Solve the equation with fractions 5. Special cases of solving equations

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

1 way. Bringing fractions to a common denominator.

Example 1

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

Solution:

1. Move 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, we recall that when moving elements to another part of the equation, the sign in front of the expressions changes to the opposite. So, if on the right side there was a “+” sign before the fraction, then on the left side there will be a “-” sign in front of it. Then on the left side we get the difference of the fractions.

2. Now we 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 the transformation in the numerator of the first fraction - we will multiply the polynomials. Recall that for this it is necessary to multiply the first term of the first polynomial, multiply 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

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

We present similar terms in the resulting expression

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

Perform a similar transformation in the numerator of the second fraction - we will multiply the polynomials

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

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 the same denominator, so you can subtract. Recall that when subtracting fractions with the same denominator from the numerator of the first fraction, it is necessary to 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 in the numerator. In order to open the brackets preceded by the “-” sign, all signs in front of the terms in brackets must be reversed

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

We present like 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 into $0$ when the roots are found.

We set the condition that the denominators are not equal to $0$

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

This means that all values ​​of the variables are allowed, except for $-3$ and $0.5$.

The root we found is a valid value, so it can be safely 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 write an algorithm for solving an equation that contains a variable in the denominator

An 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 side. 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 one using the main property of the fraction. Perform transformations using identical transformations and get the 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 variable values ​​that do not turn the denominator to $0$.

2 way. Using the basic property of proportion

The main property of a 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 members of the proportion.

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

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

Solving the resulting equation, we find the roots of the original

2. Let's find admissible values ​​of a variable.

From the previous solution (1st way) we have already found that any values ​​are allowed 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.

"Solution of fractional rational equations"

Lesson Objectives:

Tutorial:

formation of the concept of fractional rational equations; to consider various ways of solving fractional rational equations; consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero; to teach the solution of fractional rational equations according to the algorithm; checking the level of assimilation of the topic by conducting test work.

Developing:

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

Nurturing:

education of cognitive interest in the subject; education of independence in decision learning objectives; education of will and perseverance to achieve the final results.

Lesson type: lesson - explanation of new material.

During the classes

1. Organizational moment.

Hello guys! Equations are written on the blackboard, 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 today in the lesson? Formulate the topic of the lesson. So, we open notebooks and write down the topic of the lesson “Solution of fractional rational equations”.

2. Actualization of 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 Equation #1 called? ( Linear.) Method of solution linear equations. (Move everything with the unknown to the left side of the equation, all numbers to the right. Bring like terms. Find the unknown multiplier).

3. What is Equation #3 called? ( Square.) Ways to solve quadratic equations. (Selection of the full square, by formulas, using the Vieta theorem and its consequences.)

4. What is a proportion? ( Equality of two relations.) The main property of proportion. ( If the proportion is true, then the product of its extreme terms is equal to the product of the middle terms.)

5. What properties are used in solving equations? ( 1. If in the equation we transfer the term from one part to another, changing its sign, then we get an equation equivalent to the given one. 2. If both parts of the equation are multiplied or divided by the same non-zero number, then an equation will be obtained that is equivalent to the given.)

6. When is a fraction equal to zero? ( The fraction is zero when the numerator zero, and the denominator is not equal to zero.)

3. Explanation of new material.

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

Answer: 10.

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

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

x2-4x-2x+8 = x2+3x+2x+6

x2-6x-x2-5x = 6-8

Solve equation No. 4 in 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).

D=1>0, x1=3, x2=4.

Answer: 3;4.

Now try to solve equation #7 in one of the ways.

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 met the concept of an extraneous root, it is really 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 in the denominator of the number, No. 5-7 - expressions with a variable.) What is the root of the equation? ( The value of the variable at which the equation becomes a true equality.) How to find out if the number is the root of the equation? ( Make a check.)

When doing a test, 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 given error? Yes, this method is based on the condition that the fraction is equal to zero.

x2-3x-10=0, D=49, x1=5, x2=-2.

If x=5, then x(x-5)=0, so 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 themselves formulate the algorithm.

Algorithm for solving fractional rational equations:

1. Move everything to the left side.

2. Bring fractions to a common denominator.

3. Make a system: the 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 the inequality to exclude extraneous roots.

6. Write down the answer.

Discussion: how to formulate a solution if the basic property of proportion is used and the multiplication of both sides of the equation by a common denominator. (Supplement the solution: exclude from its roots those that turn the common denominator to zero).

4. Primary comprehension of new material.

Work in pairs. Students choose how to solve the equation on their own, depending on the type of equation. Tasks from the textbook "Algebra 8", 2007: No. 000 (b, c, i); No. 000(a, e, g). The teacher controls the performance of the task, answers the questions that have arisen, and provides assistance to poorly performing students. Self-test: Answers are written on the board.

b) 2 is an extraneous root. Answer:3.

c) 2 is an extraneous root. Answer: 1.5.

a) Answer: -12.5.

g) Answer: 1; 1.5.

5. Statement of homework.

2. Learn the algorithm for solving fractional rational equations.

3. Solve in notebooks No. 000 (a, d, e); No. 000(g, h).

4. Try to solve No. 000(a) (optional).

6. Fulfillment of the control task on the studied topic.

The work is done on sheets.

Job example:

A) Which of the equations are fractional rational?

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

Q) Is the number -3 the root of Equation #6?

D) Solve equation No. 7.

Task evaluation criteria:

"5" is given if the student completed more than 90% of the task correctly. "4" - 75% -89% "3" - 50% -74% "2" is given to the student who completed less than 50% of the task. Grade 2 is not put in the journal, 3 is optional.

7. Reflection.

On the leaflets with independent work, put:

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 training independent work. You will learn the results of independent work in the next lesson, at home you will have the opportunity to consolidate the knowledge gained.

What method of solving fractional rational equations, in your opinion, is easier, more accessible, more rational? Regardless of the method of solving fractional rational equations, what should not be forgotten? What is the "cunning" of fractional rational equations?

Thank you all, the lesson is over.

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 the unknown in the denominators: such equations are called fractional.

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

Multiplying both sides of it 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:

Hence, is also the 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 unknown divisor must be equal to the dividend 1 divided by the quotient 2, i.e.

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

2. We now solve the following equation:

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

After reduction we get:

Let's expand the brackets:

Bringing like terms, we have:

Solving this equation, we find:

Substituting into equation (1), we get:

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

Hence, the root of equation (1) is not. This implies that equations (1) and are not equivalent.

In this case, we 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 such operations that had not been seen before: first, we multiplied both sides of the equation by an expression containing the unknown (common denominator), and, second, we reduced algebraic fractions by factors containing the unknown .

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

It is the numbers 1 and 3 that are not admissible values ​​of the unknown for equation (1), and as a result of the transformation they became admissible 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 multiplying both parts of the equation by a factor containing the unknown, and when reducing algebraic fractions, an equation can be obtained that is not equivalent to the given one, namely: extraneous roots can appear.

Hence 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.

Solving equations with fractions let's look at examples. The examples are simple and illustrative. With their help, you can understand in the most understandable way,.
For example, you need to solve a 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 is reduced.

For example, how to solve a fractional equation:
x/5+4=9
We multiply both parts by 5. We get:
x+20=45
x=45-20=25

Another example where 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 one, which is solved in the usual way. You should only take into account the following points:

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

This is where the concept of area comes into play. allowed values(ODZ) - these are the values ​​of the roots of the equation for which the equation makes sense.

Thus, 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 DHS 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 solve the usual equation

5x - 2x = 1
3x=1
x = 1/3

Answer: x = 1/3

Let's solve the equation more complicated:

ODZ is also present here: x -2.

Solving this equation, we will not transfer everything in one direction and bring fractions to a common denominator. We immediately multiply both sides of the equation by an expression that will reduce 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. So, 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.

We write the same equation, but in a slightly different way.

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

x \u003d 4 - 2 \u003d 2, which corresponds to our ODZ

Answer: x = 2.

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