Methods for solving systems of equations with two variables. Systems of equations - basic information

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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. A general view of a system of two linear equations with two unknowns is shown 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, use equivalent transformations to 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.

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 a non-degenerate square matrix A. In this case, there exists 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 view. 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 when its solutions become easy to find (if any).

This is a view in which the upper left part of the system matrix 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 defined.

Now consider the case when r basic(main), and all the rest - non-basic(non-core, free). The last equation of the system will be:

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 deleted (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, equating non-basic variables to zero, and basic ones to 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 resolution was subtracted, multiplied by the element that was in the given row in the resolution column, to get 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

With this video I begin a series of lessons dedicated to systems of equations. Today we will talk about solving systems of linear equations addition method- This is one of the simplest methods, but at the same time one of the most effective.

The addition method consists of three simple steps:

Look at the system and choose a variable that has identical (or opposite) coefficients in each equation;
Perform algebraic subtraction (for opposite numbers - addition) of equations from each other, and then bring similar terms;
Solve the new equation obtained after the second step.

If everything is done correctly, then at the output we will get a single equation with one variable— it won’t be difficult to solve it. Then all that remains is to substitute the found root into the original system and get the final answer.

However, in practice everything is not so simple. There are several reasons for this:

Solving equations using the addition method implies that all lines must contain variables with equal/opposite coefficients. What to do if this requirement is not met?
Not always, after adding/subtracting equations in the indicated way, we get a beautiful construction that can be easily solved. Is it possible to somehow simplify the calculations and speed up the calculations?

To get the answer to these questions, and at the same time understand a few additional subtleties that many students fail at, watch my video lesson:

With this lesson we begin a series of lectures devoted to systems of equations. And we will start from the simplest of them, namely those that contain two equations and two variables. Each of them will be linear.

Systems is 7th grade material, but this lesson will also be useful for high school students who want to brush up on their knowledge of this topic.

In general, there are two methods for solving such systems:

Addition method;
A method of expressing one variable in terms of another.

Today we will deal with the first method - we will use the method of subtraction and addition. But to do this, you need to understand the following fact: once you have two or more equations, you can take any two of them and add them to each other. They are added member by member, i.e. “X’s” are added to “X’s” and similar ones are given, “Y’s” with “Y’s” are similar again, and what is to the right of the equal sign is also added to each other, and similar ones are also given there.

The results of such machinations will be a new equation, which, if it has roots, they will certainly be among the roots of the original equation. Therefore, our task is to do the subtraction or addition in such a way that either $x$ or $y$ disappears.

How to achieve this and what tool to use for this - we’ll talk about this now.

Solving easy problems using addition

So, we learn to use the addition method using the example of two simple expressions.

Task No. 1

\[\left\( \begin(align)& 5x-4y=22 \\& 7x+4y=2 \\\end(align) \right.\]

Note that $y$ has a coefficient of $-4$ in the first equation, and $+4$ in the second. They are mutually opposite, so it is logical to assume that if we add them up, then in the resulting sum the “games” will be mutually destroyed. Add it up and get:

Let's solve the simplest construction:

Great, we found the "x". What should we do with it now? We have the right to substitute it into any of the equations. Let's substitute in the first:

\[-4y=12\left| :\left(-4 \right) \right.\]

Answer: $\left(2;-3 \right)$.

Problem No. 2

\[\left\( \begin(align)& -6x+y=21 \\& 6x-11y=-51 \\\end(align) \right.\]

The situation here is completely similar, only with “X’s”. Let's add them up:

We have the simplest linear equation, let's solve it:

Now let's find $x$:

Answer: $\left(-3;3 \right)$.

Important points

So, we have just solved two simple systems of linear equations using the addition method. Key points again:

If there are opposite coefficients for one of the variables, then it is necessary to add all the variables in the equation. In this case, one of them will be destroyed.
We substitute the found variable into any of the system equations to find the second one.
The final response record can be presented in different ways. For example, like this - $x=...,y=...$, or in the form of coordinates of points - $\left(...;... \right)$. The second option is preferable. The main thing to remember is that the first coordinate is $x$, and the second is $y$.
The rule of writing the answer in the form of point coordinates is not always applicable. For example, it cannot be used when the variables are not $x$ and $y$, but, for example, $a$ and $b$.

In the following problems we will consider the technique of subtraction when the coefficients are not opposite.

Solving easy problems using the subtraction method

Task No. 1

\[\left\( \begin(align)& 10x-3y=5 \\& -6x-3y=-27 \\\end(align) \right.\]

Note that there are no opposite coefficients here, but there are identical ones. Therefore, we subtract the second from the first equation:

Now we substitute the value $x$ into any of the system equations. Let's go first:

Answer: $\left(2;5\right)$.

Problem No. 2

\[\left\( \begin(align)& 5x+4y=-22 \\& 5x-2y=-4 \\\end(align) \right.\]

We again see the same coefficient of $5$ for $x$ in the first and second equation. Therefore, it is logical to assume that you need to subtract the second from the first equation:

We have calculated one variable. Now let's find the second one, for example, by substituting the value $y$ into the second construction:

Answer: $\left(-3;-2 \right)$.

Nuances of the solution

So what do we see? Essentially, the scheme is no different from the solution of previous systems. The only difference is that we do not add equations, but subtract them. We are doing algebraic subtraction.

In other words, as soon as you see a system consisting of two equations in two unknowns, the first thing you need to look at is the coefficients. If they are the same anywhere, the equations are subtracted, and if they are opposite, the addition method is used. This is always done so that one of them disappears, and in the final equation, which remains after subtraction, only one variable remains.

Of course, that's not all. Now we will consider systems in which the equations are generally inconsistent. Those. There are no variables in them that are either the same or opposite. In this case, to solve such systems, an additional technique is used, namely, multiplying each of the equations by a special coefficient. How to find it and how to solve such systems in general, we’ll talk about this now.

Solving problems by multiplying by a coefficient

Example #1

\[\left\( \begin(align)& 5x-9y=38 \\& 3x+2y=8 \\\end(align) \right.\]

We see that neither for $x$ nor for $y$ the coefficients are not only mutually opposite, but also in no way correlated with the other equation. These coefficients will not disappear in any way, even if we add or subtract the equations from each other. Therefore, it is necessary to apply multiplication. Let's try to get rid of the $y$ variable. To do this, we multiply the first equation by the coefficient of $y$ from the second equation, and the second equation by the coefficient of $y$ from the first equation, without touching the sign. We multiply and get a new system:

\[\left\( \begin(align)& 10x-18y=76 \\& 27x+18y=72 \\\end(align) \right.\]

Let's look at it: at $y$ the coefficients are opposite. In such a situation, it is necessary to use the addition method. Let's add:

Now we need to find $y$. To do this, substitute $x$ into the first expression:

\[-9y=18\left| :\left(-9 \right) \right.\]

Answer: $\left(4;-2 \right)$.

Example No. 2

\[\left\( \begin(align)& 11x+4y=-18 \\& 13x-6y=-32 \\\end(align) \right.\]

Again, the coefficients for none of the variables are consistent. Let's multiply by the coefficients of $y$:

\[\left\( \begin(align)& 11x+4y=-18\left| 6 \right. \\& 13x-6y=-32\left| 4 \right. \\\end(align) \right .\]

\[\left\( \begin(align)& 66x+24y=-108 \\& 52x-24y=-128 \\\end(align) \right.\]

Our new system is equivalent to the previous one, but the coefficients of $y$ are mutually opposite, and therefore it is easy to apply the addition method here:

Now let's find $y$ by substituting $x$ into the first equation:

Answer: $\left(-2;1 \right)$.

Nuances of the solution

The key rule here is the following: we always multiply only by positive numbers - this will save you from stupid and offensive mistakes associated with changing signs. In general, the solution scheme is quite simple:

We look at the system and analyze each equation.
If we see that neither $y$ nor $x$ the coefficients are consistent, i.e. they are neither equal nor opposite, then we do the following: we select the variable that we need to get rid of, and then we look at the coefficients of these equations. If we multiply the first equation by the coefficient from the second, and the second, correspondingly, multiply by the coefficient from the first, then in the end we will get a system that is completely equivalent to the previous one, and the coefficients of $y$ will be consistent. All our actions or transformations are aimed only at getting one variable in one equation.
We find one variable.
We substitute the found variable into one of the two equations of the system and find the second.
We write the answer in the form of coordinates of points if we have variables $x$ and $y$.

But even such a simple algorithm has its own subtleties, for example, the coefficients of $x$ or $y$ can be fractions and other “ugly” numbers. We will now consider these cases separately, because in them you can act somewhat differently than according to the standard algorithm.

Solving problems with fractions

Example #1

\[\left\( \begin(align)& 4m-3n=32 \\& 0.8m+2.5n=-6 \\\end(align) \right.\]

First, notice that the second equation contains fractions. But note that you can divide $4$ by $0.8$. We will receive $5$. Let's multiply the second equation by $5$:

\[\left\( \begin(align)& 4m-3n=32 \\& 4m+12.5m=-30 \\\end(align) \right.\]

We subtract the equations from each other:

We found $n$, now let's count $m$:

Answer: $n=-4;m=5$

Example No. 2

\[\left\( \begin(align)& 2.5p+1.5k=-13\left| 4 \right. \\& 2p-5k=2\left| 5 \right. \\\end(align )\right.\]

Here, as in the previous system, there are fractional coefficients, but for none of the variables the coefficients do not fit into each other an integer number of times. Therefore, we use the standard algorithm. Get rid of $p$:

\[\left\( \begin(align)& 5p+3k=-26 \\& 5p-12.5k=5 \\\end(align) \right.\]

We use the subtraction method:

Let's find $p$ by substituting $k$ into the second construction:

Answer: $p=-4;k=-2$.

Nuances of the solution

That's all optimization. In the first equation, we did not multiply by anything at all, but multiplied the second equation by $5$. As a result, we received a consistent and even identical equation for the first variable. In the second system, we followed a standard algorithm.

But how do you find the numbers by which to multiply equations? After all, if we multiply by fractions, we get new fractions. Therefore, the fractions must be multiplied by a number that would give a new integer, and after that the variables must be multiplied by coefficients, following the standard algorithm.

In conclusion, I would like to draw your attention to the format for recording the response. As I already said, since here we have not $x$ and $y$, but other values, we use a non-standard notation of the form:

Solving complex systems of equations

As a final note to today's video tutorial, let's look at a couple of really complex systems. Their complexity will consist in the fact that they will have variables on both the left and right. Therefore, to solve them we will have to apply preprocessing.

System No. 1

\[\left\( \begin(align)& 3\left(2x-y \right)+5=-2\left(x+3y \right)+4 \\& 6\left(y+1 \right )-1=5\left(2x-1 \right)+8 \\\end(align) \right.\]

Each equation carries a certain complexity. Therefore, let's treat each expression as with a regular linear construction.

In total, we get the final system, which is equivalent to the original one:

\[\left\( \begin(align)& 8x+3y=-1 \\& -10x+6y=-2 \\\end(align) \right.\]

Let's look at the coefficients of $y$: $3$ fits into $6$ twice, so let's multiply the first equation by $2$:

\[\left\( \begin(align)& 16x+6y=-2 \\& -10+6y=-2 \\\end(align) \right.\]

The coefficients of $y$ are now equal, so we subtract the second from the first equation: $$

Now let's find $y$:

Answer: $\left(0;-\frac(1)(3) \right)$

System No. 2

\[\left\( \begin(align)& 4\left(a-3b \right)-2a=3\left(b+4 \right)-11 \\& -3\left(b-2a \right )-12=2\left(a-5 \right)+b \\\end(align) \right.\]

Let's transform the first expression:

Let's deal with the second one:

\[-3\left(b-2a \right)-12=2\left(a-5 \right)+b\]

\[-3b+6a-12=2a-10+b\]

\[-3b+6a-2a-b=-10+12\]

In total, our initial system will take the following form:

\[\left\( \begin(align)& 2a-15b=1 \\& 4a-4b=2 \\\end(align) \right.\]

Looking at the coefficients of $a$, we see that the first equation needs to be multiplied by $2$:

\[\left\( \begin(align)& 4a-30b=2 \\& 4a-4b=2 \\\end(align) \right.\]

Subtract the second from the first construction:

Now let's find $a$:

Answer: $\left(a=\frac(1)(2);b=0 \right)$.

That's it. I hope this video tutorial will help you understand this difficult topic, namely solving systems of simple linear equations. There will be many more lessons on this topic in the future: we will look at more complex examples, where there will be more variables, and the equations themselves will be nonlinear. See you again!

Lesson and presentation on the topic: "Systems of equations. Substitution method, addition method, method of introducing a new variable"

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Methods for solving systems of inequalities

Guys, we have studied systems of equations and learned how to solve them using graphs. Now let's see what other ways to solve systems exist?
Almost all the methods for solving them are no different from those we studied in 7th grade. Now we need to make some adjustments according to the equations that we have learned to solve.
The essence of all the methods described in this lesson is to replace the system with an equivalent system with a simpler form and solution. Guys, remember what an equivalent system is.

Substitution method

The first way to solve systems of equations with two variables is well known to us - this is the substitution method. We used this method to solve linear equations. Now let's see how to solve equations in the general case?

How should you proceed when making a decision?
1. Express one of the variables in terms of another. The variables most often used in equations are x and y. In one of the equations we express one variable in terms of another. Tip: Look at both equations carefully before you start solving, and choose the one where it is easier to express the variable.
2. Substitute the resulting expression into the second equation, instead of the variable that was expressed.
3. Solve the equation that we got.
4. Substitute the resulting solution into the second equation. If there are several solutions, then you need to substitute sequentially so as not to lose a couple of solutions.
5. As a result, you will receive a pair of numbers $(x;y)$, which must be written down as an answer.

Example.
Solve a system with two variables using the substitution method: $\begin(cases)x+y=5, \\xy=6\end(cases)$.

Solution.
Let's take a closer look at our equations. Obviously, expressing y in terms of x in the first equation is much simpler.
$\begin(cases)y=5-x, \\xy=6\end(cases)$.
Let's substitute the first expression into the second equation $\begin(cases)y=5-x, \\x(5-2x)=6\end(cases)$.
Let's solve the second equation separately:
$x(5-x)=6$.
$-x^2+5x-6=0$.
$x^2-5x+6=0$.
$(x-2)(x-3)=0$.
We obtained two solutions to the second equation $x_1=2$ and $x_2=3$.
Substitute sequentially into the second equation.
If $x=2$, then $y=3$. If $x=3$, then $y=2$.
The answer will be two pairs of numbers.
Answer: $(2;3)$ and $(3;2)$.

Algebraic addition method

We also studied this method in 7th grade.
It is known that we can multiply a rational equation in two variables by any number, not forgetting to multiply both sides of the equation. We multiplied one of the equations by a certain number so that when adding the resulting equation to the second equation of the system, one of the variables was destroyed. Then the equation was solved for the remaining variable.
This method still works, although it is not always possible to destroy one of the variables. But it allows you to significantly simplify the form of one of the equations.

Example.
Solve the system: $\begin(cases)2x+xy-1=0, \\4y+2xy+6=0\end(cases)$.

Solution.
Let's multiply the first equation by 2.
$\begin(cases)4x+2xy-2=0, \\4y+2xy+6=0\end(cases)$.
Let's subtract the second from the first equation.
$4x+2xy-2-4y-2xy-6=4x-4y-8$.
As you can see, the form of the resulting equation is much simpler than the original one. Now we can use the substitution method.
$\begin(cases)4x-4y-8=0, \\4y+2xy+6=0\end(cases)$.
Let's express x in terms of y in the resulting equation.
$\begin(cases)4x=4y+8, \\4y+2xy+6=0\end(cases)$.
$\begin(cases)x=y+2, \\4y+2(y+2)y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\4y+2y^2+4y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\2y^2+8y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\y^2+4y+3=0\end(cases)$.
$\begin(cases)x=y+2, \\(y+3)(y+1)=0\end(cases)$.
We got $y=-1$ and $y=-3$.
Let's substitute these values sequentially into the first equation. We get two pairs of numbers: $(1;-1)$ and $(-1;-3)$.
Answer: $(1;-1)$ and $(-1;-3)$.

Method for introducing a new variable

We also studied this method, but let's look at it again.

Example.
Solve the system: $\begin(cases)\frac(x)(y)+\frac(2y)(x)=3, \\2x^2-y^2=1\end(cases)$.

Solution.
Let us introduce the replacement $t=\frac(x)(y)$.
Let's rewrite the first equation with a new variable: $t+\frac(2)(t)=3$.
Let's solve the resulting equation:
$\frac(t^2-3t+2)(t)=0$.
$\frac((t-2)(t-1))(t)=0$.
We got $t=2$ or $t=1$. Let us introduce the reverse change $t=\frac(x)(y)$.
We got: $x=2y$ and $x=y$.

For each of the expressions, the original system must be solved separately:
$\begin(cases)x=2y, \\2x^2-y^2=1\end(cases)$. $\begin(cases)x=y, \\2x^2-y^2=1\end(cases)$.
$\begin(cases)x=2y, \\8y^2-y^2=1\end(cases)$. $\begin(cases)x=y, \\2y^2-y^2=1\end(cases)$.
$\begin(cases)x=2y, \\7y^2=1\end(cases)$. $\begin(cases)x=2y, \\y^2=1\end(cases)$.
$\begin(cases)x=2y, \\y=±\frac(1)(\sqrt(7))\end(cases)$. $\begin(cases)x=y, \\y=±1\end(cases)$.
$\begin(cases)x=±\frac(2)(\sqrt(7)), \\y=±\frac(1)(\sqrt(7))\end(cases)$. $\begin(cases)x=±1, \\y=±1\end(cases)$.
We received four pairs of solutions.
Answer: $(\frac(2)(\sqrt(7));\frac(1)(\sqrt(7)))$; $(-\frac(2)(\sqrt(7));-\frac(1)(\sqrt(7)))$; $(1;1)$; $(-1;-1)$.

Example.
Solve the system: $\begin(cases)\frac(2)(x-3y)+\frac(3)(2x+y)=2,\\\frac(8)(x-3y)-\frac(9 )(2x+y)=1\end(cases)$.

Solution.
Let us introduce the replacement: $z=\frac(2)(x-3y)$ and $t=\frac(3)(2x+y)$.
Let's rewrite the original equations with new variables:
$\begin(cases)z+t=2, \\4z-3t=1\end(cases)$.
Let's use the algebraic addition method:
$\begin(cases)3z+3t=6, \\4z-3t=1\end(cases)$.
$\begin(cases)3z+3t+4z-3t=6+1, \\4z-3t=1\end(cases)$.
$\begin(cases)7z=7, \\4z-3t=1\end(cases)$.
$\begin(cases)z=1, \\-3t=1-4\end(cases)$.
$\begin(cases)z=1, \\t=1\end(cases)$.
Let's introduce the reverse substitution:
$\begin(cases)\frac(2)(x-3y)=1, \\\frac(3)(2x+y)=1\end(cases)$.
$\begin(cases)x-3y=2, \\2x+y=3\end(cases)$.
Let's use the substitution method:
$\begin(cases)x=2+3y, \\4+6y+y=3\end(cases)$.
$\begin(cases)x=2+3y, \\7y=-1\end(cases)$.
$\begin(cases)x=2+3(\frac(-1)(7)), \\y=\frac(-1)(7)\end(cases)$.
$\begin(cases)x=\frac(11)(7), \\x=-\frac(11)(7)\end(cases)$.
Answer: $(\frac(11)(7);-\frac(1)(7))$.

Problems on systems of equations for independent solution

Solve systems:
1. $\begin(cases)2x-2y=6,\\xy =-2\end(cases)$.
2. $\begin(cases)x+y^2=3, \\xy^2=4\end(cases)$.
3. $\begin(cases)xy+y^2=3,\\y^2-xy=5\end(cases)$.
4. $\begin(cases)\frac(2)(x)+\frac(1)(y)=4, \\\frac(1)(x)+\frac(3)(y)=9\ end(cases)$.
5. $\begin(cases)\frac(5)(x^2-xy)+\frac(4)(y^2-xy)=-\frac(1)(6), \\\frac(7 )(x^2-xy)-\frac(3)(y^2-xy)=\frac(6)(5)\end(cases)$.