Equations with fractions, solution rules. "solving fractional rational equations"

So far we have only solved integer equations with respect to the unknown, that is, equations in which the denominators (if any) did not contain the unknown.

Often you have to solve equations that contain an unknown in the denominators: such equations are called fractional equations.

To solve this equation, we multiply both sides by that is, by the polynomial containing the unknown. Will the new equation be equivalent to this one? To answer the question, let's solve this equation.

Multiplying both sides by , we get:

Solving this equation of the first degree, we find:

So, equation (2) has a single root

Substituting it into equation (1), we get:

This means that it is also a root of equation (1).

Equation (1) has no other roots. In our example, this can be seen, for example, from the fact that in equation (1)

How the unknown divisor must be equal to the dividend 1 divided by the quotient 2, that is

So, equations (1) and (2) have a single root. This means they are equivalent.

2. Let us now solve the following equation:

The simplest common denominator: ; multiply all terms of the equation by it:

After reduction we get:

Let's expand the brackets:

Having brought similar members, will have:

Solving this equation, we find:

Substituting into equation (1), we get:

On the left side we received expressions that do not make sense.

This means that equation (1) is not a root. It follows that equations (1) and are not equivalent.

In this case, they say that equation (1) has acquired an extraneous root.

Let us compare the solution of equation (1) with the solution of the equations we considered earlier (see § 51). In solving this equation, we had to perform two operations that had not been encountered before: first, we multiplied both sides of the equation by an expression containing the unknown (common denominator), and second, we canceled algebraic fractions into factors containing the unknown.

Comparing equation (1) with equation (2), we see that not all values ​​of x that are valid for equation (2) are valid for equation (1).

It is the numbers 1 and 3 that are not acceptable values ​​of the unknown for equation (1), but as a result of the transformation they became acceptable for equation (2). One of these numbers turned out to be a solution to equation (2), but, of course, it cannot be a solution to equation (1). Equation (1) has no solutions.

This example shows that when both sides of an equation are multiplied by a factor containing an unknown, and when algebraic fractions are reduced, an equation may be obtained that is not equivalent to the given one, namely: extraneous roots may appear.

From here we draw the following conclusion. When solving an equation containing an unknown in the denominator, the resulting roots must be checked by substitution into the original equation. Extraneous roots must be discarded.

Application

Solving any type of equations online on the site for students and schoolchildren to consolidate the studied material. Solving equations online. Equations online. There are algebraic, parametric, transcendental, functional, differential and other types of equations. Some classes of equations have analytical solutions, which are convenient because they not only give exact value root, but allow you to write the solution in the form of a formula, which may include parameters. Analytical expressions allow not only to calculate the roots, but also to analyze their existence and their quantity depending on the parameter values, which is often even more important for practical application, than the specific values ​​of the roots. Solving equations online.. Equations online. Solving an equation is the task of finding such values ​​of the arguments at which this equality is achieved. On possible values arguments can be imposed additional conditions(integer, real, etc.). 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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].

Root of the equation- this is the value of the letter at which the correct numerical equality is obtained from the equation.

Solve the equation- means to find all its roots or make sure that there are no roots.

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.”

Because 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:

bring all the fractions to a common denominator and add them as algebraic fractions (only one fraction should remain on the left and right sides);

Solve the resulting equation using the rule of proportion.

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 can be 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 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.

Equations containing a variable in the denominator can be solved in two ways:

Reducing fractions to a common denominator

Using the basic property of proportion

Regardless of the chosen method, after finding the roots of the equation, it is necessary to select from the found valid values, that is, those that do not turn the denominator to $0$.

1 way. Reducing fractions to a common denominator.

Example 1

$\frac(2x+3)(2x-1)=\frac(x-5)(x+3)$

Solution:

1. Let's transfer the fraction from the right side of the equation to the left

\[\frac(2x+3)(2x-1)-\frac(x-5)(x+3)=0\]

In order to do this correctly, remember that when moving elements to another part of the equation, the sign in front of the expressions changes to the opposite. This means that if there was a “+” sign in front of the fraction on the right side, then there will be a “-” sign in front of it on the left side. Then on the left side we get the difference of the fractions.

2. Now let us note that the fractions have different denominators, which means that in order to make up the difference it is necessary to bring the fractions to a common denominator. The common denominator will be the product of the polynomials in the denominators of the original fractions: $(2x-1)(x+3)$

In order to obtain an identical expression, the numerator and denominator of the first fraction must be multiplied by the polynomial $(x+3)$, and the second by the polynomial $(2x-1)$.

\[\frac((2x+3)(x+3))((2x-1)(x+3))-\frac((x-5)(2x-1))((x+3)( 2x-1))=0\]

Let's perform a transformation in the numerator of the first fraction - multiply polynomials. Let us remember that for this it is necessary to multiply the first term of the first polynomial by each term of the second polynomial, then multiply the second term of the first polynomial by each term of the second polynomial and add the results

\[\left(2x+3\right)\left(x+3\right)=2x\cdot x+2x\cdot 3+3\cdot x+3\cdot 3=(2x)^2+6x+3x +9\]

Let us present similar terms in the resulting expression

\[\left(2x+3\right)\left(x+3\right)=2x\cdot x+2x\cdot 3+3\cdot x+3\cdot 3=(2x)^2+6x+3x +9=\] \[(=2x)^2+9x+9\]

Let's carry out a similar transformation in the numerator of the second fraction - multiply polynomials

$\left(x-5\right)\left(2х-1\right)=х\cdot 2х-х\cdot 1-5\cdot 2х+5\cdot 1=(2х)^2-х-10х+ 5=(2x)^2-11x+5$

Then the equation will take the form:

\[\frac((2x)^2+9x+9)((2x-1)(x+3))-\frac((2x)^2-11x+5)((x+3)(2x- 1))=0\]

Now fractions with same denominator, which means you can subtract. Recall that when subtracting fractions with the same denominator from the numerator of the first fraction, you must subtract the numerator of the second fraction, leaving the denominator the same

\[\frac((2x)^2+9x+9-((2x)^2-11x+5))((2x-1)(x+3))=0\]

Let's transform the expression into the numerator. In order to open brackets preceded by a “-” sign, you need to change all the signs in front of the terms in brackets to the opposite

\[(2x)^2+9x+9-\left((2x)^2-11x+5\right)=(2x)^2+9x+9-(2x)^2+11x-5\]

Let us present similar terms

$(2x)^2+9x+9-\left((2x)^2-11x+5\right)=(2x)^2+9x+9-(2x)^2+11x-5=20x+4 $

Then the fraction will take the form

\[\frac((\rm 20x+4))((2x-1)(x+3))=0\]

3. A fraction is equal to $0$ if its numerator is 0. Therefore, we equate the numerator of the fraction to $0$.

\[(\rm 20x+4=0)\]

Let's solve the linear equation:

4. Let's sample the roots. This means that it is necessary to check whether the denominators of the original fractions turn to $0$ when the roots are found.

Let us set the condition that the denominators are not equal to $0$

x$\ne 0.5$ x$\ne -3$

This means that all variable values ​​are acceptable except $-3$ and $0.5$.

The root we found is valid value, which means it can safely be considered the root of the equation. If the found root were not a valid value, then such a root would be extraneous and, of course, would not be included in the answer.

Answer:$-0,2.$

Now we can create an algorithm for solving an equation that contains a variable in the denominator

Algorithm for solving an equation that contains a variable in the denominator

Move all elements from the right side of the equation to the left. To obtain an identical equation, it is necessary to change all the signs in front of the expressions on the right side to the opposite

If on the left side we get an expression with different denominators, then we bring them to a common value using the basic property of a fraction. Perform transformations using identity transformations and obtain a final fraction equal to $0$.

Equate the numerator to $0$ and find the roots of the resulting equation.

Let's sample the roots, i.e. find valid values ​​of variables that do not make the denominator $0$.

Method 2. We use the basic property of proportion

The main property of proportion is that the product of the extreme terms of the proportion is equal to the product of the middle terms.

Example 2

We use this property to solve this task

\[\frac(2x+3)(2x-1)=\frac(x-5)(x+3)\]

1. Let’s find and equate the product of the extreme and middle terms of the proportion.

$\left(2x+3\right)\cdot(\ x+3)=\left(x-5\right)\cdot(2x-1)$

\[(2x)^2+3x+6x+9=(2x)^2-10x-x+5\]

Having solved the resulting equation, we will find the roots of the original

2. Let's find the acceptable values ​​of the variable.

From the previous solution (method 1) we have already found that any values ​​are acceptable except $-3$ and $0.5$.

Then, having established that the found root is a valid value, we found out that $-0.2$ will be the root.

Lesson objectives:

Educational:

• formation of the concept of fractional rational equations;
• consider different ways to solve fractions rational equations;
• consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero;
• teach solving fractional rational equations using an algorithm;
• checking the level of mastery of the topic by conducting a test.

Developmental:

• developing the ability to correctly operate with acquired knowledge and think logically;
• development of intellectual skills and mental operations - analysis, synthesis, comparison and generalization;
• development of initiative, the ability to make decisions, and not stop there;
• development of critical thinking;
• development of research skills.

Educating:

• fostering cognitive interest in the subject;
• fostering independence in solving educational problems;
• nurturing will and perseverance to achieve final results.

Lesson type: lesson - explanation of new material.

During the classes

1. Organizational moment.

Hello guys! There are equations written on the board, look at them carefully. Can you solve all of these equations? Which ones are not and why?

Equations in which the left and right sides are fractional rational expressions are called fractional rational equations. What do you think we will study in class today? Formulate the topic of the lesson. So, open your notebooks and write down the topic of the lesson “Solving fractional rational equations.”

2. Updating knowledge. Frontal survey, oral work with the class.

And now we will repeat the main theoretical material that we need to study new topic. Please answer the following questions:

1. What is an equation? ( Equality with a variable or variables.)
2. What is the name of equation number 1? ( Linear.) Solution linear equations. (Move everything with the unknown to the left side of the equation, all numbers to the right. Give similar terms. Find unknown factor).
3. What is the name of equation number 3? ( Square.) Solutions quadratic equations. (Isolating a complete square using formulas using Vieta’s theorem and its corollaries.)
4. What is proportion? ( Equality of two ratios.) The main property of proportion. ( If the proportion is correct, then the product of its extreme terms is equal to the product of the middle terms.)
5. What properties are used when solving equations? ( 1. If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one. 2. If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.)
6. When does a fraction equal zero? ( A fraction is equal to zero when the numerator is zero and the denominator is not zero..)

3. Explanation of new material.

Solve equation No. 2 in your notebooks and on the board.

Answer: 10.

Which fractional rational equation Can you try to solve using the basic property of proportion? (No. 5).

(x-2)(x-4) = (x+2)(x+3)

x 2 -4x-2x+8 = x 2 +3x+2x+6

x 2 -6x-x 2 -5x = 6-8

Solve equation No. 4 in your notebooks and on the board.

Answer: 1,5.

What fractional rational equation can you try to solve by multiplying both sides of the equation by the denominator? (No. 6).

x 2 -7x+12 = 0

D=1›0, x 1 =3, x 2 =4.

Answer: 3;4.

Now try to solve equation number 7 using one of the following methods.

(x 2 -2x-5)x(x-5)=x(x-5)(x+5)
(x 2 -2x-5)x(x-5)-x(x-5)(x+5)=0			x 2 -2x-5=x+5
x(x-5)(x 2 -2x-5-(x+5))=0			x 2 -2x-5-x-5=0
x(x-5)(x 2 -3x-10)=0
x=0 x-5=0 x 2 -3x-10=0
x 1 =0 x 2 =5 D=49
x 3 =5 x 4 =-2			x 3 =5 x 4 =-2
Answer: 0;5;-2.			Answer: 5;-2.

Explain why this happened? Why are there three roots in one case and two in the other? What numbers are the roots of this fractional rational equation?

Until now, students have not encountered the concept of an extraneous root; it is indeed very difficult for them to understand why this happened. If no one in the class can give a clear explanation of this situation, then the teacher asks leading questions.

• How do equations No. 2 and 4 differ from equations No. 5,6,7? ( In equations No. 2 and 4 there are numbers in the denominator, No. 5-7 are expressions with a variable.)
• What is the root of an equation? ( The value of the variable at which the equation becomes true.)
• How to find out if a number is the root of an equation? ( Make a check.)

When testing, some students notice that they have to divide by zero. They conclude that the numbers 0 and 5 are not the roots of this equation. The question arises: is there a way to solve fractional rational equations that allows us to eliminate this error? Yes, this method is based on the condition that the fraction is equal to zero.

x 2 -3x-10=0, D=49, x 1 =5, x 2 =-2.

If x=5, then x(x-5)=0, which means 5 is an extraneous root.

If x=-2, then x(x-5)≠0.

Answer: -2.

Let's try to formulate an algorithm for solving fractional rational equations in this way. Children formulate the algorithm themselves.

Algorithm for solving fractional rational equations:

1. Move everything to the left side.
2. Reduce fractions to a common denominator.
3. Create a system: a fraction is equal to zero when the numerator is equal to zero and the denominator is not equal to zero.
4. Solve the equation.
5. Check inequality to exclude extraneous roots.
6. Write down the answer.

Discussion: how to formalize the solution if you use the basic property of proportion and multiplying both sides of the equation by a common denominator. (Add to the solution: exclude from its roots those that make the common denominator vanish).

4. Initial comprehension of new material.

Work in pairs. Students choose how to solve the equation themselves depending on the type of equation. Assignments from the textbook “Algebra 8”, Yu.N. Makarychev, 2007: No. 600(b,c,i); No. 601(a,e,g). The teacher monitors the completion of the task, answers any questions that arise, and provides assistance to low-performing students. Self-test: answers are written on the board.

b) 2 – extraneous root. Answer: 3.

c) 2 – extraneous root. Answer: 1.5.

a) Answer: -12.5.

g) Answer: 1;1.5.

5. Setting homework.

1. Read paragraph 25 from the textbook, analyze examples 1-3.
2. Learn an algorithm for solving fractional rational equations.
3. Solve in notebooks No. 600 (a, d, e); No. 601(g,h).
4. Try to solve No. 696(a) (optional).

6. Completing a control task on the topic studied.

The work is done on pieces of paper.

Example task:

A) Which of the equations are fractional rational?

B) A fraction is equal to zero when the numerator is ______________________ and the denominator is _______________________.

Q) Is the number -3 the root of equation number 6?

D) Solve equation No. 7.

Assessment criteria for the assignment:

• “5” is given if the student completed more than 90% of the task correctly.
• "4" - 75%-89%
• "3" - 50%-74%
• “2” is given to a student who has completed less than 50% of the task.
• A rating of 2 is not given in the journal, 3 is optional.

7. Reflection.

On the independent work sheets, write:

• 1 – if the lesson was interesting and understandable to you;
• 2 – interesting, but not clear;
• 3 – not interesting, but understandable;
• 4 – not interesting, not clear.

8. Summing up the lesson.

So, today in the lesson we got acquainted with fractional rational equations, learned how to solve these equations different ways, tested their knowledge with the help of a training independent work. You will learn the results of your independent work in the next lesson, and at home you will have the opportunity to consolidate your knowledge.

Which method of solving fractional rational equations, in your opinion, is easier, more accessible, and more rational? Regardless of the method for solving fractional rational equations, what should you remember? What is the “cunning” of fractional rational equations?

Thanks everyone, lesson is over.

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