A linear equation in two variables is any equation that has the following form: a*x + b*y =с. 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 a pair of numbers x=8 and y=3 satisfies the linear equation x-y=5.
Solving a Linear Equation
Thus, the solution to the linear equation a*x + b*y = c is any pair of numbers (x,y) that satisfies this equation, that is, turns the equation with variables x and y into a correct numerical equality. Notice how the pair of numbers x and y are written here. This entry is shorter and more convenient. You just need to remember 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 with two unknowns is not the only one. Every linear equation in two unknowns has infinitely many different solutions. That is, there is infinitely many different two numbers x and y that convert a linear equation into a true identity.
If several equations with two variables have identical 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 with two unknowns
1. Any of the terms in the equation can be transferred from one part to another, but it is necessary to change its sign to the opposite one. The resulting equation will be equivalent to the original one.
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.
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)
Algebraic equations have been most fully studied; 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). There are a large number of specific Diophantine equations that can be solved using 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 a school mathematics course, the quadratic equation ax2 + bx + c = 0 is studied, where a, b, c are a number 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 of equation (1) satisfy the 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 us simplify the right side of the identity.
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 that 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 the theory of differential equations (reducing a second-order linear equation to canonical form).
When solving these various problems, one essentially has 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 under consideration 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.
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 series of problems with limited conditions on the coefficient of the equation, as well as methods for solving them, fall out of the sight of the teacher and, therefore, the student.
We are talking about solving an equation with two unknowns in integers or natural numbers.
At school, natural numbers and integers are studied in grades 4-6. By the time they graduate from school, not all students remember the differences between the sets of these numbers.
However, a problem like “solve an equation of the form ax + by=c in integers” is increasingly found on entrance exams to universities and in Unified State Examination materials.
Solving uncertain equations develops logical thinking, intelligence, and attention to analysis.
I propose developing several lessons on this topic. I do not have clear recommendations on the timing of these lessons. Some elements can also be used in 7th grade (for a strong class). These lessons can be taken as a basis and developed a small elective course on pre-vocational training 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 “First and second order equations”
- nurturing cognitive interest in the subject
- developing the ability to analyze, make generalizations, transfer knowledge to a new situation
Lesson 1.
During the classes.
1) Org. moment.
2) Updating basic knowledge.
Definition. A linear equation in 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 variables that turns the equation into a true equality.
Equations with two variables that have 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 to equation (1).
This equation has infinitely many solutions.
3) Historical background
Indefinite (Diophantine) equations are equations containing more than one variable.
In the 3rd century. AD – Diophantus of Alexandria wrote “Arithmetic”, in which he expanded the set of numbers to rational ones and introduced algebraic symbolism.
Diophantus also considered the problems of solving indefinite equations and he gave methods for solving indefinite equations of the second and third degree.
4) Studying new material.
Definition: A first-order inhomogeneous 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 evenly divisible by 17, there is no solution in integers.
Let k be divided by gcd (m, n). By dividing all coefficients, we can ensure that m and n become relatively prime.
Statement 2.
If m and n of equation (1) are relatively prime numbers, then this equation has at least one solution.
Statement 3.
If the coefficients m and n of equation (1) are coprime numbers, then this equation has infinitely many solutions:
Where (; ) is any solution to equation (1), t Z
Definition. A first-order homogeneous 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 coprime numbers, then any solution to equation (2) has the form 
5) Homework. Solve the equation in whole numbers:
- 9x – 18y = 5
- x + y= xy
- Several children were picking apples. Each boy collected 21 kg, and the girl collected 15 kg. In total they collected 174 kg. How many boys and how many girls picked apples?
Comment. This lesson does not provide examples of solving equations in integers. Therefore, children solve 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 whole numbers.
Using the selection method you can find a solution
Answer: (0;0), (2;2)
3) Let's make an equation:
Let the boys be x, x Z, and the girls y, y Z, then we can create the equation 21x + 15y = 174
Many students, having written an equation, will not be able to solve it.
Answer: 4 boys, 6 girls.
3) Learning new material
Having encountered difficulties with homework, students were convinced of the need to learn their methods for solving uncertain equations. Let's look at some of them.
I. Method for considering division remainders.
Example. Solve the equation in whole numbers 3x – 4y = 1.
The left side of the equation is divisible by 3, therefore the right side must be divisible. Let's consider three cases.
Answer: where m Z.
The described method is convenient to use if the numbers m and n are not small, but can be decomposed into simple factors.
Example: Solve equations in whole numbers.
Let y = 4n, then 16 - 7y = 16 – 7 4n = 16 – 28n = 4*(4-7n) is divided 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. Uncertain equations of 2nd degree
Today in the lesson we will only touch on solving second-order Diophantine equations.
And of all types of equations, we will consider the case when we can apply the difference of squares formula or another method of factorization.
Example: Solve an equation in whole numbers.
13 is a prime number, so it can only be factored in four ways: 13 = 13 1 = 1 13 = (-1)(-13) = (-13)(-1)
Let's consider these cases
Answer: (7;-3), (7;3), (-7;3), (-7;-3).
4) Homework.
Examples. Solve the equation in whole numbers:
(x - y)(x + y)=4
| 2x = 4 | 2x = 5 | 2x = 5 |
| x = 2 | x = 5/2 | x = 5/2 |
| y = 0 | doesn't fit | doesn't fit |
| 2x = -4 | doesn't fit | doesn't fit |
| 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 whole numbers?
What methods for solving uncertain 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, n Z |
| e) 7x – 4y = 29 | x = 3 + 4n, y = -2 + 7n, n Z |
| g) 19x – 5y = 119 | x = 1 + 5p, y = -20 + 19p, p Z |
| h) 28x – 40y = 60 | x = 45 + 10t, y = 30 + 7t, t Z |
2) Find integer non-negative solutions to the equation.
In the 7th grade mathematics course, we encounter for the first time equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a whole series of problems in which certain conditions are introduced on the coefficients of the equation that limit them fall out of sight. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are found more and more often in the Unified State Examination materials and in 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 equations in two variables.
Consider the equation 2x – y = 1. It becomes true when x = 2 and y = 3, so this pair of variable values is a solution to the equation in question.
Thus, the solution to any equation with two variables is a set of ordered pairs (x; y), values of the variables that turn this equation 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 equal to 3. The set of solutions to this equation can be written in the form (k; 3 – k), where k is any real number.
The main methods for solving equations with two variables are methods based on factoring expressions, isolating a complete square, using the properties of a quadratic equation, limited expressions, and estimation methods. The equation is usually transformed into a form from which a system for finding the unknowns can be obtained.
Factorization
Example 1.
Solve the equation: xy – 2 = 2x – y.
Solution.
We group the terms for the purpose of factorization:
(xy + y) – (2x + 2) = 0. From each bracket we take out a common factor:
y(x + 1) – 2(x + 1) = 0;
(x + 1)(y – 2) = 0. We have:
y = 2, x – any real number or x = -1, y – any real number.
Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.
Equality of non-negative numbers to zero
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 bracket can be folded using the squared 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.
This means x = 2/3 and y = 3/2.
Answer: (2/3; 3/2).
Estimation method
Example 3.
Solve the equation: (x 2 + 2x + 2)(y 2 – 4y + 6) = 2.
Solution.
In each bracket we select a complete square:
((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Let’s estimate 
the meaning of the expressions in parentheses.
(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, which means 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 consists of treating the equation 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 equation for 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, that is, 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 they indicate restrictions on variables.
Example 5.
Solve the equation in whole numbers: 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 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 highlight the complete 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 provided |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.
Answer: (2; -3) and (-2; -3).
Example 7.
For every pair of negative integers (x;y) satisfying the equation
x 2 – 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Please indicate the smallest amount in your answer.
Solution.
Let's select complete 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. We get the sum of the squares of two integers equal to 37 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.
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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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