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:
Solution:Z(2;-1)
Literature.
- Children's encyclopedia "Pedagogy", Moscow, 1972
- Algebra-8, N.Ya. Vilenkin, VO Nauka, Novosibirsk, 1992
- Competition problems based on number theory. V.Ya. Galkin, D.Yu. Sychugov. Moscow State University, VMK, Moscow, 2005
- Tasks of increased difficulty in the course of algebra grades 7-9. N.P. Kosrykin. “Enlightenment”, Moscow, 1991
- Algebra 7, Makarychev Yu.N., “Enlightenment”.
In 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.
Do you have any questions? Don't know how to solve equations with two variables?
To get the help of a tutor - register.
The first lesson is free!
site, with full or partial copying of the material, a link to the source is required.
Instruction
Substitution method Express one variable and substitute it into another equation. You can express any variable you like. For example, express "y" from the second equation:
x-y=2 => y=x-2 Then plug everything into the first equation:
2x+(x-2)=10 Move everything without x to the right side and count:
2x+x=10+2
3x=12 Next, for "x, divide both sides of the equation by 3:
x=4. So, you have found "x. Find "at. To do this, substitute "x" into the equation from which you expressed "y:
y=x-2=4-2=2
y=2.
Make a check. To do this, substitute the resulting values into the equations:
2*4+2=10
4-2=2
Unknown found correctly!
How to add or subtract equations Get rid of any variable at once. In our case, this is easier to do with "y.
Since in the equation “y has a sign” + , and in the second “-”, then you can perform an addition operation, i.e. We add the left side to the left, and the right side to the right:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4 Substitute "x" into any equation and find "y:
2*4+y=10
8+y=10
y=10-8
y=2 By 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 just "x. In order for x to be reduced when adding or subtracting, multiply the second equation by 2:
x-y=2
2x-2y=4 Then subtract the second equation from the first equation:
2x+y-(2x-2y)=10-4
2x+y-2x+2y=6
3y=6
find y \u003d 2 "x by expressing from any equation, i.e.
x=4
Related videos
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 setting a differential equation, which often helps to simplify the search for its integral.
Instruction
Consider a physics problem that leads to a differential equation that lacks an argument t. This is the problem of vibrations of mass m, suspended on a thread of length r, located in a vertical plane. The equation of motion of the pendulum is required if the initial one was stationary and deviated from the equilibrium state by an angle α. 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: gravity G \u003d mg and thread tension N. Both of these forces lie in a vertical plane. Therefore, to solve the problem, you can apply the equation of the rotational motion of a point around the horizontal axis passing through the point O. The equation for the 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, counted from the vertical axis counterclockwise; M is the moment of forces applied to the material point.
Calculate these quantities. I=mr^2, M=M(G)+M(N). But M(N)=0, since the line of action of the force passes through the 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 relation arises (see Fig. 1d). There is no t argument here.
We have already learned how to solve quadratic equations. Let us now extend the studied methods to rational equations.
What is a rational expression? We have already encountered this concept. Rational expressions called expressions made up of numbers, variables, their degrees and signs of mathematical operations.
Accordingly, rational equations are equations of the form: , where 
- rational expressions.
Previously, we considered only those rational equations that reduce to linear ones. Now let's consider those rational equations that can be reduced to quadratic ones.
Example 1
Solve the equation: .
Solution:
A fraction is 0 if and only if its numerator is 0 and its denominator is not 0.
We get the following system:
The first equation of the system is a quadratic equation. Before solving it, we divide all its coefficients by 3. We get:
We get two roots: ; .
Since 2 is never equal to 0, two conditions must be met: 
. Since none of the roots of the equation obtained above matches the invalid values of the variable that were obtained when solving the second inequality, they are both solutions to this equation.
Answer:.
So, let's formulate an algorithm for solving rational equations:
1. Move all terms to the left side so that 0 is obtained on the right side.
2. Transform and simplify the left side, bring all fractions to a common denominator.
3. Equate the resulting fraction to 0, according to the following algorithm: 
.
4. Write down those roots that are obtained in the first equation and satisfy the second inequality in response.
Let's look at another example.
Example 2
Solve the equation: 
.
Solution
At the very beginning, we transfer all the terms to the left side so that 0 remains on the right. We get:
Now we bring the left side of the equation to a common denominator:
This equation is equivalent to the system:
The first equation of the system is a quadratic equation.
The coefficients of this equation: . We calculate the discriminant:
We get two roots: ; .
Now we solve the second inequality: the product of factors is not equal to 0 if and only if none of the factors is equal to 0.
Two conditions must be met: 
. We get that of the two roots of the first equation, only one is suitable - 3.
Answer:.
In this lesson, we remembered what a rational expression is, and also learned how to solve rational equations, which are reduced to quadratic equations.
In the next lesson, we will consider rational equations as models of real situations, and also consider motion problems.
Bibliography
- Bashmakov M.I. Algebra, 8th grade. - M.: Enlightenment, 2004.
- Dorofeev G.V., Suvorova S.B., Bunimovich E.A. et al. Algebra, 8. 5th ed. - M.: Education, 2010.
- Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Textbook for educational institutions. - M.: Education, 2006.
- Festival of pedagogical ideas "Open Lesson" ().
- School.xvatit.com().
- Rudocs.exdat.com().
Homework
Loading...
