Different systems of equations. Systems of linear equations

More reliable than the graphical method discussed in the previous paragraph.

Substitution method

We used this method in 7th grade to solve systems 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 double digit number led to 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 into a more in simple form: Let's solve this equation for 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 received simple equations need to be considered one by one 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 this system in a much simpler form, but relatively 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 to solve this system algebraic addition:

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 different 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 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 graphical method solutions.

The method of solving systems of equations graphically is the construction of a graph for each of the specific equations that are included in a given system and are in one coordinate plane, and also 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 for graphics system equations tend to have either one single the right decision, or infinite set solutions, or have no solutions at all.

Now let’s look at each of these solutions in more detail. And so, the system of equations can have only decision in case the straight lines, which 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 schedule given 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).

A system of linear equations with two unknowns is 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.

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 of nonlinear equations; if necessary, the number of equations and unknowns is specified. 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 into true equalities all equations of the system.

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)

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).

Video on the topic

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:

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 a unique solution 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. More accurate results are provided 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.

Video on the topic

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 system matrix 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.

Video on the topic

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 ones (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 most often they do not dare. 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 determinant, and ∆i the auxiliary one. For every unknown auxiliary qualifier found by replacing the i-th column of the main determinant with the 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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