System of equations. Detailed theory with examples (2019). Systems of rational equations in mathematics

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased.

The concept of solving a system of equations means determining all the roots, that is, the values ​​that, after substituting them into the system, form an equation into an identity. When solving systems of equations, the following methods can be used:

* Substitution method. This method is that to solve the equation it is necessary to express 1 of the variables and substitute the resulting expression in place of this variable in equation 2. Having received an equation with 1 unknown, you can easily solve it and find out the value of the other variable;

* System splitting method. This method consists in factoring one of the equations of the system in such a way that on the right is \, since then each factor is equated to \ and, adding the remaining equations of the original system, we obtain several systems, each of which will be simpler than the original ones;

* Addition and subtraction method. The name itself speaks volumes about the essence of the method. Adding or subtracting 2 system equations, we obtain a new one in order to replace one of the equations of the original system;

* Division and multiplication method. The essence of the method is to divide/multiply the left and right sides of two equations of the system, respectively, to obtain a new equation and replace one of the equations of the original system with it.

Where can I solve systems of rational equations online?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

In this article I will show you seven types of solution algorithms rational equations , which can be reduced to quadratic by changing variables. In most cases, the transformations that lead to replacement are very non-trivial, and it is quite difficult to guess about them on your own.

For each type of equation, I will explain how to make a change of variable in it, and then show a detailed solution in the corresponding video tutorial.

You have the opportunity to continue solving the equations yourself, and then check your solution with the video lesson.

So, let's begin.

1 . (x-1)(x-7)(x-4)(x+2)=40

Note that on the left side of the equation there is a product of four brackets, and on the right side there is a number.

1. Let's group the brackets by two so that the sum of the free terms is the same.

2. Multiply them.

3. Let's introduce a change of variable.

In our equation, we will group the first bracket with the third, and the second with the fourth, since (-1)+(-4)=(-7)+2:

At this point the variable replacement becomes obvious:

We get the equation

Answer:

2 .

An equation of this type is similar to the previous one with one difference: on the right side of the equation is the product of the number and . And it is solved in a completely different way:

1. We group the brackets by two so that the product of the free terms is the same.

2. Multiply each pair of brackets.

3. We take x out of each factor.

4. Divide both sides of the equation by .

5. We introduce a change of variable.

In this equation, we group the first bracket with the fourth, and the second with the third, since:

Note that in each bracket the coefficient at and the free term are the same. Let's take a factor out of each bracket:

Since x=0 is not a root of the original equation, we divide both sides of the equation by . We get:

We get the equation:

Answer:

3 .

Note that the denominators of both fractions are square trinomials, for which the leading coefficient and the free term are the same. Let us take x out of the bracket, as in the equation of the second type. We get:

Divide the numerator and denominator of each fraction by x:

Now we can introduce a variable replacement:

We obtain an equation for the variable t:

4 .

Note that the coefficients of the equation are symmetrical with respect to the central one. This equation is called returnable .

To solve it,

1. Divide both sides of the equation by (We can do this since x=0 is not a root of the equation.) We get:

2. Let’s group the terms in this way:

3. In each group, let’s take the common factor out of brackets:

4. Let's introduce the replacement:

5. Express through t the expression:

From here

We get the equation for t:

Answer:

5. Homogeneous equations.

Equations that have a homogeneous structure can be encountered when solving exponential, logarithmic and trigonometric equations, so you need to be able to recognize it.

Homogeneous equations have the following structure:

In this equality, A, B and C are numbers, and the square and circle denote identical expressions. That is, on the left side of a homogeneous equation there is a sum of monomials having the same degree (in in this case the degree of the monomials is 2), and there is no free term.

To solve a homogeneous equation, divide both sides by

Attention! When dividing the right and left sides of an equation by an expression containing an unknown, you can lose roots. Therefore, it is necessary to check whether the roots of the expression by which we divide both sides of the equation are the roots of the original equation.

Let's go the first way. We get the equation:

Now we introduce variable replacement:

Let's simplify the expression and get bi quadratic equation relative to t:

Answer: or

7 .

This equation has the following structure:

To solve it, you need to select a complete square on the left side of the equation.

To select a full square, you need to add or subtract twice the product. Then we get the square of the sum or difference. This is crucial for successful variable replacement.

Let's start by finding twice the product. This will be the key to replacing the variable. In our equation, twice the product is equal to

Now let's figure out what is more convenient for us to have - the square of the sum or the difference. Let's first consider the sum of expressions:

Great! This expression is exactly equal to twice the product. Then, in order to get the square of the sum in brackets, you need to add and subtract the double product:

The topic offered for study is “ Rational inequalities and their systems. Systems of equations". To more confidently solve systems of rational inequalities and systems of equations, students are encouraged to consider solving systems of equations. The solution to the system is a pair of numbers such that, when substituted, we obtain the correct equalities from the system. The first solution of the systems is carried out by the substitution method, the second - by the graphical method.

Topic: Final review of 9th grade algebra course

Lesson: Rational inequalities and their systems. Systems of equations

Decide system of equationswith two unknowns means finding such a pair of numbers (x 0 ; y 0 ), which turns each equation of the system into a correct numerical equality. Let's consider the substitution method and the graphical method for solving systems of equations.

	Let's find the variable "y" from the second equation and substitute it into the first. Note that the first equation of the system now depends only on the variable x.
x 2 +(x+4) 2 =16 x 2 +x 2 +8x+16=16	Let's solve the first equation. Let's open the brackets. Let us present similar terms. The result is an incomplete quadratic equation. Let’s take “2x” out of brackets and get a product equal to 0. Obviously, either the first factor “2x” is equal to 0, or the second factor “x+4” is equal to 0.
First stage substitution method - From an equation, express one variable in terms of another Substitute the resulting expression for the variable into another equation and solve it
	So, we have found possible values ​​for the variable "x". Let's return to the original system of equations and find the corresponding values ​​for the second variable “y”.
	Let's substitute the found values ​​of “x”.
	So, by means of equivalent transformations of the original system of equations, we obtained two pairs of numbers. This is the answer.
Answer: (0;4) or (-4;0)	The answer can be written as two sets of two numbers. It is important that in each pair the value of the variable “x” is in first place, and the value of the variable “y” is in second place.
Second phase substitution method - Substitute the found value of the variable and calculate the value of the second variable Write down answer

Let's remember.

To solve a system of equations graphically

Let's plot the first equation

Let's plot the second equation

Let's find the coordinates of the intersection points of the graphs

Chapter 4. Systems of rational equations

The fourth chapter is devoted to the study of ways to solve systems of rational equations. It uses concepts learned in 7th grade and previously applied to systems linear equations, which makes it possible to repeat what has been learned and learn to act in a new situation. These are concepts: solutions to an equation with two (three) unknowns, systems of equations with two (three) unknowns, the concept of equivalence of equations, systems of equations.

The purpose of studying Chapter 4: to master the listed concepts, learn to solve systems of rational equations and apply them to solving word problems.

§ 9. Systems of rational equations

The main purpose of the ninth paragraph is that, based on known concepts related to equations and systems of linear equations, will learn to solve systems of rational equations, learn to apply them to solving word problems.

9.1. The concept of a system of rational equations

This paragraph introduces the concepts of a rational equation with two (three) unknowns and its solution, defines what it means to solve a system of equations, and provides statements about the equivalence of systems of equations.

The main tasks of this paragraph are tasks to establish that a given pair (three) of numbers is a solution to the system. An additional task accustoms students to solving problems with parameters.

Revision task. 805–807.

Solutions and comments

500. Is the solution to the system of equations a pair of numbers:

a) (0; 3); b) (–3; 2).

Solution. a) Since 0 + 5 3, then the pair of numbers (0; 3) is not a solution to the second equation of the system, and therefore is not a solution to the system of equations.

b) Since –3 + 5 = 2, (–3) 2 + (–3)2 – 3 = 0, then the pair of numbers (–3; 2) is a solution to the system of equations.

501. Is the solution to the system of equations
triple of numbers:

a) (1; –1; 1); b) (1; 1; 1).

Solution. a) Since 1 – 1 + 1 3, then the triple of numbers (1; –1; 1) is not a solution to the first equation of the system, and therefore is not a solution to the system of equations.

b) Since 1 + 1 + 1 = 3, 1 –1 – 1 –2, then the triple of numbers (1; 1; 1) is not a solution to the second equation of the system, and therefore is not a solution to the system of equations.

Additional task

1. At what value a a pair of numbers (2; –1) is a solution to the system of equations

Solution. Let a- a certain number for which a pair of numbers (2; –1) is a solution to a system of equations, then two numerical equalities are true:

1) 2a 2 + a= 21 and 2) 10 + a = a 2 + 4,

which can be considered as equations for a. Equation 2) has two roots: a 1 = 3 or a 2 = –2. Number a 1 is the root of equation 1), and the number a 2 = –2 - no, therefore, when a= 3 pair of numbers (2; –1) is a solution to a system of equations. And other meanings A, satisfying the conditions of the problem, there are no.

9.2. Method for substituting solutions to systems of rational equations

At this point on three examples it is shown how it can be solved by substituting rational equations in which there is at least one equation of the first.

Revision task. When studying this item, you can use the task 810.

Solutions and comments

512. Solve the system of equations:

G)
d)

Solution. d) Expressing x through y from the second equation of the system and substituting y+ 1 instead x

(1)

Now, having solved the first equation of system (1), we find its two roots y 1 = –4 and y 2 = 3. From the second equation of system (1) we obtain the corresponding values x: x 1 = –3 and x 2 = 4.

d) Expressing y through x from the second equation of the system and substituting 3 – 3 x instead of y into the first equation, we rewrite the system in the form:

(2)

Now, having solved the first equation of system (2), we find its two roots x 1 = and
x 2 = . From the second equation of system (2) we obtain the corresponding values y: y 1 = – and y 2 = 2.

Answer. d) (–3; –4), (4; 3); D 2).

Intermediate control. S-21.

9.3. Other ways to solve systems of rational equations

In this paragraph, examples of solving systems of rational equations are analyzed - by the method of adding equations, by the method of introducing new unknowns, by the method of isolating perfect squares, by the method of factorization. In this case, equivalent transformations of equations are used. Sometimes solving a system is helped by knowing that the sum of the squares of two numbers is zero if and only if those numbers are zero.

Revision task. When studying this item, you can use the task 820.

Solutions and comments

517. Solve the system of equations:

V)
d)

Solution. c) Let us replace the first equation in the system with the sum of two equations of this system. We obtain a system equivalent to the original system:

(1)

Now let’s select the perfect squares in the first equation of system (1):

(2)

Since the sum of the squares of two numbers is equal to zero if and only if these numbers are zero, then the first equation of system (2) has a unique solution (2; –6). This pair of numbers is a solution to the second equation of system (2), therefore, it is a solution to system (2) and the original system equivalent to it.

e) Let's make a change of unknowns: a= and b= . Let's rewrite the system in the form:

(3)

System (3) has a unique solution: a 1 = 1, b 1 = . Consequently, system e) also has a unique solution: x 1 = 1, y 1 = 2.

Answer. c) (2; –6); e) (1; 2).

512. g) Solve the system of equations

Solution. Usually the solution to such a system is written by replacing this system systems equivalent to it:

(4)

Equivalence signs () are set for the teacher, but in a class with in-depth study of mathematics it can be used.

The solutions to the second equation of the last of systems (4) are the following pairs of numbers ( x; y), which are solutions to at least one of the equations:

1) x + y= 1 and 2) x + y = –1.

Therefore, all solutions of the original system are the union of all solutions of two systems:

3)
and 4)

Having solved systems 3) and 4) we obtain all solutions of the original system: (–1; 2), (2; –1), (1; –2), (–2; 1).

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

518. Solve the system of equations:

A)
V)
and)

Solution. a) By introducing a new unknown a = x 2 – 4y
. It has a single root a= 1. This means that this system is equivalent to the system

(5)

Adding the equations of system (5) and replacing the first equation of the system with the resulting equation, we obtain new system, equivalent to system (5), and therefore to the original system:

(6)

Having isolated the complete squares in the first equation of system (6), we rewrite system (6) in the form:

(7)

Now it is obvious that the first equation of system (7) has a unique solution: x 1 = 3, y 1 = 2. Checking shows that this pair of numbers is a solution to the second equation of system (7), which means it is a solution to system (7) and the original system equivalent to it.

So, the original system has a unique solution (3; 2).

c) By introducing a new unknown a =
, we rewrite the first equation of the system in the form:
. It has two roots: a 1 = 1 and a 2 = –4. Therefore, all solutions of the original system are the union of all solutions of two systems:

1)
and 2)

Using substitution y = 9 – x, we solve each of the systems and find that system 1) has a unique solution (6; 3), and system 2) has a unique solution (14; –5).

So, the original system has two solutions: (6; 3), (14; –5).

g) Let us rewrite the system in the form:

(8)

If a pair of numbers ( x 0 ; y 0) is a solution to system (8), then the following numerical equalities are true: x 0 (9x 0 + 4y 0) = 1 and y 0 (9x 0 + 4y 0) = –2. Note that both sides of these numerical equalities are not zero, therefore, dividing the first equality by the second termwise, we obtain a new numerical equality:
. Whence it follows that y 0 = –2x 0 . That is, the sought solutions of system (8) are solutions of the system

(9)

Having solved system (9), we obtain two of its solutions: (1; –2), (–1; 2).

By checking we are convinced that both of these pairs of numbers are indeed solutions to the original system.

Answer. a) (3; 2); c) (6; 3), (14; –5); g) (1; –2), (–1; 2).

Comment. Note that in the process of solving problem g) we did not prove the equivalence of system (9) to the original system, but from the above reasoning it follows that any solution to the original system is a solution to system (9) (i.e., system (9) is a consequence of the original system ), therefore it is necessary to check whether each solution to system (9) is a solution to the original system. And this check is a mandatory part of the system's solution.

In fact, system (9) is equivalent to the original system, as follows from the statement proved below.

Additional tasks

1. Solve the system of equations

A)
b)

V)
G)

Solution. a) Having isolated the perfect squares in the first equation, we rewrite it in the form:

(x – 3) 2 + (y – 1) 2 = 0. (1)

Now it is obvious that the first equation of the system has a unique solution: x 1 = 3, y 1 = 1. By checking we are convinced that this pair is a solution to the second equation, and therefore a solution to the system of equations.

b) Arguing similarly, we obtain a unique solution to the system (–2, 0.5).

c) Let us factorize the left side of the first equation of the system:

x 2 – 7xy + 12y 2 = x 2 – 3xy – 4xy + 12y 2 = x(x – 3y) – 4y(x– 3y) = (x – 3y)(x – 4y).

Let us rewrite this system in the form

(2)

Now it is obvious that all solutions of system (2) are the union of all solutions of two systems:

1)
and 2)

System 1) has two solutions: (3; 1), (–3; –1). System 2) also has two solutions: (12; 3), (–12; –3). Consequently, the original system has four solutions: (3; 1), (–3; –1), (12; 3), (–12; –3).

d) Let us rewrite the original system in the form:

(3)

Obviously, the first equation of system (3) has a unique solution:
(3; –2). Checking shows that it is also a solution to the second equation of system (3), therefore, system (3), and therefore the original system, have a unique solution (3; –2).

Answer. a) (3; 1); b) (–2, 0.5); c) (3; 1), (–3; –1), (12; 3), (–12; –3); d) (3; –2).

2. Prove the statement: if f (x, y) And g (x, y) - polynomials with respect to x And y, a And b- numbers, b 0, then systems 1 are equivalent)
and 2)

Proof. 1. Let a pair of numbers ( x 0 ; y 0) is a solution to system 1), then the following numerical equalities are true: f(x 0 , y 0) = a And g(x 0 , y 0) = b. Because b 0, then g(x 0 , y 0) 0, so the numerical equality is true:
. This means that any solution to system 1) is a solution to system 2).

2. Let now a pair of numbers ( x 0 ; y 0) is the solution to system 2), then the numerical equalities are true: and g(x 0 , y 0) = b. Because b 0, then g(x 0 , y 0) 0, therefore multiplying both sides of the first numerical equality by equal non-zero numbers g(x 0 , y 0) and b, we get a new correct numerical equality: f(x 0 , y 0) = a. This means that any solution to system 2) is a solution to system 1).

3. Suppose that system 1) has no solution, and system 2) has a solution. Then from point 2 of the proof above, it follows that system 1) has a solution. The resulting contradiction shows that the assumption made is incorrect. This means that if system 1) does not have a solution, then system 2) does not have a solution.

It is similarly proven that if system 2) does not have a solution, then system 1) does not have a solution.

From the above proof it follows that systems 1) and 2) are equivalent, which is what needed to be proven.

Let's give an example of solving the system 518, and with this statement.

Having solved the last system, we obtain two of its solutions: (1; –2), (–1; 2), therefore, the original system has two solutions: (1; –2), (–1; 2).

3. Solve the system of equations:

A)
b) c)

Solution. a) The original system is equivalent to the system

which we rewrite as:

(4)

System (4) has a unique solution (1; 2). Consequently, the original system also has a unique solution (1; 2).

b) We rewrite the original system in the form

This system is equivalent to the system:

(5)

System (5) has a unique solution (–1; –5). Consequently, the original system also has a unique solution (–1; –5).

c) The original system is equivalent to the system

or system

(6)

System (6) has two solutions (1; 2; –2), (–1; –2; 2). Consequently, the original system also has two solutions (1; 2; –2), (–1; –2; –2).

Answer. a) (1; 2); b) (–1; –5); c) two solutions (1; 2; –2), (–1; –2; –2).

Intermediate control. S-22, S-23, S–24*.

9.4. Solving problems using systems of rational equations

In this paragraph, solutions to word problems leading to systems of rational equations are analyzed. You can start explaining new material with simpler tasks 513, 514, 519, 520 .

Revision task. When studying this item, you can use the task 820, 952.

Solutions and comments

513. a) Divide the number 171 into two factors, the sum of which would be equal to 28.

Solution. Let x- first factor, y - second multiplier. Let's create a system of equations:

Having solved the system, we obtain two solutions: x 1 = 9, y 1 = 19 and x 2 = 19, y 2 = 9. The order of the factors is not important here, so the required factors are 9 and 19.

Answer. 9 and 19.

519. a) If you add twice the second number to the square of the first number, you get (–7), and if you subtract the second number from the first number, you get 11. Find these numbers.

Solution. Let x- first number, y- second number. Based on the conditions of the problem, let’s create two equations: x 2 + 2y= –7 and x – y= 11. Having solved the system of these equations, we obtain two of its solutions: (–5; –16), (3; –8).x = 6 and y= 4, that is, the required number is 64.

Answer. 64.

522. b) Two workers, working together, completed all the work in 5 days. If the first worker worked twice as fast, and the second worker worked twice as slow, then they would complete all the work in 4 days. In how many days would the first worker complete this work?

Solution. Iway. Let for x And y days, the first and second workers will complete all the work, respectively. If they work together, they will complete the job in 5 days. Let's make the first equation:
.

If the first one worked 2 times faster, and the second one 2 times slower, then per day they would complete of all the work, respectively, and all the work would be completed in 4 days. Let's create the second equation:

.

952. If you sell 20 cows, then the harvested hay will last ten days longer, but if you buy 30 cows, then the supply of hay will be exhausted ten days earlier. How many cows were there and for how many days was the hay stored?

Solution. Let for x cows have prepared hay for y days. Let us briefly write down the condition of the problem:

number of cows number of days

Since with a constant supply of hay the number of days is inversely proportional to the number of cows, we will compose the first equation:
.

Let's create the second equation in the same way:
.

The system of these equations has a unique solution: x = 120, y= 50. That is, for 120 cows, hay was stored for 50 days.

Answer. For 120 cows, for 50 days.

Mathematic teacher

Municipal educational institution "Gymnasium No. 5 of Belgorod"

Lesson topic: Rational equations.

Grade: 10th grade.

UMK : Algebra and the beginnings of analysis: textbook. For 10kl. general education institutions/[S.M. Nikolsky, M.K. Potapov].-5th ed., additional-M.: Education, 2006.-432 p. pp.65-74., 45-47.

Lesson objectives:

Educational: systematize and summarize information about rational expressions known from basic school; show ways to solve rational equations;

Educational: expand and deepen learning various types rational equations using a variety of methods.

Educational: show the significance of the topic being studied in the mathematics section.

Lesson type: lesson-lecture.

Lesson structure:

1. Setting the lesson goal (1 min).
2. Preparing to study new material (2 min).
3. 3.Introduction to new material (38 min).
4. 4. Lesson summary (2 min)
5. 5.Homework (2 min)

Lesson equipment: interactive board, projector, computer.

During the classes:

Plan.

1. Rational expressions.

2. Rational equations.

3. Systems of rational equations.

I. Repetition.

Algebra originated from solving practical problems using equations. The goals of algebra remained unchanged for thousands of years - equations were solved: first linear, then quadratic, and then equations of even higher degrees. But the form in which algebraic results were presented changed beyond recognition.

An equation is the most common form of mathematical problem. The study of equations is the main content of the school algebra course. To solve equations, you need to be able to perform operations on monomials, polynomials, algebraic fractions, be able to factorize, open parentheses, etc. You need to put your knowledge in order. We will begin the review with the concept of “rational expressions”. Student's report about rational expressions known from basic school. Thus, the study of equations is impossible without the study of the laws of action.

II. Main part.

The main thing in the concept of an equation is the formulation of the question of its solution. An equation whose left and right sides are rational expressions for x is called a rational equation with unknown x.

For example, the equations 5x 6 - 9x 5 + 4x - 3x + 1 = 0, are rational.

The root (or solution) of an equation with an unknown x is a number that, when substituted into the equation instead of x, produces a true numerical equality.

Solving an equation means finding all its roots or showing that there are none. When solving rational equations, you have to multiply and divide both sides of the equation by a non-zero number, transfer terms of the equation from one part to another, apply the rules of addition and subtraction algebraic fractions. The result will be an equation equivalent to the previous one, that is, an equation that has the same roots, and only them.

Let us list the standard equations that we have studied. Student answers (linear equation, quadratic equation, simplest power equation X n =a). Converting equations to one of the standard ones is the main step in solving an equation. It is impossible to completely algorithmize the conversion process, but it is useful to remember some techniques common to all types of equations.

1).An equation of the form A(x) B(x) = O, where A(x) and B(x) are polynomials with respect to x, is calleddecaying equation.

The set of all roots of a decaying equation is the union of the sets of all roots of two equations A(x)=0 and B(x)=0. The factorization method is applied to equations of the form A(x) = 0. The essence of this method: you need to solve the equation A(x)=0, where A(x)=A 1 (x)A 2 (x)A 3 (X). The equation A(x) = 0 is replaced by a set of simple equations: A 1 (x)=0.A 2 (x)=0.A 3 (x)=0. Find the roots of the equations of this set and do a check. The factorization method is used mainly for rational and trigonometric equations.

EXAMPLE 1.

Let's solve the equation (x 2 - 5x + 6) (x 2 + x - 2) = 0.

The equation breaks down into two equations.

x 2 - 5x + 6 = 0 x 1 = 2 and x 2 = 3

x 2 + x - 2 = 0. x 3 = -2 and x 4 = 1

This means that the original equation has roots x 1 = 2, x 2 = 3, x 3 = -2, x 4 =1.

Answer. -2; 1; 2; 3.

EXAMPLE. Let's solve the equation x 3 -7x+6=0.

x 3 -x-6x+6=0

x(x 2 -1)-6(x-1)=0

x(x-1)(x+1)-6(x-1)=0

(x-1)(x(x+1)-6)=0

(x-1)(x 2 +x-6)=0

x-1=0, x 1 =1; x 2 + x-6 = 0, x 2 = 2, x 3 = -3.

Answer:1;2;-3.

2).Equation of the form, where A(x) and B(x) are polynomials relative to x.

EXAMPLE 2.

Let's solve the equation

First let's solve the equation

x 2 + 4x - 21 = 0. x 1 = 3 and x 2 = -7

Substituting these numbers into the denominator of the left side of the original equation, we get

x 1 2 - x 1 -6 = 9-3-6 = 0,

x 2 2 - x 2 - 6 = 49 + 7 - 6 = 50 ≠0.

This shows that the number x 1 = 3 is not the root of the original equation, but the number x 2 =- 7 is the root of this equation.

Answer. -7.

3).Equation of the form

where A(x), B(x), C(x) and D(x) are polynomials with respect to x, usually solved according to the following rule.

The equation A(x) D(x) - C(x)·B(x) = 0 is solved and those that do not make the denominator of the equation vanish are selected from its roots.

EXAMPLE 3.

Let's solve the equation

Let's solve the equation

x 2 - 5x + 6 - (2x + 3) (x - 3) = 0.

x 2 + 2x - 15 = 0

x 1 = -5 and x 2 = 3.

Number x 1 does not make the denominator x - 3 vanish, but the number x 2 converts. Therefore, the equation has a single root = -5.

Answer. -5.

Finding the roots of a rational equation often helps by replacing the unknown. The ability to successfully introduce a new variable is an important element of mathematical culture. The successful choice of a new variable makes the structure of the equation more transparent.

EXAMPLE 4.

Let's solve the equation x 8 + 4x 6 -10x 4 + 4x 2 + 1 = 0.

Number x 0 = 0 is not a root of the equation, so the equation is equivalent to the equation

x 4 + 4x 2 - 10 + + =0

Let us denote t =, then x 4 + =t 2 -2,

we get t 2 + 4t - 12 = 0, x 1 = 2 and x 2 = -6.

Therefore, we find the roots of the equation by combining all the roots of the two equations:=2, and =-6,

The first equation has two roots -1 and 1, and the second equation has no real roots, so the equation has only two roots: -1 and 1. Answer. -1; 1.

4). Symmetric equations.

A polynomial in several variables is called a symmetric polynomial if its form does not change with any permutation of these variables.

For example, polynomials x + y, a 2 + b 2 - 1, zt and 5a 3 + 6ab + 5b 3 - symmetric polynomials in two variables, a polynomials x + y + z, a 3 + b 3 + c 3 , - symmetric polynomials in three variables.

At the same time, the polynomials x - y, a 2 –b 2 and a 3 + ab – b 3 - non-symmetric polynomials.

Equation ax 4 +bx 3 +cx 2 +bx+a=0, where a R/ ,b R, c R is called a symmetric fourth degree equation. To solve this equation you need:

1).Divide both sides of the equation by x 2 and group the resulting expressions:.

2).Introduction of a variablethe equation is reduced to quadratic.

Example.

Solve equation x 4 +5x 3 +4x 2 -5x+1=0.

The number 0 is not the root of the equation. Divide both sides of the equation by x 2 ≠0.

Answer. .

Systems of rational equations.

Systems of equations appear when solving problems in which several quantities are unknown. These quantities are related by a certain relationship, which are written in the form of equations.

An equation whose left and right sides are rational expressions for x and y is called a rational equation with two unknowns x and y.

If we need to find all pairs of numbers x and y, each of which is a solution to each of the given equations with two unknowns x and y, then we say that we need to solve a system of equations with two unknowns x and y, and each such pair is called a solution to this system.

Unknowns can also be denoted by other letters. A system of equations in which the number of unknowns is greater than two is determined in a similar way.

If each solution to the first system of equations is a solution to the second system, and each solution to the second system of equations is a solution to the first system, then such systems are called equivalent. In particular, two systems that have no solutions are considered equivalent.

For example, systems are equivalent

1). Substitution method.

EXAMPLE 1. Let's solve the system of equations

Expressing y through x from the first equation of the system, we obtain the equation:

y = 3x - 1.

Solving the 5x equation 2 -4(3x-1)+3(3x-1) 2 =9, find its roots x 1 = 1 and x 2 = . Substituting the found numbers x 1 and x 2 into the equation y = 3x - 1, we get y 1 = 2

and y = Consequently, the system has two solutions: (1; 2) and (; )

Answer. (12), (; )

2). Algebraic addition method.

EXAMPLE 2. Let's solve the system of equations

Leaving the first equation of the system unchanged and adding the first equation with the second, we obtain a system equivalent to the system.

All solutions of the system are the union of all solutions of two systems:

(2; 1), (-2; -1),

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

3). Method of introducing new unknowns.

EXAMPLE 3. Let's solve the system of equations

Denoting u = xy, v = x - y, we rewrite the system in the form

Let's find its solutions: u 1 = 1, v 1 = 0 and u 2 = 5, v 2 = 4. Consequently, all solutions of the system are the union of all solutions of two systems:

Having solved each of these systems using the substitution method, we find its solutions to the system: (1; 1), (-1; -1), (5; 1), (-1; -5).

Answer. (1; 1), (-1; -1), (5; 1), (-1; -5).

4). Equation of the form ah 2 + bxy + su 2 = 0, where a, b, c are given non-zero numbers, is called a homogeneous equation with respect to the unknowns x and y.

Consider a system of equations in which there is a homogeneous equation.

EXAMPLE 4. Let's solve the system of equations

Designating t = , we rewrite the first equation of the system in the form t 2 +4t+3=0.

The equation has two roots t 1 = -1 and t 2 = -3, therefore all solutions of the system are the union of all solutions of two systems:

Having solved each of these systems, we find all solutions of the system:

(2,5; -2,5), (0,5; -0,5), ,(1,5;-0,5).

Answer. (2.5; -2.5), (0.5; -0.5),,(1,5;-0,5).

When solving some systems, knowledge of the properties of symmetric polynomials helps.

Example.

Let us introduce new unknowns α = x + y and β = xy, then x 4 +у 4 = α 4 -4 α 2 β+2 β 2

Therefore, the system can be rewritten in the form

Let's solve the quadratic equation for β: β 1 =6, β 2 =44.

Therefore, all solutions of the system are the union

all solutions of two systems:

The first system has two solutions x 1 = 2, y 1 = 3 and x 2 = 3, y 2 =2, and the second system has no real solutions. Therefore, the system has two solutions: (x: 1 ; y 1) and (x 2;y 2)

Answer. (2; 3), (3; 2).

Today we summed up the results of studying the topic of rational equations. We talked about general ideas, general methods, on which the entire school line of equations is based.

Methods for solving equations have been identified:

1) factorization method;

2) method of introducing new variables.

We expanded our understanding of methods for solving systems of equations.

In the next 4 lessons we will conduct practical exercises. To do this you need to learn theoretical material, and select 2 examples from the textbook on the considered methods for solving equations and systems of equations, in lesson 6 a seminar will be held on this topic, for this you need to prepare questions: Newton’s binomial formula, solving symmetric equations of degree 3.5. The final lesson on this topic is a test.

Literature.

1. Algebra and the beginnings of analysis: textbook. For 10kl. general education institutions/[S.M. Nikolsky, M.K. Potapov].-5th ed., additional-M.: Education, 2006.-432 p. pp.65-74., 45-47.
2. Mathematics: training thematic tasks of increased complexity with answers for preparation for the Unified State Exam and other forms of final and entrance exams/comp. G.I. Kovaleva, T.I. Buzulina - Volgograd: Teacher, 2009.-494 p. – pp. 62-72,194-199.
3. Titarenko A.M. Mathematics: grades 9-11: 6000 problems and examples/A.M. Titarenko.-M.: Eksmo, 2007.-336 p.

There is a lot to be said about equations. There are questions in this area of ​​mathematics that mathematicians have not yet answered. Perhaps some of you will find answers to these questions.

Albert Einstein said: “I have to divide my time between politics and equations. However, the equations, in my opinion, are much more important. Politics exists only for at this moment. And the equations will exist forever.”

Lessons 2-5 are devoted to practical exercises. The main type of activity in these lessons is the independent work of students to consolidate and deepen the theoretical material presented in the lecture. At each of them, theory questions are repeated and students are surveyed. Based independent work in the classroom and at home, repetition and assimilation of theoretical questions are ensured, targeted work is carried out to develop skills in solving problems of various levels of complexity, and students are surveyed. Goal: to consolidate and deepen the theoretical material presented in the lecture, learn to apply it in practice, master algorithms for solving typical examples and problems, and ensure that all students understand the main content of the section being studied at the level of program requirements.

The 6th and 7th lessons are allocated for the seminar, and it is advisable to conduct a seminar in the 6th lesson, and a test in the 7th lesson.

Lesson-seminar plan.

Goal: repetition, deepening and generalization of the material covered, practicing basic methods, methods and techniques for solving mathematical problems, acquiring new knowledge, learning to independently apply knowledge in non-standard situations.

1. At the beginning of the lesson, program control is organized. Purpose of the event work-check development of skills and abilities to perform simple exercises. In the process of frontal questioning of students who incorrectly indicated the answer number, the teacher finds out which of the tasks caused difficulty. Next, oral or written work is carried out to eliminate errors. No more than 10 minutes are allotted for carrying out programmed control.

2. Differentiated survey of several students on theory issues.

3. Historical reference about the emergence and development of the concept of an equation (student message). Newton's binomial formula. Solving symmetric equations of the third degree, fourth degree, fifth degree.

x 4 -2x 3 -x 2 -2x+1=0

2x 4 +x 3 -11x 2 +x+2=0

x 5 -x 4 -3x 3 -3x 2 -x+1=0

2x 5 +3x 4 -5x 3 -5x 2 +3x+2=0

4. Solving examples, checking students’ readiness to perform test work– this is one of the main objectives of the seminar.

Carrying out the test.

Carrying out a test does not mean abandoning the ongoing monitoring of students’ knowledge. Grades are given in practical and seminar classes. Some typical exercises will be tested. Students are informed in advance what theoretical material and exercises will be presented during the test. Let us present the contents of one of the cards for testing on the topic under consideration.

Level 1.

Solve the equations: (x+3) 4 +(x 2 +x-6) 2 =2(x-2) 4

X 2 +25 =24

(2x 2 -3x+1)(2x 2 -5x+1)=8x 2

Level 2.

Solve the equations: x 4 +8x 3 +8x 2 -32x-9=0

8x 3 -12x 2 +x-7=0

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