Symmetric systems of equations how to solve. Solution of symmetric systems of equations

− 4 1+ 4

−6

27≡ 0,

−4 x +4 y +27

+(y +6 )

x = 1, x

(x − 1 )

= −6.

y = −6

Note that the solution of the second equation is not yet the solution of the system. The resulting numbers must be substituted into the remaining first equation of the system. In this case, after substitution, we obtain an identity.

Answer: (1, - 6).♦

§5. Homogeneous equations and systems
Function f (x ,y )	called	homogeneous		k if
f (tx, ty) = tk f(x, y) .	For example, the function f (x ,y ) = 4x 3 y − 5xy 3 + x 2 y 2
is homogeneous of degree 4, since		f(tx, ty) = 4	(tx )3 (ty )− 5 (tx )(ty )3 +
+ (tx ) 2 (ty ) 2 = t 4 (4x 3 y − 5xy 3 + x 2 y 2 ) . Equation f (x, y) = 0, where				f (x, y) -

homogeneous function is called homogeneous. It reduces to the equation

with one unknown, if we introduce a new variable t = x y .

f (x, y) = a,

System with two variables g (x, y) \u003d b, where f (x, y), g (x, y) -

homogeneous functions of the same degree is called homogeneous. If ab ≠ 0, multiply the first equation by b, the second by a and you-

we compare one from the other - we get an equivalent system

bf (x, y) − ag(x, y) = 0, g(x, y) = b.

The first equation by change of variables t =

(or t =

) reduces to

equation with one unknown.

If a = 0

(b = 0) , then the equation f (x ,y ) = 0(g (x ,y ) = 0) by replacing

variables t =

(or t =

) reduces to an equation with one unknown

−xy+y

21 ,

Example 20. (Moscow State University, 2001, chemistry department) Solve the system

− 2xy + 15= 0.

2012-2013 academic year year, No. 1, 11 cells. Mathematics. Algebraic equations, inequalities, systems

− xy + y 2 =21,

− xy +y 2

y2 − 2xy

-2xy = -15

2xy = − 15

x ≠ 0, y ≠ 0;

19 ± 11

5x2 - 19xy + 12y2 = 0 5

− 19

12 = 0

-2xy = -15

x=3y,

y = ±5.

3 ) ,

(− 3 3;−

3 ) ,(4; 5) ,

(− 4;− 5) .♦

§6. Symmetric systems

f(x, y)

called

symmetrical,

f (x, y) = f(y, x) .

f(x, y) = a

System of equations of the form

where f (x ,y ) ,g (x ,y ) – symmet-

g (x, y) = b,

ric, is called a symmetric system. Such systems

more often	only through the introduction of new	variables
x + y= u, xy

x 3+ x 3y 3+ y 3= 17,

Example 21. Solve System of Equations

x + xy+ y= 5 .

♦ This is an algebraic (symmetric) system, usually solved by changing x + y = u ,xy = v . Noticing that

x 3+ x 3y 3+ y 3= (x + y ) (x 2− xy + y 2) + x 3y 3=

= (x+ y) ((x+ y) 2 − 3 xy) + x3 y3 = u(u2 − 3 v) + v3 ,

rewrite the system in the form

2012-2013 academic year year, No. 1, 11 cells. Mathematics. Algebraic equations, inequalities, systems

− 3uv+v

u = 5 − v,

6 =0

V=5

−5v

v=3, u=2

(in old variables)

x+y=2,

x=2-y ,

xy = 3,

y 2 − 2y + 3= 0

x+y=3,

x = 3 − y,

x=2,y=1,

y −3 y +2 =0

x=1,y=2.

xy = 2,

Answer: (2;1) ,

(1; 2) .♦

Literature

1. S. I. Kolesnikova "Intensive course of preparation for the Unified State Exam." Moscow, Iris - Press;

2. "Solving complex problems of the Unified State Examination" Moscow, Iris - Press or "Wako", 2011;

3. Magazine "Potential" №№1-2 for 2005 - articles by S. I. Kolesnikova "Irrational equations" and "Irrational inequalities";

4. S. I. Kolesnikov "Irrational Equations", Moscow, 2010,

OOO "Azbuka";

5. S. I. Kolesnikova “Irrational inequalities”, Moscow, 2010, Azbuka LLC;

6. S. I. Kolesnikova “Equations and inequalities containing modules”, Moscow, 2010, Azbuka LLC.

test questions

1(2). Find the smallest length of the interval that contains all solutions of the inequality 5x + 1≥ 2(x − 1) .

2(2). Solve the inequality x 3 + 8x 2 − 20x ≤ 2x − 4 (no need to solve the cubic equation, since there is a factor x − 2 on the right and left).

3(2). Solve the inequality 2− x ≥ x − 3.

4(2). Find the smallest length of the gap that belongs to

harvest all solutions of the inequality	x2 + 5 x− 84	≤ 0 .
	(x + 13 )(x + 14 )

5(3). Find the sum of squares of integer solutions of the inequality

2012-2013 academic year year, No. 1, 11 cells. Mathematics. Algebraic equations, inequalities, systems

4 −x −8 +x ≤x +6 .

6(3). Solve the inequality 5+ x − 8− x ≤ 3− x .

7(3). Solve the inequality	-x3 -x -1		≤x.
				9 − 4x − (x + 3) )

8(3). Solve the inequality	4 −x −(x +2 ) )(				≤ 0.
		(x + 1 )(x − 2 )(x − 3 )

9(4). Find the smallest length of the gap that belongs to

harvest all solutions of the inequality
	x+5	x+2		144-x< 0.


	X-2	4 x −5	6x − 6
	X-2	4 x −5	6x − 6

10(2). Find the smallest length of the interval that contains all solutions of the inequality 8x − 8≤ 32+ 4x − x 2 .

11(4). Find the sum of squares of all integer solutions of the non-

2(2). Find the shortest interval that contains
	(x − 1 )3 (x + 3 )
all solutions of the inequality				≤ 0 .
	2x − 1	x − 2	) (x − 1 )

3(2). Solve the inequality

4 (x− 3 ) 4 ≥ 4 (x− 7 ,5 ) 4 .

4(4). Solve the inequality

x2 + 3 x− 4

x2−16

2x 2 + 3x − 20

5(3). Solve the inequality(x 2

X +1 ) 2 −2 x 3 +x 2 +x −3 x 2

≥ 0 .

4 − 2x − 1≤ 3.

Tasks

- 5x + 6+ 9 - 2x - 5

2012-2013 academic year year, No. 1, 11 cells. Mathematics. Algebraic equations, inequalities, systems

7(4). Find all parameter values

a , for each of which

function f (x) \u003d x 2 + 4x +

x2−

x − 1

− a accepts only

non-negative

solid values.

8(4). Solve the equation 4 x − 3

x − 1

5x + 14− 3

5x + 14 - 1

9(4). Solve the Equation

x 2− 5 +

x 2 −3 \u003d x +1 +

x + 3 .

24 - x2

9 2 x

10(3). Solve the inequality

≥ 0 .

x2 − 4 7 x − 10

11(3). Three riders start at the same time from the same point on the circuit and drive at constant speeds in the same direction. The first racer caught up with the second for the first time, making his fifth lap, at a point diametrically opposite to the start, and half an hour after that he caught up with the third racer for the second time, not counting the start moment. The second rider caught up with the third for the first time 3 hours after the start. How many laps per hour does the first rider make if the second completes the lap in at least twenty minutes?

Introduction The problem of my project is that the ability to solve various systems of equations is required for the successful passing of the exam, and in the course of high school they are not given enough time to get to know this issue more deeply. The purpose of the work: to prepare for the successful delivery of the exam. Tasks of the work: Expand your knowledge in the field of mathematics related to the concept of "symmetry". Improve your mathematical culture, using the concept of "symmetry" when solving systems of equations, called symmetric, as well as other problems of mathematics.

The concept of symmetry. Symmetry - (ancient Greek συμμετρία), in a broad sense - immutability under any transformations. So, for example, the spherical symmetry of a body means that the appearance of the body will not change if it is rotated in space at arbitrary angles. Bilateral symmetry means that right and left look the same with respect to some plane.

Problem solving using symmetry. Problem 1 Two people take turns putting identical coins on a round table, and the coins should not cover each other. The one who cannot make a move loses. Who wins when played correctly? (In other words, which player has a winning strategy?)

Methods for solving symmetric systems. Symmetric systems can be solved by the change of variables, which are the main symmetric polynomials. A symmetric system of two equations with two unknowns x and y is solved by substituting u = x + y, v = xy.

Example No. 2 3 x 2y - 2xy + 3xy 2 \u003d 78, 2x - 3xy + 2y + 8 \u003d 0 Using the basic symmetric polynomials, the system can be written in the following form 3uv - 2v \u003d 78, 2u - 3v \u003d -8. Expressing u = from the second equation and substituting it into the first equation, we obtain 9v2– 28v – 156 = 0. The roots of this equation v 1 = 6 and v 2 = - allow us to find the corresponding values u1 = 5, u2= - from the expression u = .

Let us now solve the following set of systems Let us now solve the following set of systems x + y = 5, and x + y = - , xy = 6 xy = - . x \u003d 5 - y, and y \u003d -x -, xy \u003d 6 xy \u003d -. x \u003d 5 - y, and y \u003d -x -, y (5 - y) \u003d 6 x (-x -) \u003d -. x \u003d 5 - y, and y \u003d -x -, y 1 \u003d 3, y 2 \u003d 2 x 1 \u003d, x 2 \u003d - x 1 \u003d 2, x 2 \u003d 3, and x 1 \u003d, x 2 \u003d - y 1= 3, y 2 =2 y 1 = -, y 2= Answer: (2; 3), (3; 2), (; -), (- ;).

Theorems used in solving symmetric systems. Theorem 1. (on symmetric polynomials) Any symmetric polynomial in two variables can be represented as a function of two basic symmetric polynomials In other words, for any symmetric polynomial f (x, y) there exists a function of two variables φ (u, v) such that

Theorem 2. (on symmetric polynomials) Theorem 2. (on symmetric polynomials) Any symmetric polynomial in three variables can be represented as a function of three basic symmetric polynomials: In other words, for any symmetric polynomial f (x, y) there is such a function of three variables θ (u, v, w) such that

More complex symmetrical systems - systems containing the module: | x – y | + y2 = 3, | x – 1 | + | y-1 | = 2. Consider this system separately for x< 1 и при х ≥ 1. Если х < 1, то: а) при у < х система принимает вид х – у + у 2 = 3, - х + 1 – у + 1 = 2, или х – у + у 2 = 3, х + у = 0, откуда находим х 1 = 1, у 1 = - 1, х 2 = - 3, у2 = 3. Эти пары чисел не принадлежат к рассматриваемой области;

b) for x ≤ y< 1 система принимает вид б) при х ≤ у < 1 система принимает вид - х + у + у 2 = 3, - х + 1 – у + 1 = 2, или - х + у + у 2 = 3, х + у = 0, откуда находим х 1 = 3, у 1 = - 3; х 2 = - 1, у 2 = 1. Эти пары чисел не принадлежат к рассматриваемой области; в) при у ≥ 1 (тогда у >x) the system takes the form - x + y + y 2 \u003d 3, - x + 1 + y - 1 \u003d 2, or - x + y + y 2 \u003d 3, x - y \u003d - 2, from where we find x 1 \u003d - 3, y 1 \u003d - 1, x 2 \u003d - 1, y 2 \u003d 1. The second pair of numbers belongs to the area under consideration, that is, it is a solution to this system.

If x ≥ 1, then: If x ≥ 1, then: a) x > y and y< 1 система принимает вид х – у + у 2 = 3, х – 1 – у = 1 = 2, или х – у + у 2= 3, х – у = 2, откуда находим х 1 = 1, у 1 = - 1, х 2 = 4, у 2 = 2. Первая пара чисел принадлежит рассматриваемой области, т. Е. является решением данной системы; б) при х >y and y ≥ 1 the system takes the form x - y + y 2 = 3, x - 1 + y - 1 = 2, or x - y + y 2 = 3, x + y = 4, from which we find x = 1, y = 3. This pair of numbers does not belong to the area under consideration;

c) for x ≤ y (then y ≥ 1), the system takes the form c) for x ≤ y (then y ≥ 1), the system takes the form - x + y + y 2 = 3, x - 1 + y - 1 = 2, or - x + y + y 2 = 3, x + y = 4, from where we find x 1 = 5 + √8, y 1 = - 1 - √8; x 2 = 5 - √8, y 2 = - 1 + √8. These pairs of numbers do not belong to the area under consideration. Thus, x 1 \u003d - 1, y 1 \u003d 1; x 2 \u003d 1, y 2 \u003d - 1. Answer: (- 1; 1); (eleven).

Conclusion Mathematics develops human thinking, teaches through logic to find different solutions. So, having learned how to solve symmetric systems, I realized that they can be used not only to complete specific examples, but I can also solve various kinds of problems. I think that the project can benefit not only me. For those who also want to get acquainted with this topic, my work will be a good helper.

List of used literature: Bashmakov M.I., "Algebra and the beginnings of analysis", 2nd edition, Moscow, "Prosveshchenie", 1992, 350 pages. Rudchenko P.A., Yaremchuk F.P., "Algebra and elementary functions ", directory; third edition, revised and enlarged; Kyiv, Naukova, Dumka, 1987, 648 pp. Sharygin I. F., “Mathematics for high school students”, Moscow, Drofa publishing house, 1995, 490 pp. Internet resources: http://www.college. en/

The work can be used for lessons and reports on the subject "Mathematics"

Ready-made math presentations are used as visual aids that allow a teacher or parent to demonstrate the topic being studied from the textbook using slides and tables, show examples for solving problems and equations, and test knowledge. In this section of the site, you can find and download a lot of ready-made presentations in mathematics for students in grades 1,2,3,4,5,6, as well as presentations in higher mathematics for university students.

Studying additional literature on solving systems of equations, I met with a new type of systems - symmetric. And I set myself a goal:

Summarize scientific information on the topic "Systems of Equations".

Understand and learn how to solve the way of introducing new variables;

3) Consider the main theories related to symmetric systems of equations

4) Learn to solve symmetric systems of equations.

History of solving systems of equations.

The elimination of unknowns from linear equations has long been used. In the 17-18 century. in. exclusion techniques were developed by Fermat, Newton, Leibniz, Euler, Bezout, Lagrange.

In modern notation, the system of two linear equations with two unknowns has the form: a1x + b1y = c1, a2x + b2x = c2 x = c1b1 - c2b; y = а1с2 – а2с1 Solutions of this system are expressed by formulas.

a1b2 – a2b1 a1b2 – a2b1

Thanks to the coordinate method created in the 17th century. Fermat and Descartes, it became possible to solve systems of equations graphically.

In ancient Babylonian texts written in 3-2 millennia BC. e. , contains many problems solved by compiling systems of equations, in which equations of the second degree are also introduced.

Example #1:

I added up the areas of my two squares: 25. The side of the second square is equal to the side of the first and 5 more. The corresponding system of equations in the corresponding notation looks like: x2 + y2 = 25, y = x = 5

Diophantus, who had no notation for many unknowns, took great pains to choose the unknown in such a way as to reduce the solution of the system to the solution of a single equation.

Example #2:

"Find two natural numbers, knowing that their sum is 20 and the sum of their squares is 208."

The problem was also solved by compiling a system of equations, x + y = 20, but solved x2 + y2 = 208

Diophantus, choosing as the unknown half the difference of the desired numbers, i.e.

(x - y) \u003d z, + (x + y) \u003d 10

2z2 + 200 = 208 z = + 2z = -2- does not satisfy the condition of the problem, therefore, if z = 2x = 12, and y = 8

Concepts of a system of algebraic equations.

In many problems, it may be necessary to find several unknown quantities, knowing that other quantities formed with their help (functions of unknowns) are equal to each other or to some given quantities. Let's consider a simple example.

A rectangular plot of land with an area of 2400 m2 is fenced with a 200 m long fence. find the length and width of the segment. In fact, the "algebraic model" of this problem is a system of two equations and one inequality.

Possible limitations-inequalities should always be kept in mind. When you solve problems for compiling systems of equations. But still the main thing is to solve the equations themselves. I will tell you about the methods that are used.

Let's start with definitions.

A system of equations is a set of several (more than one) equations connected by a curly bracket.

The curly bracket means that all the equations of the system must be executed simultaneously, and shows that you need to find a pair of numbers (x; y) that turns each equation into a true equality.

A solution to a system is a pair of numbers x and y that, when substituted into this system, turn each of its equations into a true numerical equality.

To solve a system of equations means to find all its solutions or establish that there are none.

Substitution method.

The substitution method is that in one of the equations one variable is expressed in terms of another. The resulting expression is substituted into another equation, which then turns into an equation with one variable, and then it is solved. The resulting values of this variable are substituted into any equation of the original system and the second variable is found.

Algorithm.

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).

Example No. 1 y \u003d x - 1,

Substitute in the second equation y \u003d x - 1, we get 5x + 2 (x - 1) \u003d 16, from which x \u003d 2. we substitute the resulting expression in the first equation: y \u003d 2 - 1 \u003d 1.

Answer: (2; 1).

Example #2:

8y - x \u003d 4, 1) 2 (8y - 4) - 21y \u003d 2

2x - 21y \u003d 2 16y - 8 - 21y \u003d 2

5y \u003d 10 x \u003d 8y - 4, y \u003d -2

2x - 21y \u003d 2

2) x \u003d 8 * (-2) - 4 x \u003d 8y - 4, x \u003d -20

2 (8y - 4) - 21y \u003d 2 x \u003d 8y - 4, y \u003d -2 x \u003d -20, y \u003d -2

Answer: (-20; -2).

Example #3: x2 + y +8 = xy, 1) x2 + 2x + 8 = x * 2x y - 2x = 0 x2 + 2x + 8 = 2x2

X2 + 2x + 8 = 0 x2 + y + 8 = xy, x2 - 2x - 8 = 0 - quadratic equation y = 2x x1 = -2 x2 = 4 x2 + 2x + 8 = x * 2x 2) y1 = 2 * (-2) y = 2x y1 = -4 y2 = 2 * 4 x1 = -2 y2 = 8 x2 = 4 y = 2x x1 = -2, x2 = 4 y1 = -4, y2 = 8

Hence (-2; -4); (4; 8) are solutions of this system.

Addition method.

The addition method consists in the fact that if a given system consists of equations that, when added together, form an equation with one variable, then by solving this equation, we will obtain the values of one of the variables. The value of the second variable is found, as in the substitution method.

Algorithm for solving systems by the addition method.

1. Equalize modules of coefficients for one of the unknowns.

2. Adding or subtracting the resulting equations, find one unknown.

3. Substituting the found value into one of the equations of the original system, find the second unknown.

Example #1. Solve the system of equations by adding: x + y \u003d 20, x - y \u003d 10

Subtracting the second equation from the first equation, we get

We express from the second expression x \u003d 20 - y

Substitute y \u003d 5 into this expression: x \u003d 20 - 5 x \u003d 15.

Answer: (15; 5).

Example #2:

Let us represent the equations of the proposed system as a difference, we obtain

7y = 21, whence y = 3

Substitute this value into the value expressed from the second equation of the system x = , we get x = 4.

Answer: (4; 3).

Example #3:

2x + 11y = 15,

10x - 11y = 9

Adding these equations, we have:

2x + 10x = 15 + 9

12x \u003d 24 x \u003d 2, substituting this value into the second equation, we get:

10 * 2 - 11y \u003d 9, from where y \u003d 1.

The solution of this system is the pair: (2; 1).

Graphical way to solve systems of equations.

Algorithm.

1. Construct graphs of each of the equations of the system.

2. Finding the coordinates of the point of intersection of the constructed lines.

The case of mutual arrangement of lines on the plane.

1. If the lines intersect, i.e., have one common point, then the system of equations has one solution.

2. If the lines are parallel, that is, they do not have common points, then the system of equations has no solutions.

3. If the lines coincide, i.e., have many points, then the system of equations has an infinite number of solutions.

Example #1:

Solve graphically the system of equations x - y \u003d -1,

We express from the first and second equations y: y \u003d 1 + x, y \u003d 4 - 2x x

Let's build graphs of each of the equations of the system:

1) y \u003d 1 + x - the graph of the function is a straight line x 0 1 (1; 2) y 1 2

2) y \u003d 4 - 2x - the graph of the function is a straight line x 0 1 y 4 2

Answer: (1; 2).

Example #2: y x + 2y = 6,

4y \u003d 8 - 2x x y \u003d, y \u003d y \u003d - the graph of the function is a straight line x 0 2 y 3 2 y \u003d - the graph of the function is a straight line x 0 2 y 2 1

Answer: There are no solutions.

Example No. 3: y x - 2y \u003d 2,

3x - 6y \u003d 6 x - 2y \u003d 2, x - 2y \u003d 2 x y \u003d - the graph of the function is a straight line x 0 2 y -1 0

Answer: The system has an infinite number of solutions.

Method for introducing new variables.

The method of introducing new variables is that a new variable is introduced into only one equation or two new variables for both equations at once, then the equation or equations are solved with respect to the new variables, after which it remains to solve a simpler system of equations, from which we find the desired solution.

Example #1:

x + y = 5

Denote = z, then =.

The first equation will take the form z + = , it is equivalent to 6z - 13 + 6 = 0. Having solved the resulting equation, we have z = ; z=. Then = or = , in other words, the first equation split into two equations, therefore, we have two systems:

x + y = 5 x + y = 5

The solutions of these systems are the solutions of the given system.

The solution of the first system is the pair: (2; 3), and the second is the pair (3; 2).

Therefore, the solutions of the system + = , x + y = 5

The pairs are (2; 3); (3; 2)

Example #2:

Let = X, a = Y.

X \u003d, 5 * - 2Y \u003d 1

5X - 2Y \u003d 1 2.5 (8 - 3Y) - 2Y \u003d 1

20 - 7.5U - 2U \u003d 1

X \u003d, -9.5Y \u003d -19

5 * - 2Y = 1 Y = 2

Let's make a replacement.

2 x = 1, y = 0.5

Answer: (1; 0.5).

Symmetric systems of equations.

A system with n unknowns is called symmetric if it does not change when the unknowns are rearranged.

A symmetric system of two equations with two unknowns x and y is solved by substituting u = x + y, v = xy. Note that the expressions encountered in symmetric systems are expressed in terms of u and v. Let us give several such examples that are of undoubted interest for solving many symmetric systems: x2 + y2 = (x + y)2 - 2xy = u2 - 2v, x3 + y3 = (x + y)(x2 - xy + y2) = u ( u2 - 2v - v) = u3 - 3uv, x4 + y4 = (x2 + y2)2 - 2x2y2 = (u2 - 2v)2 - 2v2 = u4 - 4u2v + 2v2, x2 + xy + y2 = u2 - 2v + v = u2 - v, etc.

The symmetric system of three equations for the unknowns x y, z are solved by substituting x + y + z = u, xy + yz + xz = w. If u, v, w are found, then a cubic equation t2 – ut2 + vt – w = 0 is formed, whose roots t1, t2, t3 in various permutations are solutions of the original system. The most common expressions in such systems are expressed in terms of u, v, w as follows: x2 + y2 + z2 = u2 - 2v x3 + y3 + z3 = u3 - 3uv + 3w

Example #1: x2 + xy + y2 = 13, x + y = 4

Let x + y = u, xy = v.

u2 – v = 13, u = 4

16 - v = 13, u = 4 v = 3, u = 4

Let's make a replacement.

Answer: (1; 3); (3; 1).

Example #2: x3 + y3 = 28, x + y = 4

Let x + y = u, xy = v.

u3 – 3uv = 28, u = 4

64 - 12 v = 28, u = 4

12v = -36 u = 4 v = 3, u = 4

Let's make a replacement.

x + y = 4, xy = 3 x = 4 - y xy = 3 x = 4 - y,

(4 – y) y = 3 x = 4 – y, y1 = 3; y2 = 1 x1 = 1, x2 = 3, y1 = 3, y2 = 1

Answer: (1; 3); (3; 1).

Example #3: x + y + xy = 7, x2 + y2 + xy = 13

Let x = y = u, xy = v.

u + v = 7, u2 – v = 13 u2 – v = 13 u2 – 7 + u =13 u2 + u = 20 v = 7 – u, u (u + 1) =20 u2 – v =13 u = 4 v = 7 – u, u = 4 v = 3, u = 4

Let's make a replacement.

x + y = 4, xy = 3 x = 4 - y xy = 3 x = 4 - y,

(4 – y) y = 3 x = 4 – y, y1 = 3; y2 = 1 x1 = 1, x2 = 3, y1 = 3, y2 = 1

Answer: (1; 3); (3; 1).

Example #4: x + y = 5, x3 + y3 = 65

Let x + y = u, xy = v.

u = 5, u3 – 3uv = 65 u3 – 3uv = 65 125 – 15v = 65

15v = -60 u = 5, v = 4 v = 4

Let's make a replacement.

x + y = 5, xy = 4 x = 5 - y, xy = 4 x = 5 - y, y (5 - y) = 4 x = 5 - y y1 = 1, y2 = 4 x1 = 4, x2 = 1, y1 = 1, y2 = 4

Answer: (4; 1); (fourteen).

Example #5: x2 + xy + y2 = 49, x + y + xy = 23

Let's make a change of unknowns, the system will take the form u2 + v = 49, u + v = 23

Adding these equations, we get u2 + u - 72 = 0 with roots u1 = 8, u2 = -9. Accordingly, v1 = 15, v2 = 32. It remains to solve the set of systems x + y = 8, x + y = -9, xy = 15 xy = 32

The system x + y = 8 has solutions x1 = 3, y1 = 5; x2=5, y2=3.

The system x + y = -9 has no real solutions.

Answer: (3; 5), (5; 3).

Example number 6. Solve the system of equations.

2x2 - 3xy + 2y2 = 16, x + xy + y + 3 = 0

Using the basic symmetric polynomials u = y + x and v = xy, we obtain the following system of equations

2u2 - 7v = 16, u + v = -3

Substituting the expression v = -3 – u from the second equation of the system into the first equation, we obtain the following equation 2u2 + 7u + 5 = 0, whose roots are u1 = -1 and u2 = -2.5; and, accordingly, the values v1 = -2 and v2 = -0.5 are obtained from v = -3 - u.

Now it remains to solve the following set of systems x + y \u003d -1, and x + y \u003d -2.5, xy \u003d -2 xy \u003d -0.5

The solutions of this set of systems, and therefore of the original system (due to their equivalence), are as follows: (1; -2), (-2; 1), (;).

Example #7:

3x2y - 2xy + 3x2 \u003d 78,

2x - 3xy + 2y + 8 = 0

Using the basic symmetric polynomials, the system can be written in the following form

3uv - 2v = 78,

Expressing u = from the second equation and substituting it into the first equation, we get 9v2 – 28v – 156 = 0. The roots of this equation v1 = 6 and v2 = - allow us to find the corresponding values u1 = 5, u2 = - from the expression u =.

We now solve the following set of systems x + y \u003d 5, and x + y \u003d - , xy \u003d 6 xy \u003d -.

x \u003d 5 - y, and y \u003d -x -, xy \u003d 6 xy \u003d -.

x \u003d 5 - y, and y \u003d -x -, y (5 - y) \u003d 6 x (-x -) \u003d -.

x = 5 – y, and y = -x - , y1= 3, y2 =2 x1 = , x2 = - x1 = 2, x2 = 3, and x1 = , x2 = - y1= 3, y2 =2 y1 = -, y2 =

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

Conclusion.

In the process of writing the article, I got acquainted with different types of systems of algebraic equations. Summarized scientific information on the topic "Systems of Equations".

Understood and learned how to solve by introducing new variables;

Reviewed the main theories related to symmetric systems of equations

Learned how to solve symmetric systems of equations.

Home > Solution

Rational equations and inequalities

I. Rational equations.

Linear equations.

Systems of linear equations.

Return equations.

Vieta's formula for polynomials of higher degrees.

Systems of equations of the second degree.

Method for introducing new unknowns in solving equations and systems of equations.

Homogeneous equations.

Solution of symmetric systems of equations.

Equations and systems of equations with parameters.

Graphical method for solving systems of nonlinear equations.

Equations containing the modulus sign.

Basic methods for solving rational equations

II. Rational inequalities.

Properties of equivalent inequalities.

Algebraic inequalities.

interval method.

Fractional-rational inequalities.

Inequalities containing the unknown under the absolute value sign.

Inequalities with parameters.

Systems of rational inequalities.

Graphical solution of inequalities.

III. Verification test.

Rational Equations

view function

P(x) \u003d a 0 x n + a 1 x n - 1 + a 2 x n - 2 + ... + a n - 1 x + a n,

where n is a natural number, a 0 , a 1 ,…, a n are some real numbers, is called an entire rational function.

An equation of the form P(x) = 0, where P(x) is an entire rational function, is called an entire rational equation.

Type equation

P 1 (x) / Q 1 (x) + P 2 (x) / Q 2 (x) + ... + P m (x) / Q m (x) = 0,

where P 1 (x), P 2 (x), …, P m (x), Q 1 (x), Q 2 (x), …, Q m (x) are entire rational functions, is called a rational equation.

Solving the rational equation P (x) / Q (x) = 0, where P (x) and Q (x) are polynomials (Q (x)  0), reduces to solving the equation P (x) = 0 and checking whether that the roots satisfy the condition Q (x)  0.

Linear equations.

An equation of the form ax+b=0, where a and b are some constants, is called a linear equation.

If a0, then the linear equation has a single root: x = -b /a.

If a=0; b0, then the linear equation has no solutions.

If a=0; b=0, then, rewriting the original equation in the form ax = -b, it is easy to see that any x is a solution to a linear equation.

The straight line equation has the form: y = ax + b.

If the line passes through a point with coordinates X 0 and Y 0, then these coordinates satisfy the equation of the line, i.e. Y 0 = aX 0 + b.

Example 1.1. solve the equation

2x - 3 + 4(x - 1) = 5.

Decision. Let's expand the brackets one by one, give like terms and find x: 2x - 3 + 4x - 4 = 5, 2x + 4x = 5 + 4 + 3,

Example 1.2. solve the equation

2x - 3 + 2(x - 1) = 4(x - 1) - 7.

Decision. 2x + 2x - 4x = 3 +2 - 4 - 7, 0x = - 6.

Answer: .

Example 1.3. Solve the equation.

2x + 3 - 6(x - 1) = 4(x - 1) + 5.

Decision. 2x - 6x + 3 + 6 = 4 - 4x + 5,

– 4x + 9 = 9 – 4x,

4x + 4x = 9 - 9,

Answer: Any number.

Systems of linear equations.

Type equation

a 1 x 1 + a 2 x 2 + … + a n x n = b,

where a 1 , b 1 , … ,a n , b are some constants, is called a linear equation with n unknowns x 1 , x 2 , …, x n .

A system of equations is called linear if all the equations in the system are linear. If the system consists of n unknowns, then the following three cases are possible:

the system has no solutions;

the system has exactly one solution;

The system has infinitely many solutions.

Example 2.4. solve the system of equations

Decision. A system of linear equations can be solved by the substitution method, which consists in expressing one unknown in terms of other unknowns of any equation of the system, and then substituting the value of this unknown into the rest of the equations.

From the first equation we express: x = (8 - 3y) / 2. We substitute this expression into the second equation and get a system of equations

X \u003d (8 - 3y) / 2, 3 (8 - 3y) / 2 + 2y \u003d 7. From the second equation we get y \u003d 2. Taking this into account, from the first equation x \u003d 1. Answer: (1; 2). Example 2.5. Solve a system of equations

Decision. The system has no solutions, since two equations of the system cannot be satisfied simultaneously (from the first equation x + y = 3, and from the second x + y = 3.5).

Answer: There are no solutions.

Example 2.6. solve the system of equations

Decision. The system has infinitely many solutions, since the second equation is obtained from the first by multiplying by 2 (i.e., in fact, there is only one equation with two unknowns).

Answer: Infinitely many solutions.

Example 2.7. solve the system of equations

x + y - z = 2,

2x – y + 4z = 1,

Decision. When solving systems of linear equations, it is convenient to use the Gauss method, which consists in transforming the system to a triangular form.

We multiply the first equation of the system by - 2 and, adding the result obtained with the second equation, we get - 3y + 6z \u003d - 3. This equation can be rewritten as y - 2z \u003d 1. Adding the first equation with the third one, we get 7y \u003d 7, or y = 1.

Thus, the system acquired a triangular form

x + y - z = 2,

Substituting y = 1 into the second equation, we find z = 0. Substituting y =1 and z = 0 into the first equation, we find x = 1. Answer: (1; 1; 0). Example 2.8. for what values of the parameter a the system of equations

2x + ay = a + 2,

(a + 1)x + 2ay = 2a + 4

has infinitely many solutions? Decision. From the first equation we express x:

x = - (a / 2)y + a / 2 +1.

Substituting this expression into the second equation, we get

(a + 1)(– (a / 2)y + a / 2 +1) + 2ay = 2a + 4.

(a + 1)(a + 2 – ay) + 4ay = 4a + 8,

4ay – a(a + 1)y = 4(a + 2) – (a + 1)(a + 2),

ya(4 – a – 1) = (a + 2)(4 – a – 1),

ya(3 – a) = (a + 2)(3 – a).

Analyzing the last equation, we note that for a = 3 it has the form 0y = 0, i.e. it is satisfied for any values of y. Answer: 3.

Quadratic equations and equations reducing to them.

An equation of the form ax 2 + bx + c = 0, where a, b and c are some numbers (a0);

x is a variable, called a quadratic equation.

The formula for solving a quadratic equation.

First, we divide both sides of the equation ax 2 + bx + c = 0 by a - this will not change its roots. To solve the resulting equation

x 2 + (b / a) x + (c / a) = 0

select a full square on the left side

x 2 + (b / a) + (c / a) = (x 2 + 2(b / 2a)x + (b / 2a) 2) - (b / 2a) 2 + (c / a) =

= (x + (b / 2a)) 2 - (b 2) / (4a 2) + (c / a) = (x + (b / 2a)) 2 - ((b 2 - 4ac) / (4a 2 )).

For brevity, we denote the expression (b 2 - 4ac) by D. Then the resulting identity takes the form

Three cases are possible:

if the number D is positive (D > 0), then in this case it is possible to take the square root of D and write D as D = (D) 2 . Then

D / (4a 2) = (D) 2 / (2a) 2 = (D / 2a) 2 , therefore the identity takes the form

x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 - (D / 2a) 2 .

According to the formula for the difference of squares, we derive from here:

x 2 + (b / a)x + (c / a) = (x + (b / 2a) – (D / 2a))(x + (b / 2a) + (D / 2a)) =

= (x - ((-b + D) / 2a)) (x - ((- b - D) / 2a)).

Theorem: If the identity holds

ax 2 + bx + c \u003d a (x - x 1) (x - x 2),

then the quadratic equation ax 2 + bx + c \u003d 0 for X 1  X 2 has two roots X 1 and X 2, and for X 1 \u003d X 2 - only one root X 1.

By virtue of this theorem, it follows from the identity derived above that the equation

x 2 + (b / a)x + (c / a) = 0,

and thus the equation ax 2 + bx + c = 0 has two roots:

X 1 \u003d (-b +  D) / 2a; X 2 \u003d (-b -  D) / 2a.

Thus x 2 + (b / a)x + (c / a) = (x - x1)(x - x2).

Usually these roots are written in one formula:

where b 2 - 4ac \u003d D.

if the number D is equal to zero (D = 0), then the identity

x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 - (D / (4a 2))

takes the form x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 .

It follows that for D = 0, the equation ax 2 + bx + c = 0 has one root of multiplicity 2: X 1 = - b / 2a

3) If the number D is negative (D< 0), то – D >0, and therefore the expression

x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 - (D / (4a 2))

is the sum of two terms, one of which is non-negative and the other positive. Such a sum cannot be equal to zero, so the equation

x 2 + (b / a) x + (c / a) = 0

has no real roots. Neither does the equation ax 2 + bx + c = 0.

Thus, to solve the quadratic equation, one should calculate the discriminant

D \u003d b 2 - 4ac.

If D = 0, then the quadratic equation has a unique solution:

If D > 0, then the quadratic equation has two roots:

X 1 \u003d (-b + D) / (2a); X 2 \u003d (-b - D) / (2a).

If D< 0, то квадратное уравнение не имеет корней.

If one of the coefficients b or c is equal to zero, then the quadratic equation can be solved without calculating the discriminant:

b = 0; c  0; c/a<0; X1,2 = (-c / a)

b  0; c = 0; X1 = 0, X2= -b / a.

The roots of a general quadratic equation ax 2 + bx + c = 0 are found by the formula

A quadratic equation in which the coefficient at x 2 is equal to 1 is called reduced. Usually the given quadratic equation is denoted as follows:

x 2 + px + q = 0.

Vieta's theorem.

We have derived the identity

x 2 + (b / a)x + (c / a) \u003d (x - x1) (x - x2),

where X 1 and X 2 are the roots of the quadratic equation ax 2 + bx + c =0. Let us expand the brackets on the right side of this identity.

x 2 + (b / a)x + (c / a) \u003d x 2 - x 1 x - x 2 x + x 1 x 2 \u003d x 2 - (x 1 + x 2) x + x 1 x 2.

It follows that X 1 + X 2 = - b / a and X 1 X 2 = c / a. We have proved the following theorem, first established by the French mathematician F. Viet (1540 - 1603):

Theorem 1 (Vieta). The sum of the roots of the quadratic equation is equal to the coefficient at X, taken with the opposite sign and divided by the coefficient at X 2; the product of the roots of this equation is equal to the free term divided by the coefficient at X 2 .

Theorem 2 (reverse). If the equalities

X 1 + X 2 \u003d - b / a and X 1 X 2 \u003d c / a,

then the numbers X 1 and X 2 are the roots of the quadratic equation ax 2 + bx + c = 0.

Comment. Formulas X 1 + X 2 \u003d - b / a and X 1 X 2 \u003d c / a remain true even in the case when the equation ax 2 + bx + c \u003d 0 has one root X 1 of multiplicity 2, if we put in the indicated formulas X 2 = X 1 . Therefore, it is generally accepted that for D = 0, the equation ax 2 + bx + c = 0 has two roots that coincide with each other.

When solving problems related to the Vieta theorem, it is useful to use the relations

(1 / X 1) + (1 / X 2) \u003d (X 1 + X 2) / X 1 X 2;

X 1 2 + X 2 2 \u003d (X 1 + X 2) 2 - 2 X 1 X 2;

X 1 / X 2 + X 2 / X 1 \u003d (X 1 2 + X 2 2) / X 1 X 2 \u003d ((X 1 + X 2) 2 - 2X 1 X 2) / X 1 X 2;

X 1 3 + X 2 3 = (X 1 + X 2)(X 1 2 - X 1 X 2 + X 2 2) =

\u003d (X 1 + X 2) ((X 1 + X 2) 2 - 3X 1 X 2).

Example 3.9. Solve the equation 2x 2 + 5x - 1 = 0.

Decision. D = 25 – 42(– 1) = 33 >0;

X 1 \u003d (- 5 + 33) / 4; X 2 \u003d (- 5 -33) / 4.

Answer: X 1 \u003d (- 5 + 33) / 4; X 2 \u003d (- 5 -33) / 4.

Example 3.10. Solve the equation x 3 - 5x 2 + 6x = 0

Decision. Let's factorize the left side of the equation x(x 2 - 5x + 6) = 0,

hence x \u003d 0 or x 2 - 5x + 6 \u003d 0.

Solving the quadratic equation, we get X 1 \u003d 2, X 2 \u003d 3.

Answer: 0; 2; 3.

Example 3.11.

x 3 - 3x + 2 = 0. Solution. Let's rewrite the equation, writing -3x \u003d - x - 2x, x 3 - x - 2x + 2 \u003d 0, and now we group x (x 2 - 1) - 2 (x - 1) \u003d 0, (x - 1) (x( x + 1) - 2) = 0,x - 1 = 0, x 1 = 1,x 2 + x - 2 = 0, x 2 = - 2, x 3 = 1. Answer: x 1 = x 3 = 1 , x 2 = - 2. Example 3.12. Solve Equation7

Lesson Objectives:

educational: learning to solve systems of equations containing a homogeneous equation, symmetric systems of equations;
developing: development of thinking, attention, memory, ability to highlight the main thing;
educational: development of communication skills.

Lesson type: lesson learning new material.

Used learning technologies:

work in groups;
design method.

Equipment: computer, multimedia projector.

A week before the lesson, students receive topics for creative assignments (according to options).
I option. Symmetric systems of equations. Solutions.
II option. Systems containing a homogeneous equation. Solutions.

Each student, using additional educational literature, must find the appropriate educational material, select a system of equations and solve it.
One student from each option creates multimedia presentations on the topic of the creative task. The teacher provides guidance to students as needed.

I. Motivation for learning activities of students

Introductory speech of the teacher
In the previous lesson, we considered the solution of systems of equations by the method of replacing unknowns. There is no general rule for choosing new variables. However, two types of systems of equations can be distinguished when there is a reasonable choice of variables:

symmetric systems of equations;
systems of equations, one of which is homogeneous.

II. Learning new material

Students of the second option report on their homework.

1. Slideshow of a multimedia presentation "Systems containing a homogeneous equation" (presentation 1).

2. Work in pairs of students sitting at the same desk: a student of the second option explains to a neighbor in the desk the solution to a system containing a homogeneous equation.

Report of students of the 1st option.

1. Slideshow of the multimedia presentation "Symmetric systems of equations" (presentation 2).

Students write in their notebooks:

2. Work in pairs of students sitting at the same desk: a student of option I explains to a neighbor in the desk the solution of a symmetric system of equations.

III. Consolidation of the studied material

Work in groups (in a group of 4 students unite students sitting at adjacent desks).
Each of the 6 groups performs the following task.

Determine the type of system and solve it: