Method idea. An equation is chosen in which one of the variables is most simply expressed in terms of the other variables. The resulting expression for this variable is substituted into the remaining equations of the system.

1. b) Combination with other methods.

Method idea. If the direct substitution method is not applicable at the initial stage of the solution, then equivalent system transformations are used (term by term addition, subtraction, multiplication, division), and then direct substitution is carried out directly.

2) Method of independent solution of one of the equations.

Method idea. If the system contains an equation in which there are mutually inverse expressions, then a new variable is introduced and the equation is solved with respect to it. The system then breaks down into several simpler systems.

Solve a system of equations

Consider the first equation of the system:

Making the substitution , where t ≠ 0, we obtain

Whence t 1 = 4, t 2 = 1/4.

Returning to the old variables, consider two cases.

The roots of the equation 4y 2 - 15y - 4 \u003d 0 are y 1 \u003d 4, y 2 \u003d - 1/4.

The roots of the equation 4x 2 + 15x - 4 \u003d 0 are x 1 \u003d - 4, x 2 \u003d 1/4.

3) Reduction of the system to the union of simpler systems.

1. a) Factorization by taking out the common factor.

Method idea. If one of the equations has a common factor, then this equation is decomposed into factors and, taking into account the equality of the expression to zero, they proceed to solving simpler systems.

1. b) Factoring through the solution of the homogeneous equation.

Method idea. If one of the equations is a homogeneous equation (, then, having solved it with respect to one of the variables, we factor it, for example: a (x-x 1) (x-x 2) and, given the equality of the expression to zero, we proceed to solving simpler systems.

Let's solve the first system

1. c) Using homogeneity.

Method idea. If the system has an expression that is a product of variables, then using the method of algebraic addition, a homogeneous equation is obtained, and then the factorization method is used through the solution of a homogeneous equation.

4) Method of algebraic addition.

Method idea. In one of the equations, we get rid of one of the unknowns, for this we equalize the modules of the coefficients for one of the variables, then we perform either term-by-term addition of equations, or subtraction.

5) Method of multiplication of equations.

Method idea. If there are no such pairs (x; y) for which both parts of one of the equations vanish simultaneously, then this equation can be replaced by the product of both equations of the system.

Let's solve the second equation of the system.

Let = t, then 4t 3 + t 2 -12t -12 = 0. Applying the corollary from the polynomial root theorem, we have t 1 = 2.

Р(2) = 4∙2 3 + 2 2 - 12∙2 - 12 = 32 + 4 - 24 - 12 = 0. We lower the degree of the polynomial using the method of indefinite coefficients.

4t 3 + t 2 -12t -12 = (t - 2) (at 2 + bt + c).

4t 3 + t 2 -12t -12 = at 3 + bt 2 + ct - 2at 2 -2bt - 2c.

4t 3 + t 2 - 12t -12 = at 3 + (b - 2a) t 2 + (c -2b) t - 2c.

We get the equation 4t 2 + 9t + 6 = 0, which has no roots, since D = 9 2 - 4∙4∙6 = -15<0.

Returning to the variable y, we have = 2, whence y = 4.

Answer. (1;4).

6) The method of division of equations.

Method idea. If there are no such pairs (x; y) for which both parts of one of the equations vanish simultaneously, then this equation can be replaced by an equation that is obtained by dividing one equation of the system by another.

7) The method of introducing new variables.

Method idea. Some expressions from the original variables are taken as new variables, which leads to a simpler system than the original one from these variables. After the new variables are found, it is necessary to find the values ​​of the original variables.

Returning to the old variables, we have:

We solve the first system.

8) Application of the Vieta theorem.

Method idea. If the system is composed in this way, one of the equations is presented as a sum, and the second as a product of some numbers that are the roots of some quadratic equation, then using the Vieta theorem we compose a quadratic equation and solve it.

Answer. (1;4), (4;1).

Substitution is used to solve symmetric systems: x + y = a; xy = in. When solving symmetric systems, the following transformations are used:

x 2 + y 2 \u003d (x + y) 2 - 2xy \u003d a 2 - 2c; x 3 + y 3 \u003d (x + y) (x 2 - xy + y 2) \u003d a (a 2 -3c);

x 2 y + xy 2 \u003d xy (x + y) \u003d av; (x + 1) ∙ (y + 1) \u003d xy + x + y + 1 \u003d a + b + 1;

Answer. (1;1), (1;2), (2;1).

10) "Boundary problems".

Method idea. The solution of the system is obtained by logical reasoning related to the structure of the domain of definition or the set of values ​​of functions, the study of the sign of the discriminant of the quadratic equation.

The peculiarity of this system is that the number of variables in it is greater than the number of equations. For nonlinear systems, such a feature is often a sign of a "boundary problem". Based on the type of equations, we will try to find the set of values ​​of the function that occurs in both the first and second equations of the system. Since x 2 + 4 ≥ 4, it follows from the first equation that

The answer is (0;4;4), (0;-4;-4).

11) Graphical method.

Method idea. Build graphs of functions in one coordinate system and find the coordinates of their intersection points.

1) Having rewritten the first equation of the systems in the form y \u003d x 2, we come to the conclusion: the graph of the equation is a parabola.

2) Having rewritten the second equation of the systems in the form y \u003d 2 / x 2, we come to the conclusion: the graph of the equation is a hyperbola.

3) The parabola and the hyperbola intersect at point A. There is only one point of intersection, since the right branch of the parabola serves as a graph of an increasing function, and the right branch of the hyperbola is a decreasing one. Judging by the constructed geometric model, point A has coordinates (1; 2). Verification shows that the pair (1;2) is a solution to both equations of the system.

Linear Equation - an equation of the form a x = b, where x is a variable, a and b are some numbers, and a ≠ 0.

Examples of linear equations:

1. 3x=2

1. 2 7 x = − 5

Linear equations are called not only equations of the form a x \u003d b, but also any equations that, with the help of transformations and simplifications, are reduced to this form.

How to solve equations that are reduced to the form a x \u003d b? It is enough to divide the left and right sides of the equation by the value a. As a result, we get the answer: x = b a .

How to recognize if an arbitrary equation is linear or not? It is necessary to pay attention to the variable that is present in it. If the highest power of the variable is equal to one, then such an equation is a linear equation.

To solve the linear equation , it is necessary to open the brackets (if any), move the "x" to the left side, the numbers to the right, bring like terms. An equation of the form a x \u003d b will be obtained. Solution of this linear equation: x = b a .

Examples of solving linear equations:

1. 2x + 1 = 2(x − 3) + 8

This is a linear equation, since the variable is in the first power.

Let's try to convert it to the form a x = b:

Let's open the parentheses first:

2x + 1 = 4x - 6 + 8

All terms with x are transferred to the left side, numbers to the right:

2x - 4x = 2 - 1

Now let's divide the left and right parts by the number (-2) :

− 2 x − 2 = 1 − 2 = − 1 2 = − 0.5

Answer: x \u003d - 0.5

1. x 2 − 1 = 0

This equation is not a linear equation because the highest power of x is two.

1. x (x + 3) - 8 = x - 1

This equation looks linear at first glance, but after opening the parentheses, the highest power becomes equal to two:

x 2 + 3 x - 8 = x - 1

This equation is not a linear equation.

Special cases(in task 4 of the OGE they did not meet, but it is useful to know them)

Examples:

1. 2x - 4 = 2 (x - 2)

2x-4 = 2x-4

2x − 2x = − 4 + 4

And how to look for x here if it is not there? After performing the transformations, we got the correct equality (identity), which does not depend on the value of the variable x . Whatever value of x we ​​substitute into the original equation, the result is always the correct equality (identity). So x can be any number. Let's write down the answer to this linear equation.

Answer: x ∈ (− ∞ ;   + ∞)

1. 2x - 4 = 2 (x - 8)

This is a linear equation. Let's open the brackets, move the x's to the left, the numbers to the right:

2x-4 = 2x-16

2x - 2x = - 16 + 4

As a result of the transformations, x was reduced, but as a result, an incorrect equality was obtained, since. Whatever value of x we ​​substitute into the original equation, the result will always be an incorrect equality. And this means that there are no such values ​​of x at which the equality would become true. Let's write down the answer to this linear equation.

Answer: x ∈ ∅

Quadratic equations

Quadratic equation - an equation of the form a x 2 + b x + c \u003d 0, where x is a variable, a, b and c are some numbers, and a ≠ 0.

Algorithm for solving a quadratic equation:

1. Open brackets, move all terms to the left side so that the equation takes the form: a x 2 + b x + c = 0
2. Write out what the coefficients are equal in numbers: a = ... b = ... c = ...
3. Calculate the discriminant using the formula: D = b 2 − 4 a c
4. If D > 0, there will be two different roots, which are found by the formula: x 1,2 = − b ± D 2 a
5. If D = 0, there will be one root, which is found by the formula: x = − b 2 a
6. If D< 0, решений нет: x ∈ ∅

Examples of solving a quadratic equation:

1. − x 2 + 6 x + 7 = 0

a = − 1, b = 6, c = 7

D = b 2 − 4 a c = 6 2 − 4 ⋅ (− 1) ⋅ 7 = 36 + 28 = 64

D > 0 - there will be two different roots:

x 1,2 = − b ± D 2 a = − 6 ± 64 2 ⋅ (− 1) = − 6 ± 8 − 2 = [ − 6 + 8 − 2 = 2 − 2 = − 1 − 6 − 8 − 2 = − 14 − 2 = 7

Answer: x 1 \u003d - 1, x 2 \u003d 7

1. − x 2 + 4 x − 4 = 0

a = − 1, b = 4, c = − 4

D = b 2 − 4 a c = 4 2 − 4 ⋅ (− 1) ⋅ (− 4) = 16 − 16 = 0

D = 0 - there will be one root:

x = − b 2 a = − 4 2 ⋅ (− 1) = − 4 − 2 = 2

Answer: x = 2

1. 2 x 2 − 7 x + 10 = 0

a = 2, b = − 7, c = 10

D = b 2 − 4 a c = (− 7) 2 − 4 ⋅ 2 ⋅ 10 = 49 − 80 = − 31

D< 0 – решений нет.

Answer: x ∈ ∅

There are also incomplete quadratic equations (these are quadratic equations in which either b \u003d 0, or c \u003d 0, or b \u003d c \u003d 0). Watch the video on how to solve such quadratic equations!

Factorization of a square trinomial

The square trinomial can be factored as follows:

A x 2 + b x + c = a ⋅ (x − x 1) ⋅ (x − x 2)

where a is the number, coefficient before the highest coefficient,

x is a variable (that is, a letter),

x 1 and x 2 - numbers, roots of the quadratic equation a x 2 + b x + c \u003d 0, which are found through the discriminant.

If the quadratic equation has only one root, then the decomposition looks like this:

a x 2 + b x + c = a ⋅ (x − x 0) 2

Examples of factoring a square trinomial:

1. − x 2 + 6 x + 7 = 0 ⇒ x 1 = − 1,   x 2 = 7

− x 2 + 6 x + 7 = (− 1) ⋅ (x − (− 1)) (x − 7) = − (x + 1) (x − 7) = (x + 1) (7 − x)

1. − x 2 + 4 x − 4 = 0; ⇒ x0 = 2

− x 2 + 4 x − 4 = (− 1) ⋅ (x − 2) 2 = − (x − 2) 2

If the square trinomial is incomplete, ((b = 0 or c = 0) then it can be factored in the following ways:

• c = 0 ⇒ a x 2 + b x = x (a x + b)

• b = 0 ⇒ apply for difference of squares.

Fractionally rational equations

Let f (x) and g (x) be some functions depending on the variable x .

Fractionally rational equation is an equation of the form f (x) g (x) = 0 .

In order to solve a fractionally rational equation, one must remember what the ODZ is and when it arises.

ODZ– range of admissible values ​​of a variable.

In an expression like f(x) g(x) = 0

ODZ: g (x) ≠ 0 (the denominator of a fraction cannot be equal to zero).

Algorithm for solving a fractionally rational equation:

1. Write out ODZ: g (x) ≠ 0.
2. Equate the numerator of the fraction to zero f (x) = 0 and find the roots.

An example of solving a fractional rational equation:

Solve the fractionally rational equation x 2 − 4 2 − x = 1.

Solution:

We will act in accordance with the algorithm.

1. Bring the expression to the form f (x) g (x) = 0 .

We move the unity to the left side, write an additional factor to it in order to bring both terms to the same common denominator:

x 2 − 4 2 − x − 1 \ 2 − x = 0

x 2 − 4 2 − x − 2 − x 2 − x = 0

x 2 − 4 − (2 − x) 2 − x = 0

x 2 - 4 - 2 + x 2 - x = 0

x 2 + x - 6 2 - x = 0

The first step of the algorithm was successfully completed.

1. Write out ODZ:

We circle the ODZ, do not forget about it: x ≠ 2

1. Equate the numerator of the fraction to zero f (x) = 0 and find the roots:

x 2 + x - 6 = 0 - Quadratic equation. We solve through the discriminant.

a = 1, b = 1, c = − 6

D = b 2 − 4 a c = 1 2 − 4 ⋅ 1 ⋅ (− 6) = 1 + 24 = 25

D > 0 - there will be two different roots.

x 1,2 = − b ± D 2 a = − 1 ± 25 2 ⋅ 1 = − 1 ± 5 2 = [ − 1 + 5 2 = 4 2 = 2 − 1 − 5 2 = − 6 2 = − 3

[ x 1 = 2 x 2 = − 3

1. Indicate in the answer the roots from the numerator, excluding those roots that fell into the ODZ.

Roots obtained in the previous step:

[ x 1 = 2 x 2 = − 3

This means that the answer is only one root, x = − 3.

Answer: x = − 3.

Systems of equations

System of equations name two equations with two unknowns (as a rule, the unknowns are denoted by x and y), which are combined into a common system by a curly bracket.

Example of a system of equations

( x + 2 y = 8 3 x − y = − 4

Solve a system of equations – find a pair of numbers x and y that, when substituted into the system of equations, form the correct equality in both equations of the system.

There are two methods for solving systems of linear equations:

1. Substitution method.
2. Addition method.

Algorithm for solving the system of equations by the substitution method:

1. Find the remaining unknown.

Example:

Solve a system of equations using the substitution method

( x + 2 y = 8 3 x − y = − 4

Solution:

1. Express one variable from any equation in terms of another.

( x = 8 − 2 y 3 x − y = − 4

1. Substitute the resulting value in another equation instead of the expressed variable.

( x = 8 − 2 y 3 x − y = − 4

( x = 8 − 2 y 3 (8 − 2 y) − y = − 4

1. Solve an equation with one unknown.

3 (8 − 2 y) − y = − 4

24 − 6 y − y = − 4

− 7 y = − 4 − 24

− 7 y = − 28

y = − 28 − 7 = 28 7 = 4

1. Find the remaining unknown.

x = 8 − 2 y = 8 − 2 ⋅ 4 = 8 − 8 = 0

The answer can be written in one of three ways:

1. x=0, y=4
2. ( x = 0 y = 4
3. (0 ;   4)

Solving the system of equations by the addition method.

The addition method is based on the following property:

(a + c) = (b + d)

The idea behind the addition method is to get rid of one of the variables by adding the equations.

Example:

Solve a system of equations using the addition method

( x + 2 y = 8 3 x − y = − 4

Let's get rid of x in this example. The essence of the method is that in the first and second equations, opposite coefficients are placed in front of the variable x. In the second equation, x is preceded by a factor of 3. In order for the addition method to work, it is necessary that the coefficient (− 3) appears in front of the variable x. To do this, multiply the left and right sides of the first equation by (− 3) .

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

The addition method consists of three simple steps:

1. Look at the system and choose a variable that has the same (or opposite) coefficients in each equation;
2. Perform algebraic subtraction (for opposite numbers - addition) of equations from each other, and then bring like terms;
3. 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 hard to solve. Then it remains only to substitute the found root in the original system and get the final answer.

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

• Solving equations by addition implies that all rows must contain variables with the same/opposite coefficients. What if this requirement is not met?
• Not always, after adding / subtracting equations in this way, we will get a beautiful construction that is easily solved. Is it possible to somehow simplify the calculations and speed up the calculations?

To get an answer to these questions, and at the same time to deal with a few additional subtleties that many students “fall over”, watch my video tutorial:

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

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

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

1. Addition method;
2. 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 for 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 together. They are added term by term, i.e. "Xs" are added to "Xs" and similar ones are given;

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. So 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 will talk about this now.

Solving easy problems using the addition method

So, we are learning to apply the addition method using the example of two simple expressions.

Task #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 amount, the “games” will mutually annihilate. We add and get:

We solve the simplest construction:

Great, we found the X. What to do with him now? We can substitute it into any of the equations. Let's put it in the first one:

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

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

Task #2

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

Here, the situation is completely similar, only with the Xs. Let's put them together:

We got 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. Once again the key points:

1. 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.
2. We substitute the found variable into any of the equations of the system to find the second one.
3. The final record of the answer 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 $y$.
4. The rule to write the answer in the form of point coordinates is not always applicable. For example, it cannot be used when the role of variables is not $x$ and $y$, but, for example, $a$ and $b$.

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

Solving easy problems using the subtraction method

Task #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 equation from the first equation:

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

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

Task #2

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

We again see the same coefficient $5$ for $x$ in the first and second equations. 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 of $y$ into the second construct:

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

Nuances of the solution

So what do we see? In essence, 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 with 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 applied. This is always done so that one of them disappears, and in the final equation that remains after subtraction, only one variable would remain.

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

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 in general they do not correlate in any way with another 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 changing 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: for $y$, opposite coefficients. In such a situation, it is necessary to apply the addition method. Let's add:

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

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

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

Example #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 at $y$:

\[\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 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: 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:

1. We look at the system and analyze each equation.
2. If we see that neither for $y$ nor for $x$ the coefficients are consistent, i.e. they are neither equal nor opposite, then we do the following: select the variable to get rid of, and then look at the coefficients in these equations. If we multiply the first equation by the coefficient from the second, and multiply the second corresponding one 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 at $y$ will be consistent. All our actions or transformations are aimed only at getting one variable in one equation.
3. We find one variable.
4. We substitute the found variable into one of the two equations of the system and find the second one.
5. 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 in a slightly different way than according to the standard algorithm.

Solving problems with fractional numbers

Example #1

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

First, note that the second equation contains fractions. But note that you can divide $4$ by $0.8$. We get $5$. 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:

$n$ we found, now we calculate $m$:

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

Example #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, however, for none of the variables, the coefficients do not fit into each other by 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.\]

Let's use the subtraction method:

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

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, and the second equation was multiplied by $5$. As a result, we have obtained a consistent and even the same equation for the first variable. In the second system, we acted according to the standard algorithm.

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

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

Solving complex systems of equations

As a final touch 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 contain variables both on the left and on the right. Therefore, to solve them, we will have to apply preprocessing.

System #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, with each expression, let's do as with a normal 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 we 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 #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:

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

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

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

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.\]

We subtract the second from the first construction:

Now find $a$:

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

That's all. 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 further: we will analyze more complex examples, where there will be more variables, and the equations themselves will already be nonlinear. See you soon!

1. Substitution Method: from any equation of the system we express one unknown in terms of another and substitute it into the second equation of the system.

Task. Solve the system of equations:

Solution. From the first equation of the system, we express at through X and substitute into the second equation of the system. Let's get the system equivalent to the original.

After bringing such terms, the system will take the form:

From the second equation we find: . Substituting this value into the equation at = 2 - 2X, we get at= 3. Therefore, the solution of this system is a pair of numbers .

2. Algebraic addition method: by adding two equations, get an equation with one variable.

Task. Solve the system equation:

Solution. Multiplying both sides of the second equation by 2, we get the system equivalent to the original. Adding the two equations of this system, we arrive at the system

After reducing similar terms, this system will take the form: From the second equation we find . Substituting this value into Equation 3 X + 4at= 5, we get , where . Therefore, the solution of this system is a pair of numbers .

3. Method for introducing new variables: we are looking for some repeated expressions in the system, which we will denote by new variables, thereby simplifying the form of the system.

Task. Solve the system of equations:

Solution. Let's write this system differently:

Let x + y = u, hu = v. Then we get the system

Let's solve it by the substitution method. From the first equation of the system, we express u through v and substitute into the second equation of the system. Let's get the system those.

From the second equation of the system we find v 1 = 2, v 2 = 3.

Substituting these values ​​into the equation u = 5 - v, we get u 1 = 3,
u 2 = 2. Then we have two systems

Solving the first system, we get two pairs of numbers (1; 2), (2; 1). The second system has no solutions.

Exercises for independent work

1. Solve systems of equations using the substitution method.

In this lesson, we will consider methods for solving a system of linear equations. In the course of higher mathematics, systems of linear equations are required to be solved both in the form of separate tasks, for example, "Solve the system using Cramer's formulas", and in the course of solving other problems. One has to deal with systems of linear equations in almost all branches of higher mathematics.

First, a little theory. What does the mathematical word "linear" mean in this case? This means that in the equations of the system All variables are included in the first degree: no fancy stuff like etc., from which only participants of mathematical Olympiads are delighted.

In higher mathematics, not only letters familiar from childhood are used to designate variables.
A fairly popular option is variables with indices: .
Or the initial letters of the Latin alphabet, small and large:
It is not so rare to find Greek letters: - well-known to many "alpha, beta, gamma". And also a set with indices, say, with the letter "mu":

The use of one or another set of letters depends on the branch of higher mathematics in which we are faced with a system of linear equations. So, for example, in systems of linear equations encountered in solving integrals, differential equations, it is traditionally customary to use the notation

But no matter how the variables are designated, the principles, methods and methods for solving a system of linear equations do not change from this. Thus, if you come across something terrible like, do not rush to close the problem book in fear, after all, instead you can draw the sun, instead - a bird, and instead - a face (of a teacher). And, oddly enough, a system of linear equations with these notations can also be solved.

Something I have such a premonition that the article will turn out to be quite long, so a small table of contents. So, the sequential "debriefing" will be as follows:

– Solving a system of linear equations by the substitution method (“school method”);
– Solution of the system by the method of term-by-term addition (subtraction) of the equations of the system;
– Solution of the system by Cramer's formulas;
– Solution of the system using the inverse matrix;
– Solution of the system by the Gauss method.

Everyone is familiar with systems of linear equations from the school mathematics course. In fact, we start with repetition.

Solving a system of linear equations by the substitution method

This method can also be called the "school method" or the method of eliminating unknowns. Figuratively speaking, it can also be called the "half-finished Gauss method."

Example 1

Here we have a system of two equations with two unknowns. Note that the free terms (numbers 5 and 7) are located on the left side of the equation. Generally speaking, it doesn't matter where they are, on the left or on the right, it's just that in problems in higher mathematics they are often located that way. And such a record should not be confusing, if necessary, the system can always be written "as usual":. Do not forget that when transferring a term from part to part, you need to change its sign.

What does it mean to solve a system of linear equations? Solving a system of equations means finding the set of its solutions. The solution of the system is a set of values ​​of all variables included in it, which turns EVERY equation of the system into a true equality. In addition, the system can be incompatible (have no solutions).Do not be shy, this is a general definition =) We will have only one value of "x" and one value of "y", which satisfy each equation with-we.

There is a graphical method for solving the system, which can be found in the lesson. The simplest problems with a straight line. There I talked about geometric sense systems of two linear equations with two unknowns. But now in the yard is the era of algebra, and numbers-numbers, actions-actions.

We decide: from the first equation we express:
We substitute the resulting expression into the second equation:

We open the brackets, give like terms and find the value:

Next, we recall what they danced from:
We already know the value, it remains to find:

Answer:

After ANY system of equations has been solved in ANY way, I strongly recommend checking (orally, on a draft or calculator). Fortunately, this is done quickly and easily.

1) Substitute the found answer in the first equation:

- the correct equality is obtained.

2) We substitute the found answer in the second equation:

- the correct equality is obtained.

Or, to put it more simply, "everything came together"

The considered method of solution is not the only one; from the first equation it was possible to express , but not .
You can vice versa - express something from the second equation and substitute it into the first equation. By the way, note that the most disadvantageous of the four ways is to express from the second equation:

Fractions are obtained, but why is it? There is a more rational solution.

However, in some cases, fractions are still indispensable. In this regard, I draw your attention to HOW I wrote the expression. Not like this: and by no means like this: .

If in higher mathematics you are dealing with fractional numbers, then try to carry out all calculations in ordinary improper fractions.

Precisely, not or!

The comma can only be used occasionally, in particular if - this is the final answer to some problem, and no further actions need to be performed with this number.

Many readers probably thought “why such a detailed explanation, as for a correction class, and everything is clear”. Nothing of the kind, it seems to be such a simple school example, but how many VERY important conclusions! Here's another one:

Any task should be strived to be completed in the most rational way.. If only because it saves time and nerves, and also reduces the likelihood of making a mistake.

If in a task in higher mathematics you come across a system of two linear equations with two unknowns, then you can always use the substitution method (unless it is indicated that the system needs to be solved by another method) ".
Moreover, in some cases, the substitution method is advisable to use with a larger number of variables.

Example 2

Solve a system of linear equations with three unknowns

A similar system of equations often arises when using the so-called method of indefinite coefficients, when we find the integral of a rational fractional function. The system in question was taken by me from there.

When finding the integral - the goal fast find the values ​​of the coefficients, and not be sophisticated with Cramer's formulas, the inverse matrix method, etc. Therefore, in this case, the substitution method is appropriate.

When any system of equations is given, first of all it is desirable to find out, but is it possible to somehow simplify it IMMEDIATELY? Analyzing the equations of the system, we notice that the second equation of the system can be divided by 2, which we do:

Reference: a mathematical symbol means "from this follows this", it is often used in the course of solving problems.

Now we analyze the equations, we need to express some variable through the rest. Which equation to choose? You probably already guessed that the easiest way for this purpose is to take the first equation of the system:

Here, it doesn't matter which variable to express, one could just as well express or .

Next, we substitute the expression for into the second and third equations of the system:

Open the brackets and add like terms:

We divide the third equation by 2:

From the second equation, we express and substitute into the third equation:

Almost everything is ready, from the third equation we find:
From the second equation:
From the first equation:

Check: Substitute the found values ​​of the variables in the left side of each equation of the system:

1)
2)
3)

The corresponding right-hand sides of the equations are obtained, so the solution is found correctly.

Example 3

Solve a system of linear equations with 4 unknowns

This is an example for self-solving (answer at the end of the lesson).

Solution of the system by term-by-term addition (subtraction) of the equations of the system

In the course of solving systems of linear equations, one should try to use not the “school method”, but the method of term-by-term addition (subtraction) of the equations of the system. Why? This saves time and simplifies calculations, however, now it will become clearer.

Example 4

Solve the system of linear equations:

I took the same system as the first example.
Analyzing the system of equations, we notice that the coefficients of the variable are identical in absolute value and opposite in sign (–1 and 1). In this situation, the equations can be added term by term:

Actions circled in red are performed MENTALLY.
As you can see, as a result of termwise addition, we have lost the variable . This, in fact, is the essence of the method is to get rid of one of the variables.