This article is devoted to techniques for solving various equations and inequalities containing
variable under the module sign.

If you come across an equation or inequality with a modulus on the exam, you can solve it by
without knowing any special methods at all and using only the module definition. Truth,
it can take an hour and a half of precious exam time.

Therefore, we want to tell you about the techniques that simplify the solution of such problems.

First of all, remember that

Consider the different types equations with modulus... (We will move on to inequalities later.)

On the left is the module, on the right is the number

This is the simplest case. Let's solve the equation

There are only two numbers whose modules are equal to four. These are 4 and −4. Therefore, the equation
is equivalent to a combination of two simple ones:

The second equation has no solutions. Solutions to the first: x = 0 and x = 5.

Answer: 0; 5.

Variable both under the module and outside the module

Here you have to expand the module by definition. ... ... or to think!

The equation falls into two cases, depending on the sign of the expression under the modulus.
In other words, it is tantamount to a combination of two systems:

Solution of the first system:. The second system has no solutions.
Answer: 1.

First case: x ≥ 3. Remove the module:

The number, being negative, does not satisfy the condition x ≥ 3 and therefore is not a root of the original equation.

Let us find out whether the number satisfies this condition. To do this, compose the difference and determine its sign:

Hence, it is more than three and therefore is the root of the original equation

Second case: x< 3. Снимаем модуль:

Number . greater than, and therefore does not satisfy the condition x< 3. Проверим :

Means, . is the root of the original equation.

Remove the module by definition? It's scary to even think about it, because the discriminant is not a complete square. Let's better use the following consideration: an equation of the form | A | = B is equivalent to the combination of two systems:

The same, but slightly different:

In other words, we solve two equations, A = B and A = −B, and then select roots that satisfy the condition B ≥ 0.

Let's get started. First, we solve the first equation:

Then we solve the second equation:

Now, in each case, we check the sign of the right side:

Therefore, only and are suitable.

Quadratic equations with the replacement | x | = t

Let's solve the equation:

Since, it is convenient to make the replacement | x | = t. We get:

Answer: ± 1.

Module is equal to module

We are talking about equations of the form | A | = | B |. This is a gift of fate. No module disclosures by definition! It's simple:

For example, consider the equation:. It is tantamount to the following aggregate:

It remains to solve each of the equations of the set and write down the answer.

Two or more modules

Let's solve the equation:

Let's not bother with each module separately and expand it by definition - there will be too many options. There is a more rational way - the method of intervals.

The modulus expressions vanish at the points x = 1, x = 2 and x = 3. These points divide the number line into four intervals (intervals). We mark these points on the number line and arrange the signs for each of the expressions under the modules at the intervals obtained. (The order of the signs is the same as the order of the corresponding modules in the equation.)

Thus, we need to consider four cases - when x is in each of the intervals.

Case 1: x ≥ 3. All modules are removed "with a plus":

The resulting value x = 5 satisfies the condition x ≥ 3 and therefore is the root of the original equation.

Case 2: 2 ≤ x ≤ 3. The last module is now removed "with a minus":

The resulting value of x is also good - it belongs to the interval under consideration.

Case 3: 1 ≤ x ≤ 2. The second and third modules are removed "with a minus":

We got the correct numerical equality for any x from the considered interval serve as solutions to this equation.

Case 4: x ≤ 1 ≤ 1. The second and third modules are removed "with a minus":

Nothing new. We already know that x = 1 is a solution.

Answer: ∪ (5).

Module in module

Let's solve the equation:

We start by expanding the internal module.

1) x ≤ 3. We get:

The expression under the modulus vanishes at. This point belongs to the considered
interval. Therefore, we have to analyze two subcases.

1.1) We get in this case:

This value of x is not valid because it does not belong to the interval under consideration.

1.2). Then:

This x value is also not valid.

So, for x ≤ 3 there are no solutions. Let's move on to the second case.

2) x ≥ 3. We have:

Here we are in luck: the expression x + 2 is positive in the interval under consideration! Therefore, there will be no more sub-cases: the module is removed "with a plus":

This value of x is in the considered interval and therefore is the root of the original equation.

This is how all tasks of this type are solved - we open the nested modules one by one, starting with the internal one.

MBOU Secondary School No. 17 Ivanov

« Equations with modulus "
Methodical development

Compiled by

math teacher

N.V. Lebedeva

20010 g.

Explanatory note

Chapter 1. Introduction

Section 2. Basic properties Section 3. Geometric interpretation of the concept of the modulus of a number Section 4. Graph of the function y = | x | Section 5. Conventions

Chapter 2. Solving equations containing a module

Section 1. Equations of the form | F (x) | = m (simplest) Section 2. Equations of the form F (| x |) = m Section 3. Equations of the form | F (x) | = G (x) Section 4. Equations of the form | F (x) | = ± F (x) (beautiful) Section 5. Equations of the form | F (x) | = | G (x) | Section 6. Examples of solving non-standard equations Section 7. Equations of the form | F (x) | + | G (x) | = 0 Section 8. Equations of the form | a 1 x ± b 1 | ± | a 2 x ± b 2 | ±… | a n x ± in n | = m Section 9. Equations Containing Multiple Modules

Chapter 3. Examples of solving various equations with a module.

Section 1. Trigonometric equations Section 2. Exponential equations Section 3. Logarithmic equations Section 4. Irrational equations Section 5. Tasks of increased complexity Answers to exercises Bibliography

Explanatory note.

The concept of the absolute value (modulus) of a real number is one of its essential characteristics. This concept is widespread in various branches of physical, mathematical and technical sciences. In the practice of teaching a mathematics course in secondary school in accordance with the Program of the Ministry of Defense of the Russian Federation, the concept of "absolute value of a number" occurs repeatedly: in the 6th grade, the definition of a module, its geometric meaning, is introduced; in the 8th grade, the concept of absolute error is formed, the solution of the simplest equations and inequalities containing the module is considered, the properties of the arithmetic square root are studied; in the 11th grade, the concept is found in the section "Root n-th degree ". Teaching experience shows that students often face difficulties in solving tasks that require knowledge of this material, and often skip before starting to complete. In the texts of the examination tasks for the course of the 9th and 11th grades, similar tasks are also included. In addition, the requirements that universities place on school graduates differ, namely, at a higher level than the requirements of the school curriculum. For life in modern society, it is very important to form a mathematical style of thinking, manifested in certain mental skills. In the process of solving problems with modules, the ability to apply such techniques as generalization and concretization, analysis, classification and systematization, analogy is required. The solution of such tasks allows you to check the knowledge of the main sections of the school course, the level of logical thinking, the initial skills of research activities. This work is devoted to one of the sections - solving equations containing a module. It consists of three chapters. The first chapter introduces the basic concepts and the most important theoretical calculations. In the second chapter, nine basic types of equations containing a module are proposed, methods for their solution are considered, examples of different levels of complexity are analyzed. The third chapter offers more complex and non-standard equations (trigonometric, exponential, logarithmic and irrational). Each type of equation has exercises for independent solution (answers and directions are attached). The main purpose of this work is to provide methodological assistance to teachers in preparing for lessons and in organizing optional courses. The material can also be used as a teaching aid for high school students. The tasks offered in the work are interesting and not always easy to solve, which makes it possible to make the educational motivation of students more conscious, test their abilities, and improve the level of preparation of school graduates for entering universities. Differentiated selection of the proposed exercises involves the transition from the reproductive level of mastering the material to the creative one, as well as the opportunity to teach how to apply your knowledge in solving non-standard problems.

Chapter 1. Introduction.

Section 1. Determination of the absolute value .

Definition : The absolute value (modulus) of a real number a a non-negative number is called: a or -a. Designation: │ a │ The record is read as follows: "the module of the number a" or "the absolute value of the number a"

│ a, if a> 0

│a│ = │ 0 if a = 0 (1)

│ - a, if a
Examples: 1) │2,5│ = 2,5 2) │-7│ = 7 3) │1 - √2│ = √2 – 1

Expand Expression Module:

a) │x - 8│, if x> 12 b) │2x + 3│, if x ≤ -2 │x - 8│ = x - 8 │ 2x + 3│ = - 2x - 3

Section 2. Basic properties.

Let's consider the main properties of the absolute value. Property # 1: Opposite numbers have equal modules, i.e. │а│ = │- a│ Let us show that the equality is correct. Let's write the definition of the number - a : │- a│= (2) Let us compare collections (1) and (2). Obviously, the definitions of the absolute values ​​of numbers a and - a match up. Hence, │а│ = │- a│
When considering the following properties, we restrict ourselves to their formulation, since their proof is given in Property # 2: The absolute value of the sum of a finite number of real numbers does not exceed the sum of the absolute values ​​of the terms: │а 1 + а 2 + ... + а n │ ≤│а 1 │ + │а 2 │ + ... + │а n │ Property number 3: The absolute value of the difference between two real numbers does not exceed the sum of their absolute values: │а - в│ ≤│а│ + │в│ Property # 4: The absolute value of the product of a finite number of real numbers is equal to the product of the absolute values ​​of the factors: Property # 5: The absolute value of the quotient of real numbers is equal to the quotient of their absolute values:

Section 3. Geometric interpretation of the concept of the modulus of a number.

Each real number can be associated with a point on the number line, which will be the geometric image of the given real number. Each point on the number line corresponds to its distance from the origin, i.e. the length of the segment from the origin to the given point. This distance is always considered as a non-negative value. Therefore, the length of the corresponding segment will be the geometric interpretation of the absolute value of the given real number

The presented geometric illustration clearly confirms property No. 1, i.e. modules of opposite numbers are equal. Hence, the validity of the equality is easily understood: │x - a│ = │a - x│. Also, the solution of the equation │х│ = m, where m ≥ 0, namely х 1,2 = ± m, becomes more obvious. Examples: 1) │х│ = 4 x 1,2 = ± 4 2) │х - 3│ = 1
x 1.2 = 2; 4

Section 4. Graph of the function y = │х│

The scope of this function is all real numbers.

Section 5. Conventions.

In the future, when considering examples of solving equations, the following conventions will be used: (- sign of the system [- sign of the totality When solving the system of equations (inequalities), the intersection of the solutions included in the system of equations (inequalities) is found. When solving a set of equations (inequalities), the union of solutions included in the set of equations (inequalities) is found.

Chapter 2. Solving equations containing a module.

In this chapter, we will look at algebraic ways to solve equations containing one or more modules.

Section 1. Equations of the form │F (x) │ = m

An equation of this type is called the simplest. It has a solution if and only if m ≥ 0. By the definition of the modulus, the original equation is equivalent to a combination of two equations: │ F(x) │ =m
Examples:
№1. Solve the equation: │7x - 2│ = 9


Answer: x 1 = - 1; X 2 = 1 4 / 7 №2
│x 2 + 3x + 1│ = 1

x 2 + 3x + 2 = 0 x 2 + 3x = 0 x 1 = -1; x 2 = -2 x (x + 3) = 0 x 1 = 0; x 2 = -3 Answer: the sum of the roots is - 2.№3
│x 4 -5x 2 + 2│ = 2 x 4 - 5x 2 = 0 x 4 - 5x 2 + 4 = 0 x 2 (x 2 - 5) = 0 we denote x 2 = m, m ≥ 0 x = 0 ; ± √5 m 2 - 5m + 4 = 0 m = 1; 4 - both values ​​satisfy the condition m ≥ 0 x 2 = 1 x 2 = 4 x = ± 1 x = ± 2 Answer: the number of roots of the equation is 7. Exercises:
№1. Solve the equation and indicate the sum of the roots: │х - 5│ = 3 №2 ... Solve the equation and indicate the smaller root: │x 2 + x│ = 0 №3 ... Solve the equation and indicate the larger root: │x 2 - 5x + 4│ = 4 №4 .Solve the equation and indicate the whole root: │2x 2 - 7x + 6│ = 1 №5 .Solve the equation and indicate the number of roots: │x 4 - 13x 2 + 50│ = 14

Section 2. Equations of the form F (│х│) = m

The function argument on the left side is under the modulus sign, and the right side is independent of the variable. Consider two ways to solve equations of this type. Method 1: By definition of the absolute value, the original equation is equivalent to the combination of two systems. In each of which a condition is imposed on a submodule expression. F(│х│) =m
Since the function F (│х│) is even on the entire domain of definition, the roots of the equations F (x) = m and F (- x) = m are pairs of opposite numbers. Therefore, it is enough to solve one of the systems (when considering examples in this way, the solution of one system will be given). Method 2: Applying the method of introducing a new variable. In this case, the designation │х│ = a is introduced, where a ≥ 0. This method is less voluminous in design.
Examples: №1 ... Solve the equation: 3x 2 - 4│x│ = - 1 Let's use the introduction of a new variable. We denote │x│ = a, where a ≥ 0. We obtain the equation 3a 2 - 4a + 1 = 0 A = 16 - 12 = 4 a 1 = 1 a 2 = 1/3 Returning to the original variable: │x│ = 1 and │х│ = 1/3. Each equation has two roots. Answer: x 1 = 1; X 2 = - 1; X 3 = 1 / 3 ; X 4 = - 1 / 3 . №2. Solve the equation: 5x 2 + 3│x│- 1 = 1/2 │x│ + 3x 2
Let us find the solution of the first system of the set: 4x 2 + 5x - 2 = 0 D = 57 x 1 = -5 + √57 / 8 x 2 = -5-√57 / 8 Note that x 2 does not satisfy the condition x ≥ 0. Solution the second system will be the opposite of x 1. Answer: x 1 = -5+√57 / 8 ; X 2 = 5-√57 / 8 .№3 . Solve the equation: x 4 - │х│ = 0 Denote │х│ = a, where a ≥ 0. We obtain the equation a 4 - a = 0 a · (a 3 - 1) = 0 a 1 = 0 a 2 = 1 Return to the original variable: │х│ = 0 and │х│ = 1 х = 0; ± 1 Answer: x 1 = 0; X 2 = 1; X 3 = - 1.
Exercises: №6. Solve the equation: 2│x│ - 4.5 = 5 - 3/8 │x│ №7 ... Solve the equation, in the answer indicate the number of roots: 3x 2 - 7│x│ + 2 = 0 №8 ... Solve the equation, in the answer indicate the whole solutions: x 4 + │x│ - 2 = 0

Section 3. Equations of the form │F (x) │ = G (x)

The right-hand side of an equation of this form depends on the variable and, therefore, has a solution if and only if the right-hand side is a function G (x) ≥ 0. The original equation can be solved in two ways: Method 1: Standard, is based on the disclosure of a module based on its definition and consists in an equivalent transition to a combination of two systems. │ F(x) │ =G(X)

It is rational to use this method in the case of a complex expression for the function G (x) and less complex - for the function F (x), since the solution of inequalities with the function F (x) is assumed. Method 2: It consists in the transition to an equivalent system in which a condition is imposed on the right-hand side. │ F(x)│= G(x)

This method is more convenient to use if the expression for the function G (x) is less complicated than for the function F (x), since it is assumed that the inequality G (x) ≥ 0. In addition, in the case of several modules, this method is recommended to apply the second option. Examples: №1. Solve the equation: │x + 2│ = 6 -2x
(1 way) Answer: x = 1 1 / 3 №2.
│x 2 - 2x - 1│ = 2 (x + 1)
(2 way) Answer: The product of the roots is 3.
№3. Solve the equation, in the answer indicate the sum of the roots:
│x - 6│ = x 2 - 5x + 9

Answer: the sum of the roots is 4.
Exercises: №9. │x + 4│ = - 3x №10. Solve the equation, in the answer indicate the number of solutions: │x 2 + x - 1│ = 2x - 1 №11 ... Solve the equation, in the answer indicate the product of the roots: │х + 3│ = х 2 + х - 6

Section 4. Equations of the form │F (x) │ = F (x) and │F (x) │ = - F (x)

Equations of this kind are sometimes called "prettiest". Since the right-hand side of equations depends on a variable, solutions exist if and only if the right-hand side is non-negative. Therefore, the original equations are equivalent to the inequalities:
│F (x) │ = F (x) F (x) ≥ 0 and │F (x) │ = - F (x) F (x) Examples: №1 ... Solve the equation, in the answer indicate the smaller whole root: │5x - 3│ = 5x - 3 5x - 3 ≥ 0 5x ≥ 3 x ≥ 0.6 Answer: x = 1№2. Solve the equation, in the answer indicate the length of the gap: │х 2 - 9│ = 9 - х 2 х 2 - 9 ≤ 0 (х - 3) (х + 3) ≤ 0 [- 3; 3] Answer: The length of the gap is 6.№3 . Solve the equation, in the answer indicate the number of integer solutions: │2 + х - х 2 │ = 2 + х - х 2 2 + х - х 2 ≥ 0 х 2 - х - 2 ≤ 0 [- 1; 2] Answer: 4 whole solutions.№4 . Solve the equation, in the answer indicate the largest root:
│4 - x -
│ = 4 - x -
x 2 - 5x + 5 = 0 D = 5 x 1.2 =
≈ 1,4

Answer: x = 3.

Exercises: №12. Solve the equation, in the answer indicate the whole root: │x 2 + 6x + 8│ = x 2 + 6x + 8 №13. Solve the equation, in the answer indicate the number of integer solutions: │13x - x 2 - 36│ + x 2 - 13x + 36 = 0 №14. Solve the equation, in the answer write an integer that is not the root of the equation:

Section 5. Equations of the form │F (x) │ = │G (x) │

Since both sides of the equation are non-negative, the solution involves considering two cases: the submodule expressions are equal or opposite in sign. Therefore, the original equation is equivalent to a combination of two equations: │ F(x)│= │ G(x)│
Examples: №1. Solve the equation, in the answer indicate the whole root: │x + 3│ = │2x - 1│
Answer: whole root x = 4.№2. Solve the equation: │ x - x 2 - 1│ = │2x - 3 - x 2 │
Answer: x = 2.№3 . Solve the equation, in the answer indicate the product of the roots:




Equation roots 4x 2 + 2x - 1 = 0 x 1.2 = - 1 ± √5 / 4 Answer: the product of the roots is equal to - 0.25. Exercises: №15 ... Solve the equation, write the whole solution in your answer: │x 2 - 3x + 2│ = │x 2 + 6x - 1│ №16. Solve the equation, in the answer indicate the smaller root: │5x - 3│ = │7 - x│ №17 ... Solve the equation, in the answer indicate the sum of the roots:

Section 6. Examples of solving non-standard equations

In this section, we will consider examples of non-standard equations, when solving which the absolute value of an expression is revealed by definition. Examples:

№1. Solve the equation, in the answer indicate the sum of the roots: x │x│- 5x - 6 = 0
Answer: the sum of the roots is 1 №2. . Solve the equation, in the answer indicate the smaller root: x 2 - 4x
- 5 = 0
Answer: smaller root x = - 5. №3. Solve the equation:

Answer: x = -1. Exercises: №18. Solve the equation and indicate the sum of the roots: x 3x + 5│ = 3x 2 + 4x + 3
№19. Solve the equation: x 2 - 3x =

№20. Solve the equation:

Section 7. Equations of the form │F (x) │ + │G (x) │ = 0

It is easy to see that on the left side of an equation of this type is the sum of non-negative values. Consequently, the original equation has a solution if and only if both terms are simultaneously equal to zero. The equation is equivalent to a system of equations: │ F(x)│+│ G(x)│=0
Examples: №1 ... Solve the equation:
Answer: x = 2. №2. Solve the equation: Answer: x = 1. Exercises: №21. Solve the equation: №22 ... Solve the equation, in the answer indicate the sum of the roots: №23 ... Solve the equation, in the answer indicate the number of solutions:

Section 8. Equations of the form │а 1 х + в 1 │ ± │а 2 х + в 2 │ ± ... │а n х + в n │ = m

To solve equations of this type, the method of intervals is used. If we solve it by sequential expansion of modules, then we get n sets of systems, which is very cumbersome and inconvenient. Let's consider the algorithm of the method of intervals: 1). Find variable values X at which each module is equal to zero (zeros of submodule expressions):
2). Mark the found values ​​on the number line, which is divided into intervals (the number of intervals, respectively, is n+1 ) 3). Determine the sign with which each module is revealed at each of the intervals obtained (when making a solution, you can use a number line by marking signs on it) 4). The original equation is equivalent to the totality n+1 systems, each of which indicates the belonging of a variable X one of the intervals. Examples: №1 ... Solve the equation, in the answer indicate the largest root:
one). Find the zeros of the submodule expressions: x = 2; x = -3 2). Let us mark the found values ​​on the number line and determine the sign with which each module is expanded on the obtained intervals:
x - 2 x - 2 x - 2 - - + - 3 2 x 2x + 6 2x + 6 2x + 6 - + + 3)
- no solutions The equation has two roots. Answer: the largest root x = 2. №2. Solve the equation, in the answer indicate the whole root:
one). Find the zeros of the submodule expressions: x = 1.5; x = - 1 2). Let us mark the found values ​​on the number line and determine with what sign each module is revealed on the obtained intervals: х + 1 х + 1 х + 1 - + +
-1 1.5 x 2x - 3 2x - 3 2x - 3 - - +
3).
The last system has no solutions, therefore, the equation has two roots. In the course of solving the equation, you should pay attention to the "-" sign in front of the second module. Answer: whole root x = 7. №3. Solve the equation, in the answer indicate the sum of the roots: 1). Find the zeros of the submodule expressions: x = 5; x = 1; x = - 2 2). Let us mark the found values ​​on the number line and determine with what sign each module is revealed on the obtained intervals: x - 5 x - 5 x - 5 x - 5 - - - +
-2 1 5 x x - 1 x - 1 x - 1 x - 1 - - + + x + 2 x + 2 x + 2 x + 2 - + + +
3).
The equation has two roots x = 0 and 2. Answer: the sum of the roots is 2. №4 . Solve the equation: 1). Find the zeros of the submodule expressions: x = 1; x = 2; x = 3.2). Let us determine the sign with which each module is revealed on the obtained intervals. 3).
Let's combine the solutions of the first three systems. Answer: ; x = 5.
Exercises: №24. Solve the equation:
№25. Solve the equation, in the answer indicate the sum of the roots: №26. Solve the equation, in the answer indicate the smaller root: №27. Solve the equation, in the answer indicate the larger root:

Section 9. Equations Containing Multiple Modules

Equations containing multiple modules assume absolute values ​​in submodule expressions. The main principle for solving equations of this type is the sequential disclosure of modules, starting with the "external" one. In the course of the solution, the techniques discussed in sections №1, №3 are used.

Examples: №1. Solve the equation:
Answer: x = 1; - eleven. №2. Solve the equation:
Answer: x = 0; 4; - 4. №3. Solve the equation, in the answer indicate the product of the roots:
Answer: the product of the roots is - 8. №4. Solve the equation:
Let us denote the equations of the set (1) and (2) and consider the solution of each of them separately for the convenience of design. Since both equations contain more than one module, it is more convenient to carry out an equivalent transition to sets of systems. (1)

(2)


Answer:
Exercises: №36. Solve the equation, in the answer write the sum of the roots: 5 │3x-5│ = 25 x №37. Solve the equation, if there are more than one roots, in the answer indicate the sum of the roots: │x + 2│x - 3x - 10 = 1 №38. Solve the equation: 3 │2x -4│ = 9 │x│ №39. Solve the equation, in the answer indicate the number of roots by: 2 │ sin х│ = √2 №40 ... Solve the equation, in the answer indicate the number of roots:

Section 3. Logarithmic equations.

Before solving the following equations, it is necessary to repeat the properties of the logarithms and the logarithmic function. Examples: №1. Solve the equation, in the answer indicate the product of the roots: log 2 (x + 1) 2 + log 2 │x + 1│ = 6 ODZ. x + 1 ≠ 0 x ≠ - 1

1 case: if x ≥ - 1, then log 2 (x + 1) 2 + log 2 (x + 1) = 6 log 2 (x + 1) 3 = log 2 2 6 (x + 1) 3 = 2 6 x + 1 = 4 x = 3 - satisfies the condition х ≥ - 1 2 case: if х log 2 (x + 1) 2 + log 2 (-x-1) = 6 log 2 (x + 1) 2 + log 2 (- (x + 1)) = 6 log 2 (- (x + 1) 3) = log 2 2 6- (x + 1) 3 = 2 6- (x + 1) = 4 x = - 5 - satisfies condition x - 1
Answer: the product of the roots is - 15.
№2. Solve the equation, in the answer indicate the sum of the roots: lg
O.D.Z.

Answer: the sum of the roots is 0.5.
№3. Solve the equation: log 5
O.D.Z.

Answer: x = 9. №4. Solve the equation: │2 + log 0.2 x│ + 3 = │1 + log 5 x│ O.D.Z. x> 0 Let's use the formula for transition to another base. │2 - log 5 x│ + 3 = │1 + log 5 x│
│2 - log 5 x│- │1 + log 5 x│ = - 3 Find the zeros of the submodule expressions: x = 25; x = These numbers divide the range of permissible values ​​into three intervals, so the equation is equivalent to the combination of three systems.
Answer: )