system, because by substituting these triples and minus ones into the system instead of \(x\) and \(y\), both equations become into valid equalities \(\begin(cases)3-2\cdot (-1)=5 \\3 \cdot 3+2 \cdot (-1)=7 \end(cases)\)
But \(x=1\); \(y=-2\) - is not a solution to the first system, because after substitution the second equation "does not converge" \(\begin(cases)1-2\cdot(-2)=5 \\3\cdot1+2 \cdot(-2)≠7 \end(cases)\)
Note that such pairs are often written shorter: instead of "\(x=3\); \(y=-1\)" they write like this: \((3;-1)\).
How to solve a system of linear equations?
There are three main ways to solve systems linear equations:
- Substitution method.
-
\(\begin(cases)13x+9y=17\\12x-2y=26\end(cases)\)
In the second equation, each term is even, so we simplify the equation by dividing it by \(2\).
\(\begin(cases)13x+9y=17\\6x-y=13\end(cases)\)
This system can be solved in any of the ways, but it seems to me that the substitution method is the most convenient here. Let's express y from the second equation.
\(\begin(cases)13x+9y=17\\y=6x-13\end(cases)\)
Substitute \(6x-13\) for \(y\) in the first equation.
\(\begin(cases)13x+9(6x-13)=17\\y=6x-13\end(cases)\)
The first equation has become normal. We solve it.
Let's open the parentheses first.
\(\begin(cases)13x+54x-117=17\\y=6x-13\end(cases)\)
Let's move \(117\) to the right and give like terms.
\(\begin(cases)67x=134\\y=6x-13\end(cases)\)
Divide both sides of the first equation by \(67\).
\(\begin(cases)x=2\\y=6x-13\end(cases)\)
Hooray, we found \(x\)! Substitute its value into the second equation and find \(y\).
\(\begin(cases)x=2\\y=12-13\end(cases)\)\(\Leftrightarrow\)\(\begin(cases)x=2\\y=-1\end(cases )\)
Let's write down the answer.
\(\begin(cases)x-2y=5\\3x+2y=7 \end(cases)\)\(\Leftrightarrow\) \(\begin(cases)x=5+2y\\3x+2y= 7\end(cases)\)\(\Leftrightarrow\)
Substitute the resulting expression instead of this variable into another equation of the system.
\(\Leftrightarrow\) \(\begin(cases)x=5+2y\\3(5+2y)+2y=7\end(cases)\)\(\Leftrightarrow\)
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:
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 method2x+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 system of equations 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 method3x-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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More reliable than the graphical method discussed in the previous paragraph.
Substitution method
We used this method in the 7th grade to solve systems of linear equations. The algorithm that was developed in the 7th grade is quite suitable for solving systems of any two equations (not necessarily linear) with two variables x and y (of course, the variables can be denoted by other letters, which does not matter). In fact, we used this algorithm in the previous section, when the problem of double digit led to mathematical model, which is a system of equations. We solved this system of equations above by the substitution method (see example 1 from § 4).
Algorithm for using the substitution method when solving a system of two equations with two variables x, y.
1. Express y in terms of x from one equation of the system.
2. Substitute the resulting expression instead of y into another equation of the system.
3. Solve the resulting equation for x.
4. Substitute in turn each of the roots of the equation found at the third step instead of x into the expression y through x obtained at the first step.
5. Write down the answer in the form of pairs of values (x; y), which were found, respectively, in the third and fourth steps.
4) Substitute in turn each of the found values of y into the formula x \u003d 5 - Zy. If then 
5) Pairs (2; 1) and solutions of a given system of equations.
Answer: (2; 1);
Algebraic addition method
This method, like the substitution method, is familiar to you from the 7th grade algebra course, where it was used to solve systems of linear equations. Let us recall the essence of the method following example.
Example 2 Solve a system of equations
We multiply all the terms of the first equation of the system by 3, and leave the second equation unchanged: 
Subtract the second equation of the system from its first equation:
As a result of algebraic addition of two equations original system the resulting equation is simpler than the first and second equations of the given system. With this simpler equation, we have the right to replace any equation of a given system, for example, the second one. Then the given system of equations will be replaced by a simpler system:
This system can be solved by the substitution method. From the second equation we find Substituting this expression instead of y into the first equation of the system, we obtain
It remains to substitute the found values \u200b\u200bof x into the formula
If x = 2 then
Thus, we have found two solutions to the system: 
Method for introducing new variables
You got acquainted with the method of introducing a new variable when solving rational equations with one variable in the 8th grade algebra course. The essence of this method in solving systems of equations is the same, but with technical point vision, there are some features that we will discuss in the following examples.
Example 3 Solve a system of equations
Let us introduce a new variable Then the first equation of the system can be rewritten in more simple form: Let's solve this equation for the variable t:
Both of these values satisfy the condition , and therefore are roots rational equation with variable t. But that means either from where we find that x = 2y, or
Thus, with the help of the method of introducing a new variable, we were able, as it were, to “stratify” the first equation of the system, which is quite complex in appearance, into two simpler equations:
x = 2 y; y - 2x.
What's next? And then each of the two received simple equations it is necessary to consider in turn in the system with the equation x 2 - y 2 \u003d 3, which we have not yet remembered. In other words, the problem is reduced to solving two systems of equations:
It is necessary to find solutions for the first system, the second system, and include all the resulting pairs of values in the answer. Let's solve the first system of equations:
Let's use the substitution method, especially since everything is ready for it here: we substitute the expression 2y instead of x into the second equation of the system. Get
Since x \u003d 2y, we find x 1 \u003d 2, x 2 \u003d 2, respectively. Thus, two solutions to the given system are obtained: (2; 1) and (-2; -1). Let's solve the second system of equations:
Let's use the substitution method again: we substitute the expression 2x instead of y in the second equation of the system. Get
This equation has no roots, which means that the system of equations has no solutions. Thus, only the solutions of the first system should be included in the answer.
Answer: (2; 1); (-2;-1).
The method of introducing new variables when solving systems of two equations with two variables is used in two versions. First option: one new variable is introduced and used in only one equation of the system. This is exactly what happened in example 3. The second option: two new variables are introduced and used simultaneously in both equations of the system. This will be the case in example 4.
Example 4 Solve a system of equations
Let's introduce two new variables:
We learn that then
This will allow you to rewrite this system in a much simpler form, but with respect to the new variables a and b:
Since a \u003d 1, then from the equation a + 6 \u003d 2 we find: 1 + 6 \u003d 2; 6=1. Thus, for the variables a and b, we got one solution:
Returning to the variables x and y, we obtain the system of equations
To solve this system, we apply the method algebraic addition:
Since then from the equation 2x + y = 3 we find:
Thus, for the variables x and y, we got one solution:
Let us conclude this section with a brief but rather serious theoretical discussion. Have you already gained some experience in solving various equations: linear, square, rational, irrational. You know that the main idea of solving an equation is to gradually move from one equation to another, simpler but equivalent to the given one. In the previous section, we introduced the notion of equivalence for equations with two variables. This concept is also used for systems of equations.
Definition.
Two systems of equations with variables x and y are said to be equivalent if they have the same solutions or if both systems have no solutions.
All three methods (substitution, algebraic addition, and introduction of new variables) that we have discussed in this section are absolutely correct from the point of view of equivalence. In other words, using these methods, we replace one system of equations with another, simpler, but equivalent to the original system.
Graphical method for solving systems of equations
We have already learned how to solve systems of equations in such common and reliable ways as the method of substitution, algebraic addition and the introduction of new variables. And now let's remember the method that you already studied in the previous lesson. So let's recap what you know about graphic method solutions.
Method for solving systems of equations graphically is the construction of a graph for each of the specific equations that are included in this system and are in one coordinate plane, and also where it is required to find the intersections of the points of these graphs. To solve this system of equations are the coordinates of this point (x; y).
It should be remembered that for graphics system equations tend to have either one unique the right decision, or infinite set solutions, or have no solutions at all.
Now let's take a closer look at each of these solutions. And so, the system of equations can have only decision if the lines, which are the graphs of the equations of the system, intersect. If these lines are parallel, then such a system of equations has absolutely no solutions. In the case of the coincidence of the direct graphs of the equations of the system, then such a system allows you to find many solutions.
Well, now let's take a look at the algorithm for solving a system of two equations with 2 unknowns using a graphical method:
First, at first we build a graph of the 1st equation;
The second step will be to plot a graph that relates to the second equation;
Thirdly, we need to find the intersection points of the graphs.
And as a result, we get the coordinates of each intersection point, which will be the solution to the system of equations.
Let's look at this method in more detail with an example. We are given a system of equations to be solved:
Solving Equations
1. First, we will build a schedule given equation: x2+y2=9.
But it should be noted that this graph of equations will be a circle centered at the origin, and its radius will be equal to three.
2. Our next step will be to plot an equation such as: y = x - 3.
In this case, we must build a line and find the points (0;−3) and (3;0).
3. Let's see what we got. We see that the line intersects the circle at two of its points A and B.
Now we are looking for the coordinates of these points. We see that the coordinates (3;0) correspond to point A, and the coordinates (0;−3) correspond to point B.
And what do we get as a result?
The numbers (3;0) and (0;−3) obtained at the intersection of a straight line with a circle are precisely the solutions of both equations of the system. And from this it follows that these numbers are also solutions of this system of equations.
That is, the answer of this solution is the numbers: (3;0) and (0;−3).
Systems of equations received wide application in the economic sector mathematical modeling various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) 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 right part which 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 common analytical method solutions of similar 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 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 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 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 Y value. The last step is to check 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. ultimate goal 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:
- Multiply both sides of the equation by some number. As a result arithmetic operation one of the coefficients of the variable must 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.
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.
The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to the standard square trinomial. You can solve a polynomial by finding the discriminant.
It is necessary to find the value of the discriminant by well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the multipliers of the polynomial. AT 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 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 is to build on coordinate axis graphs of each equation included in the system. The coordinates of the points of intersection of the curves and will be common solution systems.
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.
The following example needs to find graphic solution systems 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 table is called a matrix. special kind 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. Matrix with units along one of the diagonals and others zero elements called singular.
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 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 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 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| - 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 notations when solving systems with large quantity 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
AT 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 solution method. 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. way algebraic transformations and substitutions is the value of one variable 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.
AT 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, says 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 program in-depth study 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 actions 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 the enumeration of 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.
