Lesson and presentation on the topic: "Systems of equations. The substitution method, the addition method, the method of introducing a new variable"
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Ways to solve 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 ways to solve them do not differ from those that we studied in the 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 the replacement of the system with an equivalent system with a simpler form and method of 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 one proceed when making a decision?
1. Express one of the variables in terms of the other. The most common variables used in equations are x and y. In one of the equations, we express one variable in terms of another. Tip: Take a good look at both equations before you start solving and choose the one where it will be 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 we got.
4. Substitute the resulting solution into the second equation. If there are several solutions, then it is necessary to substitute them sequentially so as not to lose a couple of solutions.
5. As a result, you will get a pair of numbers $(x;y)$, which must be written 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 easier.
$\begin(cases)y=5-x, \\xy=6\end(cases)$.
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 got two solutions of the second equation $x_1=2$ and $x_2=3$.
Substitute successively 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, remembering to multiply both sides of the equation. We multiplied one of the equations by a certain number so that when the resulting equation is added to the second equation of the system, one of the variables is destroyed. Then the equation was solved with respect to the remaining variable.
This method still works, although it is not always possible to destroy one of the variables. But it allows one 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.
Multiply the first equation by 2.
$\begin(cases)4x+2xy-2=0, \\4y+2xy+6=0\end(cases)$.
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 through y in the resulting equation.
$\begin(cases)4x=4y+8, \\4y+2xy+6=0\end(cases)$.
$\begin(cases)x=y+2, \\4y+2(y+2)y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\4y+2y^2+4y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\2y^2+8y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\y^2+4y+3=0\end(cases)$.
$\begin(cases)x=y+2, \\(y+3)(y+1)=0\end(cases)$.
Got $y=-1$ and $y=-3$.
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:
$\frac(t^2-3t+2)(t)=0$.
$\frac((t-2)(t-1))(t)=0$.
Got $t=2$ or $t=1$. Let us introduce the reverse change $t=\frac(x)(y)$.
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.
We 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 method of algebraic addition:
$\begin(cases)3z+3t=6, \\4z-3t=1\end(cases)$.
$\begin(cases)3z+3t+4z-3t=6+1, \\4z-3t=1\end(cases)$.
$\begin(cases)7z=7, \\4z-3t=1\end(cases)$.
$\begin(cases)z=1, \\-3t=1-4\end(cases)$.
$\begin(cases)z=1, \\t=1\end(cases)$.
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. $\begin(cases)\frac(5)(x^2-xy)+\frac(4)(y^2-xy)=-\frac(1)(6), \\\frac(7 )(x^2-xy)-\frac(3)(y^2-xy)=\frac(6)(5)\end(cases)$.
Let us first recall the definition of a solution to a system of equations in two variables.
Definition 1
A pair of numbers is called a solution to a system of equations with two variables if, when they are substituted into the equation, the correct equality is obtained.
In what follows, we will consider systems of two equations with two variables.
Exist four basic ways to solve systems of equations: substitution method, addition method, graphical method, new variable management method. Let's look at these methods with specific examples. To describe the principle of using the first three methods, we will consider a system of two linear equations with two unknowns:
Substitution method
The substitution method is as follows: any of these equations is taken and $y$ is expressed in terms of $x$, then $y$ is substituted into the equation of the system, from where the variable $x.$ is found. After that, we can easily calculate the variable $y.$
Example 1
Let us express from the second equation $y$ in terms of $x$:
Substitute in the first equation, find $x$:
\ \ \
Find $y$:
Answer: $(-2,\ 3)$
Addition method.
Consider this method with an example:
Example 2
\[\left\( \begin(array)(c) (2x+3y=5) \\ (3x-y=-9) \end(array) \right.\]
Multiply the second equation by 3, we get:
\[\left\( \begin(array)(c) (2x+3y=5) \\ (9x-3y=-27) \end(array) \right.\]
Now let's add both equations together:
\ \ \
Find $y$ from the second equation:
\[-6-y=-9\] \
Answer: $(-2,\ 3)$
Remark 1
Note that in this method it is necessary to multiply one or both equations by such numbers that when adding one of the variables "disappears".
Graphical way
The graphical method is as follows: both equations of the system are displayed on the coordinate plane and the point of their intersection is found.
Example 3
\[\left\( \begin(array)(c) (2x+3y=5) \\ (3x-y=-9) \end(array) \right.\]
Let us express $y$ from both equations in terms of $x$:
\[\left\( \begin(array)(c) (y=\frac(5-2x)(3)) \\ (y=3x+9) \end(array) \right.\]
Let's draw both graphs on the same plane:
Picture 1.
Answer: $(-2,\ 3)$
How to introduce new variables
We will consider this method in the following example:
Example 4
\[\left\( \begin(array)(c) (2^(x+1)-3^y=-1) \\ (3^y-2^x=2) \end(array) \right .\]
Solution.
This system is equivalent to the system
\[\left\( \begin(array)(c) ((2\cdot 2)^x-3^y=-1) \\ (3^y-2^x=2) \end(array) \ right.\]
Let $2^x=u\ (u>0)$ and $3^y=v\ (v>0)$, we get:
\[\left\( \begin(array)(c) (2u-v=-1) \\ (v-u=2) \end(array) \right.\]
We solve the resulting system by the addition method. Let's add the equations:
\ \
Then from the second equation, we get that
Returning to the replacement, we obtain a new system of exponential equations:
\[\left\( \begin(array)(c) (2^x=1) \\ (3^y=3) \end(array) \right.\]
We get:
\[\left\( \begin(array)(c) (x=0) \\ (y=1) \end(array) \right.\]
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
Consider in general terms the system of equations AX = B with a non-singular square matrix A. In this case, there is an inverse matrix A -1 . Let's multiply both sides by A -1 on the left. We get A -1 AX \u003d A -1 B. From here EX \u003d 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 the solution of the system, a check can be made by substituting the found values into the equations of the system. In this case, they must turn into true equalities.
For this example, let's check:
Method for solving systems of linear equations with a square matrix using Cramer's formulas
Let n=2:
If both parts of the first equation are multiplied by a 22, and both parts of the second by (-a 12), and then the resulting equations are added, then we will exclude 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
Denote
Then the system will take the form:
It follows from the resulting system that if the determinant of the system is 0, then the system will be consistent and definite. Its unique solution can be calculated by the formulas:
If = 0, a 1 0 and/or 2 0, then the equations of the system will take the form 0*х 1 = 2 and/or 0*х 1 = 2. In this case, the system will be inconsistent.
In the case when = 1 = 2 = 0, the system will be consistent and indefinite (it will have an infinite number of solutions), as it will take the form:
Cramer's theorem(we omit the proof). If the determinant of the matrix of the system n 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 the matrix A by replacing the j-th column with a column of free members.
The above formulas are called Cramer's 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 complexity (calculation of 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, the following method is more common in practice.
Gauss method (method of successive elimination of variables)
This method is used to solve a system of m linear equations with n variables in a general way. Its essence lies in applying a system of equivalent transformations to the expanded matrix, with the help of which the system of equations is transformed to the form when its solutions become easy to find (if any).
This is such 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 stepped matrix in order to determine the rank. In this case, elementary transformations are applied to the expanded matrix, which will allow one to obtain an equivalent system of equations. After that, the augmented matrix will take the form:
Obtaining such a matrix is called in a straight line Gauss method.
Finding the values of variables from the corresponding system of equations is called backwards 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 whole 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 (just discard the corresponding rows).
After that, the system will look like:
Consider first the case when r=n. Then the system will take the form:
From the last equation of the system one can uniquely find x r .
Knowing x r , one can uniquely express x r -1 from it. Then from the previous equation, knowing x r and x r -1 , we can express x r -2 and so on. up to x 1 .
So, in this case, the system will be collaborative and definite.
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:
Substituting the resulting expression instead of x r, it will be possible to express the basic variable x r -1 through non-basic ones. Etc. to variable x 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 obtained formulas. Thus, in this case, the system will be consistent and indeterminate (have an infinite number of solutions).
For example, let's solve the system of equations:
The set of basic variables will be called basis systems. The set of columns of coefficients for them will also be called basis(basic columns), or basic minor system matrices. That 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) (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, it is necessary to equate x 3 and x 4 (c 1 and c 2) to arbitrary numbers that are not equal to zero at the same time, and calculate the rest of the variables through them. For example, with c 1 = 1 and c 2 = 0, we get a non-basic solution - (4/5; -12/5; 1; 0). By substitution, it is easy to verify that both solutions are correct.
Obviously, in an indefinite system of non-basic solutions, there can be an infinite number of solutions. How many basic solutions can there be? Each row of the transformed matrix must correspond to one basic variable. In total, there are n variables in the problem, and r basic rows. Therefore, the number of possible sets of basic variables cannot exceed the number of combinations from n to 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 such a form when the matrix formed from the columns of the coefficients for these variables will be stepwise and, in this case, will consist of rrows. Those. the rank of the matrix of coefficients for these variables must be equal to r. It cannot be larger, since the number of columns is equal to r. If it turns out to be less than r, then this indicates a linear dependence of the columns with variables. Such columns cannot form a basis.
Let us consider what other basic solutions can be found in the above example. To do this, consider all possible combinations of four variables with two basic ones. Such combinations will 
, and one of them (x 1 and x 2) has already been considered.
Let's take variables x 1 and x 3 . Find the rank of the matrix of coefficients for them:
Since it is equal to two, they can be basic. We equate the non-basic variables x 2 and x 4 to zero: x 2 \u003d x 4 \u003d 0. Then from the formula x 1 \u003d 4/5 - (1/5) * x 4 it follows that x 1 \u003d 4/5, and from the formula x 2 \u003d -17/5 + x 3 - - (7/5) * x 4 \u003d -17/5 + x 3 it follows that x 3 \u003d x 2 + 17/5 \u003d 17/5. Thus, we get the basic solution (4/5; 0; 17/5; 0).
Similarly, you can get 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 is possible to determine whether or not it is possible to form a basis from some variables. When solving the example, as a result of transforming the system matrix to a stepped form, it took the form:
By choosing pairs of variables, it was possible to calculate the corresponding minors of this matrix. It is easy to see that for all pairs, except for 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 consider one more example. Let's solve the system of equations
So, the equation corresponding to the third row of the last matrix is inconsistent - it led to the wrong 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 the form when the coefficients of the variables form an identity matrix up to a permutation of rows or columns 4 (where is the rank of the system matrix).
Let's solve the system using this method:
Consider the augmented matrix of the system:
In this matrix, we select the identity element. For example, the coefficient at x 2 in the third constraint is 5. Let's make sure that in the remaining rows in this column there are zeros, i.e. make the column single. In the process of transformations, we will call this columnpermissive(leading, key). The third constraint (the third string) will also be called permissive. Myself element, which stands at the intersection of the allowing row and column (here it is a unit), is also called permissive.
The first line now contains the coefficient (-1). To get zero in its place, multiply the third row by (-1) and subtract the result from the first row (i.e. just add the first row to the third).
The second line contains a coefficient of 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 transformations will look like:
This matrix clearly shows that one of the first two constraints can be deleted (the corresponding rows are proportional, i.e. these equations follow from each other). Let's cross out the second one:
So, there are two equations in the new system. A single column (second) is received, and the unit here is in the second row. Let's remember that the basic variable x 2 will correspond to the second equation of the new system.
Let's choose a basic variable for the first row. It can be any variable except x 3 (because at 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 in order to get one in the first column of the first row.
Let's make sure that the rest of the rows (i.e., the second row) have zeros in the first column. Since now the second line is not zero, but 3, it is necessary to subtract from the second line the elements of the converted first line, multiplied by 3:
One basic solution can be directly extracted from the resulting matrix by equating the non-basic variables to zero, and the basic variables 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 \u003d 0.8 - 1.2 x 4; x 2 \u003d -3.4 + x 3 + 1.6x 4. These formulas describe the entire infinite set of solutions to the system (by 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 permissive string was divided by the permissive element to get a unit in its place,
2) from all other rows, the transformed resolving power multiplied by the element that was in the given line in the resolving column was subtracted to get zero in place of this element.
Consider once again the transformed augmented matrix of the system:
It can be seen from this entry that the rank of the matrix of system A is r.
In the course of the above reasoning, we have established that the system is consistent if and only if 
. This means that the augmented matrix of the system will look like:
Discarding zero rows, we get 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 matrix of the system 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 (because they are a linear combination of the others). The system of equations will be linearly independent only if the rank of the extended matrix is equal to the number of equations.
Moreover, for consistent 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, suppose there are five rows in the matrix (the initial row order is 12345). We need to change the second line and the fifth. In order for the second line to fall into the place of the fifth, to “move” down, we sequentially 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 up by only two consecutive changes: the fifth and fourth rows (13542) and the fifth and third (15342).
2Number of combinations from n to r 
the number of all different r-element subsets of an n-element set is called (different sets are those that have a different composition of elements, the selection order is not important). It is calculated by the formula: 
. Recall the meaning of the sign “!” (factorial): 
0!=1.)
3Since this method is more common than the Gaussian method discussed earlier, and in essence is a combination of the forward and reverse 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 matrix of the system, then it would be possible, for example, to divide both parts of the first equation by two, and then the first coefficient would become unity; or the like.
With this mathematical program, you can solve a system of two linear equations with two variables using the substitution method and the addition method.
The program not only gives the answer to the problem, but also provides a detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.
This program can be useful for high school students in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.
Rules for Entering Equations
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.
When entering equations you can use brackets. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of the elements.
For example: 6x+1 = 5(x+y)+2
In equations, you can use not only integers, but also fractional numbers in the form of decimals and common fractions.
Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &
Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)
Solve a system of equations
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A bit of theory.
Solving systems of linear equations. Substitution method
The sequence of actions when solving a system of linear equations by the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression in another equation of the system instead of this variable;
$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$
Let's express from the first equation y through x: y = 7-3x. Substituting the expression 7-3x instead of y into the second equation, we get the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$
It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$
Substituting the number 1 instead of x into the equation y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$
Pair (1;4) - solution of the system
Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.
Solving systems of linear equations by adding
Consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by the substitution method, we pass from a given system to another system equivalent to it, in which one of the equations contains only one variable.
The sequence of actions when solving a system of linear equations by the addition method:
1) multiply the equations of the system term by term, choosing factors so that the coefficients for one of the variables become opposite numbers;
2) add term by term the left and right parts of the equations of the system;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.
Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$
In the equations of this system, the coefficients of y are opposite numbers. Adding term by term the left and right parts of the equations, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$
From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38 \) we get an equation with the variable y: \(11-3y=38 \). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)
Thus, we found the solution to the system of equations by adding: \(x=11; y=-9 \) or \((11; -9) \)
Taking advantage of the fact that in the equations of the system the coefficients of y are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both parts of each of the equations of the original symmeme), in which one of the equations contains only one variable.
Books (textbooks) Abstracts of the Unified State Examination and OGE tests online Games, puzzles Graphing of functions Spelling dictionary of the Russian language Dictionary of youth slang Catalog of Russian schools Catalog of secondary schools in Russia Catalog of Russian universities List of tasksMore reliable than the graphical method discussed in the previous paragraph.
Substitution Method
We used this method in the 7th grade to solve systems of linear equations. The algorithm that was developed in the 7th grade is quite suitable for solving systems of any two equations (not necessarily linear) with two variables x and y (of course, the variables can be denoted by other letters, which does not matter). In fact, we used this algorithm in the previous 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 by the substitution method (see example 1 from § 4).
Algorithm for using the substitution method when solving a system of two equations with two variables x, y.
1. Express y in terms of x from one equation of the system.
2. Substitute the resulting expression instead of y into another equation of the system.
3. Solve the resulting equation for x.
4. Substitute in turn each of the roots of the equation found at the third step instead of x into the expression y through x obtained at the first step.
5. Write down the answer in the form of pairs of values (x; y), which were found, respectively, in the third and fourth steps.
4) Substitute in turn each of the found values of y into the formula x \u003d 5 - Zy. If then 
5) Pairs (2; 1) and solutions of a given system of equations.
Answer: (2; 1);
Algebraic addition method
This method, like the substitution method, is familiar to you from the 7th grade algebra course, where it was used to solve systems of linear equations. We recall the essence of the method in the following example.
Example 2 Solve a system of equations
We multiply all the terms of the first equation of the system by 3, and leave the second equation unchanged: 
Subtract the second equation of the system from its first equation:
As a result of algebraic addition of two equations of the original system, an equation was obtained that is simpler than the first and second equations of the given system. With this simpler equation, we have the right to replace any equation of a given system, for example, the second one. Then the given system of equations will be replaced by a simpler system:
This system can be solved by the substitution method. From the second equation we find Substituting this expression instead of y into the first equation of the system, we obtain
It remains to substitute the found values \u200b\u200bof x into the formula
If x = 2 then
Thus, we have found two solutions to the system: 
Method for introducing new variables
You got acquainted with the method of introducing a new variable when solving rational equations with one variable in the 8th grade algebra course. The essence of this method 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 a 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 the variable t. But that means either from where we find that x = 2y, or
Thus, using the method of introducing a new variable, we managed, as it were, to “stratify” the first equation of the system, which is quite complex in appearance, into two simpler equations:
x = 2 y; y - 2x.
What's next? And then each of the two simple equations obtained must be considered in turn in a system with the equation x 2 - y 2 \u003d 3, which we have not yet remembered. In other words, the problem is reduced to solving two systems of equations:
It is necessary to find solutions for the first system, the second system, and include all the resulting pairs of values in the answer. Let's solve the first system of equations:
Let's use the substitution method, especially since everything is ready for it here: we substitute the expression 2y instead of x into the second equation of the system. Get
Since x \u003d 2y, we find x 1 \u003d 2, x 2 \u003d 2, respectively. Thus, two solutions to the given system are obtained: (2; 1) and (-2; -1). Let's solve the second system of equations:
Let's use the substitution method again: we substitute the expression 2x instead of y in the second equation of the system. Get
This equation has no roots, which means that the system of equations has no solutions. Thus, only the solutions of the first system should be included in the answer.
Answer: (2; 1); (-2;-1).
The method of introducing new variables in solving systems of two equations with two variables is used in two versions. First option: one new variable is introduced and used in only one equation of the system. This is exactly what happened in example 3. The second option: two new variables are introduced and used simultaneously in both equations of the system. This will be the case in example 4.
Example 4 Solve a system of equations
Let's introduce two new variables:
We learn that then
This will allow us to rewrite the given system in a much simpler form, but with respect to the new variables a and b:
Since a \u003d 1, then from the equation a + 6 \u003d 2 we find: 1 + 6 \u003d 2; 6=1. Thus, for the variables a and b, we got one solution:
Returning to the variables x and y, we obtain the system of equations
We apply the algebraic addition method to solve this system:
Since then from the equation 2x + y = 3 we find:
Thus, for the variables x and y, we got one solution:
Let us conclude this section with a brief but rather serious theoretical discussion. You have already gained some experience in solving various equations: linear, square, rational, irrational. You know that the main idea of solving an equation is to gradually move from one equation to another, simpler but equivalent to the given one. In the previous section, we introduced the notion of equivalence for equations with two variables. This concept is also used for systems of equations.
Definition.
Two systems of equations with variables x and y are said to be equivalent if they have the same solutions or if both systems have no solutions.
All three methods (substitution, algebraic addition, and introduction of new variables) that we have discussed in this section are absolutely correct from the point of view of equivalence. In other words, using these methods, we replace one system of equations with another, simpler, but equivalent to the original system.
Graphical method for solving systems of equations
We have already learned how to solve systems of equations in such common and reliable ways as the method of substitution, algebraic addition and the introduction of new variables. And now let's remember the method that you already studied in the previous lesson. That is, let's repeat what you know about the graphical solution method.
The method of solving systems of equations graphically is the construction of a graph for each of the specific equations that are included in this system and are in the same coordinate plane, and also where it is required to find the intersection 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 a graphical system of equations it is common to have either one single correct solution, or an infinite number of solutions, or not to have solutions at all.
Now let's take a closer look at each of these solutions. And so, the system of equations can have a unique solution if the lines, which are the graphs of the equations of the system, intersect. If these lines are parallel, then such a system of equations has absolutely no solutions. In the case of the coincidence of the direct graphs of the equations of the system, then such a system allows you to find many solutions.
Well, now let's take a look at the algorithm for solving a system of two equations with 2 unknowns using a graphical method:
First, at first we build a graph of the 1st equation;
The second step will be to plot a graph that relates to the second equation;
Thirdly, we need to find the intersection points of the graphs.
And as a result, we get the coordinates of each intersection point, which will be the solution to the system of equations.
Let's look at this method in more detail with an example. We are given a system of equations to be solved:
Solving Equations
1. First, we will build a graph of this equation: x2+y2=9.
But it should be noted that this graph of equations will be a circle centered at the origin, and its radius will be equal to three.
2. Our next step will be to plot an equation such as: y = x - 3.
In this case, we must build a line and find the points (0;−3) and (3;0).
3. Let's see what we got. We see that the line intersects the circle at two of its points A and B.
Now we are looking for the coordinates of these points. We see that the coordinates (3;0) correspond to point A, and the coordinates (0;−3) correspond to point B.
And what do we get as a result?
The numbers (3;0) and (0;−3) obtained at the intersection of a straight line with a circle are precisely the solutions of both equations of the system. And from this it follows that these numbers are also solutions of this system of equations.
That is, the answer of this solution is the numbers: (3;0) and (0;−3).
