What is a linear equation with two variables. §1. Linear equations with two variables. Features of solving double inequalities

We have often come across equations of the form ax + b = 0, where a, b are numbers, x is a variable. For example, bx - 8 = 0, x + 4 = O, - 7x - 11 = 0, etc. The numbers a, b (equation coefficients) can be any, except for the case when a = 0.

The equation ax + b = 0, where a, is called a linear equation with one variable x (or a linear equation with one unknown x). We can solve it, that is, express x through a and b:

We noted earlier that quite often mathematical model the real situation is a linear equation with one variable or an equation that, after transformations, reduces to a linear one. Now let's look at this real situation.

From cities A and B, the distance between which is 500 km, two trains departed towards each other, each with its own constant speed. It is known that the first train left 2 hours earlier than the second. 3 hours after the second train left, they met. What are the train speeds?

Let's create a mathematical model of the problem. Let x km/h be the speed of the first train, y km/h be the speed of the second train. The first one was on the road for 5 hours and, therefore, covered a distance of bx km. The second train was on the way for 3 hours, i.e. walked a distance of 3 km.

Their meeting took place at point C. Figure 31 shows a geometric model of the situation. In algebraic language it can be described as follows:

5x + Zu = 500

or
5x + Zu - 500 = 0.

This mathematical model is called a linear equation with two variables x, y.
At all,

ax + by + c = 0,

where a, b, c are numbers, and , is linear the equation with two variables x and y (or with two unknowns x and y).

Let's return to the equation 5x + 3 = 500. We note that if x = 40, y = 100, then 5 40 + 3 100 = 500 is a correct equality. This means that the answer to the question of the problem can be as follows: the speed of the first train is 40 km/h, the speed of the second train is 100 km/h. A pair of numbers x = 40, y = 100 is called a solution to the equation 5x + 3 = 500. It is also said that this pair of values ​​(x; y) satisfies the equation 5x + 3 = 500.

Unfortunately, this solution is not the only one (we all love certainty and unambiguity). In fact, the following option is also possible: x = 64, y = 60; indeed, 5 64 + 3 60 = 500 is a correct equality. And this: x = 70, y = 50 (since 5 70 + 3 50 = 500 is a true equality).

But, say, a pair of numbers x = 80, y = 60 is not a solution to the equation, since with these values ​​a true equality does not work:

In general, a solution to the equation ax + by + c = 0 is any pair of numbers (x; y) that satisfies this equation, that is, turns the equality with the variables ax + by + c = 0 into a true numerical equality. There are infinitely many such solutions.

Comment. Let us return once again to the equation 5x + 3 = 500, obtained in the problem discussed above. Among the infinite number of its solutions there are, for example, the following: x = 100, y = 0 (indeed, 5 100 + 3 0 = 500 is a correct numerical equality); x = 118, y = - 30 (since 5,118 + 3 (-30) = 500 is a correct numerical equality). However, being solutions to the equation, these pairs cannot serve as solutions to this problem, because the speed of the train cannot be equal to zero(then he doesn’t move, but stands still); Moreover, the speed of the train cannot be negative (then it does not travel towards another train, as stated in the problem statement, but in the opposite direction).

Example 1. Draw solutions to a linear equation with two variables x + y - 3 = 0 points in coordinate plane xOy.

Solution. Let's select several solutions behind given equation, i.e. several pairs of numbers that satisfy the equation: (3; 0), (2; 1), (1; 2) (0; 3), (- 2; 5).

A. V. Pogorelov, Geometry for grades 7-11, Textbook for educational institutions

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Instructions

Substitution MethodExpress one variable and substitute it into another equation. You can express any variable at your discretion. For example, express y from the second equation:
x-y=2 => y=x-2Then substitute everything into the first equation:
2x+(x-2)=10 Move everything without “x” to the right side and calculate:
2x+x=10+2
3x=12 Next, to get x, divide both sides of the equation by 3:
x=4. So, you found “x. Find "y. To do this, substitute “x” into the equation from which you expressed “y”:
y=x-2=4-2=2
y=2.

Do a check. To do this, substitute the resulting values ​​into the equations:
2*4+2=10
4-2=2
The unknowns have been found correctly!

A way to add or subtract equations Get rid of any variable right away. In our case, this is easier to do with “y.
Since in the equation “y” has a “+” sign, and in the second one “-”, then you can perform the addition operation, i.e. fold the left side with the left, and the right with the right:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4Substitute “x” into any equation and find “y”:
2*4+y=10
8+y=10
y=10-8
y=2 Using the 1st method, you can check that the roots are found correctly.

If there are no clearly defined variables, then it is necessary to slightly transform the equations.
In the first equation we have “2x”, and in the second we simply have “x”. To reduce x when adding or subtracting, multiply the second equation by 2:
x-y=2
2x-2y=4Then subtract the second from the first equation:
2x+y-(2x-2y)=10-4 Note that if there is a minus in front of the bracket, then after opening, change the signs to the opposite ones:
2x+y-2x+2y=6
3у=6
find y=2x by expressing from any equation, i.e.
x=4

Video on the topic

When solving differential equations, the argument x (or time t in physical problems) is not always explicitly available. Nevertheless, this is a simplified special case of specifying a differential equation, which often helps to simplify the search for its integral.

Instructions

Consider a physical problem that results in differential equation, in which the argument t is missing. This is a problem about oscillations of a mass m suspended on a thread of length r located in a vertical plane. The equation of motion of the pendulum is required if it was initially motionless and tilted from the equilibrium state by an angle α. The forces should be neglected (see Fig. 1a).

Solution. A mathematical pendulum is a material point suspended on a weightless and inextensible thread at point O. Two forces act on the point: the force of gravity G=mg and the tension force of the thread N. Both of these forces lie in the vertical plane. Therefore, to solve the problem we can apply the equation rotational movement points around a horizontal axis passing through point O. The equation of rotational motion of a body has the form shown in Fig. 1b. In this case, I is the moment of inertia of the material point; j is the angle of rotation of the thread together with the point, measured from the vertical axis counterclockwise; M - moment of forces applied to material point.

Calculate these values. I=mr^2, M=M(G)+M(N). But M(N)=0, since the line of action of the force passes through point O. M(G)=-mgrsinj. The “-” sign means that the moment of force is directed in the direction opposite to the movement. Substitute the moment of inertia and the moment of force into the equation of motion and get the equation shown in Fig. 1s. By reducing the mass, a relationship emerges (see Fig. 1d). There is no t argument here.

Linear equation is an algebraic equation. In this equation, the total degree of its constituent polynomials is equal to one.

Linear equations are presented as follows:

In general form: a 1 x 1 + a 2 x 2 + … + a n x n + b = 0

In canonical form: a 1 x 1 + a 2 x 2 + … + a n x n = b.

Linear equation with one variable.

A linear equation with 1 variable is reduced to the form:

ax+ b=0.

For example:

2x + 7 = 0. Where a=2, b=7;

0.1x - 2.3 = 0. Where a=0.1, b=-2.3;

12x + 1/2 = 0. Where a=12, b=1/2.

The number of roots depends on a And b:

When a= b=0 , which means that the equation has an unlimited number of solutions, since .

When a=0 , b≠ 0 , which means the equation has no roots, since .

When a ≠ 0 , which means the equation has only one root.

Linear equation with two variables.

Equation with variable x is an equality of type A(x)=B(x), Where A(x) And B(x)- expressions from x. When substituting the set T values x into the equation we get a true numerical equality, which is called truth set this equation either solution of a given equation, and all such values ​​of the variable are roots of the equation.

Linear equations of 2 variables are presented in the following form:

In general form: ax + by + c = 0,

In canonical form: ax + by = -c,

In the shape of linear function: y = kx + m, Where .

The solution or roots of this equation is the following pair of variable values (x;y), which turns it into an identity. A linear equation with 2 variables has an unlimited number of these solutions (roots). The geometric model (graph) of this equation is a straight line y=kx+m.

If an equation contains x squared, then the equation is called

Lesson objectives:

• Educational:
  • repeat the topic: “Equations. Linear equations.
  • Equivalent equations and their properties”; ensure students understand the concept linear equations
• with two variables and their solution.:
  • Developmental
  • the ability to compare, build analogues, highlight the main thing;
  • the ability to generalize and systematize the material covered;
  • develop logical thinking, memory, imagination, mathematical speech;
  • develop active cognitive activity.
• Educational:
  • to cultivate independence, activity, and interest in students at all stages of the lesson;
  • to form such character qualities as perseverance, perseverance, determination.

Tasks that the teacher must solve in the lesson:

• teach to highlight main idea in the text;
• learn to ask questions to the teacher, yourself or students;
• learn to use acquired knowledge to solve non-standard problems;
• teach the ability to express your thoughts mathematically correctly.

Problems that students must solve in this lesson:

• know the definition of a linear equation with two variables;
• be able to write simple linear equations;
• be able to correctly find the values ​​of variables a, b and c;
• be able to identify linear equations with two variables among equations;
• answer the question: what is the solution to a linear equation in two variables?
• How do you know if a pair of numbers is a solution to an equation?
• be able to express one variable in terms of another.

Lesson type: lesson in learning new material.

DURING THE CLASSES

I. Organizational moment

II. Repetition of covered material

1) On the board: 2x, 2x + 5, 2x + 5 = 17.

2) Questions for the class:

– Define these expressions. (Expected answers: product, monomial, sum, polynomial, equation.)
-What is an equation called?
– Do you need an equation...? (Decide)
– What does it mean to “solve an equation”?
– What is the root of the equation?
– Which equations are equivalent?
– What properties of equivalence of equations do you know?

III. Updating students' knowledge

3) Assignment to the whole class:

– Convert expressions :(two people work at the board).

a) 2(x + 8) + 4(2x – 4) = b) 4(x – 2) + 2(3y + 4) =

After the transformation we got: a) 10x; b) 4x + 6y:

– Use them to create equations (students suggest - teacher writes equations on the board): 10x = 30; 4x + 6y = 28.

Questions:

– What is the name of the first equation?
– Why linear?
– Compare the second equation with the first. Try to formulate the definition of the second equation (Expected answer: an equation with two variables; students’ attention is focused on the type of equation – linear).

IV. Learning new material

1) The topic of the lesson is announced. Recording the topic in notebooks. Students’ independent formulation of the definition of an equation with two variables, a linear equation with two variables (by analogy with the definition of a linear equation with one variable), examples of equations with two variables. The discussion takes place in the form of a frontal conversation, dialogue - reasoning.

2) Class assignment:

a) Write two linear equations with two variables (the teacher and students listen to the answers of several students; at the teacher’s choice, one of them writes his equations on the board).

b) Together with the students, tasks and questions are determined that they should receive an answer to in this lesson. Each student receives cards with these questions.

c) Working with students to solve these issues and tasks:

– Determine which of these equations are linear equations with two variables a) 6x 2 = 36; b) 2x – 5y = 9: c) 7x + 3y 3; d) 1/2x + 1/3y = 6, etc. A problem may arise with the equation x: 5 – y: 4 = 3 (the division sign must be written as a fraction). What properties of equivalence of equations need to be applied? (Students' answers) Determine the coefficient values A, V And With.

– Linear equations with two variables, like all equations, need to be solved. What is the solution to linear equations in two variables? (Children give a definition).

Example: Find solutions to the equation: a) x – y = 12, write the answers in the form (x; y) or x = ...; y = .... How many solutions does the equation have?

Examples: Find solutions to the following equations a) 2x + y = 7; b) 5x – y = 4. How did you find the solutions to these equations? (Picked up).

– How do you know if a pair of numbers is a solution to a linear equation in two variables?

3) Working with the textbook.

– Find in the textbook those places where the main idea of ​​the topic of this lesson is highlighted

a) Oral performance of tasks: No. 1092, No. 1094.

b) Solution of examples No. 1096 (for weak students), No. 1097 (for the strong).

c) Repeat the properties of equivalence of equations.

Exercise: Using the properties of equivalence of equations, express the variable Y through the variable X in the equation 5x + 2y = 12 (“minute” on independent decision, then an overview of the solution on the board followed by an explanation).

d) Execution of example No. 1099 (one of the students completes the task at the board).

Historical reference

1. Guys, the equations that we met in class today are called Diophantine linear equations with two variables, named after the ancient Greek scientist and mathematician Diophantus, who lived about 3.5 thousand years ago. Ancient mathematicians first composed problems and then worked to solve them. Thus, many problems were compiled, which we become familiar with and learn to solve.

2. And also these equations are called indefinite equations. Many mathematicians worked on solving such equations. One of them is Pierre Fermat, a French mathematician. He studied the theory of solving indeterminate equations.

V. Lesson summary

1) Summarizing the material covered in the lesson. Answers to all questions posed to students at the beginning of the lesson:

– What equations are called linear with two variables?
– What is called solving a linear equation in two variables?
– How is this decision recorded?
– What equations are called equivalent?
– What are the properties of equivalence of equations?
– What problems did we solve in class, what questions did we answer?

2) Doing independent work.

For the weak:

– Find the values ​​of the variables a, b and c in the equation –1.1x + 3.6y = – 34?
– Find at least one solution to the equation x – y = 35?
– Are the pair of numbers (3; 2) a solution to a given linear equation with two variables 2x – y = 4?

For the strong:

– Write a linear equation with two variables for Diophantus’s problem: There are pheasants and rabbits walking in the yard of the house. The number of all legs turned out to be 26.
– Express the variable y in terms of x in the equation 3x – 5y = 8.

VI. Homework message

View all tasks in the textbook, a quick analysis of each task, select a task.

• For weak students: No. 1093, No. 1095b).
• For the strong: 1) No. 1101, No. 1104 (a). 2) solve Diophantus’ problem, find all natural solutions to this equation.

Additionally, at the request of students - No. 1105.

Instead of conclusion: I have been a mathematics teacher for over 40 years. And I want to note that public lesson– is not always the best lesson. It often happens that sometimes ordinary lessons bring more joy and satisfaction to the teacher. And then you think with regret that no one saw this lesson - the creation of the teacher and students.

A lesson is a single organism, a single whole; it is in the lesson that personal and moral experience of education is acquired for both students and teachers. 45 minutes of a lesson is so much and so little. A lot - because during this time you can “look” into the depths of centuries with your students and, “returning” from there, learn a lot of new, interesting things, and still have time to study new material.

Every student must be brought to the understanding that mathematics is the basis of human intellectual development. And the basis for this is the development of logical thinking. Therefore, before each lesson, I set a goal for myself and my students: to teach students to successfully work with definitions, skillfully distinguish the unknown from the known, proven from unproven, analyze, compare, classify, pose questions and learn to skillfully solve them. Use analogies, but if you can’t get out on your own, then next to you is not only a teacher, but your main assistant - a book.

Of course, an open lesson is some result of the teacher’s creative work. And teachers present at this lesson should pay attention to the main thing: the system of work, novelty, idea. Here, I think, it is not particularly important what teaching methodology the teacher uses in the lesson: old, modern or new innovative technologies, the main thing is that its use is appropriate and effective for the teacher and students.

I am very glad that in my life I have school, children, lessons and such kind colleagues. Thank you all!