Solving modular equations online. Equations with modulus

We don't choose mathematics her profession, and she chooses us.

Russian mathematician Yu.I. Manin

Equations with modulus

The most difficult problems to solve in school mathematics are equations containing variables under the modulus sign. To successfully solve such equations, you need to know the definition and basic properties of the module. Naturally, students must have the skills to solve equations of this type.

Basic concepts and properties

Modulus (absolute value) of a real number denoted by and is defined as follows:

TO simple properties module include the following relations:

Note, that the last two properties are valid for any even degree.

Moreover, if, where, then and

More complex module properties, which can be effectively used when solving equations with moduli, are formulated through the following theorems:

Theorem 1.For any analytical functions And inequality is true

Theorem 2. Equality is equivalent to inequality.

Theorem 3. Equality tantamount to inequality.

Let's look at typical examples of solving problems on the topic “Equations, containing variables under the modulus sign."

Solving equations with modulus

The most common method in school mathematics for solving equations with a modulus is the method, based on module expansion. This method is universal, however, in the general case, its use can lead to very cumbersome calculations. In this regard, students should know other, more effective methods and techniques for solving such equations. In particular, it is necessary to have skills in applying theorems, given in this article.

Example 1. Solve the equation. (1)

Solution. We will solve Equation (1) using the “classical” method – the method of revealing modules. To do this, let's split the number axis dots and into intervals and consider three cases.

1. If , then , , , and equation (1) takes the form . It follows from this. However, here , therefore the value found is not the root of equation (1).

2. If, then from equation (1) we obtain or .

Since then root of equation (1).

3. If, then equation (1) takes the form or . Let's note that.

Answer: , .

When solving subsequent equations with a module, we will actively use the properties of modules in order to increase the efficiency of solving such equations.

Example 2. Solve the equation.

Solution. Since and then from the equation it follows. In this regard, , , and the equation takes the form. From here we get. However , therefore the original equation has no roots.

Answer: no roots.

Example 3. Solve the equation.

Solution. Since, then. If , then and the equation takes the form.

From here we get .

Example 4. Solve the equation.

Solution.Let us rewrite the equation in equivalent form. (2)

The resulting equation belongs to equations of type .

Taking into account Theorem 2, it can be argued that equation (2) is equivalent to the inequality . From here we get .

Answer: .

Example 5. Solve the equation.

Solution. This equation has the form. That's why , according to Theorem 3, here we have inequality or .

Example 6. Solve the equation.

Solution. Let's assume that. Because , then the given equation takes the form of a quadratic equation, (3)

Where . Since equation (3) has a single positive root and , then . From here we get two roots of the original equation: And .

Example 7. Solve the equation. (4)

Solution. Since the equationis equivalent to the combination of two equations: And , then when solving equation (4) it is necessary to consider two cases.

1. If , then or .

From here we get , and .

2. If , then or .

Since, then.

Answer: , , , .

Example 8.Solve the equation . (5)

Solution. Since and , then . From here and from equation (5) it follows that and , i.e. here we have a system of equations

However, this system of equations is inconsistent.

Answer: no roots.

Example 9. Solve the equation. (6)

Solution. If we denote , then and from equation (6) we obtain

Or . (7)

Since equation (7) has the form , this equation is equivalent to the inequality . From here we get . Since , then or .

Answer: .

Example 10.Solve the equation. (8)

Solution.According to Theorem 1, we can write

(9)

Taking into account equation (8), we conclude that both inequalities (9) turn into equalities, i.e. there is a system of equations

However, according to Theorem 3, the above system of equations is equivalent to the system of inequalities

(10)

Solving the system of inequalities (10) we obtain . Since the system of inequalities (10) is equivalent to equation (8), the original equation has a single root.

Answer: .

Example 11. Solve the equation. (11)

Solution. Let and , then the equality follows from equation (11).

It follows that and . Thus, here we have a system of inequalities

The solution to this system of inequalities is And .

Answer: , .

Example 12.Solve the equation. (12)

Solution. Equation (12) will be solved by the method of sequential expansion of modules. To do this, let's consider several cases.

1. If , then .

1.1. If , then and , .

1.2. If, then. However , therefore, in this case, equation (12) has no roots.

2. If , then .

2.1. If , then and , .

2.2. If , then and .

Answer: , , , , .

Example 13.Solve the equation. (13)

Solution. Since the left side of equation (13) is non-negative, then . In this regard, and equation (13)

takes the form or .

It is known that the equation is equivalent to the combination of two equations And , solving which we get, . Because , then equation (13) has one root.

Answer: .

Example 14. Solve system of equations (14)

Solution. Since and , then and . Consequently, from the system of equations (14) we obtain four systems of equations:

The roots of the above systems of equations are the roots of the system of equations (14).

Answer: ,, , , , , , .

Example 15. Solve system of equations (15)

Solution. Since, then. In this regard, from the system of equations (15) we obtain two systems of equations

The roots of the first system of equations are and , and from the second system of equations we obtain and .

Answer: , , , .

Example 16. Solve system of equations (16)

Solution. From the first equation of system (16) it follows that .

Since then . Let's consider the second equation of the system. Because the, That , and the equation takes the form, , or .

If you substitute the valueinto the first equation of system (16), then , or .

Answer: , .

For a deeper study of problem solving methods, related to solving equations, containing variables under the modulus sign, can you advise teaching aids from the list of recommended literature.

1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.

2. Suprun V.P. Mathematics for high school students: problems increased complexity. – M.: CD “Librocom” / URSS, 2017. – 200 p.

3. Suprun V.P. Mathematics for high school students: non-standard methods for solving problems. – M.: CD “Librocom” / URSS, 2017. – 296 p.

Still have questions?

To get help from a tutor, register.

website, when copying material in full or in part, a link to the source is required.

Among examples per module Often there are equations where you need to find module roots in a module, that is, an equation of the form
||a*x-b|-c|=k*x+m .
If k=0, that is, the right side is equal to a constant (m), then it’s easier to look for a solution equations with modules graphically. Below is the method opening of double modules using examples common in practice. Understand the algorithm for calculating equations with modules well, so that you don’t have problems on quizzes, tests, and just to know.

Example 1. Solve the equation modulo |3|x|-5|=-2x-2.
Solution: Always start opening equations from the internal module
|x|=0 <->x=0.
At the point x=0, the equation with modulus is divided by 2.
At x< 0 подмодульная функция отрицательная, поэтому при раскрытии знак меняем на противоположный
|-3x-5|=-2x-2.
For x>0 or equal, expanding the module we get
|3x-5|=-2x-2 .
Let's solve the equation for negative variables (x< 0) . Оно разлагается на две системы уравнений. Первое уравнение получаем из условия, что функция после знака равенства неотрицательна. Второе - раскрывая модуль в одной системе принимаем, что подмодульная функция положительная, в иной отрицательная - меняем знак правой или левой части (зависит от методики преподавания).

From the first equation we get that the solution should not exceed (-1), i.e.

This limitation entirely belongs to the area in which we are solving. Let's move variables and constants to opposite sides of equality in the first and second systems

and find a solution


Both values ​​belong to the interval that is being considered, that is, they are roots.
Consider an equation with moduli for positive variables
|3x-5|=-2x-2.
Expanding the module we get two systems of equations

From the first equation, which is common to the two systems, we obtain the familiar condition

which, in intersection with the set on which we are looking for a solution, gives an empty set (there are no points of intersection). So the only roots of a module with a module are the values
x=-3; x=-1.4.

Example 2. Solve the equation with modulus ||x-1|-2|=3x-4.
Solution: Let's start by opening the internal module
|x-1|=0 <=>x=1.
A submodular function changes sign at one. For smaller values ​​it is negative, for larger values ​​it is positive. In accordance with this, when expanding the internal module, we obtain two equations with the module
x |-(x-1)-2|=3x-4;
x>=1 -> |x-1-2|=3x-4.
Be sure to check the right side of the modulus equation; it must be greater than zero.
3x-4>=0 -> x>=4/3.
This means that there is no need to solve the first equation, since it was written for x< 1, что не соответствует найденному условию. Раскроем модуль во втором уравнении
|x-3|=3x-4 ->
x-3=3x-4 or x-3=4-3x;
4-3=3x-x or x+3x=4+3;
2x=1 or 4x=7;
x=1/2 or x=7/4.
We received two values, the first of which is rejected because it does not belong to the required interval. Finally, the equation has one solution x=7/4.

Example 3. Solve the equation with modulus ||2x-5|-1|=x+3.
Solution: Let's open the internal module
|2x-5|=0 <=>x=5/2=2.5.
The point x=2.5 splits the number line into two intervals. Respectively, submodular function changes sign when passing through 2.5. Let us write down the condition for the solution with right side equations with modulus.
x+3>=0 -> x>=-3.
So the solution can be values ​​no less than (-3) . Let's expand the module for the negative value of the internal module
|-(2x-5)-1|=x+3;
|-2x+4|=x+3.
This module will also give 2 equations when expanded
-2x+4=x+3 or 2x-4=x+3;
2x+x=4-3 or 2x-x=3+4;
3x=1; x=1/3 or x=7 .
We reject the value x=7, since we were looking for a solution in the interval [-3;2.5]. Now we open the internal module for x>2.5. We get an equation with one module
|2x-5-1|=x+3;
|2x-6|=x+3.
When expanding the module, we obtain the following linear equations
-2x+6=x+3 or 2x-6=x+3;
2x+x=6-3 or 2x-x=3+6;
3x=3; x=1 or x=9 .
The first value x=1 does not satisfy the condition x>2.5. So on this interval we have one root of the equation with modulus x=9, and there are two in total (x=1/3). By substitution you can check the correctness of the calculations performed
Answer: x=1/3; x=9.

Example 4. Find solutions to the double module ||3x-1|-5|=2x-3.
Solution: Let's expand the internal module of the equation
|3x-1|=0 <=>x=1/3.
The point x=2.5 divides the number line into two intervals and the given equation into two cases. We write down the condition for the solution based on the form of the equation on the right side
2x-3>=0 -> x>=3/2=1.5.
It follows that we are interested in values ​​>=1.5. Thus modular equation consider on two intervals
,
|-(3x-1)-5|=2x-3;
|-3x-4|=2x-3.
The resulting module, when expanded, is divided into 2 equations
-3x-4=2x-3 or 3x+4=2x-3;
2x+3x=-4+3 or 3x-2x=-3-4;
5x=-1; x=-1/5 or x=-7 .
Both values ​​do not fall into the interval, that is, they are not solutions to the equation with moduli. Next, we will expand the module for x>2.5. We get the following equation
|3x-1-5|=2x-3;
|3x-6|=2x-3.
Expanding the module, we get 2 linear equations
3x-6=2x-3 or –(3x-6)=2x-3;
3x-2x=-3+6 or 2x+3x=6+3;
x=3 or 5x=9; x=9/5=1.8.
The second value found does not correspond to the condition x>2.5, we reject it.
Finally we have one root of the equation with moduli x=3.
Performing a check
||3*3-1|-5|=2*3-3 3=3 .
The root of the equation with the modulus was calculated correctly.
Answer: x=1/3; x=9.

In this article we will analyze in detail the absolute value of a number. We will give various definitions of the modulus of a number, introduce notation and provide graphic illustrations. At the same time, let's look at various examples of finding the modulus of a number by definition. After this, we will list and justify the main properties of the module. At the end of the article, we’ll talk about how the modulus of a complex number is determined and found.

Page navigation.

Number module - definition, notation and examples

First we introduce number modulus designation. We will write the modulus of the number a as , that is, to the left and right of the number we will put vertical dashes to form the modulus sign. Let's give a couple of examples. For example, module −7 can be written as ; module 4.125 is written as , and the module has a notation of the form .

The following definition of modulus refers to , and therefore to , and to integers, and to rational, and to irrational numbers, as constituent parts of the set of real numbers. We will talk about the modulus of a complex number in.

Definition.

Modulus of number a– this is either the number a itself, if a is a positive number, or the number −a, the opposite of the number a, if a is a negative number, or 0, if a=0.

The voiced definition of the modulus of a number is often written in the following form , this entry means that if a>0 , if a=0 , and if a<0 .

The record can be presented in a more compact form . This notation means that if (a is greater than or equal to 0), and if a<0 .

There is also the entry . Here we should separately explain the case when a=0. In this case we have , but −0=0, since zero is considered a number that is opposite to itself.

Let's give examples of finding the modulus of a number using a stated definition. For example, let's find the modules of the numbers 15 and . Let's start by finding . Since the number 15 is positive, its modulus, by definition, is equal to this number itself, that is, . What is the modulus of a number? Since is a negative number, its modulus is equal to the number opposite to the number, that is, the number . Thus, .

To conclude this point, we present one conclusion that is very convenient to use in practice when finding the modulus of a number. From the definition of the modulus of a number it follows that the modulus of a number is equal to the number under the modulus sign without taking into account its sign, and from the examples discussed above this is very clearly visible. The stated statement explains why the module of a number is also called absolute value of the number. So the modulus of a number and the absolute value of a number are one and the same.

Modulus of a number as a distance

Geometrically, the modulus of a number can be interpreted as distance. Let's give determining the modulus of a number through distance.

Definition.

Modulus of number a– this is the distance from the origin on the coordinate line to the point corresponding to the number a.

This definition is consistent with the definition of the modulus of a number given in the first paragraph. Let's clarify this point. The distance from the origin to the point corresponding to a positive number is equal to this number. Zero corresponds to the origin, therefore the distance from the origin to the point with coordinate 0 is equal to zero (you do not need to set aside a single unit segment and not a single segment that makes up any fraction of a unit segment in order to get from point O to a point with coordinate 0). The distance from the origin to a point with a negative coordinate is equal to the number opposite to the coordinate of this point, since it is equal to the distance from the origin to the point whose coordinate is the opposite number.

For example, the modulus of the number 9 is equal to 9, since the distance from the origin to the point with coordinate 9 is equal to nine. Let's give another example. The point with coordinate −3.25 is located at a distance of 3.25 from point O, so .

The stated definition of the modulus of a number is a special case of the definition of the modulus of the difference of two numbers.

Definition.

Modulus of the difference of two numbers a and b is equal to the distance between the points of the coordinate line with coordinates a and b.

That is, if points on the coordinate line A(a) and B(b) are given, then the distance from point A to point B is equal to the modulus of the difference between the numbers a and b. If we take point O (origin) as point B, then we get the definition of the modulus of a number given at the beginning of this paragraph.

Determining the modulus of a number using the arithmetic square root

Occasionally occurs determining modulus via arithmetic square root.

For example, let's calculate the moduli of the numbers −30 and based on this definition. We have. Similarly, we calculate the module of two thirds: .

The definition of the modulus of a number through the arithmetic square root is also consistent with the definition given in the first paragraph of this article. Let's show it. Let a be a positive number, and let −a be a negative number. Then And , if a=0 , then .

Module properties

The module has a number of characteristic results - module properties. Now we will present the main and most frequently used of them. When justifying these properties, we will rely on the definition of the modulus of a number in terms of distance.

Let's start with the most obvious property of the module - The modulus of a number cannot be a negative number. In literal form, this property has the form for any number a. This property is very easy to justify: the modulus of a number is a distance, and distance cannot be expressed as a negative number.

Let's move on to the next module property. The modulus of a number is zero if and only if this number is zero. The modulus of zero is zero by definition. Zero corresponds to the origin; no other point on the coordinate line corresponds to zero, since each real number is associated with a single point on the coordinate line. For the same reason, any number other than zero corresponds to a point different from the origin. And the distance from the origin to any point other than point O is not zero, since the distance between two points is zero if and only if these points coincide. The above reasoning proves that only the modulus of zero is equal to zero.

Go ahead. Opposite numbers have equal modules, that is, for any number a. Indeed, two points on the coordinate line, the coordinates of which are opposite numbers, are at the same distance from the origin, which means the modules of the opposite numbers are equal.

The following property of the module is: The modulus of the product of two numbers is equal to the product of the moduli of these numbers, that is, . By definition, the modulus of the product of numbers a and b is equal to either a·b if , or −(a·b) if . From the rules of multiplication of real numbers it follows that the product of the moduli of numbers a and b is equal to either a·b, , or −(a·b) if , which proves the property in question.

The modulus of the quotient of a divided by b is equal to the quotient of the modulus of a number divided by the modulus of b, that is, . Let us justify this property of the module. Since the quotient is equal to the product, then. By virtue of the previous property we have . All that remains is to use the equality , which is valid by virtue of the definition of the modulus of a number.

The following property of a module is written as an inequality: , a , b and c are arbitrary real numbers. The written inequality is nothing more than triangle inequality. To make this clear, let’s take points A(a), B(b), C(c) on the coordinate line, and consider a degenerate triangle ABC, whose vertices lie on the same line. By definition, the modulus of the difference is equal to the length of the segment AB, - the length of the segment AC, and - the length of the segment CB. Since the length of any side of a triangle does not exceed the sum of the lengths of the other two sides, then the inequality is true , therefore, the inequality is also true.

The inequality just proved is much more common in the form . The written inequality is usually considered as a separate property of the module with the formulation: “ The modulus of the sum of two numbers does not exceed the sum of the moduli of these numbers" But the inequality follows directly from the inequality if we put −b instead of b and take c=0.

Modulus of a complex number

Let's give definition of the modulus of a complex number. May it be given to us complex number, written in algebraic form, where x and y are some real numbers, representing, respectively, the real and imaginary parts of a given complex number z, and is the imaginary unit.

One of the most difficult topics for students is solving equations containing a variable under the modulus sign. Let's first figure out what this is connected with? Why, for example, do most children crack quadratic equations like nuts, but have so many problems with such a far from complex concept as a module?

In my opinion, all these difficulties are associated with the lack of clearly formulated rules for solving equations with a modulus. So, when solving a quadratic equation, the student knows for sure that he needs to first apply the discriminant formula, and then the formulas for the roots of the quadratic equation. What to do if a modulus is found in the equation? We will try to clearly describe the necessary action plan for the case when the equation contains an unknown under the modulus sign. We will give several examples for each case.

But first, let's remember module definition. So, modulo the number a this number itself is called if a non-negative and -a, if number a less than zero. You can write it like this:

|a| = a if a ≥ 0 and |a| = -a if a< 0

Speaking about the geometric meaning of the module, it should be remembered that each real number corresponds to a certain point on the number axis - its coordinate. So, the module or absolute value of a number is the distance from this point to the origin of the numerical axis. The distance is always specified as a positive number. Thus, the modulus of any negative number is a positive number. By the way, even at this stage, many students begin to get confused. The module can contain any number, but the result of using the module is always a positive number.

Now let's move directly to solving the equations.

1. Consider an equation of the form |x| = c, where c is a real number. This equation can be solved using the modulus definition.

We divide all real numbers into three groups: those that are greater than zero, those that are less than zero, and the third group is the number 0. We write the solution in the form of a diagram:

(±c, if c > 0

If |x| = c, then x = (0, if c = 0

(no roots if with< 0

1) |x| = 5, because 5 > 0, then x = ±5;

2) |x| = -5, because -5< 0, то уравнение не имеет корней;

3) |x| = 0, then x = 0.

2. Equation of the form |f(x)| = b, where b > 0. To solve this equation it is necessary to get rid of the module. We do it this way: f(x) = b or f(x) = -b. Now you need to solve each of the resulting equations separately. If in the original equation b< 0, решений не будет.

1) |x + 2| = 4, because 4 > 0, then

x + 2 = 4 or x + 2 = -4

2) |x 2 – 5| = 11, because 11 > 0, then

x 2 – 5 = 11 or x 2 – 5 = -11

x 2 = 16 x 2 = -6

x = ± 4 no roots

3) |x 2 – 5x| = -8, because -8< 0, то уравнение не имеет корней.

3. An equation of the form |f(x)| = g(x). According to the meaning of the module, such an equation will have solutions if its right-hand side is greater than or equal to zero, i.e. g(x) ≥ 0. Then we will have:

f(x) = g(x) or f(x) = -g(x).

1) |2x – 1| = 5x – 10. This equation will have roots if 5x – 10 ≥ 0. This is where the solution of such equations begins.

1. O.D.Z. 5x – 10 ≥ 0

2. Solution:

2x – 1 = 5x – 10 or 2x – 1 = -(5x – 10)

3. We combine O.D.Z. and the solution, we get:

The root x = 11/7 does not fit the O.D.Z., it is less than 2, but x = 3 satisfies this condition.

Answer: x = 3

2) |x – 1| = 1 – x 2 .

1. O.D.Z. 1 – x 2 ≥ 0. Let’s solve this inequality using the interval method:

(1 – x)(1 + x) ≥ 0

2. Solution:

x – 1 = 1 – x 2 or x – 1 = -(1 – x 2)

x 2 + x – 2 = 0 x 2 – x = 0

x = -2 or x = 1 x = 0 or x = 1

3. We combine the solution and O.D.Z.:

Only roots x = 1 and x = 0 are suitable.

Answer: x = 0, x = 1.

4. Equation of the form |f(x)| = |g(x)|. Such an equation is equivalent to the following two equations f(x) = g(x) or f(x) = -g(x).

1) |x 2 – 5x + 7| = |2x – 5|. This equation is equivalent to the following two:

x 2 – 5x + 7 = 2x – 5 or x 2 – 5x +7 = -2x + 5

x 2 – 7x + 12 = 0 x 2 – 3x + 2 = 0

x = 3 or x = 4 x = 2 or x = 1

Answer: x = 1, x = 2, x = 3, x = 4.

5. Equations solved by the substitution method (variable replacement). This method solutions are easiest to explain in specific example. So, let us be given a quadratic equation with modulus:

x 2 – 6|x| + 5 = 0. By the modulus property x 2 = |x| 2, so the equation can be rewritten as follows:

|x| 2 – 6|x| + 5 = 0. Let's make the replacement |x| = t ≥ 0, then we will have:

t 2 – 6t + 5 = 0. Solving this equation, we find that t = 1 or t = 5. Let’s return to the replacement:

|x| = 1 or |x| = 5

x = ±1 x = ±5

Answer: x = -5, x = -1, x = 1, x = 5.

Let's look at another example:

x 2 + |x| – 2 = 0. By the modulus property x 2 = |x| 2, therefore

|x| 2 + |x| – 2 = 0. Let’s make the replacement |x| = t ≥ 0, then:

t 2 + t – 2 = 0. Solving this equation, we get t = -2 or t = 1. Let’s return to the replacement:

|x| = -2 or |x| = 1

No roots x = ± 1

Answer: x = -1, x = 1.

6. Another type of equations is equations with a “complex” modulus. Such equations include equations that have “modules within a module.” Equations of this type can be solved using the properties of the module.

1) |3 – |x|| = 4. We will act in the same way as in equations of the second type. Because 4 > 0, then we get two equations:

3 – |x| = 4 or 3 – |x| = -4.

Now let us express the modulus x in each equation, then |x| = -1 or |x| = 7.

We solve each of the resulting equations. There are no roots in the first equation, because -1< 0, а во втором x = ±7.

Answer x = -7, x = 7.

2) |3 + |x + 1|| = 5. We solve this equation in a similar way:

3 + |x + 1| = 5 or 3 + |x + 1| = -5

|x + 1| = 2 |x + 1| = -8

x + 1 = 2 or x + 1 = -2. No roots.

Answer: x = -3, x = 1.

There is also a universal method for solving equations with a modulus. This is the interval method. But we will look at it later.

website, when copying material in full or in part, a link to the source is required.

The modulus is the absolute value of the expression. To somehow indicate a module, it is customary to use straight brackets. The value that is enclosed in even brackets is the value that is taken modulo. The process of solving any module consists in opening those very straight brackets, which in mathematical language are called modular brackets. Their disclosure occurs according to a certain number of rules. Also, in the order of solving the modules, the sets of values ​​of those expressions that were in the modular brackets are found. In most cases, the module is expanded in such a way that the expression that was submodular receives both positive and negative values, including the value zero. If we start from installed properties module, then in the process various equations or inequalities from the original expression are compiled, which then need to be solved. Let's figure out how to solve modules.

Solution process

Solving a module begins by writing the original equation with the module. To answer the question of how to solve equations with a modulus, you need to open it completely. To solve such an equation, the module is expanded. All modular expressions must be considered. It is necessary to determine at what values ​​of the unknown quantities included in its composition, the modular expression in brackets becomes zero. In order to do this, it is enough to equate the expression in modular brackets to zero, and then calculate the solution to the resulting equation. The found values ​​must be recorded. In the same way, you also need to determine the value of all unknown variables for all modules in this equation. Next, you need to start defining and considering all cases of the existence of variables in expressions when they are different from the value zero. To do this, you need to write down some system of inequalities corresponding to all modules in the original inequality. Inequalities must be written so that they cover all available and possible values ​​for a variable that are found on the number line. Then you need to draw this same number line for visualization, on which to later plot all the obtained values.

Almost everything can now be done on the Internet. The module is no exception to the rule. You can solve it online on one of the many modern resources. All those values ​​of the variable that are in the zero module will be a special constraint that will be used in the process of solving the modular equation. In the original equation, you need to open all the available modular brackets, while changing the sign of the expression so that the values ​​of the desired variable coincide with those values ​​that are visible on the number line. The resulting equation must be solved. The value of the variable that will be obtained during solving the equation must be checked against the limitation that is specified by the module itself. If the value of the variable fully satisfies the condition, then it is correct. All roots that will be obtained during the solution of the equation, but will not fit the restrictions, must be discarded.