Linear equation – an equation of the form a x = b, where x is a variable, a and b are some numbers, and a ≠ 0.

Examples:

1. 3 x = 2

1. 2 7 x = − 5

Linear equations are called not only equations of the form a x = b, but also any equations that, with the help of transformations and simplifications, are reduced to this form.

How to solve equations that are reduced to the form a x = b? It is enough to divide the left and right sides of the equation by the value a. As a result, we get the answer: x = b a.

How to recognize whether an arbitrary equation is linear or not? You need to pay attention to the variable that is present in it. If the leading power in which the variable stands is equal to one, then the equation is linear.

To solve the linear equation , you need to open the brackets (if any), move the “X’s” to the left side, the numbers to the right, and bring similar terms. The result is an equation of the form a x = b. The solution to this equation is: x = b a.

Examples:

1. 2 x + 1 = 2 (x − 3) + 8

This is a linear equation because the variable is to the first power.

Let's try to transform it to the form a x = b:

First, let's open the brackets:

2 x + 1 = 4 x − 6 + 8

All terms with x are transferred to the left side, and numbers to the right:

2 x − 4 x = 2 − 1

Now let's divide the left and right sides by the number (-2):

− 2 x − 2 = 1 − 2 = − 1 2 = − 0.5

Answer: x = − 0.5

1. x 2 − 1 = 0

This equation is not linear, since the highest power in which the variable x stands is two.

1. x (x + 3) − 8 = x − 1

This equation looks linear at first glance, but after opening the parentheses, the leading power becomes equal to two:

x 2 + 3 x − 8 = x − 1

This equation is not linear.

Special cases(they were not encountered in task 4 of the OGE, but it is useful to know them)

Examples:

1. 2 x − 4 = 2 (x − 2)

2 x − 4 = 2 x − 4

2 x − 2 x = − 4 + 4

And how can we look for x here if it doesn’t exist? After performing the transformations, we received the correct equality (identity), which does not depend on the value of the variable x. Whatever value of x we ​​substitute into the original equation, the result always results in a correct equality (identity). This means x can be any number.

Answer: x ∈ (− ∞ ;  + ∞)

1. 2 x − 4 = 2 (x − 8)

This is a linear equation. Let's open the brackets, move the X's to the left, numbers to the right:

2 x − 4 = 2 x − 16

2 x − 2 x = − 16 + 4

As a result of the transformations, x was reduced, but the result was an incorrect equality, since. No matter what value of x we ​​substitute into the original equation, the result will always be an incorrect equality. This means that there are no values ​​of x at which the equality would become true.

Lesson and presentation on the topic: "Systems of equations. Basic concepts"

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Rational equations with two unknowns

A rational equation in two variables is an equation of the form $f(x;y)= g(x;y)$.
Where f and g are rational expressions (numbers and any operations of subtraction, division, multiplication, addition and exponentiation) containing the variables x, y.

Let's look at examples of rational expressions:

A rational equation can always be represented as:
$u(x;y)=f(x;y)-g(x;y)$. Here $u(x;y)$ is a rational expression.
$u(x;y)=0$ is a whole rational equation.

The solution to the equation is: $u(x;y)= 0$. (x;y) – a pair of numbers that satisfy this equation.

Examples:

A) (3;2) - solution to the equation: $x+y=5$. Substitute x= 3 and y= 2, we get $3+2=5$

B) (1;4) - solution to the equation: $2x^2+y^2=18$. Substitute x= 1 and y= 4, we get $2+16=18$

C) Solve the equation: $(3x-6)^2+(2y-2)^2=0$.
Solution: For any x and y $(3x-6)^2≥0\; and \;(2y-2)^2≥0$. This means that the left side of the equality is always greater than or equal to zero, and is equal to zero only when both expressions are equal to zero. This means that the solution to the equation will be a pair of numbers (2;1).
Answer: (2;1).

D) Find all integer solutions to the equation: $x-y=12$.
Solution: Let x= z, then $y=z-12$, z is any integer. Then the solution will be a pair of numbers (z;z-12), where z is an integer.

D) Find integer solutions to the equation: $4x+7y=29$.
Solution: Express x in terms of y: $x=\frac(29-7y)(4)=\frac(28+1-7y)(4)=7+\frac(1-7y)(4)=7-\ frac(7y-1)(4)$.
x is an integer if $7y-1$ is divisible by 4 without a remainder. Let's look at the possible options for our division:
1) y is a multiple of 4. Then $y=4n$. $7y-1=7*4n-1=28n-1$ – not divisible by 4, which means it doesn’t fit.

2) y – when divided by 4, the remainder is 1. $y=4n+1$. $7y-1=28n+7-1=28n+6$ – not divisible by 4, which means it doesn’t fit.

3) y – when divided by 4, the remainder is 2. $y=4n+2$. $7y-1=28n+14-1=28n+13$ – not divisible by 4, which means it doesn’t fit.

4) y – when divided by 4, the remainder is 3. $y=4n+3$. $7y-1=28n+21-1=28n+20$ – divisible by 4, which means it’s suitable.

We got $y=4n+3$, let's find x.
$x=7-\frac(7y-1)(4)=7-\frac(28n+20)(4)=7-7n+5=2-7n$
Answer: ($2-7n;4n+3$).

Two rational equations are said to be equivalent if they have the same solutions.

Equivalent transformations of an equation are called:

A) Transfer of terms of the equation from one part of the equation to another, with a change of sign.
Example: $-3x+5y=2x+7y$ is equivalent to $-3x-2x=7y-5y$

B) Multiplying or dividing both sides of equations by a number that is not zero.
Example: $2x-0.5y=0.2xy$ is equivalent to $20x-5y=2xy$. (Multiply both sides of the equation by 10).

Graphing an Equation in Two Variables

Let the equation u(x;y)= 0 be given. The set of points (x;y) on the coordinate plane, which are a solution to the equation u(x;y)= 0, is called the graph of the function.

If the equation u(x;y)= 0 can be converted to the form y=f(x), then it is simultaneously considered a graph of the equation.

Graph the equation:
a) $y+2x=2$,
b) $yx=5$.

Solution:
a) The graph of our equation will be a straight line. Guys, do you remember how we plotted a linear function in 7th grade?
The graph of our function is built using two points:
Let's build a graph:

b) Let's transform our equation $yx=5$. We get $y=5/x$ – the graph of the hyperbola. Let's build it:

Distance between two points on a coordinate plane

Definition. The distance between two points A(x1;y1) and B(x2;y2) is calculated by the formula: $AB=\sqrt((x2-x1)^2+(y2-y1)^2)$

Example: Find the distance between points: A(10;34) and B(3;10).
Solution: $AB=\sqrt((x2-x1)^2+(y2-y1)^2)=\sqrt((3-10)^2+(10-34)^2)=\sqrt(7^ 2+24^2)=\sqrt(625)=$25.

Definition. The graph of the equation: $(x-a)^2+(y-b)^2=r^2$ is a circle on the coordinate plane with a center at point (a;b) and radius r.

Example: Graph the equation: $x^2+y^2=4$.
Solution: Let's rewrite our equation according to the definition: $(x-0)^2+(y-0)^2=4$. This is a circle with a center at the point (0;0) and a radius equal to 2. Let's draw our circle:

Example: Graph the equation: $x^2+y^2-6y=0$.
Solution. Let's rewrite it in the form: $x^2+y^2-6y+9-9=0$, $x^2+(y+3)^2=9$, $(x-0)^2+(y- 3)^2=9$.
This is a circle with a center at point (0; 3) and a radius equal to 3. Let's draw our circle:

Equation problems for independent solution

1. Find all integer solutions to the equation $2x+y=16$.
2. Find integer solutions: $3х+5y=23$.
3. Graph the equation: a) $y-5x=-5$, b) $yx=6$, c) $(y+2x)^2=0$.
4. Find the distance between points: A(5;25) and B(18;10).
5. Graph the equation: a) $x^2+y^2=36$, b) $x^2+8x+y^2+6y=0$.

When performing various algebraic transformations, it is often convenient to use abbreviated multiplication formulas. Often these formulas are used not so much to shorten the multiplication process, but rather to understand from the result that it can be represented as a product of some factors. Thus, these formulas need to be able to be applied not only from left to right, but also from right to left. Let us list the basic formulas for abbreviated multiplication. Squared sum:

Squared difference:

The previous two formulas are also sometimes written in a slightly different form, which gives us some kind of expression for the sum of squares:

You also need to understand what will happen if the signs in the brackets in the square are placed in a “non-standard” way:

Difference of cubes:

Sum of cubes:

Cube of sum:

Difference cube:

The last two formulas are also often conveniently used in the form:

Quadratic equation and quadratic trinomial

Let the quadratic equation have the form:

Then discriminant found by the formula:

If D> 0, then a quadratic equation has two roots, which are found using the formula:

If D= 0, then quadratic equation has one root(its multiplicity: 2), which is searched by the formula:

If D < 0, то квадратное уравнение не имеет корней. В случае когда квадратное уравнение имеет два корня, соответствующий a quadratic trinomial can be factorized using the following formula:

If a quadratic equation has one root, then the factorization of the corresponding quadratic trinomial is given by the following formula:

Only in case if a quadratic equation has two roots (i.e. the discriminant is strictly greater than zero), Vieta's Theorem holds. According to Vieta's Theorem, the sum of the roots of a quadratic equation is equal to:

The product of the roots of a quadratic equation according to Vieta’s theorem can be calculated using the formula:

The graph of a parabola is given by a quadratic function:

In this case, the coordinates of the vertex of the parabola can be calculated using the following formulas. X tops(or the point at which the quadratic trinomial reaches its largest or smallest value):

Igrek tops parabolas or the maximum if the branches of the parabola are directed downwards ( a < 0), либо минимальное, если ветви параболы направлены вверх (a> 0), the value of the quadratic trinomial:

Basic properties of degrees

Mathematical degrees have several important properties, we list them. When multiplying powers with the same bases, the exponents of the powers are added:

When dividing powers with the same bases, the exponent of the divisor is subtracted from the exponent of the dividend:

When raising a degree to a power, the exponents are multiplied:

If numbers with the same power, but different bases are multiplied, then you can first multiply the numbers, and then raise the product to this power. The reverse procedure is also possible: if there is a product to a power, then each of the multiplied ones can be raised to this power separately and the results multiplied:

Also, if numbers with the same power but different bases are being divided, then you can first divide the numbers and then raise the quotient to this power (the reverse procedure is also possible):

A few simple properties of degrees:

The last property is satisfied only when n> 0. Zero can only be raised to a positive power. Well, the main property negative degree is written as follows:

Basic properties of mathematical roots

The mathematical root can be represented as an ordinary degree, and then use all the properties of degrees given above. For representation of a mathematical root as a power use the following formula:

Nevertheless, it is possible to separately write down a number of properties of mathematical roots, which are based on the properties of powers described above:

For arithmetic roots, the following property holds (which can simultaneously be considered the definition of a root):

The latter is true: if n– odd, then for any a; if n– even, then only if it is non-negative a. For odd root The following equality also holds (a minus sign can be taken out from under the root of odd degree):

Because the value of an even root can only be non-negative, then such roots have the following important property:

Some additional information from algebra

If x 0 – root of the polynomial n th degree Pn(x), then the following equality holds (here Qn-1(x) – some polynomial ( n– 1st degree):

A procedure in which a quadratic trinomial is represented as a bracket in a square and some other term is called highlighting a complete square. And although it is easier to perform the operation of selecting a complete square each time “from scratch” in specific numbers, there is nevertheless a general formula with which you can immediately write down the result of selecting a complete square:

There is an operation inverse to the operation of adding fractions with like denominators, which is called term by term division. It consists in writing, on the contrary, each term from the sum in the numerator of a certain fraction separately above the denominator of this fraction. For the operation of term-by-term division, you can also write the general formula:

There is also a formula for factoring the sum of squares:

Solving rational equations

Solving an equation means finding all its roots. The main solution method is to reduce the equation to an equivalent equation, which can be solved simply (for example, to a quadratic equation) by means of algebraic transformations or substitution of variables. If you cannot reduce the equation to an equivalent equation, then side roots may arise. If in doubt, check the roots by substitution.

For many equations, the concept of a region of permissible values ​​for roots, hereinafter referred to as ODZ, is important. At this stage (in rational equations, i.e. those that do not contain arithmetic roots, trigonometric functions, logarithms, etc.), the main condition that the roots of the equation must meet is that when substituting them into the original form of the equation, the denominators of the fractions did not go to zero, because You cannot divide by zero. Thus, the ODZ includes all possible values ​​except those that make the denominators of fractions vanish.

When solving equations (and later inequalities), you cannot reduce factors with a variable on the left and right sides of the equation (inequality), in which case you will lose the roots. It is necessary to move all expressions to the left of the equal sign and put the “cancelling” factor out of brackets; in the future, you need to take into account the roots that it gives.

In order for the product of two or more brackets to be equal to zero, it is sufficient that any of them individually be equal to zero, and the rest exist. Therefore, in such cases, you need to equate all brackets to zero one by one. In the final answer you need to write down the roots of all these “branches” of the solution (if, of course, these roots are included in the ODZ).

Sometimes some of the fractions in a rational equation can be canceled out. You should definitely try to do this and not miss a single such opportunity. But when reducing a fraction, you may lose the ODZ, so fractions need to be reduced only after recording the ODZ, or at the end of the solution, substitute the resulting roots into the original equation to check the existence of the denominators.

So, to solve a rational equation it is necessary:

1. Factor all denominators of all fractions.
2. Move all the terms to the left so that you get zero on the right.
3. Write down the ODZ.
4. Reduce fractions if possible.
5. Reduce to a common denominator.
6. Simplify the expression in the numerator.
7. Equate the numerator to zero and solve the resulting equation.
8. Don’t forget to check the roots for compliance with the DZ.

One of the most common methods for solving equations is variable replacement method. Often, variable replacement is chosen individually for each specific example. It is important to remember two main criteria for introducing a replacement into equations. So, after introducing a replacement into some equation, this equation should:

• firstly, to become simpler;
• second, no longer contain the original variable.

In addition, it is important not to forget to perform a reverse replacement, i.e. after finding the values ​​for the new variable (for replacement), write down instead of the replacement what it is equal to through the original variable, equate this expression with the found values ​​for replacement and solve the equations again.

Let us separately dwell on the algorithm for solving very common homogeneous equations. Homogeneous equations have the form:

Here A, B and C are numbers that are not equal to zero, but f(x) And g(x) – some functions with a variable X. Homogeneous equations are solved this way: divide the entire equation by g 2 (x) and we get:

We replace the variables:

And we solve the quadratic equation:

Having received the roots of this equation, do not forget to perform the reverse substitution, and also check the roots for compliance with the ODZ.

Also, when solving some rational equations, it would be good to remember the following useful transformations:

Solving systems of rational equations

To solve a system of equations means to find not just a solution, but sets of solutions, that is, such values ​​of all variables that, being simultaneously substituted into the system, turn each of its equations into an identity. When solving systems of equations, you can use the following methods (don’t forget about ODZ):

• Substitution method. The method is to express one of the variables from one of the equations, substitute this expression instead of this unknown in the remaining equations, thus reducing the number of unknowns in the remaining equations. This procedure is repeated until one equation with one variable remains, which is then solved. The remaining unknowns are sequentially found from the already known values ​​of the found variables.
• System splitting method. This method consists of factoring one of the system's equations. In this case, it is necessary that there be a zero on the right side of this equation. Then, by equating each factor of this equation to zero in turn and adding the remaining equations of the original system, we will obtain several systems, but each of them will be simpler than the original one.
• Addition and subtraction method. This method consists of adding or subtracting two equations of the system (they can and often must first be multiplied by a certain coefficient) to obtain a new equation and replace one of the equations of the original system with it. Obviously, such a procedure makes sense only if the new equation turns out to be much simpler than the previously existing ones.
• Method of division and multiplication. This method consists of dividing or multiplying the left and right sides of two equations of the system, respectively, to obtain a new equation and replace it with one of the equations of the original system. Obviously, such a procedure again makes sense only if the new equation turns out to be much simpler than the previously existing ones.

There are other methods for solving systems of rational equations. Among which - replacing variables. Often, replacing variables is selected individually for each specific example. But there are two cases where you always need to introduce a very specific replacement. The first of these cases is the case when both equations of a system with two unknowns are homogeneous polynomials equated to a certain number. In this case, you need to use the replacement:

After applying this replacement, by the way, it will be necessary to use the division method to continue solving such systems. The second case is symmetrical systems with two variables, i.e. systems that do not change when replaced x on y, A y on x. In such systems it is necessary to use the following double substitution of variables:

Moreover, in order to introduce such a replacement into a symmetric system, the original equations will most likely have to be greatly transformed. Of course, we must not forget about ODZ and the obligation to perform reverse replacement in both of these methods.

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How to successfully prepare for the CT in physics and mathematics?

In order to successfully prepare for the CT in physics and mathematics, among other things, it is necessary to fulfill three most important conditions:

1. Study all topics and complete all tests and assignments given in the educational materials on this site. To do this, you need nothing at all, namely: devote three to four hours every day to preparing for the CT in physics and mathematics, studying theory and solving problems. The fact is that the CT is an exam where it is not enough just to know physics or mathematics, you also need to be able to quickly and without failures solve a large number of problems on different topics and of varying complexity. The latter can only be learned by solving thousands of problems.
2. Learn all the formulas and laws in physics, and formulas and methods in mathematics. In fact, this is also very simple to do; there are only about 200 necessary formulas in physics, and even a little less in mathematics. In each of these subjects there are about a dozen standard methods for solving problems of a basic level of complexity, which can also be learned, and thus, completely automatically and without difficulty solving most of the CT at the right time. After this, you will only have to think about the most difficult tasks.
3. Attend all three stages of rehearsal testing in physics and mathematics. Each RT can be visited twice to decide on both options. Again, on the CT, in addition to the ability to quickly and efficiently solve problems, and knowledge of formulas and methods, you must also be able to properly plan time, distribute forces, and most importantly, correctly fill out the answer form, without confusing the numbers of answers and problems, or your own last name. Also, during RT, it is important to get used to the style of asking questions in problems, which may seem very unusual to an unprepared person at the DT.

Successful, diligent and responsible implementation of these three points will allow you to show an excellent result at the CT, the maximum of what you are capable of.

Found a mistake?

If you think you have found an error in the training materials, please write about it by email. You can also report an error on the social network (). In the letter, indicate the subject (physics or mathematics), the name or number of the topic or test, the number of the problem, or the place in the text (page) where, in your opinion, there is an error. Also describe what the suspected error is. Your letter will not go unnoticed, the error will either be corrected, or you will be explained why it is not an error.

I. Rational equations.

1) Linear equations.

2) Systems of linear equations.

3) Quadratic equations and equations reducible to them.

4) Reciprocal equations.

5) Vieta's formula for polynomials of higher degrees.

6) Systems of equations of the second degree.

7) Method of introducing new unknowns when solving equations and systems of equations.

8) Homogeneous equations.

9) Solving symmetric systems of equations.

10) Equations and systems of equations with parameters.

11) Graphical method for solving systems of nonlinear equations.

12) Equations containing the modulus sign.

13) Basic methods for solving rational equations

II. Rational inequalities.

1) Properties of equivalent inequalities.

2) Algebraic inequalities.

3) Interval method.

4) Fractional rational inequalities.

5) Inequalities containing an unknown under the absolute value sign.

6) Inequalities with parameters.

7) Systems of rational inequalities.

8) Graphical solution of inequalities.

III. Screening test.

Rational equations

Function of the form

P(x) = a 0 x n + a 1 x n – 1 + a 2 x n – 2 + … + a n – 1 x + a n,

where n is a natural number, a 0, a 1,…, a n are some real numbers, called an entire rational function.

An equation of the form P(x) = 0, where P(x) is an entire rational function, is called an entire rational equation.

Equation of the form

P 1 (x) / Q 1 (x) + P 2 (x) / Q 2 (x) + … + P m (x) / Q m (x) = 0,

where P 1 (x), P 2 (x), ..., P m (x), Q 1 (x), Q 2 (x), ..., Q m (x) are entire rational functions, called a rational equation.

Solving the rational equation P (x) / Q (x) = 0, where P (x) and Q (x) are polynomials (Q (x) ¹ 0), comes down to solving the equation P (x) = 0 and checking that that the roots satisfy the condition Q (x) ¹ 0.

Linear equations.

An equation of the form ax+b=0, where a and b are some constants, is called a linear equation.

If a¹0, then the linear equation has a single root: x = -b /a.

If a=0; b¹0, then the linear equation has no solutions.

If a=0; b=0, then, rewriting the original equation in the form ax = -b, it is easy to see that any x is a solution to the linear equation.

The equation of the straight line is: y = ax + b.

If a line passes through a point with coordinates X 0 and Y 0, then these coordinates satisfy the equation of the line, i.e. Y 0 = aX 0 + b.

Example 1.1. Solve the equation

2x – 3 + 4(x – 1) = 5.

Solution. Sequentially open the brackets, add similar terms and find x: 2x – 3 + 4x – 4 = 5, 2x + 4x = 5 + 4 + 3,

Example 1.2. Solve the equation

2x – 3 + 2(x – 1) = 4(x – 1) – 7.

Solution. 2x + 2x – 4x = 3 +2 – 4 – 7, 0x = – 6.

Example 1.3. Solve the equation.

2x + 3 – 6(x – 1) = 4(x – 1) + 5.

Solution. 2x – 6x + 3 + 6 = 4 – 4x + 5,

– 4x + 9 = 9 – 4x,

4x + 4x = 9 – 9,

Answer: Any number.

Systems of linear equations.

Equation of the form

a 1 x 1 + a 2 x 2 + … + a n x n = b,

where a 1, b 1, …, a n, b are some constants, called a linear equation with n unknowns x 1, x 2, …, x n.

A system of equations is called linear if all equations included in the system are linear. If the system consists of n unknowns, then the following three cases are possible:

1) the system has no solutions;

2) the system has exactly one solution;

3) the system has infinitely many solutions.

Example 2.4. solve system of equations

2x + 3y = 8,

Solution. You can solve a system of linear equations using the substitution method, which consists of expressing one unknown in terms of other unknowns for any equation of the system, and then substituting the value of this unknown into the remaining equations.

From the first equation we express: x = (8 – 3y) / 2. We substitute this expression into the second equation and get a system of equations

Solution. The system has no solutions, since two equations of the system cannot be satisfied simultaneously (from the first equation x + y = 3, and from the second x + y = 3.5).

Answer: There are no solutions.

Example 2.6. solve system of equations

Solution. The system has infinitely many solutions, since the second equation is obtained from the first by multiplying by 2 (i.e., in fact, there is only one equation with two unknowns).

Answer: There are infinitely many solutions.

Example 2.7. solve system of equations

x + y – z = 2,

2x – y + 4z = 1,

– x + 6y + z = 5.

Solution. When solving systems of linear equations, it is convenient to use the Gauss method, which consists of transforming the system to a triangular form.

We multiply the first equation of the system by – 2 and, adding the resulting result with the second equation, we get – 3y + 6z = – 3. This equation can be rewritten as y – 2z = 1. Adding the first equation with the third, we get 7y = 7, or y = 1.

Thus, the system acquired a triangular shape

x + y – z = 2,

Substituting y = 1 into the second equation, we find z = 0. Substituting y = 1 and z = 0 into the first equation, we find x = 1.

Answer: (1; 1; 0).

Example 2.8. at what values ​​of parameter a is the system of equations

2x + ay = a + 2,

(a + 1)x + 2ay = 2a + 4

has infinitely many solutions?

Solution. From the first equation we express x:

x = – (a / 2)y + a / 2 +1.

Substituting this expression into the second equation, we get

(a + 1)(– (a / 2)y + a / 2 +1) + 2ay = 2a + 4.

(a + 1)(a + 2 – ay) + 4ay = 4a + 8,

4ay – a(a + 1)y = 4(a + 2) – (a + 1)(a + 2),

ya(4 – a – 1) = (a + 2)(4 – a – 1),

ya(3 – a) = (a + 2)(3 – a).

Analyzing the last equation, we note that for a = 3 it has the form 0y = 0, i.e. it is satisfied for any values ​​of y.

Quadratic equations and equations that can be reduced to them.

An equation of the form ax 2 + bx + c = 0, where a, b and c are some numbers (a¹0);

x is a variable called a quadratic equation.

Formula for solving a quadratic equation.

First, let's divide both sides of the equation ax 2 + bx + c = 0 by a - this will not change its roots. To solve the resulting equation

x 2 + (b / a)x + (c / a) = 0

select a complete square on the left side

x 2 + (b / a) + (c / a) = (x 2 + 2(b / 2a)x + (b / 2a) 2) – (b / 2a) 2 + (c / a) =

= (x + (b / 2a)) 2 – (b 2) / (4a 2) + (c / a) = (x + (b / 2a)) 2 – ((b 2 – 4ac) / (4a 2 )).

For brevity, we denote the expression (b 2 – 4ac) by D. Then the resulting identity takes the form

Three cases are possible:

1) if the number D is positive (D > 0), then in this case you can extract the square root of D and write D in the form D = (ÖD) 2. Then

D / (4a 2) = (ÖD) 2 / (2a) 2 = (ÖD / 2a) 2, therefore the identity takes the form

x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 – (ÖD / 2a) 2 .

Using the difference of squares formula, we derive from here:

x 2 + (b / a)x + (c / a) = (x + (b / 2a) – (ÖD / 2a))(x + (b / 2a) + (ÖD / 2a)) =

= (x – ((-b + ÖD) / 2a)) (x – ((– b – ÖD) / 2a)).

Theorem : If the identity holds

ax 2 + bx + c = a(x – x 1)(x – x 2),

then the quadratic equation ax 2 + bx + c = 0 for X 1 ¹ X 2 has two roots X 1 and X 2, and for X 1 = X 2 - only one root X 1.

By virtue of this theorem, from the identity derived above it follows that the equation

x 2 + (b / a)x + (c / a) = 0,

and thus the equation ax 2 + bx + c = 0 has two roots:

X 1 =(-b + Ö D) / 2a; X 2 = (-b - Ö D) / 2a.

Thus x 2 + (b / a)x + (c / a) = (x – x1)(x – x2).

Usually these roots are written with one formula:

where b 2 – 4ac = D.

2) if the number D is zero (D = 0), then the identity

x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 – (D / (4a 2))

takes the form x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2.

It follows that for D = 0 the equation ax 2 + bx + c = 0 has one root of multiplicity 2: X 1 = – b / 2a

3) If the number D is negative (D< 0), то – D >0, and therefore the expression

x 2 + (b / a)x + (c / a) = (x + (b / 2a)) 2 – (D / (4a 2))

is the sum of two terms, one of which is non-negative and the other is positive. Such a sum cannot equal zero, so the equation

x 2 + (b / a)x + (c / a) = 0

has no real roots. The equation ax 2 + bx + c = 0 does not have them either.

Thus, to solve a quadratic equation, one should calculate the discriminant

D = b 2 – 4ac.

If D = 0, then the quadratic equation has a unique solution:

If D > 0, then the quadratic equation has two roots:

X 1 =(-b + ÖD) / (2a); X 2 = (-b - ÖD) / (2a).

If D< 0, то квадратное уравнение не имеет корней.

If one of the coefficients b or c is zero, then the quadratic equation can be solved without calculating the discriminant:

1) b = 0; c¹0; c/a<0; X1,2 = ±Ö(-c / a)

2) b ¹ 0; c = 0; X1 = 0, X2= -b / a.

The roots of a general quadratic equation ax 2 + bx + c = 0 are found by the formula

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.

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