Online calculator. Solving a system of two linear equations in two variables

Instructions

Substitution MethodExpress one variable and substitute it into another equation. You can express any variable at your discretion. For example, express y from the second equation:
x-y=2 => y=x-2Then substitute everything into the first equation:
2x+(x-2)=10 Move everything without “x” to the right side and calculate:
2x+x=10+2
3x=12 Next, to get x, divide both sides of the equation by 3:
x=4. So, you found “x. Find "y. To do this, substitute “x” into the equation from which you expressed “y”:
y=x-2=4-2=2
y=2.

Do a check. To do this, substitute the resulting values ​​into the equations:
2*4+2=10
4-2=2
The unknowns have been found correctly!

A way to add or subtract equations Get rid of any variable right away. In our case, this is easier to do with “y.
Since in “y” there is a “+” sign, and in the second one “-”, then you can perform the addition operation, i.e. fold the left side with the left, and the right with the right:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4Substitute “x” into any equation and find “y”:
2*4+y=10
8+y=10
y=10-8
y=2By the 1st method you can see that they were 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 we simply have “x”. In order for x to be reduced during addition, multiply the second equation by 2:
x-y=2
2x-2y=4Then subtract the second from the first equation:
2x+y-(2x-2y)=10-4 Note that if there is a minus before the bracket, then after opening, change it to the opposite:
2x+y-2x+2y=6
3у=6
find y=2x by expressing from any equation, i.e.
x=4

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Tip 2: How to solve a linear equation in two variables

The equation, written in general form ax+bу+c=0, is called a linear equation with two variables. Such an equation itself contains an infinite number of solutions, so in problems it is always supplemented with something - another equation or limiting conditions. Depending on the conditions provided by the problem, solve a linear equation with two variables follows in different ways.

You will need

• - linear equation with two variables;
• - second equation or additional conditions.

Instructions

Given a system of two linear equations, solve it as follows. Choose one of the equations in which the coefficients are variables smaller and express one of the variables, for example, x. Then substitute this value containing y into the second equation. In the resulting equation there will be only one variable y, move all parts with y to the left side, and free ones to the right. Find y and substitute into any of the original equations to find x.

There is another way to solve a system of two equations. Multiply one of the equations by a number so that the coefficient of one of the variables, such as x, is the same in both equations. Then subtract one of the equations from the other (if the right-hand side is not equal to 0, remember to subtract the right-hand sides in the same way). You will see that the x variable has disappeared and only one y variable remains. Solve the resulting equation, and substitute the found value of y into any of the original equalities. Find x.

The third way to solve a system of two linear equations is graphical. Draw a coordinate system and graph two straight lines whose equations are given in your system. To do this, substitute any two x values ​​into the equation and find the corresponding y - these will be the coordinates of the points belonging to the line. The most convenient way to find the intersection with the coordinate axes is to simply substitute the values ​​x=0 and y=0. The coordinates of the point of intersection of these two lines will be the tasks.

If there is only one linear equation in the problem conditions, then you have been given additional conditions through which you can find a solution. Read the problem carefully to find these conditions. If variables x and y indicate distance, speed, weight - feel free to set the limit x≥0 and y≥0. It is quite possible that x or y hides the number of apples, etc. – then the values ​​can only be . If x is the son’s age, it is clear that he cannot be older than his father, so indicate this in the conditions of the problem.

Sources:

• how to solve an equation with one variable

By itself the equation with three unknown has many solutions, so most often it is supplemented by two more equations or conditions. Depending on what the initial data are, the course of the decision will largely depend.

You will need

• - a system of three equations with three unknowns.

Instructions

If two of the three systems have only two of the three unknowns, try to express some variables in terms of the others and substitute them into the equation with three unknown. Your goal in this case is to turn it into normal the equation with an unknown person. If this is , the further solution is quite simple - substitute the found value into other equations and find all the other unknowns.

Some systems of equations can be subtracted from one equation by another. See if it is possible to multiply one of or a variable so that two unknowns are canceled at once. If there is such an opportunity, take advantage of it; most likely, the subsequent solution will not be difficult. Remember that when multiplying by a number, you must multiply both the left side and the right side. Likewise, when subtracting equations, you must remember that the right-hand side must also be subtracted.

If the previous methods did not help, use the general method of solving any equations with three unknown. To do this, rewrite the equations in the form a11x1+a12x2+a13x3=b1, a21x1+a22x2+a23x3=b2, a31x1+a32x2+a33x3=b3. Now create a matrix of coefficients for x (A), a matrix of unknowns (X) and a matrix of free variables (B). Please note that by multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix of free terms, that is, A*X=B.

Find matrix A to the power (-1) by first finding , note that it should not be equal to zero. After this, multiply the resulting matrix by matrix B, as a result you will receive the desired matrix X, indicating all the values.

You can also find a solution to a system of three equations using Cramer's method. To do this, find the third-order determinant ∆ corresponding to the system matrix. Then successively find three more determinants ∆1, ∆2 and ∆3, substituting the values ​​of the free terms instead of the values ​​of the corresponding columns. Now find x: x1=∆1/∆, x2=∆2/∆, x3=∆3/∆.

Sources:

• solutions to equations with three unknowns

Solving a system of equations is challenging and exciting. The more complex the system, the more interesting it is to solve. Most often in secondary school mathematics there are systems of equations with two unknowns, but in higher mathematics there may be more variables. Systems can be solved using several methods.

Instructions

The most common method for solving a system of equations is substitution. To do this, you need to express one variable in terms of another and substitute it into the second the equation systems, thus leading the equation to one variable. For example, given the following equations: 2x-3y-1=0;x+y-3=0.

From the second expression it is convenient to express one of the variables, moving everything else to the right side of the expression, not forgetting to change the sign of the coefficient: x = 3-y.

Open the brackets: 6-2y-3y-1=0;-5y+5=0;y=1. We substitute the resulting value y into the expression: x=3-y;x=3-1;x=2.

In the first expression, all terms are 2, you can take 2 out of the bracket to the distributive property of multiplication: 2*(2x-y-3)=0. Now both parts of the expression can be reduced by this number, and then expressed as y, since the modulus coefficient for it is equal to one: -y = 3-2x or y = 2x-3.

Just as in the first case, we substitute this expression into the second the equation and we get: 3x+2*(2x-3)-8=0;3x+4x-6-8=0;7x-14=0;7x=14;x=2. Substitute the resulting value into the expression: y=2x -3;y=4-3=1.

We see that the coefficient for y is the same in value, but different in sign, therefore, if we add these equations, we will completely get rid of y: 4x+3x-2y+2y-6-8=0; 7x-14=0; x=2. Substitute the value of x into any of the two equations of the system and get y=1.

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Biquadratic the equation represents the equation fourth degree, the general form of which is represented by the expression ax^4 + bx^2 + c = 0. Its solution is based on the use of the method of substitution of unknowns. In this case, x^2 is replaced by another variable. Thus, the result is an ordinary square the equation, which needs to be solved.

Instructions

Solve the quadratic the equation, resulting from the replacement. To do this, first calculate the value in accordance with the formula: D = b^2? 4ac. In this case, the variables a, b, c are the coefficients of our equation.

Find the roots of the biquadratic equation. To do this, take the square root of the solutions obtained. If there was one solution, then there will be two - a positive and negative value of the square root. If there were two solutions, the biquadratic equation will have four roots.

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One of the classical methods for solving systems of linear equations is the Gauss method. It consists in the sequential elimination of variables, when a system of equations using simple transformations is transformed into a stepwise system, from which all variables are sequentially found, starting with the last ones.

Instructions

First, bring the system of equations into a form where all the unknowns are in a strictly defined order. For example, all unknown X's will appear first on each line, all Y's will come after X's, all Z's will come after Y's, and so on. There should be no unknowns on the right side of each equation. Mentally determine the coefficients in front of each unknown, as well as the coefficients on the right side of each equation.

§ 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) prompts 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:

1. 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
2. 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
3. HER. Tulchinskaya, Algebra 7th grade. Blitz survey: a manual for students of general education institutions, 4th edition, revised and expanded, Moscow, “Mnemosyne”, 2008
4. 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
5. 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

Using this mathematical program, you can solve a system of two linear equations with two variables using the substitution method and the addition method.

The program not only gives the answer to the problem, but also provides a detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.

This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.

In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.

Rules for entering equations

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.

When entering equations you can use parentheses. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of elements.
For example: 6x+1 = 5(x+y)+2

In equations, you can use not only whole numbers, but also fractions in the form of decimals and ordinary fractions.

Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a period or a comma.
For example: 2.1n + 3.5m = 55

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &

Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)

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A little theory.

Solving systems of linear equations. Substitution method

The sequence of actions when solving a system of linear equations using the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression into another equation of the system instead of this variable;

$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$

Let's express y in terms of x from the first equation: y = 7-3x. Substituting the expression 7-3x into the second equation instead of y, we obtain the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$

It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$

Substituting the number 1 instead of x into the equality y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$

Pair (1;4) - solution of the system

Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.

Solving systems of linear equations by addition

Let's consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by substitution, we move from this system to another, equivalent system, in which one of the equations contains only one variable.

The sequence of actions when solving a system of linear equations using the addition method:
1) multiply the equations of the system term by term, selecting factors so that the coefficients of one of the variables become opposite numbers;
2) add the left and right sides of the system equations term by term;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.

Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$

In the equations of this system, the coefficients of y are opposite numbers. By adding the left and right sides of the equations term by term, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$

From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38\) we get an equation with the variable y: \(11-3y=38\). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)

Thus, we found the solution to the system of equations by addition: \(x=11; y=-9\) or \((11;-9)\)

Taking advantage of the fact that in the equations of the system the coefficients for y are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both sides of each of the equations of the original system), in which one of the equations contains only one variable.

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 undetermined 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.

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.