What is the root of an equation? Is the number 2 the root of the equation x 3 - x = 6? What is the root of the equation? Is the number 2 the root of the equation x 3 - x = 6? What does it mean to solve an equation? What does it mean to solve an equation? What equations are called equivalent? Formulate the conditions for the transition from this equation to an equivalent equation. Give an example of two equivalent equations. Which equations are called equivalent? Formulate the conditions for the transition from this equation to an equivalent equation. Give an example of two equivalent equations. Which equation is called a linear equation with one variable? Which equation is called a linear equation with one variable? How many roots can a linear equation in one variable have? Give examples. How many roots can a linear equation with one variable have? Give examples.

Orally: find the root of the equation: 6x + 1 = 43; 12x + 2 = O; -x - 4 = 11; 1-27x = 0; 1.5 + x = 0; 2 = .5x; 5x - 8 = 1.5; x+ 2 = 0. 0 = 16 - x;

1 3 Find the root of the equation: 3.5 – 3x = 2.3 +x; x=0.3 x= x-1.4 + 6x = 0; x=0.2 1.2 = 2x + x- 1.5; x=0.2x = 5 - 0.3x; x=0.8 2.6 = x- 0.4x-4. x=1.1 1 3

Example 2 Solve the equation: 3 X+2 4 3X-1 - = = X+2 4 3X-1 - () 4 (3X-1). 12 -=-24 3 (X+2). 12 (X+2). 4- (3X-1). 3=-24 4x+8-9x+3=-24 X=7Answer: 7 4 3

The study of equations in the middle level begins with the introduction of the solution linear equations and equations that can be reduced to linear ones.

The equality of two functions considered in a common domain of definition is called an equation. The variables included in the equation are denoted by the Latin letters x, y, z, t... An equation with one variable x in general form is written as f(x)= g(x).

Any value of a variable at which the expressions f(x) and g(x) take on equal numerical values ​​is called the root of the equation.

Solving an equation means finding all its roots or proving that they do not exist.

For example, the equation 3+x=7 has a single root 4, since with this and only with this value of the variable 3+x=7 the equality is true.

The equation (x-1)(x-2)=0 has 2 roots 1 and 2.

The equation x 2 +1=0 has no real roots because the sum of two positive numbers does not equal 0.

In order to solve any equation with one variable, the student must know: firstly, the rules, formulas or algorithms for solving equations of this type and, secondly, the rules for performing identical and equivalent transformations, with the help of which this equation can be reduced to the simplest .

Thus, the solution to each equation consists of two main parts:

1. transforming this equation to the simplest;
2. solving simple equations using known rules, formulas or algorithms.

If the second part is algorithmic, then the first part is largely heuristic, which poses the greatest difficulty for students. In the process of solving an equation, they try to replace it with a simpler one, so it is important to know with the help of what transformations this is possible. Here it is necessary to give the concept of equivalence in a form accessible to the child.

Equations that have the same roots are called equivalent. Equations are also considered equivalent if each of them has no roots.

For example, the equations x+2=5 and x+5=8 are equivalent, since each of them has a single root - the number 3. The equations x 2 +1=0 and 2x 2 +5=0 are also equivalent - neither of them has roots.

The equations x-5=1 and x2=36 are not equivalent, since the first has only one root x=6, while the second has two roots 6 and –6.

Equivalent transformations include:

1) If the same number or the same integer algebraic expression containing an unknown is added to both sides of the equation, then the new equation will be equivalent to the given one.

2) If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.

For example, the equation is equivalent to x 2 – 1 = 6x

3) If you open the brackets in the equation and add similar terms, you will get an equation equivalent to the given one.

Learning to solve equations begins with the simplest linear equations and equations that reduce to them. A definition of a linear equation is given and cases when it has one solution are considered; has no solutions and has infinite set decisions.

A linear equation with one variable x is an equation of the form ax = b, where a and b are real numbers, a is called the coefficient of the variable, b is the free term.

For a linear equation ax = b can be represented in case:

As a result of transformations, many equations are reduced to linear ones.

So in 7th grade you can apply the following equations:

1)

This equation reduces to a linear equation.

Multiplying both sides by 12 (least common short denominators 3, 4, 6, 12), we get:

8 + 3x + 2 – 2x = 5x –12,

8 + 2 + 12 = 5x – 3x + 2x,

Answer: 5.5.

2) Let us show that the equation 2 (x + 1) – 1 = 3 - (1 - 2x) has no roots.

Let's simplify both sides of the equation:

2x + 2 – 1 = 3 – 1 + 2x,

2x + 1 = 2 + 2x,

2x - 2x = 2 - 1,

This equation has no roots, because the left side of 0 x is equal to 0 for any x, and therefore is not equal to 1.

3) Let us show that the equation 3(1 – x) + 2 = 5 – 3x has an infinite number of roots.

When covering the topic “linear equations in two variables,” you can offer students a graphical way to solve the equation. This method is based on the use of graphs of functions included in the equation. The essence of the method: find the abscissa of the intersection points of the graphs of functions on the left and right sides of the equation. Based on the following actions:

1) Transform the original equation to the form f(x) = g(x), where f(x) and g(x) are functions, graphs that can be constructed.
2) Graph the functions f(x) and g(x)
3) Determine the intersection points of the constructed graphs.
4) Determine the abscissa of the found points. They will give many solutions to the original equation.
5) Write down the answer.

The advantage of this method is that it makes it easy to determine the number of roots of an equation. The disadvantage is that the roots are generally determined approximately.

The next stage in the study of linear equations are equations with moduli, and some solutions are performed in several ways.

Solving equations containing a modulus sign and equations with parameters can be called an activity close to research. This is due to the fact that the choice of solution method, the solution process, and recording the answer presuppose a certain level of development of the skills to observe, compare, analyze, put forward and test a hypothesis, and generalize the results obtained.

Of particular interest are equations containing the modulus sign.

By definition of the modulus of a number, we have:

The number –a can be negative when a>0; -a positive when a<0. из определения видно, что модуль любого числа неотрицателен. Оно же показывает, как избавиться от модуля в алгебраических выражениях.

Therefore x=5 or x=-5.

Consider the equation.

There are two ways to solve the equation.

1 way. By definition of the modulus of a number we have:

Therefore x - 3 = 7 or –x + 3 = 7,

x = 10 or x = -4.

Answer: 10; -4.

Method 2 – graphic. The equation can be written as a system of two equations:

Let's build graphs of functions and .

The abscissas of the intersection points of these graphs are the solution to the equation.

Answer: -4; 10.

Let's solve an equation containing more than one module

Let's use the following algorithm.

1. Mark all zeros of submodular expressions on a number line, divided into intervals on which all submodular expressions have a constant sign.
2. Take an arbitrary number from each interval and, by counting, determine the sign of the submodular expression and expand the modules.
3. Solve the equation and choose a solution that belongs to the given interval.

So, submodular expressions vanish at x = -1 and x = -3.

I interval. Let x < - 3, then on this interval , and the equation will take the form

– x – 1 – x – 3 = 4,

and therefore is the root of the equation.

II interval. Let -3< х < -1, тогда , , we get the equation –x – 1 + x + 3 = 4,

This means that the equation has no roots on the interval (-3; -1).

III interval. Let x > -1, then

x + 1 + x + 3 = 4,

We see that the number 0 belongs to the interval. So it is the root. So the equation has two roots: 0 and -4.

On simple examples Let's consider an algorithm for solving equations with parameters: range of permissible values, domain of definition, general solutions, control values ​​of parameters, types of particular equations. Methods for finding them will be established in each type of equation separately.

Based on the introduced concepts, we define a general scheme for solving any equation F(a;x)=0 with parameter a (for the case of two parameters the scheme is similar):

• the range of permissible parameter values ​​and the scope of definition are established;
• control values ​​of the parameter are determined, dividing the region of permissible parameter values ​​into regions of similarity of partial equations;
• for control values ​​of the parameter, the corresponding partial equations are studied separately;
• general solutions x=f 1 (a),…, f k (a) of the equation F(a;x)=0 are found on the corresponding sets A f1 ,…, A fk of parameter values;
• a model of general solutions and control parameter values ​​is compiled;
• on the model, intervals of parameter values ​​with the same general solutions(areas of uniformity);
• for control values ​​of the parameter and selected areas of uniformity, the characteristics of all types of partial equations are recorded
• A special place in algebra is given to linear equations with parameters.

Let's look at a few examples.

1.	2x – 3 = m+1, 2х – 3 = + 4 m + 1,	where m is an unknown parameter. Multiplying both sides of the equation by 3, we get
	6x – 9 = m x + 12m +3, 6x - m x + 12m + 12,	Taking the common factor out of brackets, we get
	x (6-m) = 12(m+1), , 6 – m ? 0, m ? 6.	since it is in the denominator of the fraction.
Answer: , at m 6.

The equation 2x – 3 + m (x/3 + 4) + 1 has many solutions given by the formula for all values ​​of m except 6.

2. , with m 2, x 1, n 0.

mx – n = 2x – 2 + 2n + 3xn,

mx – 2x – 3xn = - 2 + 2n +n,

mx – 2x – 3xn = 3n – 2,

x (m – 2 – 3n) = 3n – 2, with m 2, x 1, n 0.

Consider the case where a = 0, then

m – 2 – 3n = 0,

m = 3n +2, for n 0

0 x = 3n – 2,

a) 3n ​​– 2 = 0,

x(4 – 2 – 3) = 3 - 2,

x – any number except x = 1.

0 x = b. In this case, the equation has no solutions.

m – 2 – 3n 0

x = , for x ? 1,

3n – 2 m – 2 – 3n,

3n + 3n 2 – 2 + m,

In this case, the equation has no solutions.

This means that with n = and m = 4, x is any number except 1; at n = 0, m = 6n

(n), m = 3n + 2 (n), m = 2 the equation has no solutions. For all other parameter values ​​x = .

Answer: 1. n = , m = 4 – x ? R\.

2. n = 0, m = 6n (n), m = 3n + 2 (n), m = 2 – there are no solutions.

3. n 0, m 6n, m 3n + 2, m 2 – x = .

In the future, it is proposed to consider solving problems by composing linear equations. This difficult process, where you need to be able to think, guess, and know the actual material well.

In the process of solving each problem, four stages must be clearly marked:

1. studying the conditions of the problem;
2. searching for a solution plan and drawing it up;
3. registration of the solution found;
4. critical analysis of the decision result.

Now let's look at problems that use linear equations to solve.

1. An alloy of copper and zinc contains 640 g more copper than zinc. After 6/7 of the copper and 60% of zinc contained in it were isolated from the alloy, the mass of the alloy turned out to be equal to 200 g. What was the mass of the alloy initially?

Let there be x g of zinc in the alloy, then copper (640 + x) g after 6/7 of copper and 60% of zinc have been isolated, 1/7 of copper and 40% of zinc remain, i.e. 0.4 parts. Knowing that the mass of the alloy turned out to be equal to 200 g, we will create an equation.

1/7 (x + 640) + 0.4 x = 200,

x + 640 + 2.8 x =1400,

3.8x = 1400 – 640,

This means that there was 200 g of zinc and 840 g of copper.

(200 + 640 = 840). 1) 200 + 840 = 1040 (g) – mass of the alloy. Answer: the initial mass of the alloy is 1040 g.

2. How many liters of 60% sulfuric acid must be added to 10 liters of 30% acid to obtain a 40% solution?

Let the number of liters of 60% acid that we add be x liters, then the solution of pure acid will be liters. And in 10 liters of a 30% solution of pure acid there will be liters. Knowing that the resulting (10 + x) mixture will contain pure acid l, we will create an equation.

60x + 300 = 40x + 400,

60x – 40x = 400 – 300,

This means you need to add 5 liters of 60% acid.

Answer: 5 l.

When studying the topic “Solving Linear Equations,” some historical background is recommended.

Problems for solving equations of the first degree are found in Babylonian cuneiform texts. They also contain some problems leading to quadratic and even cubic equations (the latter, apparently, were solved by selecting roots). Ancient Greek mathematicians found geometric shape solving a quadratic equation. In geometric form, the Arab mathematician Omar Khayyam (late 11th - early 12th century AD) studied the cubic equation, although he did not find general formula to solve it. The solution to the cubic equation was found at the beginning of the 16th century in Italy. After Scipian del Ferro decided one private view In 1535, the Italian Tartaglia found a general formula for such equations. He proved that the roots of the equation x 3 + px + q = 0 are x = .

This expression is usually called Cardano's formula, after the name of the scientist who learned it from Tartaglia and published it in 1545 in his book “The Great Art of Algebraic Rules.” Cardano's student, the young mathematician Ferrari, solved a general equation of the fourth degree. After this, the search for a formula for solving fifth-degree equations continued for two and a half centuries. In 1823, the remarkable Norwegian mathematician Niels Hendrik Abel (1802-1829) proved that such a formula does not exist. More precisely, he proved that the roots of a general equation of the fifth degree cannot be expressed in terms of its coefficients using arithmetic operations and root extraction operations. An in-depth study of the question of the conditions for the solvability of equations in radicals was carried out by the French mathematician Evariste Galois (1811-1832), who died in a duel at the age of 21. Some problems of Galois theory were solved by the Soviet algebraist I.T.Shafarevich.

Along with the search for a formula for solving a fifth-degree equation, other research was carried out in the field of the theory of algebraic equations. Vieta established a connection between the coefficients of an equation and its roots. He proved that if x 1 ,…,x n are the roots of the equation x n + a 1 x n-1 +…+a n =0, then the following formulas hold:

x 1 + x 2 + … + x n = -a,
x 1 x 2 + x 2 x 3 + … + x n-1 x n =a 2
……………………………
x 1 x 2 … x n = (-1) n d n .

Literature:

1. Magazine “Mathematics at school” 6, 1999
2. Supplement to the newspaper “First of September” - mathematics 20, 1999.
3. S.I. Tumanov “Algebra”, a manual for students in grades 6-8.
4. N.I. Alexandrov; I. P. Yarandai “Dictionary-reference book on mathematics.”
5. ABOUT. Episheva; IN AND. Krupich “Teaching schoolchildren to learn mathematics.”
6. E.I. Yamshchenko “Study of functions.”
7. A.I. Khudobin; M.F. Shurshalov “Collection of problems on algebra and elementary functions.”
8. Sh. A. Alimov, V.A. Ilyin “Algebra grades 6-8.”

In this video we will analyze a whole set of linear equations that are solved using the same algorithm - that’s why they are called the simplest.

First, let's define: what is a linear equation and which one is called the simplest?

A linear equation is one in which there is only one variable, and only to the first degree.

The simplest equation means the construction:

All other linear equations are reduced to the simplest using the algorithm:

1. Expand parentheses, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Give similar terms to the left and right of the equal sign;
4. Divide the resulting equation by the coefficient of the variable $x$.

Of course, this algorithm does not always help. The fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. In the video below we will look at several reasons why this situation is possible.
2. The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.

Now let's see how all this works using real-life examples.

Examples of solving equations

Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.

Such constructions are solved in approximately the same way:

1. First of all, you need to expand the parentheses, if there are any (as in our last example);
2. Then combine similar
3. Finally, isolate the variable, i.e. move everything connected with the variable—the terms in which it is contained—to one side, and move everything that remains without it to the other side.

Then, as a rule, you need to give similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “x”, and we will get the final answer.

In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Typically, errors are made either when opening brackets or when calculating the “pluses” and “minuses”.

In addition, it happens that a linear equation has no solutions at all, or that the solution is the entire number line, i.e. any number. We will look at these subtleties in today's lesson. But we will start, as you already understood, with the simplest tasks.

Scheme for solving simple linear equations

First, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the brackets, if any.
2. We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
3. We present similar terms.
4. We divide everything by the coefficient of “x”.

Of course, this scheme does not always work; it contains certain subtleties and tricks, and now we will get to know them.

Solving real examples of simple linear equations

Task No. 1

The first step requires us to open the brackets. But they are not in this example, so we skip this step. In the second step we need to isolate the variables. Note: we're talking about only about individual terms. Let's write it down:

We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:

\[\frac(6x)(6)=-\frac(72)(6)\]

So we got the answer.

Task No. 2

We can see the parentheses in this problem, so let's expand them:

Both on the left and on the right we see approximately the same design, but let's act according to the algorithm, i.e. separating the variables:

Here are some similar ones:

At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.

Task No. 3

The third linear equation is more interesting:

\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]

There are several brackets here, but they are not multiplied by anything, they are simply preceded by different signs. Let's break them down:

We perform the second step already known to us:

\[-x+x+2x=15-6-12+3\]

Let's do the math:

We carry out last step— divide everything by the coefficient of “x”:

\[\frac(2x)(x)=\frac(0)(2)\]

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, I would like to say the following:

• As I said above, not every linear equation has a solution - sometimes there are simply no roots;
• Even if there are roots, there may be zero among them - there is nothing wrong with that.

Zero is the same number as the others; you shouldn’t discriminate against it in any way or assume that if you get zero, then you did something wrong.

Another feature is related to the opening of brackets. Please note: when there is a “minus” in front of them, we remove it, but in parentheses we change the signs to opposite. And then we can open it using standard algorithms: we will get what we saw in the calculations above.

Understanding this simple fact will allow you to avoid making stupid and offensive mistakes in high school, when doing such actions is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complex and when performing various transformations a quadratic function will appear. However, we should not be afraid of this, because if, according to the author’s plan, we are solving a linear equation, then during the transformation process all monomials containing a quadratic function will certainly cancel.

Example No. 1

Obviously, the first step is to open the brackets. Let's do this very carefully:

Now let's take a look at privacy:

\[-x+6((x)^(2))-6((x)^(2))+x=-12\]

Here are some similar ones:

Obviously, this equation has no solutions, so we’ll write this in the answer:

\[\varnothing\]

or there are no roots.

Example No. 2

We perform the same actions. First step:

Let's move everything with a variable to the left, and without it - to the right:

Here are some similar ones:

Obviously, this linear equation has no solution, so we’ll write it this way:

\[\varnothing\],

or there are no roots.

Nuances of the solution

Both equations are completely solved. Using these two expressions as an example, we were once again convinced that even in the simplest linear equations, everything may not be so simple: there can be either one, or none, or infinitely many roots. In our case, we considered two equations, both simply have no roots.

But I would like to draw your attention to another fact: how to work with parentheses and how to open them if there is a minus sign in front of them. Consider this expression:

Before opening, you need to multiply everything by “X”. Please note: multiplies each individual term. Inside there are two terms - respectively, two terms and multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can you open the bracket from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the brackets, which means that everything below simply changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.

We do the same with the second equation:

It is not by chance that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and again learn to solve such simple equations.

Of course, the day will come when you will hone these skills to the point of automaticity. You will no longer have to perform so many transformations each time; you will write everything on one line. But while you are just learning, you need to write each action separately.

Solving even more complex linear equations

What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.

Task No. 1

\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]

Let's multiply all the elements in the first part:

Let's do some privacy:

Here are some similar ones:

Let's complete the last step:

\[\frac(-4x)(4)=\frac(4)(-4)\]

Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, they canceled each other out, which makes the equation linear and not quadratic.

Task No. 2

\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]

Let's carefully perform the first step: multiply each element from the first bracket by each element from the second. There should be a total of four new terms after the transformations:

Now let’s carefully perform the multiplication in each term:

Let’s move the terms with “X” to the left, and those without - to the right:

\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]

Here are similar terms:

Once again we have received the final answer.

Nuances of the solution

The most important note about these two equations is that as soon as we start multiplying parentheses containing more than one term, it does so by next rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we will have four terms.

About the algebraic sum

With this last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean simple design: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.

As soon as, when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.

Finally, let's look at a couple more examples that will be even more complex than the ones we just looked at, and to solve them we will have to slightly expand our standard algorithm.

Solving equations with fractions

To solve such tasks, we will have to add one more step to our algorithm. But first, let me remind you of our algorithm:

1. Open the brackets.
2. Separate variables.
3. Bring similar ones.
4. Divide by the ratio.

Alas, this wonderful algorithm, for all its effectiveness, turns out to be not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on both the left and the right in both equations.

How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be done both before and after the first action, namely, getting rid of fractions. So the algorithm will be as follows:

1. Get rid of fractions.
2. Open the brackets.
3. Separate variables.
4. Bring similar ones.
5. Divide by the ratio.

What does it mean to “get rid of fractions”? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numerical in their denominator, i.e. Everywhere the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, we will get rid of fractions.

Example No. 1

\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]

Let's get rid of the fractions in this equation:

\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]

Please note: everything is multiplied by “four” once, i.e. just because you have two parentheses doesn't mean you have to multiply each one by "four." Let's write down:

\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]

Now let's expand:

We seclude the variable:

We perform the reduction of similar terms:

\[-4x=-1\left| :\left(-4 \right) \right.\]

\[\frac(-4x)(-4)=\frac(-1)(-4)\]

We have received the final solution, let's move on to the second equation.

Example No. 2

\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]

Here we perform all the same actions:

\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]

\[\frac(4x)(4)=\frac(4)(4)\]

The problem is solved.

That, in fact, is all I wanted to tell you today.

Key points

Key findings are:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Don't worry if you see quadratic functions, most likely, in the process of further transformations they will decrease.
• There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and no roots at all.

I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site and solve the examples presented there. Stay tuned, many more interesting things await you!

When solving equations, in order to simplify it, we perform identical transformations of expressions. In equations with one variable, sometimes the solution of the equation can be reduced to the solution of an equivalent linear equation with one variable.

Let's look at examples. Let's solve the equation (2x+1)(3x-2)-6x(x+4)=67-2x. On the left side of the equation, multiply the polynomial 2x+1 by the polynomial 3x-2, as well as the monomial 6x by the polynomial x+4. After multiplying the polynomial 2x+1 by the polynomial 3x-2 we get the polynomial 6x 2 +3x-4x-2, and after multiplying the monomial 6x by the polynomial x+4 we get the polynomial 6x 2 +24x. Our equation will take the form (6x 2 +3x-4x-2)-(6x 2 +24x)=67-2x. After this, open the brackets and get 6x 2 +3x-4x-2-6x 2 -24x=67-2x. Let us move the terms with the unknown to the left side, and those without the unknown to the right. The new equivalent equation looks like this: 6x 2 -6x 2 +3x-4x+2x-24x=67+2. Let's give similar ones. We get -23x=69. Divide both sides of the equation by -23. We get x=-3. We successively replaced the equations with equivalent ones. This means that the original equation is equivalent to the equation -23x=69 and has a single root - the number -3.

Example two. Let's solve the equation (x+2)/3-(3x-1)/4=-2. On the left side of this equation are the fractions (x+2)/3 and (3x-1)/4. Let's multiply both sides of the equation by the least common denominator of these fractions - the number 12. [(x+2)/3-(3x-1)4].12=-2.12. Let's open the brackets and multiply each fraction by 12. We get (x+2)12/3-(3x-1)12/4+-24. In the first fraction, 12 and 3 will be reduced, and in the second, 12 and 4. After the reduction, our equation will become 4(x+2)-3(3x-1)=-24. Thus, we freed ourselves from denominators. After opening the brackets, we get 4x+8-9x+3=-24. We move everything that contains a variable to the left side, and everything that does not contain a variable - to the right. The equation becomes 4x-9x=-24-8-3. Let's bring similar ones and get -5x=-35. We divide both sides of the equation by -5 and it turns out that x = 7. Replacing the equation step by step with an equivalent parameter, we obtained a linear equation -5x=-35, equivalent to this one. This linear equation has a single root - the number 7.

In the examples considered, the solution to the original equation was reduced to solving a linear equation of the form ax=b, in which the coefficient a is not equal to 0.

However, it may also happen that by replacing one equation with another that is equivalent to it, we can obtain a linear equation of the form 0x=b, where b is not equal to 0 or 0x=0. In the first case, we can conclude that the original equation has no roots, because on the left side of the equation there is 0, and on the right the number is not equal to 0. In the second case, the equation has an infinite number of roots, because on the left side of the equation there will always be 0, and the right one is also 0. The equality will always be true, regardless of the value of the variable.

Example three. Let's solve the equation (2x-7)/2-(4x-1)/4=0. Again, our equation contains fractions, so we multiply both sides of the equation by the lowest common denominator. This number is 4. We get [(2x-7)/2-(4x-1)/4].4=0.4. Let's open the brackets: 4(2x-7)/2-4(4x-1)/4=0. Let's reduce the factors and get the equation 2(2x-7)-(4x-1)=0. Let's open the brackets again: 4x-14-4x+1=0. Let's move the terms with the unknown to the left side of the equation, and those without the unknown to the right. The equation becomes 4x-4x=14-1. We bring similar ones and get 0x=13. This equation has no roots because 0x is equal to 0 for any value of x. It turns out that equality will never be achieved, for any value of x. This means that the original equation equivalent to it has no roots.

Example four. Let's solve the equation (5x-1)-2(3x-6)=11-x. Let's open the brackets: 5x-1-6x+12=11x. Let's move the terms containing x to the left side, and those not containing x to the right side of the equation. We get 5x-6x+x=11+1-12. Let's give similar ones: 0x=0. This equation 0x=0, and therefore the equivalent original equation, has an infinite number of roots. Since 0 multiplied by any number equals 0, the equality holds for any value of x.

1. General provisions

1.1. In order to maintain business reputation and ensure compliance with federal legislation, the Federal State Institution State Research Institute of Technology "Informika" (hereinafter referred to as the Company) considers the most important task to be ensuring the legitimacy of the processing and security of personal data of subjects in the Company's business processes.

1.2. To solve this problem, the Company has introduced, operates and undergoes periodic review (monitoring) of a personal data protection system.

1.3. The processing of personal data in the Company is based on the following principles:

The legality of the purposes and methods of processing personal data and integrity;

Compliance of the purposes of processing personal data with the goals predetermined and stated when collecting personal data, as well as with the powers of the Company;

Correspondence of the volume and nature of the processed personal data, methods of processing personal data to the purposes of processing personal data;

The reliability of personal data, their relevance and sufficiency for the purposes of processing, the inadmissibility of processing personal data that is excessive in relation to the purposes of collecting personal data;

The legitimacy of organizational and technical measures to ensure the security of personal data;

Continuous improvement of the level of knowledge of Company employees in the field of ensuring the security of personal data during their processing;

Striving for continuous improvement of the personal data protection system.

2. Purposes of processing personal data

2.1. In accordance with the principles of processing personal data, the Company has determined the composition and purposes of processing.

Purposes of processing personal data:

Conclusion, support, change, termination employment contracts, which are the basis for the emergence or termination labor relations between the Company and its employees;

Providing a portal and services personal account for students, parents and teachers;

Storage of learning results;

Fulfillment of obligations provided for by federal legislation and other regulatory legal acts;

3. Rules for processing personal data

3.1. The Company processes only those personal data that are presented in the approved List of personal data processed in the Federal State Autonomous Institution State Scientific Research Institute of Information Technology "Informika"

3.2. The Company does not allow processing following categories personal data:

Race;

Political Views;

Philosophical beliefs;

About the state of health;

State intimate life;

Nationality;

Religious Beliefs.

3.3. The Company does not process biometric personal data (information that characterizes physiological and biological features person, on the basis of which his identity can be established).

3.4. The Company does not carry out cross-border transfer of personal data (transfer of personal data to the territory of a foreign state to an authority of a foreign state, a foreign to an individual or a foreign legal entity).

3.5. The Company prohibits making decisions regarding personal data subjects based solely on automated processing of their personal data.

3.6. The Company does not process data on subjects' criminal records.

3.7. The company does not publish the subject’s personal data in publicly available sources without his prior consent.

4. Implemented Requirements to ensure the security of personal data

4.1. In order to ensure the security of personal data during their processing, the Company implements the following requirements: regulatory documents Russian Federation in the field of processing and ensuring the security of personal data:

Federal Law of July 27, 2006 No. 152-FZ “On Personal Data”;

Government Decree Russian Federation dated November 1, 2012 N 1119 “On approval of requirements for the protection of personal data during their processing in information systems personal data";

Decree of the Government of the Russian Federation dated September 15, 2008 No. 687 “On approval of the Regulations on the specifics of processing personal data carried out without the use of automation tools”;

Order of the FSTEC of Russia dated February 18, 2013 N 21 “On approval of the composition and content of organizational and technical measures to ensure the security of personal data during their processing in personal data information systems”;

Basic model of threats to the security of personal data during their processing in personal data information systems (approved by the Deputy Director of the FSTEC of Russia on February 15, 2008);

Methodology for determining current threats to the security of personal data during their processing in personal data information systems (approved by the Deputy Director of the FSTEC of Russia on February 14, 2008).

4.2. The company assesses the harm that may be caused to personal data subjects and identifies threats to the security of personal data. In accordance with identified current threats, the Company applies necessary and sufficient organizational and technical measures, including the use of information security tools, detection of unauthorized access, restoration of personal data, establishment of rules for access to personal data, as well as monitoring and evaluation of the effectiveness of the measures applied.

4.3. The Company has appointed persons responsible for organizing the processing and ensuring the security of personal data.

4.4. The Company's management is aware of the need and is interested in ensuring an adequate level of security for personal data processed as part of the Company's core business, both in terms of the requirements of regulatory documents of the Russian Federation and justified from the point of view of assessing business risks.