I. Rational equations.
1) Linear equations.
2) Systems of linear equations.
3) Quadratic equations and equations reducible to them.
4) Reciprocal equations.
5) Vieta's formula for polynomials of higher degrees.
6) Systems of equations of the second degree.
7) Method of introducing new unknowns when solving equations and systems of equations.
8) Homogeneous equations.
9) Solving symmetric systems of equations.
10) Equations and systems of equations with parameters.
11) Graphical method for solving systems of nonlinear equations.
12) Equations containing the modulus sign.
13) Basic methods for solving rational equations
II. Rational inequalities.
1) Properties of equivalent inequalities.
2) Algebraic inequalities.
3) Interval method.
4) Fractional rational inequalities.
5) Inequalities containing an unknown under the absolute value sign.
6) Inequalities with parameters.
7) Systems of rational inequalities.
8) Graphical solution of inequalities.
III. Screening test.
Rational equations
Function of the form
P(x) = 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, 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.
Equation of the form
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, called a rational equation.
Solving the rational equation P (x) / Q (x) = 0, where P (x) and Q (x) are polynomials (Q (x) ¹ 0), comes down to solving the equation P (x) = 0 and checking that that the roots satisfy the condition Q (x) ¹ 0.
Linear equations.
An equation of the form ax+b=0, where a and b are some constants, is called a linear equation.
If a¹0, then the linear equation has a single root: x = -b /a.
If a=0; b¹0, 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 the linear equation.
The equation of the straight line is: y = ax + b.
If a 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.
Solution. Sequentially open the brackets, add similar 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.
Solution. 2x + 2x – 4x = 3 +2 – 4 – 7, 0x = – 6.
Example 1.3. Solve the equation.
2x + 3 – 6(x – 1) = 4(x – 1) + 5.
Solution. 2x – 6x + 3 + 6 = 4 – 4x + 5,
– 4x + 9 = 9 – 4x,
4x + 4x = 9 – 9,
Answer: Any number.
Systems of linear equations.
Equation of the form
a 1 x 1 + a 2 x 2 + … + a n x n = b,
where a 1, b 1, …, a n, b are some constants, called a linear equation with n unknowns x 1, x 2, …, x n.
A system of equations is called linear if all equations included in the system are linear. If the system consists of n unknowns, then the following three cases are possible:
1) the system has no solutions;
2) the system has exactly one solution;
3) the system has infinitely many solutions.
Example 2.4. solve system of equations
2x + 3y = 8,Solution. You can solve a system of linear equations using the substitution method, which consists of expressing one unknown in terms of other unknowns for any equation of the system, and then substituting the value of this unknown into the remaining 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
Solution. 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 system of equations
Solution. 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: There are infinitely many solutions.
Example 2.7. solve system of equations
x + y – z = 2,2x – y + 4z = 1,
– x + 6y + z = 5.
Solution. When solving systems of linear equations, it is convenient to use the Gauss method, which consists of transforming the system to a triangular form.
We multiply the first equation of the system by – 2 and, adding the resulting result with the second equation, we get – 3y + 6z = – 3. This equation can be rewritten as y – 2z = 1. Adding the first equation with the third, we get 7y = 7, or y = 1.
Thus, the system acquired a triangular shape
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. at what values of parameter a is the system of equations
2x + ay = a + 2,(a + 1)x + 2ay = 2a + 4
has infinitely many solutions?
Solution. 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.
Quadratic equations and equations that can be reduced to them.
An equation of the form ax 2 + bx + c = 0, where a, b and c are some numbers (a¹0);
x is a variable called a quadratic equation.
Formula for solving a quadratic equation.
First, let's 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 complete 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:
1) if the number D is positive (D > 0), then in this case you can extract the square root of D and write D in the form D = (ÖD) 2. Then
D / (4a 2) = (ÖD) 2 / (2a) 2 = (ÖD / 2a) 2, therefore the identity takes the form
x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 – (ÖD / 2a) 2 .
Using the difference of squares formula, we derive from here:
x 2 + (b / a)x + (c / a) = (x + (b / 2a) – (ÖD / 2a))(x + (b / 2a) + (ÖD / 2a)) =
= (x – ((-b + ÖD) / 2a)) (x – ((– b – ÖD) / 2a)).
Theorem : If the identity holds
ax 2 + bx + c = a(x – x 1)(x – x 2),
then the quadratic equation ax 2 + bx + c = 0 for X 1 ¹ X 2 has two roots X 1 and X 2, and for X 1 = X 2 - only one root X 1.
By virtue of this theorem, from the identity derived above it follows 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 =(-b + Ö D) / 2a; X 2 = (-b - Ö D) / 2a.
Thus x 2 + (b / a)x + (c / a) = (x – x1)(x – x2).
Usually these roots are written with one formula:
where b 2 – 4ac = D.
2) if the number D is 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 is positive. Such a sum cannot equal zero, so the equation
x 2 + (b / a)x + (c / a) = 0
has no real roots. The equation ax 2 + bx + c = 0 does not have them either.
Thus, to solve a quadratic equation, one should calculate the discriminant
D = 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 =(-b + ÖD) / (2a); X 2 = (-b - ÖD) / (2a).
If D< 0, то квадратное уравнение не имеет корней.
If one of the coefficients b or c is zero, then the quadratic equation can be solved without calculating the discriminant:
1) b = 0; c¹0; c/a<0; X1,2 = ±Ö(-c / a)
2) b ¹ 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
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Chapter 4. Systems of rational equations
The fourth chapter is devoted to the study of ways to solve systems of rational equations. Here, concepts learned in 7th grade and previously applied to systems of linear equations are used, 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 goal of the ninth paragraph is to, based on known concepts related to equations and systems of linear equations, learn to solve systems of rational equations and 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
In this paragraph, using three examples, we show how you can solve 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) 
e) 
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) 
e) 
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 with equivalent systems:
(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 up the equations of system (5) and replacing the first equation of the system with the resulting equation, we obtain a 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.
Davydova M.G.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., add.-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;
Developmental: expand and deepen the study of various types of 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:
- Setting the lesson goal (1 min).
- Preparing to study new material (2 min).
- 3.Introduction to new material (38 min).
- 4. Lesson summary (2 min)
- 5.Homework (2 min)
Lesson equipment: interactive whiteboard, 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 the other, and apply the rules for adding and subtracting 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 the general ideas, the 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 the theoretical material, and select 2 examples from the textbook for the considered methods of 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.
- 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.
- 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.
- 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 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 on independent work in class 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. The purpose of the work is to test the 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 information 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 and checking students’ readiness to take the test is one of the main tasks 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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We introduced the equation above in § 7. First, let us recall what a rational expression is. This is an algebraic expression made up of numbers and the variable x using the operations of addition, subtraction, multiplication, division and exponentiation with a natural exponent.
If r(x) is a rational expression, then the equation r(x) = 0 is called a rational equation.
However, in practice it is more convenient to use a slightly broader interpretation of the term “rational equation”: this is an equation of the form h(x) = q(x), where h(x) and q(x) are rational expressions.
Until now, we could not solve any rational equation, but only one that, as a result of various transformations and reasoning, was reduced to linear equation. Now our capabilities are much greater: we will be able to solve a rational equation that reduces not only to linear
mu, but also to the quadratic equation.
Let us recall how we solved rational equations before and try to formulate a solution algorithm.
Example 1. Solve the equation
Solution. Let's rewrite the equation in the form
In this case, as usual, we take advantage of the fact that the equalities A = B and A - B = 0 express the same relationship between A and B. This allowed us to move the term to the left side of the equation with the opposite sign.
Let's transform the left side of the equation. We have
Let us recall the conditions of equality fractions zero: if and only if two relations are simultaneously satisfied:
1) the numerator of the fraction is zero (a = 0); 2) the denominator of the fraction is different from zero).
Equating the numerator of the fraction on the left side of equation (1) to zero, we obtain
It remains to check the fulfillment of the second condition indicated above. The relation means for equation (1) that . The values x 1 = 2 and x 2 = 0.6 satisfy the indicated relationships and therefore serve as the roots of equation (1), and at the same time the roots of the given equation.
1) Let's transform the equation to the form
2) Let us transform the left side of this equation:
(simultaneously changed the signs in the numerator and
fractions).
Thus, the given equation takes the form
3) Solve the equation x 2 - 6x + 8 = 0. Find
4) For the found values, check the fulfillment of the condition 
. The number 4 satisfies this condition, but the number 2 does not. This means that 4 is the root of the given equation, and 2 is an extraneous root.
ANSWER: 4.
2. Solving rational equations by introducing a new variable
The method of introducing a new variable is familiar to you; we have used it more than once. Let us show with examples how it is used in solving rational equations.
Example 3. Solve the equation x 4 + x 2 - 20 = 0.
Solution. Let's introduce a new variable y = x 2 . Since x 4 = (x 2) 2 = y 2, then the given equation can be rewritten as
y 2 + y - 20 = 0.
This is a quadratic equation, the roots of which can be found using known formulas; we get y 1 = 4, y 2 = - 5.
But y = x 2, which means the problem has been reduced to solving two equations:
x 2 =4; x 2 = -5.
From the first equation we find that the second equation has no roots.
Answer: .
An equation of the form ax 4 + bx 2 + c = 0 is called a biquadratic equation (“bi” is two, i.e., a kind of “double quadratic” equation). The equation just solved was precisely biquadratic. Any biquadratic equation is solved in the same way as the equation from Example 3: introduce a new variable y = x 2, solve the resulting quadratic equation with respect to the variable y, and then return to the variable x.
Example 4. Solve the equation
Solution. Note that the same expression x 2 + 3x appears twice here. This means that it makes sense to introduce a new variable y = x 2 + 3x. This will allow us to rewrite the equation in a simpler and more pleasant form (which, in fact, is the purpose of introducing a new variable- and simplifying the recording
becomes clearer, and the structure of the equation becomes clearer):
Now let’s use the algorithm for solving a rational equation.
1) Let’s move all the terms of the equation into one part:
= 0
2) Transform the left side of the equation
So, we have transformed the given equation to the form
3) From the equation - 7y 2 + 29y -4 = 0 we find (you and I have already solved quite a lot of quadratic equations, so it’s probably not worth always giving detailed calculations in the textbook).
4) Let's check the found roots using condition 5 (y - 3) (y + 1). Both roots satisfy this condition.
So, the quadratic equation for the new variable y is solved: 
Since y = x 2 + 3x, and y, as we have established, takes two values: 4 and , we still have to solve two equations: x 2 + 3x = 4; x 2 + Zx = . The roots of the first equation are the numbers 1 and - 4, the roots of the second equation are the numbers
In the examples considered, the method of introducing a new variable was, as mathematicians like to say, adequate to the situation, that is, it corresponded well to it. Why? Yes, because the same expression clearly appeared in the equation several times and there was a reason to designate this expression with a new letter. But this does not always happen; sometimes a new variable “appears” only during the transformation process. This is exactly what will happen in the next example.
Example 5. Solve the equation
x(x-1)(x-2)(x-3) = 24.
Solution. We have
x(x - 3) = x 2 - 3x;
(x - 1)(x - 2) = x 2 -Зx+2.
This means that the given equation can be rewritten in the form
(x 2 - 3x)(x 2 + 3x + 2) = 24
Now a new variable has “appeared”: y = x 2 - 3x.
With its help, the equation can be rewritten in the form y (y + 2) = 24 and then y 2 + 2y - 24 = 0. The roots of this equation are the numbers 4 and -6.
Returning to the original variable x, we obtain two equations x 2 - 3x = 4 and x 2 - 3x = - 6. From the first equation we find x 1 = 4, x 2 = - 1; the second equation has no roots.
ANSWER: 4, - 1.
