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

Instruction

Perhaps the most obvious point here is, of course, . Numerical fractions do not pose any danger (fractional equations, where all denominators contain only numbers, will generally be linear), but if there is a variable in the denominator, then this must be taken into account and prescribed. Firstly, it is that x, which turns the denominator to 0, cannot be, and in general it is necessary to separately register the fact that x cannot be equal to this number. Even if you succeed that when substituting into the numerator, everything converges perfectly and satisfies the conditions. Secondly, we cannot multiply either or both sides of the equation by equal to zero.

After this, such an equation is reduced to transferring all its terms to the left side so that 0 remains on the right side.

It is necessary to bring all the terms to a common denominator, multiplying, where necessary, the numerators by the missing expressions.
Next, we solve the usual equation written in the numerator. We can endure common factors out of brackets, apply abbreviated multiplication, give similar ones, calculate the roots of a quadratic equation through the discriminant, etc.

The result should be a factorization in the form of a product of brackets (x-(i-th root)). This may also include polynomials that do not have roots, for example, a square trinomial with a discriminant less than zero (unless, of course, there are only real roots in the problem, as most often happens).
Be sure to factorize and denominator from the location of the brackets there, already contained in the numerator. If the denominator contains expressions like (x-(number)), then it is better, when reducing to a common denominator, not to multiply the parentheses in it “head-on”, but to leave them in the form of a product of the original simple expressions.
The same brackets in the numerator and denominator can be reduced by pre-writing, as mentioned above, the conditions on x.
The answer is written in curly braces, as a set of x values, or simply by enumeration: x1=..., x2=..., etc.

Sources:

• Fractional rational equations

Something that cannot be dispensed with in physics, mathematics, chemistry. Least. We learn the basics of their solution.

Instruction

In the most general and simplest classification, it can be divided according to the number of variables they contain, and according to the degrees in which these variables stand.

Solve the equation all its roots or prove that they do not exist.

Any equation has at most P roots, where P is the maximum of the given equation.

But some of these roots may coincide. So, for example, the equation x ^ 2 + 2 * x + 1 = 0, where ^ is the exponentiation icon, folds into the square of the expression (x + 1), that is, into the product of two identical brackets, each of which gives x = - 1 as a solution.

If there is only one unknown in the equation, this means that you will be able to explicitly find its roots (real or complex).

To do this, you will most likely need various transformations: abbreviated multiplication, calculating the discriminant and roots of a quadratic equation, transferring terms from one part to another, reducing to a common denominator, multiplying both parts of the equation by the same expression, squaring, and so on.

Transformations that do not affect the roots of the equation are identical. They are used to simplify the process of solving an equation.

You can also use instead of the traditional analytical graphic method and write this equation in the form , after conducting its study.

If there is more than one unknown in the equation, then you will only be able to express one of them in terms of the other, thereby showing a set of solutions. Such, for example, are equations with parameters in which there is an unknown x and a parameter a. To solve a parametric equation means for all a to express x through a, that is, to consider all possible cases.

If the equation contains derivatives or differentials of unknowns (see picture), congratulations, this is a differential equation, and here you cannot do without higher mathematics).

Sources:

• Identity transformations

To solve the problem with fractions gotta learn to do with them arithmetic operations. They can be decimal, but are most commonly used natural fractions with numerator and denominator. Only after that you can move on to solving mathematical problems with fractional values.

You will need

• - calculator;
• - knowledge of the properties of fractions;
• - Ability to work with fractions.

Instruction

A fraction is a record of dividing one number by another. Often this cannot be done completely, and therefore this action is left “unfinished. The number that is divisible (it is above or before the fraction sign) is called the numerator, and the second number (under or after the fraction sign) is called the denominator. If the numerator is greater than the denominator, the fraction is called an improper fraction, and an integer part can be extracted from it. If the numerator is less than the denominator, then such a fraction is called proper, and its integer part is 0.

Tasks are divided into several types. Determine which one is the task. The simplest option- finding the fraction of a number expressed as a fraction. To solve this problem, it is enough to multiply this number by a fraction. For example, 8 tons of potatoes were brought in. In the first week, 3/4 of its total was sold. How many potatoes are left? To solve this problem, multiply the number 8 by 3/4. It will turn out 8 ∙ 3/4 \u003d 6 t.

If you need to find a number by its part, multiply the known part of the number by the reciprocal of the fraction that shows what proportion of this part is in the number. For example, 8 out of 1/3 of the total number of students. How many in ? Since 8 people is the part that represents 1/3 of the total, then find the reciprocal that is 3/1 or just 3. Then to get the number of students in the class 8∙3=24 students.

When you need to find what part of a number is one number from another, divide the number that represents the part by the one that is the whole number. For example, if the distance is 300 km and the car has traveled 200 km, how much of this will be from the total journey? Divide part of the path 200 by the full path 300, after reducing the fraction you get the result. 200/300=2/3.

To find the part of the unknown fraction of a number, when there is a known one, take the integer as a conventional unit, and subtract the known fraction from it. For example, if 4/7 of the lesson has already passed, is there still left? Take the whole lesson as a conventional unit and subtract 4/7 from it. Get 1-4/7=7/7-4/7=3/7.

"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 of critical thinking; development of research skills.

Nurturing:

education of cognitive interest in the subject; education of independence in solving educational problems; 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 full square, by formulas, using the Vieta theorem and its corollaries.)

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? ( A fraction is zero when the numerator is zero and the denominator is non-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 formalize the 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.

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Equations with fractions themselves are not difficult and very interesting. Consider the types fractional equations and ways to solve them.

How to solve equations with fractions - x in the numerator

If a fractional equation is given, where the unknown is in the numerator, the solution does not require additional conditions and is solved without extra hassle. General form such an equation is x/a + b = c, where x is an unknown, a, b and c are ordinary numbers.

Find x: x/5 + 10 = 70.

In order to solve the equation, you need to get rid of the fractions. Multiply each term of the equation by 5: 5x/5 + 5x10 = 70x5. 5x and 5 is reduced, 10 and 70 are multiplied by 5 and we get: x + 50 = 350 => x = 350 - 50 = 300.

Find x: x/5 + x/10 = 90.

This example is a slightly more complicated version of the first. There are two solutions here.

• Option 1: Get rid of fractions by multiplying all terms of the equation by a larger denominator, that is, by 10: 10x/5 + 10x/10 = 90x10 => 2x + x = 900 => 3x = 900 => x=300.
• Option 2: Add the left side of the equation. x/5 + x/10 = 90. The common denominator is 10. Divide 10 by 5, multiply by x, we get 2x. 10 divided by 10, multiplied by x, we get x: 2x+x/10 = 90. Hence 2x+x = 90×10 = 900 => 3x = 900 => x = 300.

Often there are fractional equations in which x's are on opposite sides of the equal sign. In such a situation, it is necessary to transfer all fractions with x in one direction, and the numbers in another.

• Find x: 3x/5 = 130 - 2x/5.
• Move 2x/5 to the right with opposite sign: 3x/5 + 2x/5 = 130 => 5x/5 = 130.
• We reduce 5x/5 and get: x = 130.

How to solve an equation with fractions - x in the denominator

This type of fractional equations requires writing additional conditions. The indication of these conditions is a mandatory and integral part right decision. By not attributing them, you run the risk, since the answer (even if it is correct) may simply not be counted.

The general form of fractional equations, where x is in the denominator, is: a/x + b = c, where x is an unknown, a, b, c are ordinary numbers. Note that x may not be any number. For example, x cannot be zero, since you cannot divide by 0. This is precisely the additional condition that we must specify. This is called the range of acceptable values, abbreviated - ODZ.

Find x: 15/x + 18 = 21.

We immediately write the ODZ for x: x ≠ 0. Now that the ODZ is indicated, we solve the equation according to the standard scheme, getting rid of the fractions. We multiply all terms of the equation by x. 15x/x+18x = 21x => 15+18x = 21x => 15 = 3x => x = 15/3 = 5.

Often there are equations where the denominator contains not only x, but also some other operation with it, such as addition or subtraction.

Find x: 15/(x-3) + 18 = 21.

We already know that the denominator cannot be equal to zero, which means x-3 ≠ 0. We transfer -3 to the right side, while changing the “-” sign to “+” and we get that x ≠ 3. ODZ is indicated.

Solve the equation, multiply everything by x-3: 15 + 18x(x - 3) = 21x(x - 3) => 15 + 18x - 54 = 21x - 63.

Move the x's to the right, the numbers to the left: 24 = 3x => x = 8.