Given a fractional rational equation. How to solve a rational equation
§ 1 Integer and fractional rational equations
In this lesson we will look at concepts such as rational equation, rational expression, integer expression, fractional expression. Let's consider solving rational equations.
A rational equation is an equation in which the left and right sides are rational expressions.
Rational expressions are:
Fractional.
An integer expression is made up of numbers, variables, integer powers using the operations of addition, subtraction, multiplication, and division by a number other than zero.
For example:
Fractional expressions involve division by a variable or an expression with a variable. For example:
A fractional expression does not make sense for all values of the variables included in it. For example, the expression
at x = -9 it does not make sense, since at x = -9 the denominator goes to zero.
This means that a rational equation can be integer or fractional.
A whole rational equation is a rational equation in which the left and right sides are whole expressions.
For example:
A fractional rational equation is a rational equation in which either the left or right sides are fractional expressions.
For example:
§ 2 Solution of an entire rational equation
Let's consider the solution of an entire rational equation.
For example:
Let's multiply both sides of the equation by the least common denominator of the denominators of the fractions included in it.
For this:
1. find the common denominator for denominators 2, 3, 6. It is equal to 6;
2. find an additional factor for each fraction. To do this, divide the common denominator 6 by each denominator
additional factor for fraction
additional factor for fraction
3. multiply the numerators of the fractions by their corresponding additional factors. Thus, we obtain the equation
which is equivalent to the given equation
Let's open the brackets on the left, move the right part to the left, changing the sign of the term when transferred to the opposite one.
Let us bring similar terms of the polynomial and get
We see that the equation is linear.
Having solved it, we find that x = 0.5.
§ 3 Solution of a fractional rational equation
Let's consider solving a fractional rational equation.
For example:
1.Multiply both sides of the equation by the least common denominator of the denominators of the rational fractions included in it.
Let's find the common denominator for the denominators x + 7 and x - 1.
It is equal to their product (x + 7)(x - 1).
2. Let's find an additional factor for each rational fraction.
To do this, divide the common denominator (x + 7)(x - 1) by each denominator. Additional factor for fractions
equal to x - 1,
additional factor for fraction
equals x+7.
3.Multiply the numerators of the fractions by their corresponding additional factors.
We obtain the equation (2x - 1)(x - 1) = (3x + 4)(x + 7), which is equivalent to this equation
4.Multiply the binomial by the binomial on the left and right and get the following equation
5. We move the right side to the left, changing the sign of each term when transferring to the opposite:
6. Let us present similar terms of the polynomial:
7. Both parts can be divided by -1. We get quadratic equation:
8. Having solved it, we will find the roots
Since in Eq.
the left and right sides are fractional expressions, and in fractional expressions, for some values of the variables, the denominator can become zero, then it is necessary to check whether the common denominator does not go to zero when x1 and x2 are found.
At x = -27, the common denominator (x + 7)(x - 1) does not vanish; at x = -1, the common denominator is also not zero.
Therefore, both roots -27 and -1 are roots of the equation.
When solving a fractional rational equation, it is better to immediately indicate the region acceptable values. Eliminate those values at which the common denominator goes to zero.
Let's consider another example of solving a fractional rational equation.
For example, let's solve the equation
We factor the denominator of the fraction on the right side of the equation
We get the equation
Let's find the common denominator for the denominators (x - 5), x, x(x - 5).
It will be the expression x(x - 5).
Now let's find the range of acceptable values of the equation
To do this, we equate the common denominator to zero x(x - 5) = 0.
We obtain an equation, solving which we find that at x = 0 or at x = 5 the common denominator goes to zero.
This means that x = 0 or x = 5 cannot be the roots of our equation.
Additional multipliers can now be found.
Additional factor for rational fractions
additional factor for the fraction
will be (x - 5),
and the additional factor of the fraction
We multiply the numerators by the corresponding additional factors.
We get the equation x(x - 3) + 1(x - 5) = 1(x + 5).
Let's open the brackets on the left and right, x2 - 3x + x - 5 = x + 5.
Let's move the terms from right to left, changing the sign of the transferred terms:
X2 - 3x + x - 5 - x - 5 = 0
And after bringing similar terms, we obtain a quadratic equation x2 - 3x - 10 = 0. Having solved it, we find the roots x1 = -2; x2 = 5.
But we have already found out that at x = 5 the common denominator x(x - 5) goes to zero. Therefore, the root of our equation
will be x = -2.
§ 4 Brief summary lesson
Important to remember:
When solving fractional rational equations, proceed as follows:
1. Find the common denominator of the fractions included in the equation. Moreover, if the denominators of fractions can be factored, then factor them and then find the common denominator.
2.Multiply both sides of the equation by a common denominator: find additional factors, multiply the numerators by additional factors.
3.Solve the resulting whole equation.
4. Eliminate from its roots those that make the common denominator vanish.
List of used literature:
- Makarychev Yu.N., N.G. Mindyuk, Neshkov K.I., Suvorova S.B. / Edited by Telyakovsky S.A. Algebra: textbook. for 8th grade. general education institutions. - M.: Education, 2013.
- Mordkovich A.G. Algebra. 8th grade: In two parts. Part 1: Textbook. for general education institutions. - M.: Mnemosyne.
- Rurukin A.N. Lesson developments in algebra: 8th grade. - M.: VAKO, 2010.
- Algebra 8th grade: lesson plans based on the textbook by Yu.N. Makarycheva, N.G. Mindyuk, K.I. Neshkova, S.B. Suvorova / Auth.-comp. T.L. Afanasyeva, L.A. Tapilina. -Volgograd: Teacher, 2005.
We have already learned how to solve quadratic equations. Now let's extend the studied methods to rational equations.
What is a rational expression? We have already encountered this concept. Rational expressions are expressions made up of numbers, variables, their powers and symbols of mathematical operations.
Accordingly, rational equations are equations of the form: , where 
- rational expressions.
Previously, we considered only those rational equations that can be reduced to linear ones. Now let's look at those rational equations that can be reduced to quadratic equations.
Example 1
Solve the equation: .
Solution:
A fraction is equal to 0 if and only if its numerator is equal to 0 and its denominator is not equal to 0.
We get the following system:
The first equation of the system is a quadratic equation. Before solving it, let's divide all its coefficients by 3. We get:
We get two roots: ; .
Since 2 never equals 0, two conditions must be met: 
. Since none of the roots of the equation obtained above coincides with the invalid values of the variable that were obtained when solving the second inequality, they are both solutions given equation.
Answer:.
So, let's formulate an algorithm for solving rational equations:
1. Transfer all terms to left side, so that the right side turns out to be 0.
2. Transform and simplify the left side, reduce all fractions to common denominator.
3. Equate the resulting fraction to 0 using the following algorithm: 
.
4. Write down those roots that were obtained in the first equation and satisfy the second inequality in the answer.
Let's look at another example.
Example 2
Solve the equation: 
.
Solution
At the very beginning, we move all the terms to the left so that 0 remains on the right. We get:
Now let's 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.
Coefficients of this equation: . We calculate the discriminant:
We get two roots: ; .
Now let's 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 find 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 reduce to quadratic equations.
In the next lesson we will look at rational equations as models of real situations, and also look at motion problems.
Bibliography
- Bashmakov M.I. Algebra, 8th grade. - M.: Education, 2004.
- Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra, 8. 5th ed. - M.: Education, 2010.
- Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Tutorial for educational institutions. - M.: Education, 2006.
- Festival pedagogical ideas "Public lesson" ().
- School.xvatit.com ().
- Rudocs.exdat.com ().
We have already learned how to solve quadratic equations. Now let's extend the studied methods to rational equations.
What is a rational expression? We have already encountered this concept. Rational expressions are expressions made up of numbers, variables, their powers and symbols of mathematical operations.
Accordingly, rational equations are equations of the form: , where - rational expressions.
Previously, we considered only those rational equations that can be reduced to linear ones. Now let's look at those rational equations that can be reduced to quadratic equations.
Example 1
Solve the equation: .
Solution:
A fraction is equal to 0 if and only if its numerator is equal to 0 and its denominator is not equal to 0.
We get the following system:
The first equation of the system is a quadratic equation. Before solving it, let's divide all its coefficients by 3. We get:
We get two roots: ; .
Since 2 never equals 0, two conditions must be met: . Since none of the roots of the equation obtained above coincides with 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 terms to the left side so that the right side ends up with 0.
2. Transform and simplify the left side, bring all fractions to a common denominator.
3. Equate the resulting fraction to 0 using the following algorithm: .
4. Write down those roots that were obtained in the first equation and satisfy the second inequality in the answer.
Let's look at another example.
Example 2
Solve the equation: .
Solution
At the very beginning, we move all the terms to the left so that 0 remains on the right. We get:
Now let's 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.
Coefficients of this equation: . We calculate the discriminant:
We get two roots: ; .
Now let's 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 find 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 reduce to quadratic equations.
In the next lesson we will look at rational equations as models of real situations, and also look at motion problems.
Bibliography
- Bashmakov M.I. Algebra, 8th grade. - M.: Education, 2004.
- Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra, 8. 5th ed. - M.: Education, 2010.
- Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Textbook for general education institutions. - M.: Education, 2006.
- Festival of pedagogical ideas "Open Lesson" ().
- School.xvatit.com ().
- Rudocs.exdat.com ().
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\(\bullet\) A rational equation is an equation represented in the form \[\dfrac(P(x))(Q(x))=0\] where \(P(x), \Q(x)\) - polynomials (the sum of “X’s” in various powers, multiplied by various numbers).
The expression on the left side of the equation is called a rational expression.
The EA (range of acceptable values) of a rational equation is all the values of \(x\) at which the denominator does NOT vanish, that is, \(Q(x)\ne 0\) .
\(\bullet\) For example, equations \[\dfrac(x+2)(x-3)=0,\qquad \dfrac 2(x^2-1)=3, \qquad x^5-3x=2\] are rational equations.
In the first equation, the ODZ are all \(x\) such that \(x\ne 3\) (write \(x\in (-\infty;3)\cup(3;+\infty)\)); in the second equation – these are all \(x\) such that \(x\ne -1; x\ne 1\) (write \(x\in (-\infty;-1)\cup(-1;1)\cup(1;+\infty)\)); and in the third equation there are no restrictions on the ODZ, that is, the ODZ is all \(x\) (they write \(x\in\mathbb(R)\)). \(\bullet\) Theorems:
1) The product of two factors is equal to zero if and only if one of them is equal to zero, and the other does not lose meaning, therefore, the equation \(f(x)\cdot g(x)=0\) is equivalent to the system \[\begin(cases) \left[ \begin(gathered)\begin(aligned) &f(x)=0\\ &g(x)=0 \end(aligned) \end(gathered) \right.\\ \ text(ODZ equations)\end(cases)\] 2) A fraction is equal to zero if and only if the numerator is equal to zero and the denominator is not equal to zero, therefore, the equation \(\dfrac(f(x))(g(x))=0\) is equivalent to a system of equations \[\begin(cases) f(x)=0\\ g(x)\ne 0 \end(cases)\]\(\bullet\) Let's look at a few examples.
1) Solve the equation \(x+1=\dfrac 2x\) . Let us find the ODZ of this equation - this is \(x\ne 0\) (since \(x\) is in the denominator).
This means that the ODZ can be written as follows: .
Let's move all the terms into one part and bring them to a common denominator: \[\dfrac((x+1)\cdot x)x-\dfrac 2x=0\quad\Leftrightarrow\quad \dfrac(x^2+x-2)x=0\quad\Leftrightarrow\quad \begin( cases) x^2+x-2=0\\x\ne 0\end(cases)\] The solution to the first equation of the system will be \(x=-2, x=1\) . We see that both roots are non-zero. Therefore, the answer is: \(x\in \(-2;1\)\) .
2) Solve the equation \(\left(\dfrac4x - 2\right)\cdot (x^2-x)=0\). Let's find the ODZ of this equation. We see that the only value of \(x\) for which the left side does not make sense is \(x=0\) . So, the ODZ can be written like this: \(x\in (-\infty;0)\cup(0;+\infty)\).
Thus, this equation is equivalent to the system:
\[\begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x-2=0\\ &x^2-x=0 \end(aligned) \end(gathered) \right. \\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x=2\\ &x(x-1)= 0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &x =2\\ &x=1\\ &x=0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \left[ \begin(gathered) \begin(aligned) &x=2\\ &x=1 \end(aligned) \end(gathered) \right.\] Indeed, despite the fact that \(x=0\) is the root of the second factor, if you substitute \(x=0\) into the original equation, then it will not make sense, because expression \(\dfrac 40\) is not defined.
Thus, the solution to this equation is \(x\in \(1;2\)\) .
3) Solve the equation \[\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1)\] In our equation \(4x^2-1\ne 0\) , from which \((2x-1)(2x+1)\ne 0\) , that is, \(x\ne -\frac12; \frac12\) .
Let's move all the terms to the left side and bring them to a common denominator:
\(\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1) \quad \Leftrightarrow \quad \dfrac(x^2+4x- 3+x+x^2)(4x^2-1)=0\quad \Leftrightarrow \quad \dfrac(2x^2+5x-3)(4x^2-1)=0 \quad \Leftrightarrow\)
\(\Leftrightarrow \quad \begin(cases) 2x^2+5x-3=0\\ 4x^2-1\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) (2x-1 )(x+3)=0\\ (2x-1)(2x+1)\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered) \begin( aligned) &x=\dfrac12\\ &x=-3 \end(aligned)\end(gathered) \right.\\ x\ne \dfrac 12\\ x\ne -\dfrac 12 \end(cases) \quad \ Leftrightarrow \quad x=-3\)
Answer: \(x\in \(-3\)\) .
Comment. If the answer consists of a finite set of numbers, then they can be written separated by semicolons in curly braces, as shown in the previous examples.
Problems that require solving rational equations are encountered every year in the Unified State Examination in mathematics, so when preparing to pass the certification test, graduates should definitely repeat the theory on this topic on their own. Graduates taking both the basic and specialized level of the exam must be able to cope with such tasks. Having mastered the theory and dealt with practical exercises on the topic “Rational Equations”, students will be able to solve problems with any number of actions and count on receiving competitive scores on the Unified State Examination.
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Our specialists prepared and presented all the necessary theory on the topic “Rational Equations” in the most accessible form. After studying the information presented, students will be able to fill gaps in knowledge.
For successful preparation For the Unified State Exam, graduates need not only to refresh their memory of basic theoretical material on the topic “Rational Equations”, but also to practice completing tasks on specific examples. A large selection of tasks is presented in the “Catalog” section.
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