An equation is an equality containing a letter whose value must be found.

In equations, the unknown is usually represented by a lowercase letter. The most commonly used letters are “x” [ix] and “y” [y].

Having solved the equation, we always write down a check after the answer.

Information for parents

Dear parents, we draw your attention to the fact that primary school and in 5th grade, children DO NOT know the topic “Negative Numbers”.

Therefore, they must solve equations using only the properties of addition, subtraction, multiplication, and division. Methods for solving equations for grade 5 are given below.

Do not try to explain the solution of equations by transferring numbers and letters from one part of the equation to another with a change in sign.

You can brush up on concepts related to addition, subtraction, multiplication and division in the lesson “Laws of Arithmetic”.

Solving addition and subtraction equations

How to find the unknown
term

How to find the unknown
minuend

How to find the unknown
subtrahend

To find the unknown term, you need to subtract the known term from the sum.

To find the unknown minuend, you need to add the subtrahend to the difference.

To find the unknown subtrahend, you need to subtract the difference from the minuend.

x + 9 = 15
x = 15 − 9
x=6
Examination

x − 14 = 2
x = 14 + 2
x = 16
Examination

16 − 2 = 14
14 = 14

5 − x = 3
x = 5 − 3
x = 2
Examination

Solving multiplication and division equations

How to find an unknown
factor

How to find the unknown
dividend

How to find an unknown
divider

To find an unknown factor, you need to divide the product by the known factor.

To find the unknown dividend, you need to multiply the quotient by the divisor.

To find an unknown divisor, you need to divide the dividend by the quotient.

y 4 = 12
y=12:4
y=3
Examination

y: 7 = 2
y = 2 7
y=14
Examination

8:y=4
y=8:4
y=2
Examination

An equation is an equality containing a letter whose sign must be found. The solution to an equation is the set of letter values ​​that turns the equation into a true equality:

Recall that to solve equation you need to transfer the terms with the unknown to one part of the equality, and the numerical terms to the other, bring similar ones and get the following equality:

From the last equality we determine the unknown according to the rule: “one of the factors is equal to the quotient divided by the second factor.”

Since rational numbers a and b can have the same and different signs, then the sign of the unknown is determined by the rules for dividing rational numbers.

Procedure for solving linear equations

The linear equation must be simplified by opening the brackets and performing the second step operations (multiplication and division).

Move the unknowns to one side of the equal sign, and the numbers to the other side of the equal sign, obtaining an equality identical to the given one,

Bring similar ones to the left and right of the equal sign, obtaining an equality of the form ax = b.

Calculate the root of the equation (find the unknown X from equality x = b : a),

Check by substituting the unknown into the given equation.

If we obtain an identity in a numerical equality, then the equation is solved correctly.

Special cases of solving equations

1. If the equation given a product equal to 0, then to solve it we use the property of multiplication: “the product is equal to zero if one of the factors or both factors are equal to zero.”

27 (x - 3) = 0
27 is not equal to 0, which means x - 3 = 0

The second example has two solutions to the equation, since
this is a second degree equation:

If the coefficients of the equation are ordinary fractions, then first of all we need to get rid of the denominators. For this:

Find the common denominator;

Determine additional factors for each term of the equation;

Multiply the numerators of fractions and integers by additional factors and write all terms of the equation without denominators (the common denominator can be discarded);

Move the terms with unknowns to one side of the equation, and the numerical terms to the other from the equal sign, obtaining an equivalent equality;

Bring similar members;

Basic properties of equations

In any part of the equation, you can add similar terms or open a parenthesis.

Any term of the equation can be transferred from one part of the equation to another by changing its sign to the opposite.

Both sides of the equation can be multiplied (divided) by the same number, except 0.

In the example above, all its properties were used to solve the equation.

How to solve an equation with an unknown in a fraction

Sometimes linear equations take the form when unknown appears in the numerator of one or more fractions. Like in the equation below.

In such cases, such equations can be solved in two ways.

I solution method
Reducing an equation to a proportion

When solving equations using the proportion method, you must perform the following steps:

So let's go back to our equation. On the left side we already have only one fraction, so no transformations are needed in it.

We will work with the right side of the equation. Let's simplify right side equations so that only one fraction remains. To do this, remember the rules for adding a number with an algebraic fraction.

Now we use the rule of proportion and solve the equation to the end.

II method of solution
Reduction to a linear equation without fractions

Let's look at the equation above again and solve it in a different way.

We see that there are two fractions in the equation "

How to solve equations with fractions. Exponential solution of equations with fractions.

Solving equations with fractions Let's look at examples. The examples are simple and illustrative. With their help, you will be able to understand in the most understandable way.
For example, you need to solve the simple equation x/b + c = d.

An equation of this type is called linear, because The denominator contains only numbers.

The solution is performed by multiplying both sides of the equation by b, then the equation takes the form x = b*(d – c), i.e. the denominator of the fraction on the left side cancels.

For example, how to solve a fractional equation:
x/5+4=9
We multiply both sides by 5. We get:
x+20=45

Another example when the unknown is in the denominator:

Equations of this type are called fractional-rational or simply fractional.

We would solve a fractional equation by getting rid of fractions, after which this equation, most often, turns into a linear or quadratic equation, which is solved in the usual way. You just need to consider the following points:

• the value of a variable that turns the denominator to 0 cannot be a root;
• You cannot divide or multiply an equation by the expression =0.

This is where the concept of the region of permissible values ​​(ADV) comes into force - these are the values ​​of the roots of the equation for which the equation makes sense.

Thus, when solving the equation, it is necessary to find the roots, and then check them for compliance with the ODZ. Those roots that do not correspond to our ODZ are excluded from the answer.

For example, you need to solve a fractional equation:

Based on the above rule, x cannot be = 0, i.e. ODZ in this case: x – any value other than zero.

We get rid of the denominator by multiplying all terms of the equation by x

And we solve the usual equation

5x – 2x = 1
3x = 1
x = 1/3

Let's solve a more complicated equation:

ODZ is also present here: x -2.

When solving this equation, we will not move everything to one side and bring the fractions to a common denominator. We will immediately multiply both sides of the equation by an expression that will cancel out all the denominators at once.

To reduce the denominators, you need to multiply the left side by x+2, and the right side by 2. This means that both sides of the equation must be multiplied by 2(x+2):

This is the most common multiplication of fractions, which we have already discussed above.

Let's write the same equation, but slightly differently

The left side is reduced by (x+2), and the right by 2. After the reduction, we obtain the usual linear equation:

x = 4 – 2 = 2, which corresponds to our ODZ

Solving equations with fractions not as difficult as it might seem. In this article we have shown this with examples. If you have any difficulties with how to solve equations with fractions, then unsubscribe in the comments.

Solving equations with fractions grade 5

Solving equations with fractions. Solving fraction problems.

View document contents
“Solving equations with fractions, grade 5”

— Addition of fractions with the same denominators.

— Subtraction of fractions with the same denominators.

Adding fractions with like denominators.

To add fractions with the same denominators, you need to add their numerators and leave the denominator the same.

Subtracting fractions with like denominators.

To subtract fractions with the same denominators, you need to subtract the numerator of the minuend from the numerator of the minuend, but leave the denominator the same.

When solving equations, it is necessary to use the rules for solving equations, the properties of addition and subtraction.

Solving equations using properties.

Solving equations using rules.

The expression on the left side of the equation is the sum.

term + term = sum.

To find the unknown term, you need to subtract the known term from the sum.

minuend – subtrahend = difference

To find the unknown subtrahend, you need to subtract the difference from the minuend.

The expression on the left side of the equation is the difference.

To find the unknown minuend, you need to add the subtrahend to the difference.

USING RULES FOR SOLVING EQUATIONS.

On the left side of the equation, the expression is the sum.

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).

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 known at this point linear equations And quadratic equations 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 doing 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, the solution to the original integer equation is reduced to the solution 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 the equation has two real roots, which we find by 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 more high degrees Often you have 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 entire 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 theorem converse 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 root formula for 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 arbitrary type. 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.
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