Systems of rational equations in mathematics. Video lesson “Rational equations Rational equations and systems of rational equations

Maintaining your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please review our privacy practices and let us know if you have any questions.

Collection and use of personal information

Personal information refers to data that can be used to identify or contact a specific person.

You may be asked to provide your personal information at any time when you contact us.

Below are some examples of the types of personal information we may collect and how we may use such information.

What personal information do we collect:

• When you submit an application on the site, we may collect various information, including your name, phone number, email address, etc.

How we use your personal information:

• The personal information we collect allows us to contact you with unique offers, promotions and other events and upcoming events.
• From time to time, we may use your personal information to send important notices and communications.
• We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
• If you participate in a prize draw, contest or similar promotion, we may use the information you provide to administer such programs.

Disclosure of information to third parties

We do not disclose the information received from you to third parties.

Exceptions:

• If necessary - in accordance with the law, judicial procedure, in legal proceedings, and/or on the basis of public requests or requests from government authorities in the territory of the Russian Federation - to disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public importance purposes.
• In the event of a reorganization, merger, or sale, we may transfer the personal information we collect to the applicable successor third party.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as unauthorized access, disclosure, alteration and destruction.

Respecting your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security standards to our employees and strictly enforce privacy practices.

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.

Let's continue talking about solving equations. In this article we will go into detail about rational equations and principles of solving rational equations with one variable. First, let's figure out what type of equations are called rational, give a definition of whole rational and fractional rational equations, and give examples. Next, we will obtain algorithms for solving rational equations, and, of course, we will consider solutions to typical examples with all the necessary explanations.

Page navigation.

Based on the stated definitions, we give several examples of rational equations. For example, x=1, 2·x−12·x 2 ·y·z 3 =0, , are all rational equations.

From the examples shown, it is clear that rational equations, as well as equations of other types, can be with one variable, or with two, three, etc. variables. In the following paragraphs we will talk about solving rational equations with one variable. Solving equations in two variables and their large number deserve special attention.

In addition to dividing rational equations by the number of unknown variables, they are also divided into integer and fractional ones. Let us give the corresponding definitions.

Definition.

The rational equation is called whole, if both its left and right sides are integer rational expressions.

Definition.

If at least one of the parts of a rational equation is a fractional expression, then such an equation is called fractionally rational(or fractional rational).

It is clear that whole equations do not contain division by a variable; on the contrary, fractional rational equations necessarily contain division by a variable (or a variable in the denominator). So 3 x+2=0 and (x+y)·(3·x 2 −1)+x=−y+0.5– these are whole rational equations, both of their parts are whole expressions. A and x:(5 x 3 +y 2)=3:(x−1):5 are examples of fractional rational equations.

Concluding this point, let us pay attention to the fact that the linear equations and quadratic equations known to this point are entire rational equations.

Solving whole equations

One of the main approaches to solving entire equations is to reduce them to equivalent ones algebraic equations. This can always be done by performing the following equivalent transformations of the equation:

• first, the expression from the right side of the original integer equation is transferred to the left side with the opposite sign to obtain zero on the right side;
• after this, on the left side of the equation the resulting standard form.

The result is an algebraic equation that is equivalent to the original integer equation. Thus, in the simplest cases, solving entire equations is reduced to solving linear or quadratic equations, and in the general case, to solving an algebraic equation of degree n. For clarity, let's look at the solution to the example.

Example.

Find the roots of the whole equation 3·(x+1)·(x−3)=x·(2·x−1)−3.

Solution.

Let us reduce the solution of this entire equation to the solution of an equivalent algebraic equation. To do this, firstly, we transfer the expression from the right side to the left, as a result we arrive at the equation 3·(x+1)·(x−3)−x·(2·x−1)+3=0. And, secondly, we transform the expression formed on the left side into a standard form polynomial by completing the necessary: 3·(x+1)·(x−3)−x·(2·x−1)+3= (3 x+3) (x−3)−2 x 2 +x+3= 3 x 2 −9 x+3 x−9−2 x 2 +x+3=x 2 −5 x−6. Thus, solving the original integer equation is reduced to solving the quadratic equation x 2 −5·x−6=0.

We calculate its discriminant D=(−5) 2 −4·1·(−6)=25+24=49, it is positive, which means that the equation has two real roots, which we find using the formula for the roots of a quadratic equation:

To be completely sure, let's do it checking the found roots of the equation. First we check the root 6, substitute it instead of the variable x in the original integer equation: 3·(6+1)·(6−3)=6·(2·6−1)−3, which is the same, 63=63. This is a valid numerical equation, therefore x=6 is indeed the root of the equation. Now we check the root −1, we have 3·(−1+1)·(−1−3)=(−1)·(2·(−1)−1)−3, from where, 0=0 . When x=−1, the original equation also turns into a correct numerical equality, therefore, x=−1 is also a root of the equation.

Answer:

6 , −1 .

Here it should also be noted that the term “degree of the whole equation” is associated with the representation of an entire equation in the form of an algebraic equation. Let us give the corresponding definition:

Definition.

The power of the whole equation is called the degree of an equivalent algebraic equation.

According to this definition, the entire equation from the previous example has the second degree.

This could have been the end of solving entire rational equations, if not for one thing…. As is known, solving algebraic equations of degree above the second is associated with significant difficulties, and for equations of degree above the fourth there are no general root formulas at all. Therefore, to solve entire equations of the third, fourth and higher degrees, it is often necessary to resort to other solution methods.

In such cases, an approach to solving entire rational equations based on factorization method. In this case, the following algorithm is adhered to:

• first, they ensure that there is a zero on the right side of the equation; to do this, they transfer the expression from the right side of the whole equation to the left;
• then, the resulting expression on the left side is presented as a product of several factors, which allows us to move on to a set of several simpler equations.

The given algorithm for solving an entire equation through factorization requires a detailed explanation using an example.

Example.

Solve the whole equation (x 2 −1)·(x 2 −10·x+13)= 2 x (x 2 −10 x+13) .

Solution.

First, as usual, we transfer the expression from the right side to the left side of the equation, not forgetting to change the sign, we get (x 2 −1)·(x 2 −10·x+13)− 2 x (x 2 −10 x+13)=0 . Here it is quite obvious that it is not advisable to transform the left-hand side of the resulting equation into a polynomial of the standard form, since this will give an algebraic equation of the fourth degree of the form x 4 −12 x 3 +32 x 2 −16 x−13=0, the solution of which is difficult.

On the other hand, it is obvious that on the left side of the resulting equation we can x 2 −10 x+13 , thereby presenting it as a product. We have (x 2 −10 x+13) (x 2 −2 x−1)=0. The resulting equation is equivalent to the original whole equation, and it, in turn, can be replaced by a set of two quadratic equations x 2 −10·x+13=0 and x 2 −2·x−1=0. Finding their roots using known root formulas through a discriminant is not difficult; the roots are equal. They are the desired roots of the original equation.

Answer:

Also useful for solving entire rational equations method for introducing a new variable. In some cases, it allows you to move to equations whose degree is lower than the degree of the original whole equation.

Example.

Find the real roots of a rational equation (x 2 +3 x+1) 2 +10=−2 (x 2 +3 x−4).

Solution.

Reducing this whole rational equation to an algebraic equation is, to put it mildly, not a very good idea, since in this case we will come to the need to solve a fourth-degree equation that does not have rational roots. Therefore, you will have to look for another solution.

Here it is easy to see that you can introduce a new variable y and replace the expression x 2 +3·x with it. This replacement leads us to the whole equation (y+1) 2 +10=−2·(y−4) , which, after moving the expression −2·(y−4) to the left side and subsequent transformation of the expression formed there, is reduced to a quadratic equation y 2 +4·y+3=0. The roots of this equation y=−1 and y=−3 are easy to find, for example, they can be selected based on the theorem inverse to Vieta’s theorem.

Now we move on to the second part of the method of introducing a new variable, that is, to performing a reverse replacement. After performing the reverse substitution, we obtain two equations x 2 +3 x=−1 and x 2 +3 x=−3, which can be rewritten as x 2 +3 x+1=0 and x 2 +3 x+3 =0 . Using the formula for the roots of a quadratic equation, we find the roots of the first equation. And the second quadratic equation has no real roots, since its discriminant is negative (D=3 2 −4·3=9−12=−3 ).

Answer:

In general, when we are dealing with entire equations of high degrees, we must always be prepared to search for a non-standard method or an artificial technique for solving them.

Solving fractional rational equations

First, it will be useful to understand how to solve fractional rational equations of the form , where p(x) and q(x) are integer rational expressions. And then we will show how to reduce the solution of other fractionally rational equations to the solution of equations of the indicated type.

One approach to solving the equation is based on the following statement: the numerical fraction u/v, where v is a non-zero number (otherwise we will encounter , which is undefined), is equal to zero if and only if its numerator is equal to zero, then is, if and only if u=0 . By virtue of this statement, solving the equation is reduced to fulfilling two conditions p(x)=0 and q(x)≠0.

This conclusion corresponds to the following algorithm for solving a fractional rational equation. To solve a fractional rational equation of the form , you need

• solve the whole rational equation p(x)=0 ;
• and check whether the condition q(x)≠0 is satisfied for each root found, while
• if true, then this root is the root of the original equation;
• if it is not satisfied, then this root is extraneous, that is, it is not the root of the original equation.

Let's look at an example of using the announced algorithm when solving a fractional rational equation.

Example.

Find the roots of the equation.

Solution.

This is a fractional rational equation, and of the form , where p(x)=3·x−2, q(x)=5·x 2 −2=0.

According to the algorithm for solving fractional rational equations of this type, we first need to solve the equation 3 x−2=0. This is a linear equation whose root is x=2/3.

It remains to check for this root, that is, check whether it satisfies the condition 5 x 2 −2≠0. We substitute the number 2/3 into the expression 5 x 2 −2 instead of x, and we get . The condition is met, so x=2/3 is the root of the original equation.

Answer:

2/3 .

You can approach solving a fractional rational equation from a slightly different position. This equation is equivalent to the integer equation p(x)=0 on the variable x of the original equation. That is, you can stick to this algorithm for solving a fractional rational equation :

• solve the equation p(x)=0 ;
• find the ODZ of variable x;
• take roots belonging to the region of acceptable values ​​- they are the desired roots of the original fractional rational equation.

For example, let's solve a fractional rational equation using this algorithm.

Example.

Solve the equation.

Solution.

First, we solve the quadratic equation x 2 −2·x−11=0. Its roots can be calculated using the root formula for the even second coefficient, we have D 1 =(−1) 2 −1·(−11)=12, And .

Secondly, we find the ODZ of the variable x for the original equation. It consists of all numbers for which x 2 +3·x≠0, which is the same as x·(x+3)≠0, whence x≠0, x≠−3.

It remains to check whether the roots found in the first step are included in the ODZ. Obviously yes. Therefore, the original fractional rational equation has two roots.

Answer:

Note that this approach is more profitable than the first if the ODZ is easy to find, and is especially beneficial if the roots of the equation p(x) = 0 are irrational, for example, or rational, but with a rather large numerator and/or denominator, for example, 127/1101 and −31/59. This is due to the fact that in such cases, checking the condition q(x)≠0 will require significant computational effort, and it is easier to exclude extraneous roots using the ODZ.

In other cases, when solving the equation, especially when the roots of the equation p(x) = 0 are integers, it is more profitable to use the first of the given algorithms. That is, it is advisable to immediately find the roots of the entire equation p(x)=0, and then check whether the condition q(x)≠0 is satisfied for them, rather than finding the ODZ, and then solving the equation p(x)=0 on this ODZ . This is due to the fact that in such cases it is usually easier to check than to find DZ.

Let us consider the solution of two examples to illustrate the specified nuances.

Example.

Find the roots of the equation.

Solution.

First, let's find the roots of the whole equation (2 x−1) (x−6) (x 2 −5 x+14) (x+1)=0, composed using the numerator of the fraction. The left side of this equation is a product, and the right side is zero, therefore, according to the method of solving equations through factorization, this equation is equivalent to a set of four equations 2 x−1=0 , x−6=0 , x 2 −5 x+ 14=0 , x+1=0 . Three of these equations are linear and one is quadratic; we can solve them. From the first equation we find x=1/2, from the second - x=6, from the third - x=7, x=−2, from the fourth - x=−1.

With the roots found, it is quite easy to check whether the denominator of the fraction on the left side of the original equation vanishes, but determining the ODZ, on the contrary, is not so simple, since for this you will have to solve an algebraic equation of the fifth degree. Therefore, we will abandon finding the ODZ in favor of checking the roots. To do this, we substitute them one by one instead of the variable x in the expression x 5 −15 x 4 +57 x 3 −13 x 2 +26 x+112, obtained after substitution, and compare them with zero: (1/2) 5 −15·(1/2) 4 + 57·(1/2) 3 −13·(1/2) 2 +26·(1/2)+112= 1/32−15/16+57/8−13/4+13+112= 122+1/32≠0 ;
6 5 −15·6 4 +57·6 3 −13·6 2 +26·6+112= 448≠0 ;
7 5 −15·7 4 +57·7 3 −13·7 2 +26·7+112=0;
(−2) 5 −15·(−2) 4 +57·(−2) 3 −13·(−2) 2 + 26·(−2)+112=−720≠0 ;
(−1) 5 −15·(−1) 4 +57·(−1) 3 −13·(−1) 2 + 26·(−1)+112=0 .

Thus, 1/2, 6 and −2 are the desired roots of the original fractional rational equation, and 7 and −1 are extraneous roots.

Answer:

1/2 , 6 , −2 .

Example.

Find the roots of a fractional rational equation.

Solution.

First, let's find the roots of the equation (5 x 2 −7 x−1) (x−2)=0. This equation is equivalent to a set of two equations: square 5 x 2 −7 x−1=0 and linear x−2=0. Using the formula for the roots of a quadratic equation, we find two roots, and from the second equation we have x=2.

Checking whether the denominator goes to zero at the found values ​​of x is quite unpleasant. And determining the range of permissible values ​​of the variable x in the original equation is quite simple. Therefore, we will act through ODZ.

In our case, the ODZ of the variable x of the original fractional rational equation consists of all numbers except those for which the condition x 2 +5·x−14=0 is satisfied. The roots of this quadratic equation are x=−7 and x=2, from which we draw a conclusion about the ODZ: it consists of all x such that .

It remains to check whether the found roots and x=2 belong to the range of acceptable values. The roots belong, therefore, they are roots of the original equation, and x=2 does not belong, therefore, it is an extraneous root.

Answer:

It will also be useful to separately dwell on the cases when in a fractional rational equation of the form there is a number in the numerator, that is, when p(x) is represented by some number. Wherein

• if this number is non-zero, then the equation has no roots, since a fraction is equal to zero if and only if its numerator is equal to zero;
• if this number is zero, then the root of the equation is any number from the ODZ.

Example.

Solution.

Since the numerator of the fraction on the left side of the equation contains a non-zero number, then for any x the value of this fraction cannot be equal to zero. Therefore, this equation has no roots.

Answer:

no roots.

Example.

Solve the equation.

Solution.

The numerator of the fraction on the left side of this fractional rational equation contains zero, so the value of this fraction is zero for any x for which it makes sense. In other words, the solution to this equation is any value of x from the ODZ of this variable.

It remains to determine this range of acceptable values. It includes all values ​​of x for which x 4 +5 x 3 ≠0. The solutions to the equation x 4 +5 x 3 =0 are 0 and −5, since this equation is equivalent to the equation x 3 (x+5)=0, and it in turn is equivalent to the combination of two equations x 3 =0 and x +5=0, from where these roots are visible. Therefore, the desired range of acceptable values ​​is any x except x=0 and x=−5.

Thus, a fractional rational equation has infinitely many solutions, which are any numbers except zero and minus five.

Answer:

Finally, it's time to talk about solving fractional rational equations of arbitrary form. They can be written as r(x)=s(x), where r(x) and s(x) are rational expressions, and at least one of them is fractional. Looking ahead, let's say that their solution comes down to solving equations of the form already familiar to us.

It is known that transferring a term from one part of the equation to another with the opposite sign leads to an equivalent equation, therefore the equation r(x)=s(x) is equivalent to the equation r(x)−s(x)=0.

We also know that any , identically equal to this expression, is possible. Thus, we can always transform the rational expression on the left side of the equation r(x)−s(x)=0 into an identically equal rational fraction of the form .

So we move from the original fractional rational equation r(x)=s(x) to the equation, and its solution, as we found out above, reduces to solving the equation p(x)=0.

But here it is necessary to take into account the fact that when replacing r(x)−s(x)=0 with , and then with p(x)=0, the range of permissible values ​​of the variable x may expand.

Consequently, the original equation r(x)=s(x) and the equation p(x)=0 that we arrived at may turn out to be unequal, and by solving the equation p(x)=0, we can get roots that will be extraneous roots of the original equation r(x)=s(x) . You can identify and not include extraneous roots in the answer either by performing a check or by checking that they belong to the ODZ of the original equation.

Let's summarize this information in algorithm for solving fractional rational equation r(x)=s(x). To solve the fractional rational equation r(x)=s(x) , you need

• Get zero on the right by moving the expression from the right side with the opposite sign.
• Perform operations with fractions and polynomials on the left side of the equation, thereby transforming it into a rational fraction of the form.
• Solve the equation p(x)=0.
• Identify and eliminate extraneous roots, which is done by substituting them into the original equation or by checking their belonging to the ODZ of the original equation.

For greater clarity, we will show the entire chain of solving fractional rational equations:
.

Let's look at the solutions of several examples with a detailed explanation of the solution process in order to clarify the given block of information.

Example.

Solve a fractional rational equation.

Solution.

We will act in accordance with the solution algorithm just obtained. And first we move the terms from the right side of the equation to the left, as a result we move on to the equation.

In the second step, we need to convert the fractional rational expression on the left side of the resulting equation to the form of a fraction. To do this, we reduce rational fractions to a common denominator and simplify the resulting expression: . So we come to the equation.

In the next step, we need to solve the equation −2·x−1=0. We find x=−1/2.

It remains to check whether the found number −1/2 is not an extraneous root of the original equation. To do this, you can check or find the VA of the variable x of the original equation. Let's demonstrate both approaches.

Let's start with checking. We substitute the number −1/2 into the original equation instead of the variable x, and we get the same thing, −1=−1. The substitution gives the correct numerical equality, so x=−1/2 is the root of the original equation.

Now we will show how the last point of the algorithm is performed through ODZ. The range of permissible values ​​of the original equation is the set of all numbers except −1 and 0 (at x=−1 and x=0 the denominators of the fractions vanish). The root x=−1/2 found in the previous step belongs to the ODZ, therefore, x=−1/2 is the root of the original equation.

Answer:

−1/2 .

Let's look at another example.

Example.

Find the roots of the equation.

Solution.

We need to solve a fractional rational equation, let's go through all the steps of the algorithm.

First, we move the term from the right side to the left, we get .

Secondly, we transform the expression formed on the left side: . As a result, we arrive at the equation x=0.

Its root is obvious - it is zero.

At the fourth step, it remains to find out whether the found root is extraneous to the original fractional rational equation. When it is substituted into the original equation, the expression is obtained. Obviously, it doesn't make sense because it contains division by zero. Whence we conclude that 0 is an extraneous root. Therefore, the original equation has no roots.

7, which leads to Eq. From this we can conclude that the expression in the denominator of the left side must be equal to that of the right side, that is, . Now we subtract from both sides of the triple: . By analogy, from where, and further.

The check shows that both roots found are roots of the original fractional rational equation.

Answer:

Bibliography.

• Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Mordkovich A. G. Algebra. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
• Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.

You have already met in the 7th grade algebra course, but these were only systems of a special type - systems of two linear equations with two variables. In the 8th grade, you learned to solve rational equations with one variable, which means you can think about solving systems of rational equations with two variables, especially since such systems quite often represent mathematical models of the situations being studied. You already learned about one of these models from the Algebra-8 textbook. The example below is taken from the referenced textbook.

In practice, a broader interpretation of the term “rational equation with two variables” is more convenient: this is an equation of the form - rational expressions with two variables x and y.
Examples of rational equations with two variables:

Of course, you can consider rational equations with other variables, not necessarily with x, for example, a3 - bx = 3ab - a rational equation with two variables a, b. But according to tradition, in algebra they prefer to use the letters x and y as variables.

Definition 2.

A solution to the equation p (x, y) = 0 is any pair of numbers (x; y) that satisfies this equation, i.e. turns equality with variables p (x, y) = 0 into a true numerical equality.

For example:

1) (3; 7) - solution to the equation x 2 + y 2 = 58. Indeed, 3 2 + 7 2 = 58 is a correct numerical equality.
2) - solution to the equation x 2 + y 2 - 58. In fact, - correct numerical equality (22 + 36 = 58).

3) (0; 5) - solution to the equation 2xy + x 3 = 0. Indeed, 2 0 5 + 0+ O 2 = 0 is a correct numerical equality.
4) (1; 2) is not a solution to the equation 2xy + x 3 = 0. In fact, 2 1 2 + 3 = 0 is an incorrect equality (it turns out 5 = 0).

For equations with two variables, as well as for equations with one variable, we can introduce the concept of equivalence of equations.

Definition 3.

Two equations p(x, y) = 0 and d(x, y) = 0 are called equivalent if they have the same solutions (in particular, if both equations have no solutions).

Usually, when solving an equation, they try to replace this equation with a simpler one, but equivalent to it. Such a replacement is called an equivalent transformation of the equation. The two main equivalent conversions are listed below:

1) Transferring terms of the equation from one part of the equation to another with opposite signs.
For example, replacing the equation 2x + bу = 7x - 8у with the equation 2x - 7x - -8у - bу is an equivalent transformation of the equation.
2) Multiplying or dividing both sides of an equation by the same non-zero number or expression.
For example, replacing the equation 0.5l:2 - 0.3xy = 2y with the equation 5l:2 - 3xy = 20y (both sides of the equation were multiplied term by term by 10) is an equivalent transformation of the equation.

Inequivalent transformations of the equation, as in the case of equations with one variable, are:

1) Liberation from denominators containing variables.
2) Squaring both sides of the equation.

If one of the indicated non-equivalent transformations was used in the process of solving the equation, then all the solutions found must be checked by substitution into the original equation, since there may be extraneous solutions among them.

Sometimes it is possible to switch to a geometric (graphical) model of an equation with two variables, i.e. graph the equation. You probably remember that the graph of a linear equation with two variables ax + bу + c = 0 (a, b, c are numbers, coefficients, where at least one of the numbers a, b is different from zero) is a straight line - a geometric model linear equation. Let's try to find the corresponding graphical models for some more rational equations with two variables x and y.

Example 2. Draw a graph of the equation y - 2x2 = 0.

Solution. Let's transform the equation to the form y = 2x2. The graph of the function y - 2x2 is a parabola, which is also considered the graph of the equation y - 2x2 = 0 (Fig. 33).

Example 3. Graph the equation xy = 2.
Solution. Let's transform the equation to the form The graph of the function - is a hyperbola, it is also considered the graph of the equation xy = 2 (Fig. 34).

Thus, if the equation p(x, y) = O can be transformed to the form y = f (x), then the graph of the function y - f (x) is considered at the same time to be the graph of the equation p(x, y) - 0.

Example 4. Graph the equation x 2 + y 2 = 16.

Solution.

Let's use a theorem from the geometry course: the graph of the equation x 2 + y 2 = r 2, where r is a positive number, is a circle with a center at the origin and radius r. This means that the graph of the equation x 2 + y 2 = 16 is a circle with a center at the origin and radius 4 (Fig. 35).

The theorem mentioned above is a special case of the following theorem, which we hope is also known to you from your geometry course.

Example 5. Graph the equation:

a) (x - I) 2 + (y - 2) 2 = 9; b) x 2 + y 2 + 4x = 0.

Solution:

a) Let's rewrite the equation in the form (x - I) 2 + (y - 2) 2 = 32. The graph of this equation, according to the theorem, is a circle with a center at point (1; 2) and radius 3 (Fig. 37).

b) Let's rewrite the equation in the form (x 2 + 4x + 4) + y 2 = 4, i.e. (x + 2) 2 + y 2 = 4 and further (x - (-2)) 2 + (y - O) 2 = 22. The graph of this equation, according to the theorem, is a circle with the center at the point (-2; 0 ) and radius 2 (Fig. 38).

Definition 4.

If the task is to find pairs of values ​​(x; y) that simultaneously satisfy the equation p (x, y) = 0 and the equation q (x, y) = 0, then they say that these equations form a system of equations:

A pair of values ​​(x; y), which is simultaneously a solution to both the first and second equations of the system, is called a solution to the system of equations. Solving a system of equations means finding all its solutions or establishing that there are no solutions.
For example, pair (3; 7) - solution to the system of equations

In fact, this pair satisfies both the first and second equations of the system, which means it is its solution. Usually it is written like this: (3; 7) - a solution to the system or A pair (5; 9) is not a solution to system (1): it does not satisfy the first equation (although it satisfies the second equation of the system).

Of course, the variables in the equations that form a system of equations can be designated by other letters, for example: But in any case, when writing the answer in the form of a pair of numbers, the lexicographic method is used, i.e. The first place is given to the one of the two letters that appears earlier in the Latin alphabet.

Sometimes you can solve a system of equations using a graphical method with which you are familiar: you need to graph the first equation, then graph the second equation, and finally find the intersection points of the graphs; the coordinates of each intersection point serve as a solution to the system of equations.

Example 6. Solve system of equations

Solution.

1) Construct a graph of the equation x 2 + y 2 = 16 - a circle with a center at the origin and radius 4 (Fig. 39).
2) Let's build a graph of the equation y - x = 4. This is a straight line passing through the points (0; 4) and (-4; 0) (Fig. 39).
3) The circle and the straight line intersect at points A and B (Fig. 39). Judging by the constructed geometric model, point A has coordinates A(-4; 0), and point B has coordinates B(0; 4). The check shows that in fact the pair (-4; 0) and the pair (0; 4) are solutions to both equations of the system, and therefore solutions to the system of equations. Consequently, the given system of equations has two solutions: (-4; 0) and (0; 4).

Answer: (-4; 0); (0; 4).

Example 7. Solve system of equations

Solution.

1) Having rewritten the first equation of the system in the form y = 2x 2, we come to the conclusion: the graph of the equation is a parabola (Fig. 40).
2) Having rewritten the second equation of the system in the form we come to the conclusion: the graph of the equation is a hyperbola (Fig. 40).

3) The parabola and hyperbola intersect at point A (Fig. 40). Judging by the constructed geometric model, point A has coordinates A (1; 2). Checking shows that indeed the pair (1; 2) is a solution to both equations of the system, and therefore a solution to the system of equations. Consequently, the given system of equations has one solution: (1; 2).

Answer: (1; 2).

The graphical method for solving systems of equations, like the graphical method for solving equations, is beautiful, but unreliable: firstly, because we will not always be able to construct graphs of equations; secondly, even if it was possible to construct graphs of the equations, the intersection points may not be as “good” as in specially selected examples 6 and 7, and may even be outside the boundaries of the drawing. This means that we need to have reliable algebraic methods for solving systems of two equations in two variables. This will be discussed in the next paragraph.

A.G. Mordkovich Algebra 9th grade

Online math materials, problems and answers by grade, math lesson plans

The lowest common denominator is used to simplify this equation. This method is used when you cannot write a given equation with one rational expression on each side of the equation (and use the crisscross method of multiplication). This method is used when you are given a rational equation with 3 or more fractions (in the case of two fractions, it is better to use criss-cross multiplication).

Find the lowest common denominator of the fractions (or least common multiple). NOZ is the smallest number that is evenly divisible by each denominator.

• Sometimes NPD is an obvious number. For example, if given the equation: x/3 + 1/2 = (3x +1)/6, then it is obvious that the least common multiple of the numbers 3, 2 and 6 is 6.
• If the NCD is not obvious, write down the multiples of the largest denominator and find among them one that will be a multiple of the other denominators. Often the NOD can be found by simply multiplying two denominators. For example, if the equation is given x/8 + 2/6 = (x - 3)/9, then NOS = 8*9 = 72.
• If one or more denominators contain a variable, the process becomes somewhat more complicated (but not impossible). In this case, the NOC is an expression (containing a variable) that is divided by each denominator. For example, in the equation 5/(x-1) = 1/x + 2/(3x) NOZ = 3x(x-1), because this expression is divided by each denominator: 3x(x-1)/(x-1 ) = 3x; 3x(x-1)/3x = (x-1); 3x(x-1)/x = 3(x-1).

Multiply both the numerator and denominator of each fraction by a number equal to the result of dividing the NOC by the corresponding denominator of each fraction. Since you are multiplying both the numerator and denominator by the same number, you are effectively multiplying the fraction by 1 (for example, 2/2 = 1 or 3/3 = 1).

• So in our example, multiply x/3 by 2/2 to get 2x/6, and 1/2 multiply by 3/3 to get 3/6 (the fraction 3x +1/6 does not need to be multiplied because it the denominator is 6).
• Proceed similarly when the variable is in the denominator. In our second example, NOZ = 3x(x-1), so multiply 5/(x-1) by (3x)/(3x) to get 5(3x)/(3x)(x-1); 1/x multiplied by 3(x-1)/3(x-1) and you get 3(x-1)/3x(x-1); 2/(3x) multiplied by (x-1)/(x-1) and you get 2(x-1)/3x(x-1).

Find x. Now that you have reduced the fractions to a common denominator, you can get rid of the denominator. To do this, multiply each side of the equation by the common denominator. Then solve the resulting equation, that is, find “x”. To do this, isolate the variable on one side of the equation.

• In our example: 2x/6 + 3/6 = (3x +1)/6. You can add 2 fractions with the same denominator, so write the equation as: (2x+3)/6=(3x+1)/6. Multiply both sides of the equation by 6 and get rid of the denominators: 2x+3 = 3x +1. Solve and get x = 2.
• In our second example (with a variable in the denominator), the equation looks like (after reduction to a common denominator): 5(3x)/(3x)(x-1) = 3(x-1)/3x(x-1) + 2 (x-1)/3x(x-1). By multiplying both sides of the equation by N3, you get rid of the denominator and get: 5(3x) = 3(x-1) + 2(x-1), or 15x = 3x - 3 + 2x -2, or 15x = x - 5 Solve and get: x = -5/14.