Solution of simple linear equations. More complex examples of equations

An equation is a mathematical expression that is an equation containing an unknown. If the equality is true for any admissible values ​​of the unknowns included in it, then it is called an identity; for example: a relation like (x – 1)2 = (x – 1)(x – 1) holds for all values ​​of x.

If an equation involving an unknown x holds only for certain values ​​of x, and not for all values ​​of x, as in the case of an identity, then it may be useful to determine those values ​​of x for which the equation is valid. Such values ​​of x are called roots or solutions of the equation. For example, the number 5 is the root of the equation 2x + 7= 17.

In the branch of mathematics called the theory of equations, the main subject of study is the methods for solving equations. AT school course algebra equations are given great attention.

The history of the study of equations goes back many centuries. The most famous mathematicians who contributed to the development of the theory of equations were:

Archimedes (circa 287-212 BC) - Ancient Greek scientist, mathematician and mechanic. In the study of one problem, which reduces to cubic equation, Archimedes found out the role of the characteristic, which later became known as the discriminant.

François Viet lived in the 16th century. He made a great contribution to the study various problems mathematics. In particular, he introduced the literal notation for the coefficients of an equation and established a connection between the roots of a quadratic equation.

Leonhard Euler (1707 - 1783) - mathematician, mechanic, physicist and astronomer. The author of St. 800 papers on mathematical analysis, differential equations, geometry, number theory, approximate calculations, celestial mechanics, mathematics, optics, ballistics, shipbuilding, music theory, etc. He had a significant impact on the development of science. He derived formulas (Euler formulas) expressing trigonometric functions variable x through an exponential function.

Lagrange Joseph Louis (1736 - 1813), French mathematician and mechanic. He owns outstanding research, among them research on algebra (the symmetric function of the roots of an equation, on differential equations (the theory of singular solutions, the method of variation of constants).

J. Lagrange and A. Vandermonde - French mathematicians. In 1771, for the first time, the method of solving systems of equations (the substitution method) was used.

Gauss Karl Friedrich (1777 -1855) - German mathematician. Wrote a book that outlines the theory of circle division equations (i.e., equations xn - 1 = 0), which in many ways was a prototype of the Galois theory. Apart from common methods solving these equations, established a connection between them and the construction of regular polygons. He, for the first time after the ancient Greek scientists, made a significant step forward in this matter, namely: he found all those values ​​of n for which regular n-gon can be built with a compass and a ruler. Learned how to add. He concluded that systems of equations can be added, divided, and multiplied among themselves.

O. I. Somov - enriched various parts of mathematics with important and numerous works, among them the theory of certain algebraic equations higher degrees.

Galois Evariste (1811-1832), French mathematician. His main merit is the formulation of a set of ideas, to which he came in connection with the continuation of research on the solvability of algebraic equations, begun by J. Lagrange, N. Abel and others, created the theory of algebraic equations of higher degrees with one unknown.

A. V. Pogorelov (1919 - 1981) - In his work, geometric methods are associated with analytical methods theory of differential equations with partial derivatives. His works also had a significant impact on the theory of non-linear differential equations.

P. Ruffini - Italian mathematician. He devoted a number of works to the proof of the unsolvability of the equation of the 5th degree, systematically uses the closedness of the set of substitutions.

Despite the fact that scientists have been studying equations for a long time, science does not know how and when people got the need to use equations. It is only known that problems leading to the solution of the simplest equations have been solved by people since the time they became people. Another 3 - 4 thousand years BC. e. the Egyptians and Babylonians knew how to solve equations. The rule for solving these equations coincides with the modern one, but it is not known how they got to this point.

AT Ancient Egypt and Babylon, the false position method was used. An equation of the first degree with one unknown can always be reduced to the form ax + b = c, in which a, b, c are integers. According to the rules arithmetic operations ax \u003d c - b,

If b > c, then c b is a negative number. Negative numbers were unknown to the Egyptians and many other later peoples (on an equal footing with positive numbers they began to be used in mathematics only in the seventeenth century). To solve the problems that we now solve with equations of the first degree, the false position method was invented. In the papyrus of Ahmes, 15 problems are solved by this method. The Egyptians had a special sign to denote unknown number, which, until recently, was read “how” and translated by the word “heap” (“heap” or “unknown number” of units). Now they read a little less inaccurately: "aha." The solution method used by Ahmes is called the method of one false position. Using this method, equations of the form ax = b are solved. This method consists in dividing each side of the equation by a. It was used by both the Egyptians and the Babylonians. At different peoples the method of two false positions was used. The Arabs mechanized this method and obtained the form in which it passed into the textbooks of European peoples, including Magnitsky's Arithmetic. Magnitsky calls the method of solving the "false rule" and writes in the part of his book expounding this method:

Zelo bo cunning is this part, Like you can put everything with it. Not only what is in citizenship, But also the higher sciences in space, Even are listed in the sphere of heaven, Like the wise there is a need.

The content of Magnitsky's poems can be summarized as follows: this part of arithmetic is very tricky. With its help, you can calculate not only what is needed in everyday practice, but it also solves the “higher” questions that confront the “wise”. Magnitsky uses a "false rule" in the form given to it by the Arabs, calling it the "arithmetic of two errors" or the "method of weights." Indian mathematicians often gave problems in verse. Lotus challenge:

Above the quiet lake, half a measure above the water, Lotus color was visible. He grew up alone, and the wind in a wave bent him to the side, and no longer

Flowers above the water. Found his fisherman's eye Two measures from where he grew up. How many lakes here are the water deep? I will offer you a question.

Types of equations

Linear equations

Linear equations are equations of the form: ax + b = 0, where a and b are some constants. If a is not equal to zero, then the equation has one single root: x \u003d - b: a (ax + b; ax \u003d - b; x \u003d - b: a.).

For example: solve a linear equation: 4x + 12 = 0.

Solution: T. to a \u003d 4, and b \u003d 12, then x \u003d - 12: 4; x = - 3.

Check: 4 (- 3) + 12 = 0; 0 = 0.

Since k 0 = 0, then -3 is the root of the original equation.

Answer. x = -3

If a is zero and b is zero, then the root of the equation ax + b = 0 is any number.

For example:

0 = 0. Since 0 is 0, then the root of the equation 0x + 0 = 0 is any number.

If a is zero and b is not zero, then the equation ax + b = 0 has no roots.

For example:

0 \u003d 6. Since 0 is not equal to 6, then 0x - 6 \u003d 0 has no roots.

Systems of linear equations.

A system of linear equations is a system in which all equations are linear.

To solve a system means to find all its solutions.

Before solving a system of linear equations, you can determine the number of its solutions.

Let the system of equations be given: (а1х + b1y = с1, (а2х + b2y = c2.

If a1 divided by a2 is not equal to b1 divided by b2, then the system has one unique solution.

If a1 divided by a2 is equal to b1 divided by b2, but equal to c1 divided by c2, then the system has no solutions.

If a1 divided by a2 is equal to b1 divided by b2, and equal to c1 divided by c2, then the system has infinitely many solutions.

A system of equations that has at least one solution is called consistent.

A joint system is called definite if it has finite number solutions, and indefinite if the set of its solutions is infinite.

A system that does not have a single solution is called inconsistent or inconsistent.

Ways to solve linear equations

There are several ways to solve linear equations:

1) Selection method. This is the most simplest way. It consists in the fact that they select all allowed values unknown by enumeration.

For example:

Solve the equation.

Let x = 1. Then

4 = 6. Since 4 is not equal to 6, then our assumption that x = 1 was wrong.

Let x = 2.

6 = 6. Since 6 equals 6, then our assumption that x = 2 was correct.

Answer: x = 2.

2) Way to simplify

This method lies in the fact that all members containing the unknown are transferred to the left side, and known to the right with opposite sign, give similar ones, and divide both sides of the equation by the coefficient of the unknown.

For example:

Solve the equation.

5x - 4 \u003d 11 + 2x;

5x - 2x \u003d 11 + 4;

3x = 15; : (3) x = 5.

Answer. x = 5.

3) Graphical way.

It consists in the fact that a graph of functions is built given equation. Because in the linear equation y \u003d 0, then the graph will be parallel to the y-axis. The point of intersection of the graph with the x-axis will be the solution to this equation.

For example:

Solve the equation.

Let y = 7. Then y = 2x + 3.

Let's build a graph of the functions of both equations:

Ways to solve systems of linear equations

In the seventh grade, three ways to solve systems of equations are studied:

1) Substitution method.

This method consists in the fact that in one of the equations one unknown is expressed in terms of another. The resulting expression is substituted into another equation, which then turns into an equation with one unknown, then it is solved. The resulting value of this unknown is substituted into any equation of the original system and the value of the second unknown is found.

For example.

Solve the system of equations.

5x - 2y - 2 = 1.

3x + y = 4; y \u003d 4 - 3x.

Substitute the resulting expression into another equation:

5x - 2 (4 - 3x) -2 \u003d 1;

5x - 8 + 6x \u003d 1 + 2;

11x = 11; : (11) x = 1.

Substitute the resulting value into the equation 3x + y \u003d 4.

3 1 + y = 4;

3 + y = 4; y \u003d 4 - 3; y = 1.

Examination.

/3 1 + 1 = 4,

\5 1 - 2 1 - 2 = 1;

Answer: x = 1; y = 1.

2) Method of addition.

This method is that if this system consists of equations that, when added term by term, form an equation with one unknown, then by solving this equation, we will obtain the value of one of the unknowns. The resulting value of this unknown is substituted into any equation of the original system and the value of the second unknown is found.

For example:

Solve the system of equations.

/ 3y - 2x \u003d 5,

\5x - 3y \u003d 4.

Let's solve the resulting equation.

3x = 9; : (3) x = 3.

Let's substitute the obtained value into the equation 3y - 2x = 5.

3y - 2 3 = 5;

3y = 11; : (3) y = 11/3; y = 3 2/3.

So x = 3; y = 3 2/3.

Examination.

/3 11/3 - 2 3 = 5,

\5 3 - 3 11/ 3 = 4;

Answer. x = 3; y = 3 2/3

3) Graphical way.

This method is based on the fact that graphs of equations are plotted in one coordinate system. If the graphs of the equation intersect, then the coordinates of the intersection point are the solution of this system. If the graphs of an equation are parallel lines, then the given system has no solutions. If the graphs of the equations merge into one straight line, then the system has infinitely many solutions.

For example.

Solve the system of equations.

18x + 3y - 1 = 8.

2x - y \u003d 5; 18x + 3y - 1 = 8;

Y \u003d 5 - 2x; 3y \u003d 9 - 18x; : (3) y = 2x - 5. y = 3 - 6x.

We construct graphs of functions y \u003d 2x - 5 and y \u003d 3 - 6x on the same coordinate system.

The graphs of the functions y \u003d 2x - 5 and y \u003d 3 - 6x intersect at point A (1; -3).

Therefore, the solution to this system of equations will be x = 1 and y = -3.

Examination.

2 1 - (- 3) = 5,

18 1 + 3 (-3) - 1 = 8.

18 - 9 – 1 = 8;

Answer. x = 1; y = -3.

Conclusion

Based on all of the above, we can conclude that equations are necessary in modern world not only for solving practical problems, but also as a scientific tool. Therefore, so many scientists have studied this issue and continue to study.

The equation that represents square trinomial, is commonly referred to as a quadratic equation. From the point of view of algebra, it is described by the formula a*x^2+b*x+c=0. In this formula, x is the unknown to be found (it is called the free variable); a, b and c are numerical coefficients. With regard to the components of this, there are a number of restrictions: for example, the coefficient a should not be equal to 0.

Solving the equation: the concept of the discriminant

The value of the unknown x, at which quadratic equation turns into a true equality, is called the root of such an equation. In order to solve a quadratic equation, you must first find the value of a special coefficient - the discriminant, which will show the number of roots of the considered equality. The discriminant is calculated by the formula D=b^2-4ac. In this case, the result of the calculation can be positive, negative or equal to zero.

In this case, it should be borne in mind that the concept requires that only the coefficient a be strictly different from 0. Therefore, the coefficient b can be equal to 0, and the equation itself in this case is a*x^2+c=0. In such a situation, the coefficient value equal to 0 should be used in the formulas for calculating the discriminant and roots. So, the discriminant in this case will be calculated as D=-4ac.

Solution of the equation with a positive discriminant

If the discriminant of the quadratic equation turned out to be positive, we can conclude from this that this equality has two roots. These roots can be calculated using the following formula: x=(-b±√(b^2-4ac))/2a=(-b±√D)/2a. Thus, to calculate the value of the roots of the quadratic equation for positive value discriminant used known values coefficients available in . Thanks to the use of the sum and difference in the formula for calculating the roots, the result of the calculations will be two values ​​that turn the equality in question into the correct one.

Solution of the equation with zero and negative discriminant

If the discriminant of the quadratic equation turned out to be equal to 0, we can conclude that said equation has one root. Strictly speaking, in this situation, the equation still has two roots, but due to the zero discriminant, they will be equal to each other. In this case x=-b/2a. If, in the course of calculations, the value of the discriminant turns out to be negative, it should be concluded that the considered quadratic equation has no roots, that is, such values ​​of x at which it turns into a true equality.

And so on, it is logical to get acquainted with equations of other types. Next in line are linear equations, the purposeful study of which begins in algebra lessons in grade 7.

It is clear that first you need to explain what a linear equation is, give a definition of a linear equation, its coefficients, show it general form. Then you can figure out how many solutions a linear equation has depending on the values ​​of the coefficients, and how the roots are found. This will allow you to move on to solving examples, and thereby consolidate the studied theory. In this article we will do this: we will dwell in detail on all theoretical and practical points regarding linear equations and their solution.

Let's say right away that here we will consider only linear equations with one variable, and in a separate article we will study the principles of solving linear equations in two variables.

Page navigation.

What is a linear equation?

The definition of a linear equation is given by the form of its notation. Moreover, in different textbooks of mathematics and algebra, the formulations of the definitions of linear equations have some differences that do not affect the essence of the issue.

For example, in an algebra textbook for grade 7 by Yu. N. Makarycheva and others, a linear equation is defined as follows:

Definition.

Type equation ax=b, where x is a variable, a and b are some numbers, is called linear equation with one variable.

Let us give examples of linear equations corresponding to the voiced definition. For example, 5 x=10 is a linear equation with one variable x , here the coefficient a is 5 , and the number b is 10 . Another example: −2.3 y=0 is also a linear equation, but with the variable y , where a=−2.3 and b=0 . And in the linear equations x=−2 and −x=3.33 a are not explicitly present and are equal to 1 and −1, respectively, while in the first equation b=−2 , and in the second - b=3.33 .

And a year earlier, in the textbook of mathematics by N. Ya. Vilenkin, linear equations with one unknown, in addition to equations of the form a x = b, were also considered equations that can be reduced to this form by transferring terms from one part of the equation to another with the opposite sign, as well as by reducing like terms. According to this definition, equations of the form 5 x=2 x+6 , etc. are also linear.

In turn, the following definition is given in the algebra textbook for 7 classes by A. G. Mordkovich:

Definition.

Linear equation with one variable x is an equation of the form a x+b=0 , where a and b are some numbers, called the coefficients of the linear equation.

For example, linear equations of this kind are 2 x−12=0, here the coefficient a is equal to 2, and b is equal to −12, and 0.2 y+4.6=0 with coefficients a=0.2 and b =4.6. But at the same time, there are examples of linear equations that have the form not a x+b=0 , but a x=b , for example, 3 x=12 .

Let's, so that we do not have any discrepancies in the future, under a linear equation with one variable x and coefficients a and b we will understand an equation of the form a x+b=0 . This type of linear equation seems to be the most justified, since linear equations are algebraic equations first degree. And all the other equations above, as well as equations that, using equivalent transformations are reduced to the form a x+b=0 , we will call equations reducing to linear equations. With this approach, the equation 2 x+6=0 is a linear equation, and 2 x=−6 , 4+25 y=6+24 y , 4 (x+5)=12, etc. are linear equations.

How to solve linear equations?

Now it's time to figure out how the linear equations a x+b=0 are solved. In other words, it's time to find out if the linear equation has roots, and if so, how many and how to find them.

The presence of roots of a linear equation depends on the values ​​of the coefficients a and b. In this case, the linear equation a x+b=0 has

• the only root at a≠0 ,
• has no roots for a=0 and b≠0 ,
• has infinitely many roots for a=0 and b=0 , in which case any number is a root of a linear equation.

Let us explain how these results were obtained.

We know that in order to solve equations, it is possible to pass from the original equation to equivalent equations, that is, to equations with the same roots or, like the original one, without roots. To do this, you can use the following equivalent transformations:

• transfer of a term from one part of the equation to another with the opposite sign,
• and also multiplying or dividing both sides of the equation by the same non-zero number.

So, in a linear equation with one type variable a x+b=0 we can move the term b from the left side to right side with the opposite sign. In this case, the equation will take the form a x=−b.

And then the division of both parts of the equation by the number a suggests itself. But there is one thing: the number a can be equal to zero, in which case such a division is impossible. To deal with this problem, we will first assume that the number a is different from zero, and consider the case of zero a separately a bit later.

So, when a is not equal to zero, then we can divide both parts of the equation a x=−b by a , after that it is converted to the form x=(−b): a , this result can be written using a solid line as .

Thus, for a≠0, the linear equation a·x+b=0 is equivalent to the equation , from which its root is visible.

It is easy to show that this root is unique, that is, the linear equation has no other roots. This allows you to do the opposite method.

Let's denote the root as x 1 . Suppose that there is another root of the linear equation, which we denote x 2, and x 2 ≠ x 1, which, due to definitions equal numbers through the difference is equivalent to the condition x 1 − x 2 ≠0 . Since x 1 and x 2 are the roots of the linear equation a x+b=0, then the numerical equalities a x 1 +b=0 and a x 2 +b=0 take place. We can subtract the corresponding parts of these equalities, which the properties of numerical equalities allow us to do, we have a x 1 +b−(a x 2 +b)=0−0 , whence a (x 1 −x 2)+( b−b)=0 and then a (x 1 − x 2)=0 . And this equality is impossible, since both a≠0 and x 1 − x 2 ≠0. So we have come to a contradiction, which proves the uniqueness of the root of the linear equation a·x+b=0 for a≠0 .

So we have solved the linear equation a x+b=0 with a≠0 . The first result given at the beginning of this subsection is justified. There are two more that meet the condition a=0 .

For a=0 the linear equation a·x+b=0 becomes 0·x+b=0 . From this equation and the property of multiplying numbers by zero, it follows that no matter what number we take as x, when we substitute it into the equation 0 x+b=0, we get the numerical equality b=0. This equality is true when b=0 , and in other cases when b≠0 this equality is false.

Therefore, with a=0 and b=0, any number is the root of the linear equation a x+b=0, since under these conditions, substituting any number instead of x gives the correct numerical equality 0=0. And for a=0 and b≠0, the linear equation a x+b=0 has no roots, since under these conditions, substituting any number instead of x leads to an incorrect numerical equality b=0.

The above justifications make it possible to form a sequence of actions that allows solving any linear equation. So, algorithm for solving a linear equation is:

• First, by writing a linear equation, we find the values ​​of the coefficients a and b.
• If a=0 and b=0 , then this equation has infinitely many roots, namely, any number is a root of this linear equation.
• If a is different from zero, then
• the coefficient b is transferred to the right side with the opposite sign, while the linear equation is transformed to the form a x=−b ,
• after which both parts of the resulting equation are divided by a non-zero number a, which gives the desired root of the original linear equation.

The written algorithm is an exhaustive answer to the question of how to solve linear equations.

In conclusion of this paragraph, it is worth saying that a similar algorithm is used to solve equations of the form a x=b. Its difference lies in the fact that when a≠0, both parts of the equation are immediately divided by this number, here b is already in the desired part of the equation and it does not need to be transferred.

To solve equations of the form a x=b, the following algorithm is used:

• If a=0 and b=0 , then the equation has infinitely many roots, which are any numbers.
• If a=0 and b≠0 , then the original equation has no roots.
• If a is non-zero, then both sides of the equation are divided by a non-zero number a, from which the only root of the equation equal to b / a is found.

Examples of solving linear equations

Let's move on to practice. Let us analyze how the algorithm for solving linear equations is applied. Let us present solutions of typical examples corresponding to different meanings coefficients of linear equations.

Example.

Solve the linear equation 0 x−0=0 .

Solution.

In this linear equation, a=0 and b=−0 , which is the same as b=0 . Therefore, this equation has infinitely many roots, any number is the root of this equation.

Answer:

x is any number.

Example.

Does the linear equation 0 x+2.7=0 have solutions?

Solution.

AT this case the coefficient a is equal to zero, and the coefficient b of this linear equation is equal to 2.7, that is, it is different from zero. Therefore, the linear equation has no roots.

Linear equations. Solution, examples.

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

Linear equations.

Linear equations are not the best difficult topic school mathematics. But there are some tricks there that can puzzle even a trained student. Shall we figure it out?)

A linear equation is usually defined as an equation of the form:

ax + b = 0 where a and b- any numbers.

2x + 7 = 0. Here a=2, b=7

0.1x - 2.3 = 0 Here a=0.1, b=-2.3

12x + 1/2 = 0 Here a=12, b=1/2

Nothing complicated, right? Especially if you do not notice the words: "where a and b are any numbers"... And if you notice, but carelessly think about it?) After all, if a=0, b=0(any numbers are possible?), then we get a funny expression:

But that's not all! If, say, a=0, a b=5, it turns out something quite absurd:

What strains and undermines confidence in mathematics, yes ...) Especially in exams. But of these strange expressions, you also need to find X! Which doesn't exist at all. And, surprisingly, this X is very easy to find. We will learn how to do it. In this lesson.

How to recognize a linear equation in appearance? It depends what appearance.) The trick is that linear equations are called not only equations of the form ax + b = 0 , but also any equations that are reduced to this form by transformations and simplifications. And who knows if it is reduced or not?)

A linear equation can be clearly recognized in some cases. Say, if we have an equation in which there are only unknowns in the first degree, yes numbers. And the equation doesn't fractions divided by unknown , it is important! And division by number, or a numeric fraction - that's it! For example:

This is a linear equation. There are fractions here, but there are no x's in the square, in the cube, etc., and there are no x's in the denominators, i.e. No division by x. And here is the equation

cannot be called linear. Here x's are all in the first degree, but there is division by expression with x. After simplifications and transformations, you can get a linear equation, and a quadratic one, and anything you like.

It turns out that it is impossible to find out a linear equation in some intricate example until you almost solve it. It's upsetting. But in assignments, as a rule, they don’t ask about the form of the equation, right? In tasks, equations are ordered decide. This makes me happy.)

Solution of linear equations. Examples.

The entire solution of linear equations consists of identical transformations of equations. By the way, these transformations (as many as two!) underlie the solutions all equations of mathematics. In other words, the decision any The equation begins with these same transformations. In the case of linear equations, it (the solution) on these transformations ends with a full-fledged answer. It makes sense to follow the link, right?) Moreover, there are also examples of solving linear equations.

Let's start with the simplest example. Without any pitfalls. Let's say we need to solve the following equation.

x - 3 = 2 - 4x

This is a linear equation. Xs are all to the first power, there is no division by X. But, actually, we don't care what the equation is. We need to solve it. The scheme here is simple. Collect everything with x's on the left side of the equation, everything without x's (numbers) on the right.

To do this, you need to transfer - 4x to the left side, with a change of sign, of course, but - 3 - to the right. By the way, this is first identical transformation of equations. Surprised? So, they didn’t follow the link, but in vain ...) We get:

x + 4x = 2 + 3

We give similar, we consider:

What do we lack for complete happiness? Yes, so that there is a clean X on the left! Five gets in the way. Get rid of the five with second identical transformation of equations. Namely, we divide both parts of the equation by 5. We get a ready-made answer:

An elementary example, of course. This is for a warm-up.) It is not very clear why I recalled identical transformations here? OK. We take the bull by the horns.) Let's decide something more impressive.

For example, here is this equation:

Where do we start? With X - to the left, without X - to the right? Could be so. In small steps long road. And you can immediately, in a universal and powerful way. Unless, of course, in your arsenal there are identical transformations of equations.

I ask you key question: What do you dislike the most about this equation?

95 people out of 100 will answer: fractions ! The answer is correct. So let's get rid of them. So we start right away with second identical transformation. What do you need to multiply the fraction on the left by so that the denominator is completely reduced? That's right, 3. And on the right? By 4. But math allows us to multiply both sides by the same number. How do we get out? Let's multiply both sides by 12! Those. on the common denominator. Then the three will be reduced, and the four. Do not forget that you need to multiply each part entirely. Here's what the first step looks like:

Expanding the brackets:

Note! Numerator (x+2) I took in brackets! This is because when multiplying fractions, the numerator is multiplied by the whole, entirely! And now you can reduce fractions and reduce:

Opening the remaining parentheses:

Not an example, but pure pleasure!) Now we recall the spell from lower grades: with x - to the left, without x - to the right! And apply this transformation:

Here are some like:

And we divide both parts by 25, i.e. apply the second transformation again:

That's all. Answer: X=0,16

Take note: to bring the original confusing equation to a pleasant form, we used two (only two!) identical transformations- translation left-right with a change of sign and multiplication-division of the equation by the same number. it universal way! We will work in this way any equations! Absolutely any. That is why I keep repeating these identical transformations all the time.)

As you can see, the principle of solving linear equations is simple. We take the equation and simplify it with identical transformations before receiving a response. The main problems here are in the calculations, and not in the principle of the solution.

But ... There are such surprises in the process of solving the most elementary linear equations that they can drive into a strong stupor ...) Fortunately, there can be only two such surprises. Let's call them special cases.

Special cases in solving linear equations.

Surprise first.

Suppose you got elementary equation, something like:

2x+3=5x+5 - 3x - 2

Slightly bored, we transfer with X to the left, without X - to the right ... With a change of sign, everything is chin-chinar ... We get:

2x-5x+3x=5-2-3

We believe, and ... oh my! We get:

In itself, this equality is not objectionable. Zero is really zero. But X is gone! And we must write in the answer, what x is equal to. Otherwise, the solution doesn't count, yes...) A dead end?

Calm! In such doubtful cases, the most general rules save. How to solve equations? What does it mean to solve an equation? This means, find all values ​​of x that, when substituted into the original equation, will give us the correct equality.

But we have the correct equality already happened! 0=0, where really?! It remains to figure out at what x's this is obtained. What values ​​of x can be substituted into initial equation if these x's still shrink to zero? Come on?)

Yes!!! Xs can be substituted any! What do you want. At least 5, at least 0.05, at least -220. They will still shrink. If you don't believe me, you can check it.) Substitute any x values ​​in initial equation and calculate. All the time the pure truth will be obtained: 0=0, 2=2, -7.1=-7.1 and so on.

Here is your answer: x is any number.

The answer can be written in different mathematical symbols, the essence does not change. This is a completely correct and complete answer.

Surprise second.

Let's take the same elementary linear equation and change only one number in it. This is what we will decide:

2x+1=5x+5 - 3x - 2

After the same identical transformations, we get something intriguing:

Like this. Solved a linear equation, got a strange equality. talking mathematical language, we got wrong equality. And speaking plain language, this is not true. Rave. But nevertheless, this nonsense is quite a good reason for right decision equations.)

Again, we think from general rules. What x, when substituted into the original equation, will give us correct equality? Yes, none! There are no such xes. Whatever you substitute, everything will be reduced, nonsense will remain.)

Here is your answer: there are no solutions.

This is also a perfectly valid answer. In mathematics, such answers often occur.

Like this. Now, I hope, the loss of x's in the process of solving any (not just linear) equation will not bother you at all. The matter is familiar.)

Now that we have dealt with all the pitfalls in linear equations, it makes sense to solve them.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

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.

To begin with, let's define: what is a linear equation and which of them should be called the simplest?

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

The simplest equation means the construction:

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

1. Open brackets, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Lead like terms to the left and right of the equals 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 you get something like $0\cdot x=8$, i.e. on the left is zero, and on the right is a non-zero number. 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.

And now let's see how it all works on the example of real problems.

Examples of solving equations

Today we deal 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 open the brackets, if any (as in our last example);
2. Then bring similar
3. Finally, isolate the variable, i.e. everything that is connected with the variable - the terms in which it is contained - is transferred to one side, and everything that remains without it is transferred to the other side.

Then, as a rule, you need to bring similar on each side of the resulting equality, and after that it remains only to divide by the coefficient at "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. Usually, mistakes are made either when opening brackets, or when counting "pluses" and "minuses".

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

Scheme for solving simple linear equations

To begin with, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the parentheses, if any.
2. Seclude variables, i.e. everything that contains "x" is transferred to one side, and without "x" - to the other.
3. We present similar terms.
4. We divide everything by the coefficient at "x".

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

Solving real examples of simple linear equations

Task #1

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

We give similar terms on the left and on the right, but this has already been done here. Therefore, we proceed to the fourth step: divide by a factor:

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

Here we got the answer.

Task #2

In this task, we can observe the brackets, so let's expand them:

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

Here are some like:

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

Task #3

The third linear equation is already more interesting:

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

There are a few brackets here, but they are not multiplied by anything, they just stand in front of them various signs. Let's break them down:

We perform the second step already known to us:

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

Let's calculate:

We perform the last step - we divide everything by the coefficient at "x":

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

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, then 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, zero can get in among them - there is nothing wrong with that.

Zero is the same number as the rest, you should not somehow discriminate it or assume that if you get zero, then you did something wrong.

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

Understanding this simple fact will keep you from making stupid and hurtful mistakes in high school when doing such things is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complicated and a quadratic function will appear when performing various transformations. However, you should not be afraid of this, because if, according to the author's intention, we solve a linear equation, then in the process of transformation all monomials containing a quadratic function will necessarily be reduced.

Example #1

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

Now let's take privacy:

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

Here are some like:

Obviously, this equation has no solutions, so in the answer we write as follows:

\[\variety \]

or no roots.

Example #2

We perform the same steps. First step:

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

Here are some like:

Obviously, this linear equation has no solution, so we write it like this:

\[\varnothing\],

or no roots.

Nuances of the solution

Both equations are completely solved. On the example of these two expressions, we once again made sure that even in the simplest linear equations, everything can be not so simple: there can be either one, or none, or infinitely many. In our case, we considered two equations, in both there are simply no roots.

But I would like to draw your attention to another fact: how to work with brackets and how to expand 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: multiply each individual term. Inside there are two terms - respectively, two terms and is multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can the bracket be opened from the point of view that there is a minus sign after it. Yes, yes: only now, when the transformations are done, we remember that there is a minus sign in front of the brackets, which means that everything below just 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 no coincidence that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence elementary transformations where the inability to clearly and competently perform simple steps leads to the fact that high school students come to me and learn to solve such simple equations again.

Of course, the day will come when you will hone these skills to automatism. You no longer have to perform so many transformations each time, you will write everything in 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 #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 a retreat:

Here are some like:

Let's do 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, however, they mutually annihilated, which makes the equation exactly linear, not square.

Task #2

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

Let's do the first step carefully: multiply every element in the first bracket by every element in the second. In total, four new terms should be obtained after transformations:

And now carefully perform the multiplication in each term:

Let's move the terms with "x" to the left, and without - to the right:

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

Here are similar terms:

We have received a definitive answer.

Nuances of the solution

The most important remark about these two equations is this: as soon as we start multiplying brackets in which there is more than a term, then this is done according to the following 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 get four terms.

On the algebraic sum

In the last example, I would like to remind students what is algebraic sum. In classical mathematics, by $1-7$ we mean a simple construction: we subtract seven from one. In algebra, we mean by this the following: to the number "one" we add another number, namely "minus seven." This algebraic sum differs from the usual 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.

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

Solving equations with a fraction

To solve such tasks, one more step will have to be added to our algorithm. But first, I will remind our algorithm:

1. Open brackets.
2. Separate variables.
3. Bring similar.
4. Divide by a factor.

Alas, this wonderful algorithm, for all its efficiency, is not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on the left and on 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 performed both before the first action and after it, namely, to get rid of fractions. Thus, the algorithm will be as follows:

1. Get rid of fractions.
2. Open brackets.
3. Separate variables.
4. Bring similar.
5. Divide by a factor.

What does it mean to "get rid of fractions"? And why is it possible to do this both after and before the first standard step? In fact, in our case, all fractions are numeric in terms of the denominator, i.e. everywhere the denominator is just a number. Therefore, if we multiply both parts of the equation by this number, then we will get rid of fractions.

Example #1

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

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

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

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

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

Now let's open it:

We perform seclusion of a variable:

We carry out the reduction of similar terms:

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

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

We got final decision, we pass to the second equation.

Example #2

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

Here we perform all the same actions:

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

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

Problem solved.

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

Key points

The key findings are as follows:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Do not worry if somewhere you have quadratic functions, most likely, in the process of further transformations, they will be reduced.
• The roots in linear equations, even the simplest ones, are of three types: one single root, the entire number line is a root, there are 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, solve the examples presented there. Stay tuned, there are many more interesting things waiting for you!