Equation with one variable. Equations with one variable Card for knowledge correction

The equation Is an equality in which one or more variables are present.
We will consider the case when there is one variable in the equation, that is, one unknown number. Essentially, an equation is a kind of mathematical model. Therefore, first of all, we need equations to solve problems.

Let's remember how a mathematical model is made to solve a problem.
For example, in the new academic year, the number of students at school # 5 has doubled. After 20 students moved to another school, a total of 720 students began to study at school No. 5. How many students were there last year?

We need to express what is said in the condition in mathematical language. Let the number of students last year be X. Then, according to the condition of the problem,
2X - 20 = 720. We have got a mathematical model, which is one-variable equation... More precisely, this is an equation of the first degree with one variable. It remains to find its root.

What is the root of an equation?

The value of the variable at which our equation turns into a true equality is called the root of the equation. There are equations that have many roots. For example, in the equation 2 * X = (5-3) * X, any value of X is a root. And the equation X = X +5 has no roots at all, since no matter what value we substitute for X, we will not get the correct equality. Solving an equation means finding all its roots, or determining that it has no roots. So, to answer our question, we need to solve the equation 2X - 20 = 720.

How to solve equations in one variable?

First, let's write down the basic definitions. Each equation has a right side and a left side. In our case, (2X - 20) is the left side of the equation (it stands to the left of the equal sign), and 720 is the right side of the equation. The terms on the right and left sides of the equation are called the terms of the equation. Our equation terms are 2X, -20, and 720.

Let's just say about 2 properties of equations:

1. Any term in the equation can be transferred from the right side of the equation to the left, and vice versa. In this case, it is necessary to change the sign of this term of the equation to the opposite. That is, records of the form 2X - 20 = 720, 2X - 20 - 720 = 0, 2X = 720 + 20, -20 = 720 - 2X are equivalent.
2. Both sides of the equation can be multiplied or divided by the same number. This number must not be zero. That is, records of the form 2X - 20 = 720, 5 * (2X - 20) = 720 * 5, (2X - 20): 2 = 720: 2 are also equivalent.

Let's use these properties to solve our equation.

Move -20 to the right side with the opposite sign. We get:

2X = 720 + 20. Add what we have on the right side. We get 2X = 740.

Now divide the left and right sides of the equation by 2.

2X: 2 = 740: 2 or X = 370. We found the root of our equation and at the same time found the answer to our problem. Last year, there were 370 students at school # 5.

Let's check if our root actually turns the equation into a true equality. Substitute 370 for X in the equation 2X - 20 = 720.

2*370-20 = 720.

Everything is correct.

So, in order to solve an equation with one variable, it must be reduced to the so-called linear equation of the form ax = b, where a and b are some numbers. Then divide the left and right sides by the number a. We get that x = b: a.

What does it mean to bring an equation to a linear equation?

Consider this equation:

5X - 2X + 10 = 59 - 7X + 3X.

It is also an equation with one unknown variable X. Our task is to bring this equation to the form ax = b.

To do this, first we collect all the terms having X as a factor on the left side of the equation, and the remaining terms on the right side. Terms that have the same letter as a factor are called similar terms.

5X - 2X + 7X - 3X = 59 - 10.

According to the distribution property of multiplication, we can take the same factor out of the brackets, and add the coefficients (factors at the variable x). This process is also called the reduction of such terms.

X (5-2 + 7-3) = 49.

7X = 49. We brought the equation to the form ax = b, where a = 7, b = 49.

And as we wrote above, the root of an equation of the form ax = b will be x = b: a.

That is, X = 49: 7 = 7.

Algorithm for finding the roots of an equation with one variable.

1. Collect similar terms on the left side of the equation, the rest of the terms on the right side of the equation.
2. Give similar terms.
3. Bring the equation to the form ax = b.
4. Find the roots by the formula x = b: a.

Note... In this article, we have not considered those cases when a variable is raised to any power. In other words, we considered equations of the first degree in one variable.

• Equality with a variable is called an equation.
• To solve an equation means to find many of its roots. An equation can have one, two, several, many roots, or none at all.
• Each value of a variable at which a given equation turns into a true equality is called the root of the equation.
• Equations that have the same roots are called equivalent equations.
• Any term in the equation can be transferred from one side of the equality to the other, while changing the sign of the term to the opposite.
• If both sides of the equation are multiplied or divided by the same nonzero number, then you get an equation that is equivalent to this equation.

Examples. Solve the equation.

1. 1.5x + 4 = 0.3x-2.

1.5x-0.3x = -2-4. We collected the terms containing the variable on the left side of the equality, and the free terms on the right side of the equality. In this case, the property was applied:

1.2x = -6. They brought similar terms according to the rule:

x = -6 : 1.2. Both sides of the equality were divided by the coefficient of the variable, since

x = -5. Divided according to the rule of dividing a decimal fraction by a decimal fraction:

to divide a number by a decimal fraction, you need to move the commas in the dividend and divisor by as many digits to the right as there are after the decimal point in the divisor, and then divide by a natural number:

6 : 1,2 = 60 : 12 = 5.

Answer: 5.

2. 3∙ (2x-9) = 4 ∙ (x-4).

6x-27 = 4x-16. Expand the parentheses using the distribution law of multiplication versus subtraction: (a-b) ∙ c = a ∙ c-b ∙ c.

6x-4x = -16 + 27. We collected the terms containing the variable on the left side of the equality, and the free terms on the right side of the equality. In this case, the property was applied: any term in the equation can be transferred from one side of the equality to the other, while changing the sign of the term to the opposite.

2x = 11. Brought similar terms according to the rule: to bring such terms down, you need to add their coefficients and multiply the result by their common letter part (that is, attribute their common letter part to the result obtained).

x = 11 : 2. Both sides of the equality were divided by the coefficient of the variable, since if both sides of the equation are multiplied or divided by the same nonzero number, then an equation is obtained that is equivalent to this equation.

Answer: 5,5.

3. 7x- (3 + 2x) = x-9.

7x-3-2x = x-9. Expanded brackets according to the parenthesis expansion rule, preceded by a "-" sign: if there is a "-" sign in front of the brackets, then remove the brackets, the "-" sign and write down the terms in parentheses with opposite signs.

7x-2x-x = -9 + 3. We collected the terms containing the variable on the left side of the equality, and the free terms on the right side of the equality. In this case, the property was applied: any term in the equation can be transferred from one side of the equality to the other, while changing the sign of the term to the opposite.

4x = -6. They brought similar terms according to the rule: to bring such terms down, you need to add their coefficients and multiply the result by their common letter part (that is, attribute their common letter part to the result obtained).

x = -6 : 4. Both sides of the equality were divided by the coefficient of the variable, since if both sides of the equation are multiplied or divided by the same nonzero number, then an equation is obtained that is equivalent to this equation.

Answer: -1,5.

3 ∙ (x-5) = 7 ∙ 12 — 4 ∙ (2x-11). Multiply both sides of the equality by 12 - the lowest common denominator for the denominators of these fractions.

3x-15 = 84-8x + 44. Expand the parentheses using the distribution law of multiplication versus subtraction: in order to multiply the difference of two numbers by the third number, you can separately reduce and separately subtract multiplied by the third number, and then subtract the second result from the first result, i.e.(a-b) ∙ c = a ∙ c-b ∙ c.

3x + 8x = 84 + 44 + 15. We collected the terms containing the variable on the left side of the equality, and the free terms on the right side of the equality. In this case, the property was applied: any term in the equation can be transferred from one side of the equality to the other, while changing the sign of the term to the opposite.

7B algebra lesson plan.

Linear equation in one variable.

(04.10.2012)

The purpose of the lesson... Formation of the skill of solving an equation with one unknown, reducing it to a linear equation using the properties of equivalence.

Lesson type: combined.

Lesson Objectives:

1) educational:

To acquaint students with the form of a linear equation and the way to solve it, to achieve the assimilation of the rule for solving linear equations, its understanding and the ability to use it when solving;

2) developing:

continue the formation of mathematical knowledge and methods of mental activity (the ability to analyze the situation and navigate in actions, learn how to perform a new action, bring it to automation). Form elements of mathematical logic.

3) educational:

the formation of the skill of step-by-step work under the guidance of a teacher (explanation of new material, initial consolidation), perception of information by ear (cards), the formation of self-esteem (reflection).

During the classes

I. Checking homework frontally.

II. Oral work (on cards)

The purpose of oral work: diagnostics of the formation of skills in solving linear equations with one variable.

1. Instead of (*) put the sign "+" or "-", and instead of dots - numbers:

a) (* 5) + (* 7) = 2;

b) (* 8) - (* 8) = (* 4) -12;

c) (* 9) + (* 4) = - 5;

d) (-15) - (* ...) = 0;

e) (* 8) + (* ...) = - 12;

e (* 10) - (* ...) = 12.

2. Make up equations that are equivalent to the equation:

a) x-7 = 5;

b) 2x-4 = 0;

c) x-11 = x-7;

d) 2 (x-12) = 2x-24.

III. Generalization of the ability to solve equations by reducing them to a linear equation.

Teamwork with the class.

Collective work form: frontal

Let's solve the equation

12 - (4x-18) = (36 + 5x) + (28 - 6x). (one)

To do this, we will perform the following transformations:

1. Let's expand the brackets. If there is a plus sign in front of the brackets, then the brackets can be omitted, keeping the sign of each term enclosed in brackets. If there is a minus sign in front of the brackets, then the brackets can be omitted by changing the sign of each term enclosed in brackets:

12 - 4x + 18 = 36 + 5x + 28 - 6x. (2)

Equations (2) and (1) are equivalent.

2. Move the unknown terms with opposite signs so that they are only in one side of the equation (either on the left or on the right). Simultaneously, we transfer the known terms with opposite signs so that they are only in the other side of the equation.

For example, we transfer the unknown terms with opposite signs to the left, and the known ones to the right side of the equation, then we get the equation

4x-5x + 6x = 36 + 28-18, (3)

which is equivalent to equation (2), and hence to equation (1).

3. Here are similar terms:

3x = 46. (4)

Equation (4) is equivalent to equation (3), and hence to equation (1).

4. Let us divide both sides of equation (4) by the coefficient of the unknown. The resulting equation x = 46 / -3 or -15 1/3 will be equivalent to equation (4), and hence to equations (3), (2), (1).

Therefore, the root of equation (1) will be the number -15 1/3.

Using this scheme (algorithm), we solve the equations in today's lesson:

1. Expand brackets.

2. Collect the terms containing unknowns in one side of the equation, and the remaining terms in the other.

3. Bring similar terms.

4. Divide both sides of the equation by the coefficient of the unknown.

Note: it should be noted that the given scheme is not obligatory, since there are often equations, for the solution of which some of the indicated steps are unnecessary. When solving other equations, it is easier to deviate from this scheme, as, for example, in the equation:

7 (x-2) = 42.

IV. Training exercises.

№№ 132 (a, d), 133 (a, d), 136 (c), 138 (d) - with a note on the board.

№132. Find the root of the equation:

a) (13x-15) - (9 + 6x) = - 3x

Let's expand the brackets:

13x-15-9-6x = -3x.

Move the unknown terms with opposite signs to the left, and the known ones to the right side of the equation, then we get the equation:

13x-6x + 3x = 15 + 9.

Here are similar terms.

10x = 24.

Let us divide both sides of the equation by the coefficient of the unknown.

x = 2.4

Answer: 2.4

d) (0.5x + 1.2) - (3.6-4.5x) = (4.8-0.3x) + (10.5x + 0.6);

0.5x + 1.2-3.6 + 4.5x = 4.8-0.3x + 10.5x + 0.6;

0.5x + 4.5x + 0.3x-10.5x = 4.8 + 0.6-1.2 + 3.6;

5.2x = 7.8;

x = -1.5

Answer: -1.5

№133 Find the root of the equation:

a) 5 (3x + 1.2) + x = 6.8,

15x + 6 + x = 6.8,

15x + x = 6.8 - 6,

16x = 0.8,

x = 0.8: 16,

x = 0.05,

Answer: 0.05

d) 5.6 - 7y = - 4 (2y - 0.9) + 2, 4,

5.6 - 7y = - 8y + 3, 6 + 2.4,

8y - 7y = 3.6 + 2.4 - 5.6,

y = 0.4,

Answer: 0.4

№ 136. Solve the equation:

c) 0.8x - (0.7x + 0.36) = 7.1,

0.8x - 0.7x - 0.36 = 7.1,

0.1x = 0.36 + 7.1,

0.1x = 7.46,

x = 7.46: 0.1,

x = 74.6

Answer: 74.6.

№ 138. Find the root of the equation:

d) -3 (y + 2.5) = 6.9 - 4.2y,

3y - 7.5 = 6.9 - 4.2y,

4.2y - 3y = 6.9 + 7.5,

1.2y = 14.4,

y = 14.4: 1.2,

y = 12,

Answer: 12

V. Independent work, taking into account the individual abilities of students.

I. Option.

1. To solve the equation 5x = -40, you need to divide -40 by 5. What is the root of this equation?

2. Underline the coefficient at x and solve the equations:

a) 7x = 49;

6) - Zx = 111;

c) 12x = 1.

3. Solving the equation 12x = -744, Kolya found what x = -62. Substituting 62 instead of x, check if the root of the equation is found correctly.

4. Solve the equations.

a) 6x = 24;

b) 13x = -39;

c) 8x = 4;

d) 6x = 7.5; e) 7x = 63;

f) - 4x = 12;

g) 9x = - 3;

h) 9x = 0,36.

5. At what value of x:

a) the value of the expression 8x is equal to -64;

b) the value of the expression 7x is equal to 1;

c) is the value of the expression -x equal to 11?

6. Move the terms containing x to the left part equations, and the rest to the right, changing at the same time their signs to the opposite:

a) 2x - 3 = 5x + 8; c) -2x - 5 = 6x - 8;

b) 4x - 12 = -3x + 3; d) -4x - 2 = - 13x+ 21.

7. Complete the solution to the equation:

a) 2x - 4 = -8x + 12; b) Zx - 2 = 7x - 14;

c) 2x + 8x = 12 + 4 d) Zx - 7x = -14 + 2

8. Solve the equation:

a) Zx + 8 = x - 12;

b) x + 4 = 3 - 2x;

c) 5y = 2y + 16;

d) -2x + 9 - 8 = x - 1.

9. Solve the equation:

a) 1.2x = -4.8; d) Zx - 4 = 11; g) 2x - 1 = 3x + 6;

b) -6x = 7.2; e) 5 - 2x = 0; h) x - 8 = 4x - 9;

B) -X = -0.6; f) -12 - x = 3; i) 5 - 6x = 0.3 - 5x.

10. At what value of a

a) the value of the expression 3 + 2a is 43,

b) the value of expression 12 - a is equal to 100;

c) the values ​​of expressions 13a + 17 and 5a + 9 are equal;

d) the values ​​of expressions 5a + 14 and 2a + 7 are the opposite positive numbers?

II. Option

1. For each equation of the form ax = b, write down what is a and what is equal to:

a) 2.3x = 6.9;

b) –x = -1;

c) - x = 6;

d) 1.2x = 0.

2.a) Finish the record: to solve the equation ax = b, in which a = 0, you need ...

b) Solve the equation 12x = -60 and check.

3. Solve the equation:

1) a) 2x = 12; b) -5x = 15; c) - x = 32; d) -11x = 0;

2) a) 3x = 5; b) - 6x = -15; c) 29x = - 27; d) 16x = - 1;

3) a) 5x = 1/3 |; b) 4x = - 2/7; c) 1 / 3x = 6; d) -2 / 7x = 14.

4) a) 0.01x = 6.5; b) - 1.4x = 0.42; c) 0, Zx = 10; d) -0.6x = - 0.5.

4. At what value of x:

a) the value of the expression 5x is equal to - 1;

b) the value of the expression -0.1x is equal to 0.5;

c) is the value of expression 16x equal to 0?

5. The solution to an equation of the form ax = b was written on the blackboard, but the right side of the equation was erased. Rebuild it:

a) 5x = ... b) Zx = ... c) 4x = ...

x = -12; x = 1/6; x = 0.8.

6. Find the value of a for which the equation ax = 114 has a root of 6.

7. Solve the equation:

a) Zx-4 = 20

b) 54 - 5x ~ -6;

c) 1.2-0.3x = 0;

d) 16-7x = 0;

e) 5/6-x = 1/6

8. Solve the equation:

a) 5x-11 = 2x + 8; d) 0.8x-4 = 0.5-7;

b) 6-7x = 11-6x; e) 2.6x + 8 = 2x;

c) 3 - x = x + 13; f) 12 + 1 / 3x = 15 - 1 / 6x

9. At what value of a:

a) the value of the expression 5-Za is 17;

b) the meaning of expressions 3-2a and 5a + 10 are equal;

c) the value of expression 5 - 9a is 4 more than the value of expression a + 1;

d) is the value of the expression 7 + 8a 5 less than the value of the expression 2a + 1?

10. Solve the equation:

a) 15 (x + 2) = 40; c) 5 (2x + 1) = 3 (2x);

b) - 2 (1-x) = x; d) -6 (2-x) -5 (1 + x).

11. Solve the equation:

a) 43 + 4x + (11-5x) = 7; d) 6 (x + 11) -7x = 73 + x;

b) 12-4x - (2 + x) = 5x; e) 8 (3-x) - 12 + 6x = 25-x;

c) 5x + 12-3 (x + 16) = - 20; f) 6-x-3 (2-5x) - 12 + 8x.

For self-control: after expanding the parentheses, the equation is obtained:

a) 43 + 4x + 11-5x = 7; d) 6x + 66-7x = 73 + x;

b) 12-4x-2-x = 5x; e) 24-8x-12 + 6x - 25x;

c) 5x + 12-Zx-48 = -20; f) 6-x-6 + 15x = 12 + 8x.

III. Option

1. Solve the equation:

a) 6x = 36; c) -x = 18; e) 49x = 0; g) 21x = - 3;

b) 5x = 5/7; d) 11x = -1/3; c) 1 / 3x = 0; e) -3 / 7x = - 1;

2. Solve the equation and check:

a) 0.08x - 1; c) - 0.1x = 1; e) 0.6x = - 5; g) - 0.3x = - 1.1;

b) 0.Zx = 1/3; d) - 1 / 7x = 0; f) 0.2x = 1/7 h) - 3.6x - - 6.

3. Make up any equation of the form ax = b, which

a) has the root number 3;

b) has the root number 0;

c) has no roots;

d) has infinitely many roots.

4. At what values ​​of x

A) the value of the expression 1 / 3x is 3;

b) the value of the expression - 0.8x is equal to 0;

c) the value of the expression 0.01x is equal to 30;

d) the value of the expression -15x is equal to - 0.1.

5. Having solved an equation of the form ax = b, the student erased the coefficient a. Rebuild it if possible:

a) ... x = 1/8 b) ... x = -4 c) ... x = 0

x = 4 x = - 1 x = 0

6 ... For what integer values ​​of a is the root of the equation ax = 8 an integer?

8. Expressions for + 2 and a-5 are given. At what values ​​of a

a) the values ​​of these expressions are equal;

b) the value of the first expression is 12 more than the value of the second;

c) the value of the first expression is 7 less than the value of the second;

d) the value of the first expression is 5 times greater than the value of the second

rogo?

9. Solve the equation:

a) - (2x + 1) = 41; d) 5 (x-1) - 3 (2x + 2) = - 1;

b) 5 (12-x) = 27; e) 12 (1-x) - 4 = 2 (4x + 6);

c) 1.2 (2x-1) = 3.6; f) 0.5 (2x-1) - x = 6.5.

10. For the equation ax-11 = 3x + 1 find

a) the values ​​of a at which the root of this equation is the number 6;

b) the values ​​of a at which this equation has no roots;

c) natural values ​​of a, for which the root of the equation is a natural number.

11. Solve the equation:

a) 5 (x - 18) - 7x = 21 + x; d) 6 (x - 1) +12 (3 - 2x) = 45 - 17x;

b) Zx + 6 (1 - x) = - 2 (2 + x); e) 15 (3 - x) - 5 (x + 11) = 1 - 19x;

c) 1.7 - 8 (x - 1) = 3.7 + 2x; f) - (5 - x) - 8 (6 + x) = 11.8 + x.

VI ... Lesson summary. Algorithm for reducing an equation to a linear equation.

Vii ... Homework: p. 3, No. 128, 129, 131.

The check showed that the students completed these tasks, that is, they mastered the given topic.

Introspection lesson

1. There are 25 students in the class. Five people can study for 4-5, 8 people for fours, the rest cannot study without guidance. When planning the lesson, this was taken into account and determined the choice of methods and techniques for presenting new material and ways to consolidate the knowledge gained.

2. This is the second lesson on the topic "Equations with one variable". In this academic year, this material was studied; at the beginning of the lesson, knowledge was updated in the form of a teacher's reminder of the necessary information. This lesson is important for the subsequent study of the topic "Linear function" in the course of algebra. Specificity - there are many concepts, models, knowledge that are better systematized and formalized in the form of a synopsis. Lesson type - combined lesson.

3. In the lesson, the following tasks were solved:

Didactic purpose of the lesson: Promote the awareness and comprehension of new educational information about geometric and analytical models of a linear equation with one variable.

Educational purpose: Form the concept of a linear equation and methods for its solution and achieve an understanding of the essence of its name, notation and algebraic notation.

Developing goal: Promote the development of the ability to model a situation and systematize knowledge in the form of a table.

Educational purpose: Formation of self-esteem, respect for intellectual work.

The complexity of their solution is thought out. The main tasks were educational tasks, while solving them, both developing and educating tasks were solved along the way. The developing problem was solved through the methods of accessible study of the material, and the upbringing one was already at the stage of choosing a class for an open lesson.

4. This structure of the lesson is dictated by the inability of students to perceive the monotonous material presented for a long time and with concentration. Therefore, the lesson in the first half is denser and more dynamic. The survey was conducted with the aim of updating existing knowledge and consolidating new ones. The links between the stages are logical. Homework contains three numbers, students can complete as many as they wish: 3-one number, 4-two, 5-three.

5. The main emphasis was on the concepts: linear equation, root of the equation. The main concepts of the topic are selected, the skills are being developed to designate, name, write down the algebraic model of the numerical interval.

6. Teaching methods selected partially search, visual, activity.

7. There was no need to apply differentiated teaching methods. It is enough to provide individual assistance.

8. Control of assimilation of knowledge was carried out by observing the independence and activity of students, as they studied new material.

9. Learning tools were used: Textbook Yu.N. Makarychev et al. - 2009, cards for oral and individual work, the board was actively used.

10. The tasks are fully implemented.

Khartsyzsk secondary school number 25 "Intellect"

with in-depth study of individual subjects

Introductory lesson in algebra in grade 7

Linear Equation

with one variable

Mathematic teacher

L.P. Nakonechnaya

Khartsyzk, 2017

Lesson topic. Linear Equation in One Variable

Lesson type: combined.

Lesson teaching method: the use of modular technology.

The purpose of the lesson. Deepen, expand and generalize previously acquired knowledge about

equation.

Lesson Objectives

Educational:

To deepen and consolidate the knowledge of students about solving equations;

Formation of the skill of solving an equation with one unknown reduction to a linear equation using the properties of equivalence;

Form the ability to solve equations with a module;

To acquaint students with solving equations with a parameter;

Build a vocabulary of terms on the topic of the equation.

Developing:

Form independence and the ability to analyze, compare and generalize;

Develop creative thinking;

Develop the ability to apply knowledge in life situations.

Develop mathematical speech;

Educational:

Contribute to the education of a conscious and interested attitude to the subject;

To instill an interest in research activities;

To cultivate a kind attitude towards comrades, the ability to offer help.

During the classes

1. Organizational stage

Check the availability of educational supplies from students.

Nature cannot part with warmth -

So let go and fall asleep….

September always comes, year after year

Similar to August a little - a little

And the forest greens have not yet faded,

And in summer fur coats the beast,

And the sun shines like summer in the sky,

Waste your warmth.

In a warm, friendly atmosphere, we will begin our journey to the world of ALGEBRA

2. Introductory conversation of the teacher

On this warm September day, we begin to study a new subject for you - algebra, with which you will be friends until graduation.

Algebra is an ancient science. The ancient Babylonians and Egyptians more than 4000 years ago already owned some algebraic concepts and general methods of solving problems. But the outstanding ancient Greek mathematician Diophantus (III century) is rightfully called the "father of algebra". Already in those distant times, he was able to solve very complex equations, using letter designations for unknown numbers.

In 825, the Arab scholar Muhammad al-Khwarizmi wrote the book Kitab al Jabr wal-muqabala, which means “The Book of Reconstruction and Opposition,” in which algebra is considered as an independent area of ​​mathematics. It was the world's first algebra textbook. The word "algebra" itself comes from the word "al-jabr", which means "transfer of negative terms from one part of the equation to another with a change in sign."

The "father of modern algebra" is considered the French mathematician Francois Vieta, who was born in 1540 in the small French town of Fontenay. He was a lawyer by profession, but mathematics was his real calling. Carried away by some mathematical problem, he could sometimes work on it for three days in a row without food or sleep.

A great contribution to the further development of algebraic symbolism was made by the outstanding French mathematician and philosopher Rene Descartes (1596 - 1650), the designations he introduced have survived to our time.

Collaboration with algebra doesn't end at school. There are special educational institutions where mathematicians are trained, for whom this science becomes a profession.

Knowledge of algebra is essential in everyday life. It allows you to solve complex problems that relate to the needs of technology and production.

To move on to the next stage of acquaintance with algebra, I suggest you guess the "Pentagon"

1. She teaches many, although she is constantly silent.

2. Some people try to teach her, but not everyone succeeds.

3. She can delight, she can make you angry, she can send you on a trip and even lock you in a room for several days.

4. She can tell you something, advise something, she can ask you a problem, but in any case she will make you think.

5. You can take it with you, even put it in a briefcase or put it in a closet.

Quite right guys this is a book. And now we will get acquainted with a textbook that will take us into the fascinating world of algebra.

(Acquaintance with the textbook Algebra. Grade 7: a textbook for general education organizations / Yu.N. Makarychev, N.G. Mindyuk, K.I.Neshkov, S.B. Suvorov; under the editorship of S.A. Telyakovsky. - 6 ed. - M.: Education, 2016.)

3. Updating basic knowledge.

Frontal poll

What is called an equation?

(An equation is an equality containing a variable to be found.)

What is called the root of the equation?

(The root of the equation is the value of the variable, when substituted into the equation, the correct equality is obtained)

What does it mean to solve an equation?

(To solve an equation means to find all its roots or to show that they do not exist);

How to open parentheses before which there is a "+" sign.

(The signs in brackets are left unchanged)

How to open parentheses before which there is a "-" sign.

(The signs in brackets are reversed)

What terms are called similar?

(Terms that have the same letter part are called similar)

How to bring such terms?

(we perform actions with coefficients and assign the letter part to the result)

What is called the modulus of a number?

(The modulus of a number is the distance from the origin to a point with a given coordinate)

4. Formulation of the goal and objectives of the lesson

In the 5th - 6th grade, we worked mainly with numerical expressions. In algebra, actions are mainly studied not with specific numbers, but with numbers that are indicated by letters and the topic of our lesson today is "Linear Equation with One Variable" (Define the tasks of today's lesson together with the students.) In today's lesson we will deepen your knowledge of the equation and continue acquaintance with equations with a modulus and equations containing a parameter.

Expected results:

Know: Definitions of the concepts "equation", "root of an equation", "linear equation", "equivalent equation", an algorithm for solving a linear equation.

Be able to: Solve linear equations, determine the number of roots of a linear equation, solve the simplest equations containing the sign of the modulus, investigate the solution of simple equations containing a parameter.

5. Motivation for educational and cognitive activities

Little is known about Diophantus, it is even impossible to accurately establish the years of his life. But he was such a famous mathematician that according to legend, even the epitaph on his gravestone was written in the form of a problem. It read: “Traveler! Under this stone lies the ashes of Diophantus, who died in extreme old age. The sixth part of his long life he was a child, the twelfth part of his youth, the seventh part he spent unmarried. Five years after his marriage, he had a son, who lived half his father's life. Four years after the death of his son, Diophantus himself, mourned by his loved ones, fell asleep in eternal sleep. Tell me, if you can count how many years Diophantus lived? "

The most common way to solve this problem is to write an equation. And I propose after our lesson to compose and solve it at home.

(Solution. Let's take for x - the age of Diophantus, then we can make the equation:

6. Deepening and systematization of knowledge(Students work with the textbook)

Definition. An equation of the form ax = b, where x is a variable, and and b are some numbers called linear equation in one variable

Definition The equations are called tantamount to if they have the same roots. Equations that have no solutions are also considered to be equivalent.

Equation properties

1. If both sides of the equation are multiplied or divided by one and the same nonzero number, then we get an equation equivalent to the given one;

2. If in the equation to transfer the term from one part to another, changing its sign, then an equation equivalent to the given one will be obtained.

To solve a linear equation with one variable you need:

1. Expand the brackets.

2. Collect the terms containing unknowns in one part of the equation, and the remaining terms in the other.

3.Reduce similar terms

on both sides of the equation.

4. Divide both sides of the equation by the coefficient of the unknown

ah = in

If a ≠ 0, the equation has a unique solution;

If a = 0 and b = 0, the equation has many roots;

If a = 0, and b ≠ 0, the equation of solutions has no

| x | = a

If a = 0, then x = 0

If a ˂ 0, there are no solutions

If a ˃ 0, x = a or x = -a

We have big houses, (hands up)
There are many smaller houses (they lower their hands a little lower)
Greens are bright around (spread their arms to the sides)
Swings in the wind (hands swing to the right and then to the left)
You are my friend and I am your friend (right hand forward, then left hand forward)
Let the friendship never end (clap their hands)

7. Consolidation of knowledge and skills.

(Teamwork and work in pairs. We carry out in each block task a, tasks b) and c) we solve independently with subsequent mutual check)

1. At what value of x:

a) the value of the expression 11x is equal to -1;

b) the value of the expression - 0.1x is equal to 0.7;

c) is the value of the expression 19x equal to 0?

2. At what value of y:

a) the value of the expression 7 - 4y is 19;

b) the meaning of expressions 3 - 2y and 5y + 10 are equal;

c) the value of the expression 5 - 9y is 4 more than the value of the expression y + 1;

2. The solution to an equation of the form ax = b was written on the board, but the right side of the equation was erased. Reconstruct the right side of the equation

a) 19x = ... b) 6x = ... c) 7x = ...

x = - 4; x =; x = 2.6.

3. Solve Equations

a) 7.2 (x + 5) = 36 + 7.2x; b) 12x - (3x +4) = 17 + 9x; c) 1.3x + 9 = 0.7x + 27;

7.2x + 36 = 36 + 7.2x; 12x - 3x - 4 = 17 + 9x; 1.3x - 0.7x = 27 - 9;

0x = 0.12x - 3x - 9x = 17 +4; 0.6x = 18;

0x = 21.x = 18: 0.6;

- (Solution of equation d) comment on the board)

d) (2 - x) (x - 7) = 0;

The product of two factors is zero if at least one of the factors is zero.

2 - x = 0 or x - 7 = 0

a) the solution is any number.

b) there are no solutions;

c) one solution x = 30.

d) two solutions x = 2, x = 7.

"Brainstorming" (Statement of a problematic question)

Does an equation always have roots? Has one root?

Can an equation have three roots, four roots, five roots? Give an example of such an equation.

Is this equation linear?

What property of multiplication is the solution of such equations based on?

(Tasks 4, 5, 6, 7 collective work)

4. Solve the equations

a) | x | = 4.5; b) | x | = - 17; c) | 3x + 2 | = 8;

x = 4.5; no solutions; 3x + 2 = 8; or 3x + 2 = - 8;

3x = 6; 3x = -10;

x = 2.x = - 3.

5. Find the value of a for which the equation ax = 156 has a root of 6.

Solution. Since the root of the equation is 6, then when substituted into the equation we get the correct equality a 6 = 156

6. Solve the equation (a - 2) · x = 4;

Solution. With a = 2, (a - 2) = 0, we get the equation 0 x = 4, which has no roots. If a - 2 ≠ 0, and ≠ 2, then x =.

7. Find all integer values ​​of a for which the root of the equation ax = 8 is an integer.

Solution. Let us find the value of x for a ≠ 0, x =. For the root of the equation to be an integer, it is necessary that a is a divisor of 8. Therefore, a = (-8; -4; -2; -1; 1; 2; 4; 8)

8. Lesson summary

What equation is called linear?

How many roots does a linear equation have?

What properties do you know for solving equations?

9. Reflection.

Proverb: A wise man walked, and three people were carrying stones for construction towards him. The sage stopped and asked each of them a question. The first one asked: "What have you been doing all day?" And he answered: "He carried the damned stones." Second: "And I conscientiously did my job." And the third smiled and replied: "And I took part in the construction of the temple."

Guys, who worked in good faith today? Who took part in the "building of the temple"?

9. Homework

Learn the definitions and properties of equations

№131 (a, b), №134 (a), №135 (a, b, c), solve the problem about the age of Diophantus.

Literature.

1. Algebra. Grade 7: textbook for general education. organizations / Yu.N. Makarychev, N.G. Mindyuk, K.I. Neshkov, S.B. Suvorova; ed. S.A. Telyakovsky. - 6th ed. - M.: Education, 2016.

2. Kostrykina N.P. Problems of increased difficulty in the course of algebra 7-9 grades. - M .: Education, 1991.

3.Bartenev F.A. Non-standard problems in algebra. - M .: Education, 1976.

4. Chervatyuk OG, Shimanskaya GD Elements of interesting mathematics in mathematics lessons. - К .: "Radianska school", 1968.

5. Perelman Ya.I. Live mathematics. - M .: "Science", 1978.

6. Shunda N.M. Collection of problems in algebra for grades 6 - 8. - K.: "Radianska school", 1987.

Linear Equation with one variable

Examination work number 1

Target:

Show the skills of mastering the topic "Linear Equation with One Variable" Be able to compose an expression with variables according to the condition of the problem. Perform transformations of expressions: bring similar terms, open parentheses. Find the value of an expression with variables given the values ​​of the variables.

Task number 1

• Solve the equation:

• Option 1

• a) 6x- 15 = 4x + 11;
• b) 9 - 7 (x + 3) = 5 - 4x.

• Option 2

• a) 9x - 8 = 4x + 12;
• b) 6 - 8 (x + 2) = 3 - 2x.

Task number 2

• Option 1

The first box contained 5 times more apples than the second. When they took 7kg of apples from the first box, and added 5kg to the second, then there were equal parts of apples in the boxes. How many kg. There were apples in each box first?

• Option 2

The first basket contained 4 times more mushrooms than the second. When 4 more mushrooms were put in the first basket, and 31 mushrooms in the second, then in the baskets of mushrooms it became equal. How many mushrooms were in each basket first?

Task number 3

• Solve the equation:

• Option 1

a) (8y - 16) * (2.1 + 0.3y) = 0;

b) 7x - (4x + 3) = 3x + 2.

• Option 2

a) (12y + 30) (1.4 - 0.7y) = 0;

b) 9x - (5x - 4) = 4x + 4.

Task number 4

• Option 1

100 kg were delivered to the first store sweets, and in the second-240kg. The first store sold 12 kg of sweets every day, and the second one - 46 kg. In how many days will there be 4 times less sweets in the second store than in the first?

• Option 2

The first warehouse had 300 tons of coal, and the second - 178 tons. 15 tons of coal were transported daily from the first warehouse, and 18 tons from the second. In how many days will there be 3 times more tons of coal in the first warehouse than in the second?

Task number 5

• Option 1

At what value of a is the equation (a + 3) x = 12

a) has a root equal to 6;

b) has no roots?

• Option 2

At what value of a is the equation (a -2) x = 35

a) has a root equal to 5;