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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Subject:Linear function
Lesson:Linear equation in two variables and its graph
We became familiar with the concepts of a coordinate axis and a 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 of 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;
§ 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 construct 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 assign some specific value to the variable y, and only then calculate the value of x.
§ 3 System of equations
Find two natural numbers whose sum is 11 and 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 a single point that lies on both lines; this 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 test papers in a new form for students of general education institutions, edited by A.G. Mordkovich, Moscow, “Mnemosyne”, 2011
- Alexandrova L.A. Algebra 7th grade. Independent works for students of general education institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, “Mnemosyne”, 2010
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.
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.
Don't despair if you have difficulty solving equations with two unknowns. With a little practice, you can handle any equation.
Still have questions? Don't know how to solve equations in two variables?
To get help from a tutor -.
The first lesson is free!
blog.site, when copying material in full or in part, a link to the original source is required.