The study of equations in the middle link begins with the introduction of the solution linear equations and equations reducing to linear ones.

The equality of two functions considered in the general domain of definition is called an equation. The variables included in the equation are denoted with Latin letters x, y, z, t ... An equation with one variable x in general form is written as follows f (x) \u003d g (x).

Any value of the variable, in which the expressions f(x) and g(x) take equal numerical values, is called the root of the equation.

Solving an equation means finding all its roots or proving that there are none.

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, since 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 ones.

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

1. transformations given equation to the simplest
2. solving the simplest equations according to known rules, formulas or algorithms.

If the second part is algorithmic, then the first part is largely heuristic, which is the most difficult for students. In the process of solving the equation, they try to replace it with a simpler one, so it is important to know what transformations this is possible with. 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, each of which 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 - none of them has roots.

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

Equivalent transformations include:

1) If we add the same number or the same whole algebraic expression containing the unknown to both parts of the equation, then the new equation will be equivalent to the given one.

2) If both parts of the equation are multiplied or divided by the same non-zero number, then an equation equivalent to the given one will be obtained.

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

3) If in the equation to expand the brackets and bring like terms, then we get an equation equivalent to the given one.

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

A linear equation with one variable x is an equation of the form ax \u003d b, where a and b are real numbers, a is called the coefficient of the variable, b is a free member.

For a linear equation ax = b can be presented on occasion:

Many equations are reduced to linear ones as a result of transformations.

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

1)

This equation reduces to a linear equation.

Multiplying both parts by 12 (lowest common denominator 3, 4, 6, 12), we get:

8 + 3x + 2 - 2x = 5x -12,

8 + 2 + 12 = 5x - 3x + 2x,

Answer: 5.5.

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

Simplify both sides of the equation:

2x + 2 - 1 = 3 - 1 + 2x,

2x + 1 = 2 + 2x,

2x - 2x \u003d 2 - 1,

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

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

When going through the topic “linear equations with 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: to find the abscissas of the intersection points of the graphs of functions on the left and right sides of the equation. Based on the following steps:

1) Convert the original equation to the form f(x) = g(x), where f(x) and g(x) are functions, graphs that can be built.
2) Construct graphs of functions f(x) and g(x)
3) Determine the intersection points of the constructed graphs.
4) Determine the abscissas of the found points. They will give a set of solutions to the original equation.
5) Write down the answer.

Advantage 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 step in the study of linear equations are equations with modules, and some solutions are performed in several ways.

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

Equations containing the modulus sign are of particular interest.

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

The number –a can be negative if a>0; -a positive for 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.

2 way - graphic. The equation can be written as a system of two equations:

We construct graphs of functions and .

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

Answer: -4; ten.

Solve an equation containing more than one module

Let's use the following algorithm.

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

So, submodule 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 \u003d 4,

and hence is the root of the equation.

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

So on the interval (-3; -1) the equation has no roots.

III interval. Let x > -1 then

x + 1 + x + 3 = 4,

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

On the simple examples consider an algorithm for solving equations with parameters: area allowed values, domain of definition, general solutions, control values ​​of parameters, types of particular equations. The ways of finding them will be established in each type of equations separately.

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

• the area of ​​admissible values ​​of the parameter and the area of ​​definition are set;
• control parameter values, dividing the region of admissible values ​​of the parameter into regions of uniformity of particular equations;
• for the 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 А f1 ,…, А fk of parameter values;
• a model of general solutions, control values ​​of the parameter is compiled;
• intervals of parameter values ​​with the same common solutions(areas of uniformity);
• for the control values ​​of the parameter and selected areas of uniformity, the characteristics of all types of partial equations are written
• Special place in algebra is assigned to linear equations with parameters.

Let's look at a few examples.

1.	2x - 3 \u003d m + 1, 2x - 3 \u003d + 4 m + 1,	where m is an unknown parameter. Multiplying both sides of the equation by 3, we get
	6x - 9 \u003d m x + 12m +3, 6x - m x + 12m + 12,	Let's take out common factor brackets, we get
	x (6-m) = 12(m+1), , 6 – m? 0,m? 6.	because it is in the denominator of a fraction.
Answer: , for 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. , for 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 \u003d 3n - 2,

a) 3n ​​- 2 = 0,

x(4 - 2 - 3) = 3 - 2,

x is any number except x = 1.

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

m – 2 – 3n 0

x = , when x ? one,

3n - 2m - 2 - 3n,

3n + 3n 2 – 2 + m,

In this case, the equation has no solutions.

Hence, for n = and m = 4, x is any number except 1; for n = 0, m = 6n

(n), m \u003d 3n + 2 (n), m \u003d 2, the equation has no solutions. For all other parameter values ​​x = .

Answer: 1. n = , m = 4 - x? R\.

2. n \u003d 0, m \u003d 6n (n), m \u003d 3n + 2 (n), m \u003d 2 - there are no solutions.

3. n 0, m 6n, m 3n + 2, m 2 – x = .

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

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

1. studying the condition of the problem;
2. search for a solution plan and its preparation;
3. execution of the found solution;
4. critical analysis decision result.

Now consider the problems in the solution of which linear equations are used.

1. An alloy of copper and zinc contains 640 g more copper than zinc. After 6/7 of the copper contained in it and 60% of zinc were isolated from the alloy, the mass of the alloy turned out to be 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. 0.4 parts. Knowing that the mass of the alloy turned out to be equal to 200 g, we make an equation.

1/7 (x + 640) + 0.4 x \u003d 200,

x + 640 + 2.8 x \u003d 1400,

3.8x \u003d 1400 - 640,

So, zinc was 200 g, and copper 840 g.

(200 + 640 = 840). 1) 200 + 840 = 1040 (g) - weight 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 get a 40% solution?

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

60x + 300 = 40x + 400,

60x - 40x \u003d 400 - 300,

So, you need to add 5 liters of 60% acid.

Answer: 5 liters.

When studying the topic “Solution of linear equations”, some historical background is recommended.

Problems for solving equations of the first degree are found in the Babylonian cuneiform texts. They also have some problems that lead to quadratic and even cubic equations (the latter, apparently, were solved using the selection of roots). Ancient Greek mathematicians found geometric shape solution of 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. Decision cubic equation was found at the beginning of the 16th century in Italy. After Scipian del Ferro decided one private view such equations in 1535, the Italian Tartaglia found a general formula. He proved that the roots of the equation x 3 + px + q = 0 have the form x = .

This expression is usually called Cardano's formula, after the scientist who learned it from Tartaglia and published it in 1545 in his book The Great Art of Algebraic Rules. A student of Cardano, a young mathematician Ferrari, solved the general equation of the fourth degree. After that, for two and a half centuries, the search for a formula for solving equations of the fifth degree continued. In 1823, the remarkable Norwegian mathematician Niels Hendrik Abel (1802-1829) proved that there was no such formula. More precisely, he proved that the roots general equation fifth degree cannot be expressed in terms of its coefficients using arithmetic and root extraction operations. A deep 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 studies were also carried out in the field of the theory of algebraic equations. Vieta established a connection between the coefficients of equations 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 formulas take place:

x 1 + x 2 + ... + x n \u003d -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. Journal “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. Epishev; IN AND. Krupych “Teaching schoolchildren to learn mathematics”.
6. E.I.Yamshchenko “Study of functions”.
7. A.I. Khudobin; M.F. Shurshalov “Collection of problems in algebra and elementary functions”.
8. Sh. A. Alimov, V.A. Ilyin "Algebra grades 6-8".

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

Let's look at examples. 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, and also 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) \u003d 67-2x. After that, we open the brackets and get 6x 2 + 3x-4x-2-6x 2 -24x \u003d 67-2x. We move the terms with the unknown to the left side, and without the unknown - to the right. The new equivalent equation looks like this: 6x 2 -6x 2 +3x-4x+2x-24x=67+2. We present 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. So the original equation is equivalent to the equation -23x=69 and has a single root - the number -3.

Second example. 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. Multiply both sides of the equation by the smallest 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) \u003d -24. Thus, we got rid of the denominators. After opening the brackets, we get 4x + 8-9x + 3 \u003d -24. Everything that contains a variable is moved to the left side, and everything that does not contain a variable is moved to the right. The equation becomes 4x-9x=-24-8-3. We give similar ones and get -5x \u003d -35. 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 have obtained a linear equation -5x=-35, which is equivalent to the given one. This linear equation has a single root - the number 7.

In the considered examples, the solution of the original equation was reduced to the solution of 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 equivalent to it, we can get 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 there is 0 on the left side of the equation, and the number is not equal to 0 on the right. In the second case, the equation has infinite number roots, because the left side of the equation will always be 0, and the right side will also be 0. 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 least 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. We reduce the factors and get the equation 2(2x-7)-(4x-1)=0. 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 without the unknown - to the right. The equation will take the form 4x-4x=14-1. We give similar ones and get 0x \u003d 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 values ​​of x. This means that the original equation equivalent to it has no roots.

Example four. Solve the equation (5x-1)-2(3x-6)=11-x. Let's open the brackets: 5x-1-6x+12=11-x. Let's move the terms containing x to the left side, and those not containing x - to right side equations. We get 5x-6x+x=11+1-12. Let's give similar ones: 0x=0. This equation 0x=0, and hence 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.

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 that correspond to the sounded 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 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 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 indicated above, as well as equations that are reduced to the form a x+b=0 with the help of equivalent transformations, will be called 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 the 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 the case equal to zero a will be considered separately 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. Here are the solutions characteristic examples corresponding different meanings coefficients of linear equations.

Example.

Solve the linear equation 0 x−0=0 .

Decision.

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?

Decision.

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.

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. Bring like 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 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 down 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 the following: as soon as we start multiplying brackets in which there is a term greater than it, then this is done according to 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 get four terms.

On the algebraic sum

With the last example, I would like to remind students what an algebraic sum is. 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:

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

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!