System linear equations with two unknowns - these are two or more linear equations for which it is necessary to find all of them general solutions. We will consider systems of two linear equations in two unknowns. The general view of 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. There are several ways to solve a system of linear equations. Let's consider one of the ways to solve a system of linear equations, namely the addition method.
Algorithm for solving by addition method
An algorithm for solving a system of linear equations with two unknowns using the addition method.
1. If required, by equivalent transformations equalize the coefficients of one of the unknown variables in both equations.
2. By adding or subtracting the resulting equations, obtain a linear equation with one unknown
3. Solve the resulting equation with one unknown and find one of the variables.
4. Substitute the resulting expression into any of the two equations of the system and solve this equation, thus obtaining the second variable.
5. Check the solution.
An example of a solution using the addition method
For greater clarity, let us solve the following system of linear equations with two unknowns using the addition method:
(3*x + 2*y = 10;
(5*x + 3*y = 12;
Since none of the variables have identical coefficients, we equalize the coefficients of the variable y. To do this, multiply the first equation by three, and the second equation by two.
(3*x+2*y=10 |*3
(5*x + 3*y = 12 |*2
We get the following system of equations:
(9*x+6*y = 30;
(10*x+6*y=24;
Now we subtract the first from the second equation. We present similar terms and solve the resulting linear equation.
10*x+6*y - (9*x+6*y) = 24-30; x=-6;
We substitute the resulting value into the first equation from our original system and solve the resulting equation.
(3*(-6) + 2*y =10;
(2*y=28; y =14;
The result is a pair of numbers x=6 and y=14. We are checking. Let's make a substitution.
(3*x + 2*y = 10;
(5*x + 3*y = 12;
{3*(-6) + 2*(14) = 10;
{5*(-6) + 3*(14) = 12;
{10 = 10;
{12=12;
As you can see, we got two correct equalities, therefore, we found the correct solution.
Instructions
Addition method.
You need to write two strictly below each other:
549+45y+4y=-7, 45y+4y=549-7, 49y=542, y=542:49, y≈11.
In an arbitrarily chosen (from the system) equation, insert the number 11 instead of the already found “game” and calculate the second unknown:
X=61+5*11, x=61+55, x=116.
The answer to this system of equations is x=116, y=11.
Graphic method.
It consists of practically finding the coordinates of the point at which the lines are mathematically written in a system of equations. The graphs of both lines should be drawn separately in the same coordinate system. General view: – y=khx+b. To construct a straight line, it is enough to find the coordinates of two points, and x is chosen arbitrarily.
Let the system be given: 2x – y=4
Y=-3x+1.
A straight line is constructed using the first one, for convenience it should be written down: y=2x-4. Come up with (easier) values for x, substituting it into the equation, solving it, and finding y. We get two points along which a straight line is constructed. (see picture)
x 0 1
y -4 -2
A straight line is constructed using the second equation: y=-3x+1.
Also construct a straight line. (see picture)
y 1 -5
Find the coordinates of the intersection point of two constructed lines on the graph (if the lines do not intersect, then the system of equations does not have - so).
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If the same system of equations is solved by three different ways, the answer will be the same (if the solution is correct).
Sources:
- 8th grade algebra
- solve an equation with two unknowns online
- Examples of solving systems of linear equations with two
System equations is a collection of mathematical records, each of which contains a number of variables. There are several ways to solve them.
You will need
- -Ruler and pencil;
- -calculator.
Instructions
Let's consider the sequence of solving the system, which consists of linear equations having the form: a1x + b1y = c1 and a2x + b2y = c2. Where x and y are unknown variables, and b,c are free terms. When applying this method, each system represents the coordinates of points corresponding to each equation. To begin, in each case, express one variable in terms of another. Then set the variable x to any number of values. Two is enough. Substitute into the equation and find y. Construct a coordinate system, mark the resulting points on it and draw a line through them. Similar calculations must be carried out for other parts of the system.
The system has only decision, if the constructed lines intersect and one common point. It is incompatible if parallel to each other. And it has infinitely many solutions when the lines merge with each other.
This method considered very visual. 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 resulting values instead of the variables. You can also find its solution using several methods. If the solution of the system is correct, then everyone should turn out the same.
Often there are equations in which one of the terms is unknown. To solve the equation, you need to remember and do it with the given numbers specific set actions.
You will need
- - paper;
- - pen or pencil.
Instructions
Imagine that there are 8 rabbits in front of you, and you only have 5 carrots. Think about it, you still need to buy more carrots so that each rabbit gets one.
Let's present this problem in the form of an equation: 5 + x = 8. Let's substitute the number 3 in place of x. Indeed, 5 + 3 = 8.
When you substituted a number for x, you did the same thing as when you subtracted 5 from 8. So, to find unknown term, subtract the known term from the sum.
Let's say you have 20 rabbits and only 5 carrots. Let's make it up. An equation is an equality that holds only for certain values of the letters included in it. The letters whose meanings need to be found are called . Write an equation with one unknown, call it x. When solving our rabbit problem, we get the following equation: 5 + x = 20.
Let's find the difference between 20 and 5. When subtracting, the number from which it is subtracted is the one being reduced. The number that is being subtracted is called , and final result called difference. So, x = 20 – 5; x = 15. You need to buy 15 carrots for the rabbits.
Check: 5 + 15 = 20. The equation is solved correctly. Of course, when we're talking about about such simple ones, it is not necessary to perform a check. However, when you have equations with three-digit, four-digit, etc. numbers, you definitely need to check to be absolutely sure of the result of your work.
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Helpful advice
To find the unknown minuend, you need to add the subtrahend to the difference.
To find unknown subtrahend, you need to subtract the difference from the minuend.
Tip 4: How to solve a system of three equations with three unknowns
A system of three equations with three unknowns may not have solutions, despite a sufficient number of equations. You can try to solve it using the substitution method or using Cramer's method. Cramer's method, in addition to solving the system, allows you to evaluate whether the system is solvable before finding the values of the unknowns.
Instructions
The substitution method consists of sequentially sequentially one unknown through two others and substituting the resulting result into the equations of the system. Let a system of three equations be given in general view:
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Express x from the first equation: x = (d1 - b1y - c1z)/a1 - and substitute into the second and third equations, then express y from the second equation and substitute into the third. You'll get linear expression for z through the coefficients of the system equations. Now go “backward”: substitute z into the second equation and find y, and then substitute z and y into the first and solve for x. The process is generally shown in the figure before finding z. Further writing in general form will be too cumbersome; in practice, by substituting , you can quite easily find all three unknowns.
Cramer's method consists of constructing a matrix of the system and calculating the determinant of this matrix, as well as three more auxiliary matrices. The system matrix is composed of coefficients for the unknown terms of the equations. A column containing the numbers on the right-hand sides of equations, a column of right-hand sides. It is not used in the system, but is used when solving the system.
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note
All equations in the system must provide additional information independent of other equations. Otherwise, the system will be underdetermined and it will not be possible to find an unambiguous solution.
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.
Instructions
If two of the three systems have only two of the three unknowns, try to express some variables in terms of the others and substitute them into the equation with three unknown. Your goal in this case is to turn it into normal the equation with an unknown person. If this is , the further solution is quite simple - substitute the found value into other equations and find all the other unknowns.
Some systems of equations can be subtracted from one equation by another. See if it is possible to multiply one of or a variable so that two unknowns are canceled at once. If there is such an opportunity, take advantage of it; most likely, the subsequent solution will not be difficult. Don't forget that when multiplying by a number you must multiply as left side, and the right one. Likewise, when subtracting equations, you must remember that the right-hand side must also be subtracted.
If the previous methods did not help, use in a general way solutions to any equations with three unknown. To do this, rewrite the equations in the form a11x1+a12x2+a13x3=b1, a21x1+a22x2+a23x3=b2, a31x1+a32x2+a33x3=b3. Now create a matrix of coefficients for x (A), a matrix of unknowns (X) and a matrix of free variables (B). Please note that by multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix of free terms, that is, A*X=B.
Find matrix A to the power (-1) by first finding , note that it should not be equal to zero. After this, multiply the resulting matrix by matrix B, as a result you will receive the desired matrix X, indicating all the values.
You can also find a solution to a system of three equations using Cramer's method. To do this, find the third-order determinant ∆ corresponding to the system matrix. Then successively find three more determinants ∆1, ∆2 and ∆3, substituting the values of the free terms instead of the values of the corresponding columns. Now find x: x1=∆1/∆, x2=∆2/∆, x3=∆3/∆.
Sources:
- solutions to equations with three unknowns
When starting to solve a system of equations, figure out what kind of equations they are. Methods for solving linear equations have been studied quite well. Nonlinear equations are most often not solved. There are only one special cases, each of which is practically individual. Therefore, the study of solution techniques should begin with linear equations. Such equations can even be solved purely algorithmically.
the denominators of the unknowns found are exactly the same. Yes, and the numerators show some patterns in their construction. If the dimension of the system of equations were greater than two, then the elimination method would lead to very cumbersome calculations. To avoid them, they are designed purely algorithmic methods solutions. The simplest of them is Cramer's algorithm (Cramer's formulas). For you should find out general system equations from n equations.
System n linear algebraic equations with n unknowns has the form (see Fig. 1a). In it, аij are the coefficients of the system,
xj – unknowns, bi – free terms (i=1, 2, ... , n; j=1, 2, ... , n). Such a system can be written compactly in matrix form AX=B. Here A is the matrix of system coefficients, X is the column matrix of unknowns, B is the column matrix of free terms (see Figure 1b). According to Cramer's method, each unknown xi =∆i/∆ (i=1,2…,n). The determinant ∆ of the coefficient matrix is called the main one, and ∆i the auxiliary one. For each unknown, the auxiliary determinant is found by replacing the i-th column of the main determinant with a column of free terms. The Cramer method for the case of second- and third-order systems is presented in detail in Fig. 2.
The system is a combination of two or more equalities, each of which contains two or more unknowns. There are two main ways to solve systems of linear equations that are used within school curriculum. One of them is called the method, the other - the addition method.
Standard form of a system of two equations
At standard form the first equation has the form a1*x+b1*y=c1, the second equation has the form a2*x+b2*y=c2 and so on. For example, in the case of two parts of the system, both given a1, a2, b1, b2, c1, c2 are some numerical coefficients represented in specific equations. In turn, x and y represent unknowns whose values need to be determined. The required values turn both equations simultaneously into true equalities.Solving the system using the addition method
In order to solve the system, that is, to find those values of x and y that will turn them into true equalities, you need to take several simple steps. The first of them is to transform either equation so that the numerical coefficients for the variable x or y in both equations are the same in magnitude, but different in sign.For example, suppose a system consisting of two equations is given. The first of them has the form 2x+4y=8, the second has the form 6x+2y=6. One of the options for completing the task is to multiply the second equation by a coefficient of -2, which will lead it to the form -12x-4y=-12. The correct choice of coefficient is one of key tasks in the process of solving a system by addition, since it determines the entire further move procedures for finding unknowns.
Now it is necessary to add the two equations of the system. Obviously, the mutual destruction of variables with coefficients equal in value but opposite in sign will lead to the form -10x=-4. After this, it is necessary to solve this simple equation, from which it clearly follows that x = 0.4.
The last step in the solution process is to substitute the found value of one of the variables into any of the original equalities available in the system. For example, substituting x=0.4 into the first equation, you can get the expression 2*0.4+4y=8, from which y=1.8. Thus, x=0.4 and y=1.8 are the roots of the example system.
In order to make sure that the roots were found correctly, it is useful to check by substituting the found values into the second equation of the system. For example, in in this case we get an equality of the form 0.4*6+1.8*2=6, which is true.
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The material in this article is intended for a first acquaintance with systems of equations. Here we will introduce the definition of a system of equations and its solutions, and also consider the most common types of systems of equations. As usual, we will give explanatory examples.
Page navigation.
What is a system of equations?
We will approach the definition of the system of equations gradually. First, let’s just say that it is convenient to give it, indicating two points: firstly, the type of recording, and, secondly, the meaning embedded in this recording. Let's look at them in turn, and then generalize the reasoning into the definition of systems of equations.
Let there be several of them in front of us. For example, let's take two equations 2 x+y=−3 and x=5. Let's write them one below the other and combine them on the left with a curly brace:
Records of this type, which are several equations arranged in a column and united on the left by a curly brace, are records of systems of equations.
What do such entries mean? They define the set of all such solutions to the equations of the system that are a solution to each equation.
It wouldn't hurt to describe it in other words. Let's say that some solutions to the first equation are solutions to all other equations of the system. So the system record just means them.
Now we are ready to adequately accept the definition of a system of equations.
Definition.
Systems of equations call records that are equations located one below the other, united on the left by a curly brace, which denote the set of all solutions to equations that are also solutions to each equation of the system.
A similar definition is given in the textbook, but it is not given there for general case, and for two rational equations with two variables.
Main types
It is clear that there are an infinite number of different equations. Naturally, there are also an infinite number of systems of equations compiled using them. Therefore, for the convenience of studying and working with systems of equations, it makes sense to divide them into groups according to similar characteristics, and then move on to considering systems of equations of individual types.
The first division suggests itself by the number of equations included in the system. If there are two equations, then we can say that we have a system of two equations, if there are three, then a system of three equations, etc. It is clear that it makes no sense to talk about a system of one equation, since in this case, in essence, we are dealing with the equation itself, and not with the system.
The next division is based on the number of variables involved in writing the equations of the system. If there is one variable, then we are dealing with a system of equations with one variable (they also say with one unknown), if there are two, then with a system of equations with two variables (with two unknowns), etc. For example, 
is a system of equations with two variables x and y.
This refers to the number of all different variables involved in the recording. They do not have to all be included in the record of each equation at once; their presence in at least one equation is sufficient. Eg, 
is a system of equations with three variables x, y and z. In the first equation, the variable x is present explicitly, and y and z are implicit (we can assume that these variables have zero), and in the second equation there are x and z, but the variable y is not explicitly presented. In other words, the first equation can be viewed as 
, and the second – as x+0·y−3·z=0.
The third point in which systems of equations differ is the type of equations themselves.
At school, the study of systems of equations begins with systems of two linear equations in two variables. That is, such systems constitute two linear equations. Here are a couple of examples: 
And 
. They learn the basics of working with systems of equations.
When deciding more complex tasks You can also encounter systems of three linear equations with three unknowns.
Further in the 9th grade, nonlinear equations are added to systems of two equations with two variables, mostly entire equations of the second degree, less often - more high degrees. These systems are called systems nonlinear equations, if necessary, clarify the number of equations and unknowns. Let us show examples of such systems of nonlinear equations: 
And .
And then in systems there are also, for example, . They are usually called simply systems of equations, without specifying which equations. It is worth noting here that most often a system of equations is simply referred to as a “system of equations,” and clarifications are added only if necessary.
In high school, as the material is studied, irrational, trigonometric, logarithmic and exponential equations
: 
, 
, 
.
If we look even further into the first-year university curriculum, the main emphasis is on the study and solution of systems of linear algebraic equations (SLAEs), that is, equations in which the left-hand sides contain polynomials of the first degree, and the right-hand sides contain certain numbers. But there, unlike school, they no longer take two linear equations with two variables, but an arbitrary number of equations with any number variables, often not matching the number of equations.
What is the solution to a system of equations?
The term “solution of a system of equations” directly refers to systems of equations. At school, the definition of solving a system of equations with two variables is given :
Definition.
Solving a system of equations with two variables is called a pair of values of these variables that turns each equation of the system into the correct one, in other words, is a solution to each equation of the system.
For example, a pair of variable values x=5, y=2 (it can be written as (5, 2)) is a solution to a system of equations by definition, since the equations of the system, when x=5, y=2 are substituted into them, become correct numerical equalities 5+2=7 and 5−2=3 respectively. But the pair of values x=3, y=0 is not a solution to this system, since when substituting these values into the equations, the first of them will turn into the incorrect equality 3+0=7.
Similar definitions can be formulated for systems with one variable, as well as for systems with three, four, etc. variables.
Definition.
Solving a system of equations with one variable there will be a value of the variable that is the root of all equations of the system, that is, turning all equations into correct numerical equalities.
Let's give an example. Consider a system of equations with one variable t of the form 
. The number −2 is its solution, since both (−2) 2 =4 and 5·(−2+2)=0 are true numerical equalities. And t=1 is not a solution to the system, since substituting this value will give two incorrect equalities 1 2 =4 and 5·(1+2)=0.
Definition.
Solving a system with three, four, etc. variables called three, four, etc. values of the variables, respectively, turning all equations of the system into true equalities.
So, by definition, a triple of values of the variables x=1, y=2, z=0 is a solution to the system 
, since 2·1=2, 5·2=10 and 1+2+0=3 are true numerical equalities. And (1, 0, 5) is not a solution to this system, since when substituting these values of variables into the equations of the system, the second of them turns into the incorrect equality 5·0=10, and the third too 1+0+5=3.
Note that systems of equations may not have solutions, they may have final number solutions, for example, one, two, ..., but can have infinitely many solutions. You will see this as you delve deeper into the topic.
Taking into account the definitions of a system of equations and their solutions, we can conclude that the solution to a system of equations is the intersection of the sets of solutions of all its equations.
To conclude, here are a few related definitions:
Definition.
non-joint, if it has no solutions, in otherwise the system is called joint.
Definition.
The system of equations is called uncertain, if it has infinitely many solutions, and certain, if it has a finite number of solutions or does not have them at all.
These terms are introduced, for example, in a textbook, but they are used quite rarely at school; they are more often heard in higher educational institutions.
Bibliography.
- Algebra: textbook for 7th grade general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 17th ed. - M.: Education, 2008. - 240 p. : ill. - ISBN 978-5-09-019315-3.
- Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
- Mordkovich A. G. Algebra. 7th grade. At 2 p.m. Part 1. Textbook for students educational institutions/ A. G. Mordkovich. - 17th ed., add. - M.: Mnemosyne, 2013. - 175 p.: ill. ISBN 978-5-346-02432-3.
- Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., erased. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
- Mordkovich A. G. Algebra and beginnings mathematical analysis. Grade 11. At 2 p.m. Part 1. Textbook for students of general education institutions ( profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.
- Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
- A. G. Kurosh. Higher algebra course.
- Ilyin V. A., Poznyak E. G. Analytic geometry: Textbook: For universities. – 5th ed. – M.: Science. Fizmatlit, 1999. – 224 p. - (Well higher mathematics and mat. physics). – ISBN 5-02-015234 – X (Issue 3)
Let us first consider the case when the number of equations is equal to the number of variables, i.e. m = n. Then the matrix of the system is square, and its determinant is called the determinant of the system.
Inverse matrix method
Let us consider in general form the system of equations AX = B with non-degenerate square matrix A. In this case there is inverse matrix A -1. Let's multiply both sides by A -1 on the left. We get A -1 AX = A -1 B. Hence EX = A -1 B and
The last equality is a matrix formula for finding solutions to such systems of equations. The use of this formula is called the inverse matrix method
For example, let’s use this method to solve the following system:
;
At the end of solving the system, you can check by substituting the found values into the system equations. In doing so, they must turn into true equalities.
For the example considered, let's check:
Method for solving systems of linear equations with a square matrix using Cramer's formulas
Let n= 2:
If we multiply both sides of the first equation by a 22, and both sides of the second by (-a 12), and then add the resulting equations, then we eliminate the variable x 2 from the system. Similarly, you can eliminate the variable x 1 (by multiplying both sides of the first equation by (-a 21), and both sides of the second by a 11). As a result, we get the system:
The expression in brackets is the determinant of the system
Let's denote
Then the system will take the form:
From the resulting system it follows that if the determinant of the system is 0, then the system will be consistent and definite. Its only solution can be calculated using the formulas:
If = 0, a 1 0 and/or 2 0, then the system equations will take the form 0*x 1 = 2 and/or 0*x 1 = 2. In this case, the system will be inconsistent.
In the case when = 1 = 2 = 0, the system will be consistent and indefinite (will have an infinite number of solutions), since it will take the form:
Cramer's theorem(we will omit the proof). If the determinant of the matrix of a system of equations is not equal to zero, then the system has a unique solution, determined by the formulas:
,
where j is the determinant of the matrix obtained from matrix A by replacing the j-th column with a column of free terms.
The above formulas are called Cramer formulas.
As an example, let’s use this method to solve a system that was previously solved using the inverse matrix method:
Disadvantages of the considered methods:
1) significant labor intensity (calculating determinants and finding the inverse matrix);
2) limited scope (for systems with a square matrix).
Real economic situations are often modeled by systems in which the number of equations and variables is quite significant, and there are more equations than variables. Therefore, in practice, the following method is more common.
Gaussian method (method of sequential elimination of variables)
This method is used to solve a system of m linear equations with n variables in general form. Its essence lies in applying a system of equivalent transformations to the extended matrix, with the help of which the system of equations is transformed to a form where its solutions become easy to find (if any).
This is the kind in which the left top part The matrix of the system will be a stepped matrix. This is achieved using the same techniques that were used to obtain a step matrix to determine the rank. In this case, elementary transformations are applied to the extended matrix, which will allow one to obtain an equivalent system of equations. After this, the expanded matrix will take the form:
Obtaining such a matrix is called straight ahead Gauss method.
Finding the values of variables from the corresponding system of equations is called in reverse Gauss method. Let's consider it.
Note that the last (m – r) equations will take the form:
If at least one of the numbers 
is not equal to zero, then the corresponding equality will be false, and the entire system will be inconsistent.
Therefore, for any joint system 
. In this case, the last (m – r) equations for any values of the variables will be identities 0 = 0, and they can be ignored when solving the system (simply discard the corresponding rows).
After this, the system will look like:
Let us first consider the case when r=n. Then the system will take the form:
From the last equation of the system, x r can be uniquely found.
Knowing x r, we can unambiguously express x r -1 from it. Then from the previous equation, knowing x r and x r -1, we can express x r -2, etc. up to x 1 .
So, in this case the system will be joint and determined.
Now consider the case when r
From this equation we can express the basic variable x r in terms of non-basic ones:
The penultimate equation will look like:
By substituting the resulting expression into it instead of x r, it will be possible to express the basic variable x r -1 in terms of non-basic ones. Etc. to variablex 1 . To obtain a solution to the system, you can equate non-basic variables to arbitrary values and then calculate the basic variables using the resulting formulas. Thus, in this case the system will be consistent and indefinite (have an infinite number of solutions).
For example, let's solve the system of equations:
We will call the set of basic variables basis systems. We will also call the set of columns of coefficients for them basis(base columns), or basic minor system matrices. The solution of the system in which all non-basic variables are equal to zero will be called basic solution.
In the previous example, the basic solution will be (4/5; -17/5; 0; 0) (the variables x 3 and x 4 (c 1 and c 2) are set to zero, and the basic variables x 1 and x 2 are calculated through them) . To give an example of a non-basic solution, we need to equate x 3 and x 4 (c 1 and c 2) to arbitrary numbers that are not simultaneously zero, and calculate the remaining variables through them. For example, with 1 = 1 and 2 = 0, we obtain a non-basic solution - (4/5; -12/5; 1; 0). By substitution it is easy to verify that both solutions are correct.
It is obvious that in an indefinite system there can be an infinite number of non-basic solutions. How many basic solutions can there be? Each row of the transformed matrix must correspond to one basis variable. There are n variables in the problem, and r base lines. Therefore, the number of all possible sets of basic variables cannot exceed the number of combinations of n by 2. It may be less than 
, because it is not always possible to transform the system to such a form that this particular set of variables is the basis.
What kind is this? This is the type when the matrix formed from columns of coefficients for these variables will be stepped, and at the same time will consist of r rows. Those. the rank of the coefficient matrix for these variables must be equal to r. It cannot be greater, since the number of columns is equal. If it turns out to be less than r, then this indicates a linear dependence of the columns on the variables. Such columns cannot form a basis.
Let's consider what other basic solutions can be found in the example discussed above. To do this, consider all possible combinations of four variables, two basic ones each. There will be such combinations 
, and one of them (x 1 and x 2) has already been considered.
Let's take the variables x 1 and x 3. Let us find the rank of the matrix of coefficients for them:
Since it is equal to two, they can be basic. Let us equate the non-basic variables x 2 and x 4 to zero: x 2 = x 4 = 0. Then from the formula x 1 = 4/5 – (1/5)*x 4 it follows that x 1 = 4/5, and from the formula x 2 = -17/5 + x 3 - - (7/5)*x 4 = -17/5 + x 3 it follows that x 3 = x 2 +17/5 = 17/5. Thus, we get the basic solution (4/5; 0; 17/5; 0).
Similarly, you can obtain basic solutions for the basic variables x 1 and x 4 – (9/7; 0; 0; -17/7); x 2 and x 4 – (0; -9; 0; 4); x 3 and x 4 – (0; 0; 9; 4).
The variables x 2 and x 3 in this example cannot be taken as basic ones, since the rank of the corresponding matrix is equal to one, i.e. less than two:
.
Another approach to determining whether or not it is possible to construct a basis from certain variables is also possible. When solving the example, as a result of converting the system matrix to a stepwise form, it took the form:
By selecting pairs of variables, it was possible to calculate the corresponding minors of this matrix. It is easy to verify that for all pairs except x 2 and x 3 they are not equal to zero, i.e. the columns are linearly independent. And only for columns with variables x 2 and x 3 
, which indicates their linear dependence.
Let's look at another example. Let's solve the system of equations
So, the equation corresponding to the third row of the last matrix is contradictory - it resulted in the incorrect equality 0 = -1, therefore, this system is inconsistent.
Jordan-Gauss method 3 is a development of the Gaussian method. Its essence is that the extended matrix of the system is transformed to a form where the coefficients of the variables form an identity matrix up to permutation of rows or columns 4 (where r is the rank of the system matrix).
Let's solve the system using this method:
Let's consider the extended matrix of the system:
In this matrix we select a unit element. For example, the coefficient for x 2 in the third constraint is 5. Let's ensure that the remaining rows in this column contain zeros, i.e. Let's make the column single. During the transformation process we will call this columnpermissive(leading, key). The third limitation (third line) we will also call permissive. Myself element, which stands at the intersection of the resolving row and column (here it is one), is also called permissive.
The first line now contains the coefficient (-1). To get a zero in its place, multiply the third line by (-1) and subtract the result from the first line (i.e. simply add the first line to the third).
The second line contains the coefficient 2. To get zero in its place, multiply the third line by 2 and subtract the result from the first line.
The result of the transformation will look like:
From this matrix it is clearly visible that one of the first two restrictions can be crossed out (the corresponding rows are proportional, i.e. these equations follow from each other). Let's cross out, for example, the second:
So, the new system has two equations. A single column (second) is obtained, and the unit here appears in the second row. Let us remember that the second equation of the new system will correspond to the basic variable x 2.
Let's choose a base variable for the first row. This can be any variable except x 3 (because for x 3 the first constraint has a zero coefficient, i.e. the set of variables x 2 and x 3 cannot be basic here). You can take the first or fourth variable.
Let's choose x 1. Then the resolving element will be 5, and both sides of the resolving equation will have to be divided by five to get one in the first column of the first row.
Let's ensure that the remaining rows (i.e., the second row) have zeros in the first column. Since now the second line contains not zero, but 3, we need to subtract from the second line the elements of the transformed first line, multiplied by 3:
From the resulting matrix, one can directly extract one basic solution by equating non-basic variables to zero, and basic ones to the free terms in the corresponding equations: (0.8; -3.4; 0; 0). You can also derive general formulas expressing basic variables through non-basic ones: x 1 = 0.8 – 1.2 x 4; x 2 = -3.4 + x 3 + 1.6x 4. These formulas describe the entire infinite set of solutions to the system (equating x 3 and x 4 to arbitrary numbers, you can calculate x 1 and x 2).
Note that the essence of the transformations at each stage of the Jordan-Gauss method was as follows:
1) the resolution line was divided by the resolution element to obtain a unit in its place,
2) from all other rows, the transformed resolving element was subtracted, multiplied by the element that was in the given line in the resolving column, to obtain a zero in place of this element.
Let us consider again the transformed extended matrix of the system:
From this record it is clear that the rank of the matrix of system A is equal to r.
In the course of our reasoning, we established that the system will be cooperative if and only if 
. This means that the extended matrix of the system will look like:
By discarding zero rows, we obtain that the rank of the extended matrix of the system is also equal to r.
Kronecker-Capelli theorem. A system of linear equations is consistent if and only if the rank of the system's matrix is equal to the rank of the extended matrix of this system.
Recall that the rank of a matrix is equal to the maximum number of its linearly independent rows. It follows from this that if the rank of the extended matrix is less than the number of equations, then the equations of the system are linearly dependent, and one or more of them can be excluded from the system (since they are a linear combination of the others). A system of equations will be linearly independent only if the rank of the extended matrix is equal to the number of equations.
Moreover, for simultaneous systems of linear equations, it can be argued that if the rank of the matrix is equal to the number of variables, then the system has a unique solution, and if it is less than the number of variables, then the system is indefinite and has infinitely many solutions.
1For example, let there be five rows in the matrix (the original row order is 12345). We need to change the second line and the fifth. In order for the second line to take the place of the fifth and “move” down, we successively change the adjacent lines three times: the second and third (13245), the second and fourth (13425) and the second and fifth (13452). Then, in order for the fifth row to take the place of the second in the original matrix, it is necessary to “shift” the fifth row upward by only two consecutive changes: the fifth and fourth rows (13542) and the fifth and third (15342).
2Number of combinations from n to r 
they call the number of all different r-element subsets of an n-element set (those that have different compositions of elements are considered different sets; the order of selection is not important). It is calculated using the formula: 
. Let us recall the meaning of the sign “!” (factorial): 
0!=1.)
3 Since this method is more common than the previously discussed Gaussian method, and is essentially a combination of the forward and backward steps of the Gaussian method, it is also sometimes called the Gaussian method, omitting the first part of the name.
4For example, 
.
5If there were no units in the system matrix, then it would be possible, for example, to divide both sides of the first equation by two, and then the first coefficient would become unity; or the like
More reliable than the graphical method discussed in the previous paragraph.
Substitution method
We used this method in 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 designated by other letters, which does not matter). In fact, we used this algorithm in the previous paragraph, when the problem of a two-digit number led to a mathematical model, which is a system of equations. We solved this system of equations above using the substitution method (see example 1 from § 4).
An 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 in the third step instead of x into the expression y through x obtained in the first step.
5. Write the answer in the form of pairs of values (x; y), which were found in the third and fourth steps, respectively.
4) Substitute one by one each of the found values of y into the formula x = 5 - 3. If then 
5) Pairs (2; 1) and solutions to 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 using the following example.
Example 2. Solve system of equations
Let's multiply all 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 the algebraic addition of two equations of the original system, an equation was obtained that was 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 using the substitution method. From the second equation we find. Substituting this expression instead of y into the first equation of the system, we get
It remains to substitute the found values of x into the formula
If x = 2 then
Thus, we found two solutions to the system: 
Method for introducing new variables
You were introduced to 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 for solving systems of equations is the same, but from a technical point of view there are some features that we will discuss in the following examples.
Example 3. Solve system of equations
Let's introduce a new variable. Then the first equation of the system can be rewritten in a simpler form: Let's solve this equation with respect to the variable t:
Both of these values satisfy the condition and therefore are the roots of a rational equation with variable t. But that means either where we find that x = 2y, or
Thus, using the method of introducing a new variable, we managed to “stratify” the first equation of the system, which was quite complex in appearance, into two simpler equations:
x = 2 y; y - 2x.
What's next? And then each of the two simple equations obtained must be considered in turn in a system with the equation x 2 - y 2 = 3, which we have not yet remembered. In other words, the problem comes down to solving two systems of equations:
We need to find solutions to 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: let's substitute the expression 2y instead of x into the second equation of the system. We get
Since x = 2y, we find, respectively, x 1 = 2, x 2 = 2. Thus, two solutions of 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: substitute the expression 2x instead of y into the second equation of the system. We get
This equation has no roots, which means the system of equations has no solutions. Thus, only the solutions of the first system need to 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. 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 system of equations
Let's introduce two new variables:
Let's take into account that then
This will allow you to rewrite the given system in a much simpler form, but with respect to the new variables a and b:
Since a = 1, then from the equation a + 6 = 2 we find: 1 + 6 = 2; 6=1. Thus, regarding the variables a and b, we got one solution:
Returning to the variables x and y, we obtain a system of equations
Let us apply the method of algebraic addition to solve this system:
Since then from the equation 2x + y = 3 we find:
Thus, regarding the variables x and y, we got one solution:
Let us conclude this paragraph with a brief but rather serious theoretical conversation. You have already gained some experience in solving various equations: linear, quadratic, 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 paragraph we introduced the concept 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 called equivalent if they have the same solutions or if both systems have no solutions.
All three methods (substitution, algebraic addition and introducing new variables) that we 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. Now let's remember the method that you already studied in the previous lesson. That is, let's repeat what you know about the graphical solution method.
The method of solving systems of equations graphically involves constructing a graph for each of the specific equations that are included in a given system and are located in the same coordinate plane, as well as where it is necessary 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 it is common for a graphical system of equations to have either one single correct solution, or an infinite number of solutions, or to have no solutions at all.
Now let’s look at each of these solutions in more detail. And so, a system of equations can have a unique solution if the lines that are the graphs of the system’s equations intersect. If these lines are parallel, then such a system of equations has absolutely no solutions. If the direct graphs of the equations of the system coincide, then such a system allows one to find many solutions.
Well, now let’s look at the algorithm for solving a system of two equations with 2 unknowns using a graphical method:
Firstly, first we build a graph of the 1st equation;
The second step will be to construct 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 using an example. We are given a system of equations that needs to be solved:
Solving equations
1. First, we will build a graph of this equation: x2+y2=9.
But it should be noted that this graph of the equations will be a circle with a center at the origin, and its radius will be equal to three.
2. Our next step will be to graph an equation such as: y = x – 3.
In this case, we must construct a straight line and find the points (0;−3) and (3;0).
3. Let's see what we got. We see that the straight 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 when the line intersects the circle are precisely the solutions to both equations of the system. And from this it follows that these numbers are also solutions to this system of equations.
That is, the answer to this solution is the numbers: (3;0) and (0;−3).