How to solve rational equations correctly. Rational Equations

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 the 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 what is additional condition, which 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 using standard scheme getting rid of 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.

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:

• upbringing cognitive interest to 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! The 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.) Methods for solving 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 to solve 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.

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

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

x 2 -4x-2x + 8 \u003d x 2 + 3x + 2x + 6

x 2 -6x-x 2 -5x \u003d 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).

x 2 -7x+12 = 0

D=1>0, x 1 =3, x 2 =4.

Answer: 3;4.

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

(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 \u003d 0 x 2 \u003d 5 D \u003d 49
x 3 \u003d 5 x 4 \u003d -2			x 3 \u003d 5 x 4 \u003d -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 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 a number is the root of an 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 roots. given equation. The question arises: is there a way to solve fractional rational equations that eliminates 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, 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.
2. Bring fractions to a common denominator.
3. Make up a system: a fraction is zero when the numerator is zero and the denominator is not zero.
4. Solve the equation.
5. Check 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", Yu.N. Makarychev, 2007: No. 600 (b, c, i); No. 601(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.

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

We have already learned how to solve quadratic equations. Let us now extend the studied methods to rational equations.

What rational expression? We have already encountered this concept. Rational expressions called expressions made up of numbers, variables, their degrees and signs of mathematical operations.

Accordingly, rational equations are equations of the form: , where - rational expressions.

Previously, we considered only those rational equations that reduce to linear ones. Now let's consider those rational equations that can be reduced to quadratic ones.

Example 1

Solve the equation: .

Decision:

A fraction is 0 if and only if its numerator is 0 and its denominator is not 0.

We get the following system:

The first equation of the system is quadratic equation. Before solving it, we divide all its coefficients by 3. We get:

We get two roots: ; .

Since 2 is never equal to 0, two conditions must be met: . Since none of the roots of the equation obtained above matches the invalid values ​​of the variable that were obtained when solving the second inequality, they are both solutions to this equation.

Answer:.

So, let's formulate an algorithm for solving rational equations:

1. Move all the terms to the left side so that 0 is obtained on the right side.

2. Transform and simplify the left side, bring all fractions to a common denominator.

3. Equate the resulting fraction to 0, according to the following algorithm: .

4. Write down those roots that are obtained in the first equation and satisfy the second inequality in response.

Let's look at another example.

Example 2

Solve the equation: .

Decision

At the very beginning, we transfer all the terms to left side so that 0 remains on the right. We get:

Now we bring the left side of the equation to a common denominator:

This equation is equivalent to the system:

The first equation of the system is a quadratic equation.

The coefficients of this equation: . We calculate the discriminant:

We get two roots: ; .

Now we solve the second inequality: the product of factors is not equal to 0 if and only if none of the factors is equal to 0.

Two conditions must be met: . We get that of the two roots of the first equation, only one is suitable - 3.

Answer:.

In this lesson, we remembered what a rational expression is, and also learned how to solve rational equations, which are reduced to quadratic equations.

In the next lesson, we will consider rational equations as models of real situations, and also consider motion problems.

Bibliography

1. Bashmakov M.I. Algebra, 8th grade. - M.: Enlightenment, 2004.
2. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. et al. Algebra, 8. 5th ed. - M.: Education, 2010.
3. Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Tutorial for educational institutions. - M.: Education, 2006.

1. Festival of Pedagogical Ideas " Public lesson" ().
2. School.xvatit.com().
3. Rudocs.exdat.com().

Homework

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.

Simply put, these are equations in which there is at least one with a variable in the denominator.

For example:

\(\frac(9x^2-1)(3x)\) \(=0\)
\(\frac(1)(2x)+\frac(x)(x+1)=\frac(1)(2)\)
\(\frac(6)(x+1)=\frac(x^2-5x)(x+1)\)

Example not fractional rational equations:

\(\frac(9x^2-1)(3)\) \(=0\)
\(\frac(x)(2)\) \(+8x^2=6\)

How are fractional rational equations solved?

The main thing to remember about fractional rational equations is that you need to write in them. And after finding the roots, be sure to check them for admissibility. Otherwise, extraneous roots may appear, and the whole solution will be considered incorrect.

Algorithm for solving a fractional rational equation:

Write out and "solve" the ODZ.

Multiply each term in the equation by a common denominator and reduce the resulting fractions. The denominators will disappear.

Write the equation without opening brackets.

Solve the resulting equation.

Check the found roots with ODZ.

Write down in response the roots that passed the test in step 7.

Do not memorize the algorithm, 3-5 solved equations - and it will be remembered by itself.

Example . Solve fractional rational equation \(\frac(x)(x-2) - \frac(7)(x+2)=\frac(8)(x^2-4)\)

Decision:

Answer: \(3\).

Example . Find the roots of the fractional rational equation \(=0\)

Decision:

\(\frac(x)(x+2) + \frac(x+1)(x+5)-\frac(7-x)(x^2+7x+10)\)\(=0\) ODZ: \(x+2≠0⇔x≠-2\) \(x+5≠0 ⇔x≠-5\) \(x^2+7x+10≠0\) \(D=49-4 \cdot 10=9\) \(x_1≠\frac(-7+3)(2)=-2\) \(x_2≠\frac(-7-3)(2)=-5\)		We write down and "solve" ODZ. Expand \(x^2+7x+10\) into the formula: \(ax^2+bx+c=a(x-x_1)(x-x_2)\). Fortunately \(x_1\) and \(x_2\) we have already found.
\(\frac(x)(x+2) + \frac(x+1)(x+5)-\frac(7-x)((x+2)(x+5))\)\(=0\)		Obviously, the common denominator of fractions: \((x+2)(x+5)\). We multiply the whole equation by it.
\(\frac(x(x+2)(x+5))(x+2) + \frac((x+1)(x+2)(x+5))(x+5)-\) \(-\frac((7-x)(x+2)(x+5))((x+2)(x+5))\)\(=0\)		We reduce fractions
\(x(x+5)+(x+1)(x+2)-7+x=0\)		Opening the brackets
\(x^2+5x+x^2+3x+2-7+x=0\)		We give like terms
\(2x^2+9x-5=0\)		Finding the roots of the equation
\(x_1=-5;\) \(x_2=\frac(1)(2).\)		One of the roots does not fit under the ODZ, so in response we write down only the second root.

Answer: \(\frac(1)(2)\).