The author's approach to this topic is not accidental. Equations with two variables are first encountered in the 7th grade course. One equation with two variables has an infinite number of solutions. This is clearly demonstrated by the graph of a linear function given as ax + by=c. In the school course, students study systems of two equations with two variables. As a result, a whole range of problems, with limited conditions on the coefficient of the equation, as well as methods for solving them, fall out of the teacher's field of vision and, therefore, the student.
We are talking about solving an equation with two unknowns in integer or natural numbers.
At school, natural and integer numbers are studied in grades 4-6. By the time they leave school, not all students remember the differences between the sets of these numbers.
However, a task like “solve an equation of the form ax + by=c in integers” is increasingly common in university entrance exams and in USE materials.
Solving indefinite equations develops logical thinking, ingenuity, and attention to analyze.
I propose the development of several lessons on this topic. I don't have clear recommendations on the timing of these lessons. Separate elements can be used in the 7th grade (for a strong class). These lessons can be taken as a basis and develop a small elective course on pre-profile preparation in the 9th grade. And, of course, this material can be used in grades 10-11 to prepare for exams.
The purpose of the lesson:
- repetition and generalization of knowledge on the topic “Equations of the first and second order”
- education of cognitive interest in the subject
- formation of skills to analyze, make generalizations, transfer knowledge to a new situation
Lesson 1.
During the classes.
1) Org. moment.
2) Actualization of basic knowledge.
Definition. A linear equation with two variables is an equation of the form
mx + ny = k, where m, n, k are numbers, x, y are variables.
Example: 5x+2y=10
Definition. A solution to an equation with two variables is a pair of values of the variables that turns this equation into a true equality.
Equations with two variables having the same solutions are called equivalent.
1.5x+2y=12 (2)y=-2.5x+6
This equation can have any number of solutions. To do this, it is enough to take any x value and find the corresponding y value.
Let x = 2, y = -2.5 2+6 = 1
x = 4, y = -2.5 4+6 =- 4
Pairs of numbers (2;1); (4;-4) - solutions of equation (1).
This equation has infinitely many solutions.
3) Historical background
Indefinite (Diophantine) equations are equations containing more than one variable.
In the III century. AD – Diophantus of Alexandria wrote “Arithmetic”, in which he expanded the set of numbers to rational ones, introduced algebraic symbolism.
Also, Diophantus considered the problems of solving indefinite equations and he gave methods for solving indefinite equations of the second and third degree.
4) Learning new material.
Definition: An inhomogeneous first-order Diophantine equation with two unknowns x, y is an equation of the form mx + ny = k, where m, n, k, x, y Z k0
Statement 1.
If the free term k in equation (1) is not divisible by the greatest common divisor (GCD) of the numbers m and n, then equation (1) has no integer solutions.
Example: 34x - 17y = 3.
GCD (34; 17) = 17, 3 is not divisible by 17, there is no solution in integers.
Let k be divisible by gcd(m, n). By dividing all the coefficients, one can achieve that m and n become coprime.
Statement 2.
If m and n of equation (1) are coprime numbers, then this equation has at least one solution.
Statement 3.
If the coefficients m and n of equation (1) are relatively prime numbers, then this equation has infinitely many solutions:
Where (; ) is any solution of equation (1), t Z
Definition. A homogeneous first-order Diophantine equation with two unknowns x, y is an equation of the form mx + ny = 0, where (2)
Statement 4.
If m and n are relatively prime numbers, then any solution to Eq. (2) has the form 
5) Homework. Solve the equation in integers:
- 9x - 18y = 5
- x+y=xy
- Several children were picking apples. Each boy collected 21 kg, and the girl 15 kg. In total, they collected 174 kg. How many boys and how many girls were picking apples?
Comment. This lesson does not provide examples of solving equations in integers. Therefore, children solve their homework based on statement 1 and selection.
Lesson 2
1) Organizational moment
2) Checking homework
1) 9x - 18y = 5
5 is not divisible by 9, there are no solutions in integers.
The selection method can find a solution
Answer: (0;0), (2;2)
3) Let's make an equation:
Let boys x, x Z, and girls y, y Z, then we can write the equation 21x + 15y = 174
Many students, having made an equation, will not be able to solve it.
Answer: 4 boys, 6 girls.
3) Learning new material
Faced with difficulties in doing homework, students became convinced of the need to study their methods for solving indefinite equations. Let's consider some of them.
I. Method of consideration of remainders from division.
Example. Solve the equation in integers 3x – 4y = 1.
The left side of the equation is divisible by 3, so the right side must also be divisible. Let's consider three cases.
Answer: where m Z.
The described method is convenient to apply if the numbers m and n are not small, but decompose into simple factors.
Example: Solve equations in integers.
Let y = 4n, then 16 - 7y = 16 - 7 4n = 16 - 28n = 4*(4-7n) is divisible by 4.
y = 4n+1, then 16 - 7y = 16 - 7 (4n + 1) = 16 - 28n - 7 = 9 - 28n is not divisible by 4.
y = 4n+2, then 16 - 7y = 16 - 7 (4n + 2) = 16 - 28n - 14 = 2 - 28n is not divisible by 4.
y = 4n+3, then 16 - 7y = 16 - 7 (4n + 3) = 16 - 28n - 21 = -5 - 28n is not divisible by 4.
Therefore, y = 4n, then
4x = 16 – 7 4n = 16 – 28n, x = 4 – 7n
Answer: , where n Z.
II. Indefinite equations of the 2nd degree
Today in the lesson we will only touch on the solution of second-order Diophantine equations.
And of all types of equations, consider the case when you can apply the difference of squares formula or another way of factoring.
Example: Solve the equation in integers.
13 is a prime number, so it can only be factored in four ways: 13 = 13 1 = 1 13 = (-1)(-13) = (-13)(-1)
Consider these cases
Answer: (7;-3), (7;3), (-7;3), (-7;-3).
4) Homework.
Examples. Solve the equation in integers:
(x - y)(x + y)=4
| 2x=4 | 2x=5 | 2x=5 |
| x=2 | x=5/2 | x=5/2 |
| y=0 | not suitable | not suitable |
| 2x = -4 | not suitable | not suitable |
| x=-2 | ||
| y=0 |
Answer: (-2;0), (2;0).
Answers: (-10;9), (-5;3), (-2;-3), (-1;-9), (1;9), (2;3), (5;-3) , (10;-9).
V) 
Answer: (2;-3), (-1;-1), (-4;0), (2;2), (-1;3), (-4;5).
Results. What does it mean to solve an equation in integers?
What methods of solving indefinite equations do you know?
Application:
Exercises for training.
1) Solve in whole numbers.
| a) 8x + 12y = 32 | x = 1 + 3n, y = 2 - 2n, n Z |
| b) 7x + 5y = 29 | x = 2 + 5n, y = 3 – 7n, n Z |
| c) 4x + 7y = 75 | x = 3 + 7n, y = 9 – 4n, n Z |
| d) 9x – 2y = 1 | x = 1 – 2m, y = 4 + 9m, m Z |
| e) 9x - 11y = 36 | x = 4 + 11n, y = 9n, nZ |
| f) 7x - 4y = 29 | x = 3 + 4n, y = -2 + 7n, n Z |
| g) 19x - 5y = 119 | x = 1 + 5p, y = -20 + 19p, pZ |
| h) 28x - 40y = 60 | x = 45 + 10t, y = 30 + 7t, t Z |
2) Find integer non-negative solutions of the equation.
Equality f(x; y) = 0 represents an equation with two variables. The solution to such an equation is a pair of variable values that turns the two-variable equation into a true equality.
If we have an equation with two variables, then in its record, by virtue of tradition, we must put x in the first place, and y in the second.
Consider the equation x - 3y \u003d 10. Pairs (10; 0), (16; 2), (-2; -4) are solutions to the equation under consideration, while the pair (1; 5) is not a solution.
To find other pairs of solutions to this equation, it is necessary to express one variable in terms of the other - for example, x through y. As a result, we get the equation
x = 10 + 3y. Compute x values by choosing arbitrary y values.
If y \u003d 7, then x \u003d 10 + 3 ∙ 7 \u003d 10 + 21 \u003d 31.
If y \u003d -2, then x \u003d 10 + 3 ∙ (-2) \u003d 10 - 6 \u003d 4.
Thus, pairs (31; 7), (4; -2) are also solutions of the given equation.
If equations with two variables have the same roots, then such equations are called equivalent.
For equations with two variables, theorems on equivalent transformations of equations are valid.
Consider the graph of an equation with two variables.
Let an equation with two variables f(x; y) = 0 be given. All its solutions can be represented by points on the coordinate plane, obtaining a certain set of points of the plane. This set of points in the plane is called the graph of the equation f(x; y) = 0.
So, the graph of the equation y - x 2 \u003d 0 is a parabola y \u003d x 2; the graph of the equation y - x \u003d 0 is a straight line; the graph of the equation y - 3 \u003d 0 is a straight line parallel to the x axis, etc.
An equation of the form ax + by = c, where x and y are variables and a, b and c are numbers, is called linear; the numbers a, b are called the coefficients of the variables, c is the free term.
The graph of the linear equation ax + by = c is:
Let's plot the equation 2x - 3y = -6.
1. Because none of the coefficients for the variables is equal to zero, then the graph of this equation will be a straight line.
2. To build a straight line, we need to know at least two of its points. Substitute the values of x into the equations and get the values of y and vice versa:
if x = 0, then y = 2; (0 ∙ x - 3y \u003d -6);
if y \u003d 0, then x \u003d -3; (2x - 3 ∙ 0 \u003d -6). 
So, we got two points of the graph: (0; 2) and (-3; 0).
3. Let's draw a straight line through the obtained points and get a graph of the equation
2x - 3y \u003d -6.
If the linear equation ax + by = c has the form 0 ∙ x + 0 ∙ y = c, then we have to consider two cases:
1. c \u003d 0. In this case, any pair (x; y) satisfies the equation, and therefore the graph of the equation is the entire coordinate plane;
2. c ≠ 0. In this case, the equation has no solution, which means that its graph does not contain a single point.
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With this mathematical program, you can solve a system of two linear equations with two variables using the substitution method and the addition method.
The program not only gives the answer to the problem, but also provides a detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.
This program can be useful for high school students in preparation for tests and exams, when testing knowledge before the Unified State Examination, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.
Rules for Entering Equations
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.
When entering equations you can use brackets. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of the elements.
For example: 6x+1 = 5(x+y)+2
In equations, you can use not only integers, but also fractional numbers in the form of decimal and ordinary fractions.
Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &
Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)
Solve a system of equations
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A bit of theory.
Solving systems of linear equations. Substitution method
The sequence of actions when solving a system of linear equations by the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression in another equation of the system instead of this variable;
$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$
Let's express from the first equation y through x: y = 7-3x. Substituting the expression 7-3x instead of y into the second equation, we get the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$
It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$
Substituting the number 1 instead of x into the equation y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$
Pair (1;4) - solution of the system
Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.
Solving systems of linear equations by adding
Consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by the substitution method, we pass from a given system to another system equivalent to it, in which one of the equations contains only one variable.
The sequence of actions when solving a system of linear equations by the addition method:
1) multiply the equations of the system term by term, choosing factors so that the coefficients for one of the variables become opposite numbers;
2) add term by term the left and right parts of the equations of the system;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.
Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$
In the equations of this system, the coefficients of y are opposite numbers. Adding term by term the left and right parts of the equations, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$
From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38 \) we get an equation with the variable y: \(11-3y=38 \). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)
Thus, we found the solution to the system of equations by adding: \(x=11; y=-9 \) or \((11; -9) \)
Taking advantage of the fact that the coefficients of y in the equations of the system are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both parts of each of the equations of the original symmeme), in which one of the equations contains only one variable.
Books (textbooks) Abstracts of the Unified State Examination and OGE tests online Games, puzzles Graphing of functions Spelling dictionary of the Russian language Dictionary of youth slang Catalog of Russian schools Catalog of secondary schools in Russia Catalog of Russian universities List of tasksIn the 7th grade mathematics course, they first meet with equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a number of problems fall out of sight, in which certain conditions are introduced on the coefficients of the equation that limit them. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are encountered more and more often in the USE materials and at entrance exams.
Which equation will be called an equation with two variables?
So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are two-variable equations.
Consider the equation 2x - y = 1. It turns into a true equality at x = 2 and y = 3, so this pair of variable values is the solution to the equation under consideration.
Thus, the solution of any equation with two variables is the set of ordered pairs (x; y), the values of the variables that this equation turns into a true numerical equality.
An equation with two unknowns can:
A) have one solution. For example, the equation x 2 + 5y 2 = 0 has a unique solution (0; 0);
b) have multiple solutions. For example, (5 -|x|) 2 + (|y| – 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);
V) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;
G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers whose sum is 3. The set of solutions to this equation can be written as (k; 3 - k), where k is any real number.
The main methods for solving equations with two variables are methods based on factoring expressions, highlighting the full square, using the properties of a quadratic equation, bounded expressions, and evaluation methods. The equation, as a rule, is transformed into a form from which a system for finding unknowns can be obtained.
Factorization
Example 1
Solve the equation: xy - 2 = 2x - y.
Solution.
We group the terms for the purpose of factoring:
(xy + y) - (2x + 2) = 0. Take out the common factor from each bracket:
y(x + 1) – 2(x + 1) = 0;
(x + 1)(y - 2) = 0. We have:
y = 2, x is any real number or x = -1, y is any real number.
Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.
Equality to zero of non-negative numbers
Example 2
Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).
Solution.
Grouping:
(9x 2 - 12x + 4) + (4y 2 - 12y + 9) = 0. Now each parenthesis can be collapsed using the square difference formula.
(3x - 2) 2 + (2y - 3) 2 = 0.
The sum of two non-negative expressions is zero only if 3x - 2 = 0 and 2y - 3 = 0.
So x = 2/3 and y = 3/2.
Answer: (2/3; 3/2).
Evaluation method
Example 3
Solve the equation: (x 2 + 2x + 2) (y 2 - 4y + 6) = 2.
Solution.
In each bracket, select the full square:
((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Estimate 
the meaning of the expressions in brackets.
(x + 1) 2 + 1 ≥ 1 and (y - 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:
(x + 1) 2 + 1 = 1 and (y - 2) 2 + 2 = 2, so x = -1, y = 2.
Answer: (-1; 2).
Let's get acquainted with another method for solving equations with two variables of the second degree. This method is that the equation is considered as square with respect to some variable.
Example 4
Solve the equation: x 2 - 6x + y - 4√y + 13 = 0.
Solution.
Let's solve the equation as a quadratic one with respect to x. Let's find the discriminant:
D = 36 - 4(y - 4√y + 13) = -4y + 16√y - 16 = -4(√y - 2) 2 . The equation will have a solution only when D = 0, i.e., if y = 4. We substitute the value of y into the original equation and find that x = 3.
Answer: (3; 4).
Often in equations with two unknowns indicate restrictions on variables.
Example 5
Solve the equation in integers: x 2 + 5y 2 = 20x + 2.
Solution.
Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of the resulting equation, when divided by 5, gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number that is not divisible by 5 gives a remainder of 1 or 4. Thus equality is impossible and there are no solutions.
Answer: no roots.
Example 6
Solve the equation: (x 2 - 4|x| + 5) (y 2 + 6y + 12) = 3.
Solution.
Let's select the full squares in each bracket:
((|x| – 2) 2 + 1)((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible if |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.
Answer: (2; -3) and (-2; -3).
Example 7
For each pair of negative integers (x; y) satisfying the equation
x 2 - 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Answer the smallest amount.
Solution.
Select full squares:
(x 2 - 2xy + y 2) + (y 2 + 4y + 4) = 37;
(x - y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. The sum of the squares of two integers, equal to 37, we get if we add 1 + 36. Therefore:
(x - y) 2 = 36 and (y + 2) 2 = 1 
(x - y) 2 = 1 and (y + 2) 2 = 36.
Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).
Answer: -17.
Do not despair if you have difficulties when solving equations with two unknowns. With a little practice, you will be able to master any equation.
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A two-variable linear equation is any equation that has the following form: a*x + b*y =c. Here x and y are two variables, a,b,c are some numbers.
Below are a few examples of linear equations.
1. 10*x + 25*y = 150;
Like equations with one unknown, a linear equation with two variables (unknowns) also has a solution. For example, the linear equation x-y=5, with x=8 and y=3, turns into the correct identity 8-3=5. In this case, the pair of numbers x=8 and y=3 is said to be a solution to the linear equation x-y=5. You can also say that the pair of numbers x=8 and y=3 satisfies the linear equation x-y=5.
Solving a linear equation
Thus, the solution of the linear equation a * x + b * y = c, is any pair of numbers (x, y) that satisfies this equation, that is, it turns the equation with the variables x and y into the correct numerical equality. Notice how the pair of numbers x and y is written here. Such a record is shorter and more convenient. It should only be remembered that the first place in such a record is the value of the variable x, and the second is the value of the variable y.
Please note that the numbers x=11 and y=8, x=205 and y=200 x= 4.5 and y= -0.5 also satisfy the linear equation x-y=5, and therefore are solutions to this linear equation.
Solving a linear equation in two unknowns is not the only one. Every linear equation in two unknowns has infinitely many different solutions. That is, there is an infinite number of different two numbers x and y that turn the linear equation into a true identity.
If several equations in two variables have the same solutions, then such equations are called equivalent equations. It should be noted that if equations with two unknowns do not have solutions, then they are also considered equivalent.
Basic properties of linear equations in two unknowns
1. Any of the terms in the equation can be transferred from one part to another, while it is necessary to change its sign to the opposite. The resulting equation will be equivalent to the original.
2. Both sides of the equation can be divided by any number that is not zero. As a result, we obtain an equation equivalent to the original one.
