Solving equations in integers is one of the oldest mathematical problems. Already at the beginning of the 2nd millennium BC. e. The Babylonians knew how to solve systems of such equations with two variables. This area of mathematics reached its greatest flourishing in Ancient Greece. Our main source is Diophantus' Arithmetic, which contains various types of equations. In it, Diophantus (after his name the name of the equations is Diophantine equations) anticipates a number of methods for studying equations of the 2nd and 3rd degrees, which developed only in the 19th century.
The simplest Diophantine equations are ax + y = 1 (equation with two variables, first degree) x2 + y2 = z2 (equation with three variables, second degree)
Most fully studied algebraic equations, their solution was one of the most important problems in algebra in the 16th and 17th centuries.
By the beginning of the 19th century, the works of P. Fermat, L. Euler, K. Gauss investigated a Diophantine equation of the form: ax2 + bxy + cy2 + dx + ey + f = 0, where a, b, c, d, e, f are numbers; x, y unknown variables.
This is a 2nd degree equation with two unknowns.
K. Gauss developed a general theory of quadratic forms, which is the basis for solving certain types of equations with two variables (Diophantine equations). Exists large number specific Diophantine equations solved by elementary methods. /p>
Theoretical material.
In this part of the work, the basic mathematical concepts will be described, terms will be defined, and the expansion theorem will be formulated using the method of indefinite coefficients, which were studied and considered when solving equations with two variables.
Definition 1: Equation of the form ax2 + bxy + cy2 + dx + ey + f = 0, where a, b, c, d, e, f are numbers; x, y unknown variables is called a second degree equation with two variables.
In the school mathematics course we study quadratic equation ax2+inx+c=0, where a, b, c numbers x variable, with one variable. There are many ways to solve this equation:
1. Finding roots using a discriminant;
2. Finding the roots for the even coefficient in (according to D1=);
3. Finding roots using Vieta’s theorem;
4. Finding roots by isolating the perfect square of a binomial.
Solving an equation means finding all its roots or proving that they do not exist.
Definition 2: The root of an equation is a number that, when substituted into an equation, forms a true equality.
Definition 3: The solution to an equation with two variables is called a pair of numbers (x, y) when substituted into the equation, it turns into a true equality.
The process of finding solutions to an equation very often usually consists of replacing the equation with an equivalent equation, but one that is simpler to solve. Such equations are called equivalent.
Definition 4: Two equations are said to be equivalent if each solution of one equation is a solution of the other equation, and vice versa, and both equations are considered in the same domain.
To solve equations with two variables, use the theorem on the decomposition of the equation into a sum of complete squares (by the method of indefinite coefficients).
For the second order equation ax2 + bxy + cy2 + dx + ey + f = 0 (1), the expansion a(x + py + q)2 + r(y + s)2 + h (2) takes place
Let us formulate the conditions under which expansion (2) takes place for equation (1) of two variables.
Theorem: If the coefficients a,b,c equations(1) satisfy conditions a0 and 4ab – c20, then expansion (2) is determined in a unique way.
In other words, equation (1) with two variables can be reduced to form (2) using the method of indefinite coefficients if the conditions of the theorem are met.
Let's look at an example of how the method of indefinite coefficients is implemented.
METHOD No. 1. Solve the equation using the method of undetermined coefficients
2 x2 + y2 + 2xy + 2x +1= 0.
1. Let’s check the fulfillment of the conditions of the theorem, a=2, b=1, c=2, which means a=2.4av – c2= 4∙2∙1- 22= 40.
2. The conditions of the theorem are met; they can be expanded according to formula (2).
3. 2 x2 + y2 + 2xy + 2x +1= 2(x + py + q)2 + r(y + s)2 + h, based on the conditions of the theorem, both parts of the identity are equivalent. Let's simplify right side identities.
4. 2(x + py + q)2 + r(y +s)2 +h =
2(x2+ p2y2 + q2 + 2pxy + 2pqy + 2qx) + r(y2 + 2sy + s2) + h =
2x2+ 2p2y2 + 2q2 + 4pxy + 4pqy + 4qx + ry2 + 2rsy + rs2 + h =
X2(2) + y2(2p2 + r) + xy(4p) + x(4q) + y(4pq + 2rs) + (2q2 + rs2 + h).
5. We equate the coefficients for identical variables with their degrees.
x2 2 = 2 y21 = 2p2 + r) xy2 = 4p x2 = 4q y0 = 4pq + 2rs x01 = 2q2 + rs2 + h
6. Let's get a system of equations, solve it and find the values of the coefficients.
7. Substitute the coefficients into (2), then the equation will take the form
2 x2 + y2 + 2xy + 2x +1= 2(x + 0.5y + 0.5)2 + 0.5(y -1)2 +0
Thus, the original equation is equivalent to the equation
2(x + 0.5y + 0.5)2 + 0.5(y -1)2 = 0 (3), this equation is equivalent to a system of two linear equations.
Answer: (-1; 1).
If you pay attention to the type of expansion (3), you will notice that it is identical in form to isolating a complete square from a quadratic equation with one variable: ax2 + inx + c = a(x +)2 +.
Let's apply this technique when solving an equation with two variables. Let us solve, using the selection of a complete square, a quadratic equation with two variables that has already been solved using the theorem.
METHOD No. 2: Solve the equation 2 x2 + y2 + 2xy + 2x +1= 0.
Solution: 1. Let's imagine 2x2 as the sum of two terms x2 + x2 + y2 + 2xy + 2x +1= 0.
2. Let's group the terms in such a way that we can fold them using the formula of a complete square.
(x2 + y2 + 2xy) + (x2 + 2x +1) = 0.
3. Select complete squares from the expressions in brackets.
(x + y)2 + (x + 1)2 = 0.
4. This equation is equivalent to a system of linear equations.
Answer: (-1;1).
If you compare the results, you can see that the equation solved by method No. 1 using the theorem and the method of indefinite coefficients and the equation solved by method No. 2 using the extraction of a complete square have the same roots.
Conclusion: A quadratic equation with two variables can be expanded into a sum of squares in two ways:
➢ The first method is the method of indefinite coefficients, which is based on the theorem and expansion (2).
➢ The second way is using identity transformations, which allow you to select sequentially complete squares.
Of course, when solving problems, the second method is preferable, since it does not require memorizing expansion (2) and conditions.
This method can also be used for quadratic equations with three variables. Isolating a perfect square in such equations is more labor-intensive. I will be doing this type of transformation next year.
It is interesting to note that a function that has the form: f(x,y) = ax2 + vxy + cy2 + dx + ey + f is called a quadratic function of two variables. Quadratic functions play an important role in various branches of mathematics:
In mathematical programming (quadratic programming)
In linear algebra and geometry (quadratic forms)
In theory differential equations(bringing a second-order linear equation to canonical form).
When solving these various problems, you essentially have to apply the procedure of isolating a complete square from a quadratic equation (one, two or more variables).
Lines whose equations are described by a quadratic equation of two variables are called second-order curves.
This is a circle, ellipse, hyperbola.
When constructing graphs of these curves, the method of sequentially isolating a complete square is also used.
Let's look at how the method of sequentially selecting a complete square works using specific examples.
Practical part.
Solve equations using the method of sequentially isolating a complete square.
1. 2x2 + y2 + 2xy + 2x + 1 = 0; x2 + x2 + y2 + 2xy + 2x + 1 = 0;
(x +1)2 + (x + y)2 = 0;
Answer:(-1;1).
2. x2 + 5y2 + 2xy + 4y + 1 = 0; x2 + 4y2 + y2 + 2xy + 4y + 1 = 0;
(x + y)2 + (2y + 1)2 = 0;
Answer:(0.5; - 0.5).
3. 3x2 + 4y2 - 6xy - 2y + 1 = 0;
3x2 + 3y2 + y2 – 6xy – 2y +1 = 0;
3x2 +3y2 – 6xy + y2 –2y +1 = 0;
3(x2 - 2xy + y2) + y2 - 2y + 1 = 0;
3(x2 - 2xy + y2)+(y2 - 2y + 1)=0;
3(x-y)2 + (y-1)2 = 0;
Answer:(-1;1).
Solve equations:
1. 2x2 + 3y2 – 4xy + 6y +9 =0
(reduce to the form: 2(x-y)2 + (y +3)2 = 0)
Answer: (-3; -3)
2. – 3x2 – 2y2 – 6xy –2y + 1=0
(reduce to the form: -3(x+y)2 + (y –1)2= 0)
Answer: (-1; 1)
3. x2 + 3y2+2xy + 28y +98 =0
(reduce to the form: (x+y)2 +2(y+7)2 =0)
Answer: (7; -7)
Conclusion.
In this scientific work equations with two variables of the second degree were studied, and methods for solving them were considered. The task has been completed, a shorter method of solution has been formulated and described, based on isolating a complete square and replacing the equation with an equivalent system of equations, as a result the procedure for finding the roots of an equation with two variables has been simplified.
An important point of the work is that the technique in question is used when solving various mathematical problems related to a quadratic function, constructing second-order curves, and finding the largest (smallest) value of expressions.
Thus, the technique of decomposing a second-order equation with two variables into a sum of squares has the most numerous applications in mathematics.
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 equation with two variables into a true equality.
If we have an equation with two variables, then, by tradition, we must put x in first place and y in second place.
Consider the equation x – 3y = 10. Pairs (10; 0), (16; 2), (-2; -4) are solutions to the equation under consideration, while 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 another - for example, x in terms of y. As a result, we get the equation
x = 10 + 3y. Let's calculate the values of x by choosing arbitrary values of y.
If y = 7, then x = 10 + 3 ∙ 7 = 10 + 21 = 31.
If y = -2, then x = 10 + 3 ∙ (-2) = 10 – 6 = 4.
Thus, pairs (31; 7), (4; -2) are also solutions to 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 on the plane. This set of points on the plane is called the graph of the equation f(x; y) = 0.
Thus, the graph of the equation y – x 2 = 0 is the parabola y = x 2; the graph of the equation y – x = 0 is a straight line; the graph of the equation y – 3 = 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 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 of the variables is equal to zero, then the graph of this equation will be a straight line.
2. To construct a straight line, we need to know at least two of its points. Substitute the x values into the equations and get the y values and vice versa:
if x = 0, then y = 2; (0 ∙ x – 3y = -6);
if y = 0, then x = -3; (2x – 3 ∙ 0 = -6). 
So, we got two points on 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 = -6.
If the linear equation ax + by = c has the form 0 ∙ x + 0 ∙ y = c, then we must consider two cases:
1. c = 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 its graph does not contain a single point.
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Subject:Linear function
Lesson:Linear equation with two variables and its graph
We got acquainted with the concepts coordinate axis and coordinate plane. We know that each point on the plane uniquely defines a pair of numbers (x; y), with the first number being the abscissa of the point, and the second being the ordinate.
We will very often encounter a linear equation in two variables, the solution of which is a pair of numbers that can be represented on the coordinate plane.
Equation of the form:
Where a, b, c are numbers, and 
It is called a linear equation with two variables x and y. The solution to such an equation will be any such pair of numbers x and y, substituting which into the equation we will obtain the correct numerical equality.
A pair of numbers will be depicted on the coordinate plane as a point.
For such equations we will see many solutions, that is, many pairs of numbers, and all the corresponding points will lie on the same straight line.
Let's look at an example:
To find solutions to this equation, you need to select the corresponding pairs of numbers x and y:
Let , then the original equation turns into an equation with one unknown:
,
That is, the first pair of numbers that is a solution to a given equation (0; 3). We got point A(0; 3)
Let . We get the original equation with one variable: 
, from here, we got point B(3; 0)
Let's put the pairs of numbers in the table:
Let's plot points on the graph and draw a straight line: 
Note that any point on a given line will be a solution to the given equation. Let's check - take a point with a coordinate and use the graph to find its second coordinate. It is obvious that at this point. Let's substitute this pair numbers into the equation. We get 0=0 - a correct numerical equality, which means a point lying on a line is a solution.
For now, we cannot prove that any point lying on the constructed line is a solution to the equation, so we accept this as true and will prove it later.
Example 2 - graph the equation:
Let's make a table; we only need two points to construct a straight line, but we'll take a third one for control:
In the first column we took a convenient one, we will find it from:
, ,
In the second column we took a convenient one, let's find x:
, , ,
Let's check and find:
, ,
Let's build a graph:
Let's multiply the given equation by two:
From such a transformation, the set of solutions will not change and the graph will remain the same.
Conclusion: we learned to solve equations with two variables and build their graphs, we learned that the graph of such an equation is a straight line and that any point on this line is a solution to the equation
1. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra 7. 6th edition. M.: Enlightenment. 2010
2. Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7. M.: VENTANA-GRAF
3. Kolyagin Yu.M., Tkacheva M.V., Fedorova N.E. and others. Algebra 7.M.: Enlightenment. 2006
2. Portal for family viewing ().
Task 1: Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7, No. 960, Art. 210;
Task 2: Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7, No. 961, Art. 210;
Task 3: Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7, No. 962, Art. 210;
Nonlinear equations with two unknowns
Definition 1. Let A be some set of pairs of numbers (x; y) . They say that the set A is given numeric function z from two variables x and y , if a rule is specified with the help of which each pair of numbers from set A is associated with a certain number.
Exercise numerical function z from two variables x and y often denote So:
Where f (x , y) – any function other than a function
f (x , y) = ax+by+c ,
where a, b, c are given numbers.
Definition 3. Solving equation (2) call a pair of numbers ( x; y) , for which formula (2) is a true equality.
Example 1. Solve the equation
Since the square of any number is non-negative, it follows from formula (4) that the unknowns x and y satisfy the system of equations
the solution to which is a pair of numbers (6; 3).
Answer: (6; 3)
Example 2. Solve the equation
Therefore, the solution to equation (6) is infinite number of pairs of numbers type
(1 + y ; y) ,
where y is any number.
linear
Definition 4. Solving a system of equations
call a pair of numbers ( x; y) , when substituting them into each of the equations of this system, the correct equality is obtained.
Systems of two equations, one of which is linear, have the form
g(x , y)
Example 4. Solve system of equations
Solution . Let us express the unknown y from the first equation of system (7) through the unknown x and substitute the resulting expression into the second equation of the system:
Solving the equation
x 1 = - 1 , x 2 = 9 .
Hence,
y 1 = 8 - x 1 = 9 ,
y 2 = 8 - x 2 = - 1 .
Systems of two equations, one of which is homogeneous
Systems of two equations, one of which is homogeneous, have the form
where a, b, c are given numbers, and g(x , y) – function of two variables x and y.
Example 6. Solve system of equations
Solution . Let's solve the homogeneous equation
3x 2 + 2xy - y 2 = 0 ,
3x 2 + 17xy + 10y 2 = 0 ,
treating it as a quadratic equation with respect to the unknown x:
.
In case x = - 5y, from the second equation of system (11) we obtain the equation
5y 2 = - 20 ,
which has no roots.
In case
from the second equation of system (11) we obtain the equation
,
whose roots are numbers y 1 = 3 , y 2 = - 3 . Finding for each of these values y the corresponding value x, we obtain two solutions to the system: (- 2 ; 3) , (2 ; - 3) .
Answer: (- 2 ; 3) , (2 ; - 3)
Examples of solving systems of equations of other types
Example 8. Solve a system of equations (MIPT)
Solution . Let us introduce new unknowns u and v, which are expressed through x and y according to the formulas:
In order to rewrite system (12) in terms of new unknowns, we first express the unknowns x and y in terms of u and v. From system (13) it follows that
Let us solve the linear system (14) by eliminating the variable x from the second equation of this system. For this purpose, we perform the following transformations on system (14):
- We will leave the first equation of the system unchanged;
- from the second equation we subtract the first equation and replace the second equation of the system with the resulting difference.
As a result, system (14) is transformed into an equivalent system
from which we find
Using formulas (13) and (15), we rewrite the original system (12) in the form
The first equation of system (16) is linear, so we can express from it the unknown u through the unknown v and substitute this expression into the second equation of the system.
§ 1 Selection of equation roots in real situations
Let's consider this real situation:
The master and apprentice together made 400 custom parts. Moreover, the master worked for 3 days, and the student for 2 days. How many parts did each person make?
Let's create an algebraic model of this situation. Let the master produce parts in 1 day. And the student is at the details. Then the master will make 3 parts in 3 days, and the student will make 2 parts in 2 days. Together they will produce 3 + 2 parts. Since, according to the condition, a total of 400 parts were manufactured, we obtain the equation:
The resulting equation is called a linear equation in two variables. Here we need to find a pair of numbers x and y for which the equation will take the form of a true numerical equality. Note that if x = 90, y = 65, then we get the equality:
3 ∙ 90 + 65 ∙ 2 = 400
Since the correct numerical equality has been obtained, the pair of numbers 90 and 65 will be a solution to this equation. But the solution found is not the only one. If x = 96 and y = 56, then we get the equality:
96 ∙ 3 + 56 ∙ 2 = 400
This is also a true numerical equality, which means that the pair of numbers 96 and 56 is also a solution to this equation. But a pair of numbers x = 73 and y = 23 will not be a solution to this equation. In fact, 3 ∙ 73 + 2 ∙ 23 = 400 will give us the incorrect numerical equality 265 = 400. It should be noted that if we consider the equation in relation to this real situation, then there will be pairs of numbers that, being a solution to this equation, will not be a solution to the problem. For example, a couple of numbers:
x = 200 and y = -100
is a solution to the equation, but the student cannot make -100 parts, and therefore such a pair of numbers cannot be the answer to the question of the problem. Thus, in each specific real situation it is necessary to take a reasonable approach to selecting the roots of the equation.
Let's summarize the first results:
An equation of the form ax + bу + c = 0, where a, b, c are any numbers, is called a linear equation with two variables.
The solution to a linear equation in two variables is a pair of numbers corresponding to x and y, for which the equation turns into a true numerical equality.
§ 2 Graph of a linear equation
The very recording of the pair (x;y) leads us to think about the possibility of depicting it as a point with coordinates xy y on a plane. This means that we can obtain a geometric model of a specific situation. For example, consider the equation:
2x + y - 4 = 0
Let's select several pairs of numbers that will be solutions to this equation and construct points with the found coordinates. Let these be points:
A(0; 4), B(2; 0), C(1; 2), D(-2; 8), E(- 1; 6).
Note that all points lie on the same line. This line is called the graph of a linear equation in two variables. It is a graphical (or geometric) model of a given equation.
If a pair of numbers (x;y) is a solution to the equation
ax + vy + c = 0, then the point M(x;y) belongs to the graph of the equation. We can say the other way around: if the point M(x;y) belongs to the graph of the equation ax + y + c = 0, then the pair of numbers (x;y) is a solution to this equation.
From the geometry course we know:
To plot a straight line, you need 2 points, so to plot a graph of a linear equation with two variables, it is enough to know only 2 pairs of solutions. But guessing the roots is not always a convenient or rational procedure. You can act according to another rule. Since the abscissa of a point (variable x) is an independent variable, you can give it any convenient value. Substituting this number into the equation, we find the value of the variable y.
For example, let the equation be given:
Let x = 0, then we get 0 - y + 1 = 0 or y = 1. This means that if x = 0, then y = 1. A pair of numbers (0;1) is the solution to this equation. Let's set another value for the variable x: x = 2. Then we get 2 - y + 1 = 0 or y = 3. The pair of numbers (2;3) is also a solution to this equation. Using the two points found, it is already possible to construct a graph of the equation x - y + 1 = 0.
You can do this: first give some specific meaning variable y, and only then calculate the value of x.
§ 3 System of equations
Find two natural numbers, the sum of which is 11 and the difference is 1.
To solve this problem, we first create a mathematical model (namely, an algebraic one). Let the first number be x and the second number y. Then the sum of the numbers x + y = 11 and the difference of the numbers x - y = 1. Since both equations deal with the same numbers, these conditions must be met simultaneously. Usually in such cases a special record is used. The equations are written one below the other and combined with a curly brace.
Such a record is called a system of equations.
Now let’s construct sets of solutions to each equation, i.e. graphs of each of the equations. Let's take the first equation:
If x = 4, then y = 7. If x = 9, then y = 2.
Let's draw a straight line through points (4;7) and (9;2).
Let's take the second equation x - y = 1. If x = 5, then y = 4. If x = 7, then y = 6. We also draw a straight line through the points (5;4) and (7;6). We obtained a geometric model of the problem. The pair of numbers we are interested in (x;y) must be a solution to both equations. In the figure we see the only point, which lies on both lines, is the point of intersection of the lines.
Its coordinates are (6;5). Therefore, the solution to the problem will be: the first required number is 6, the second is 5.
List of used literature:
- Mordkovich A.G., Algebra 7th grade in 2 parts, Part 1, Textbook for general education institutions / A.G. Mordkovich. – 10th ed., revised – Moscow, “Mnemosyne”, 2007
- Mordkovich A.G., Algebra 7th grade in 2 parts, Part 2, Problem book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, “Mnemosyne”, 2007
- HER. Tulchinskaya, Algebra 7th grade. Blitz survey: a manual for students of general education institutions, 4th edition, revised and expanded, Moscow, “Mnemosyne”, 2008
- Alexandrova L.A., Algebra 7th grade. Thematic testing work V new form for students of general education institutions, edited by A.G. Mordkovich, Moscow, “Mnemosyne”, 2011
- Alexandrova L.A. Algebra 7th grade. Independent work for students of general education institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, “Mnemosyne”, 2010