Solving a system of fractional rational equations. How to solve a rational equation

\(\bullet\) 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 go to zero, 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.

How to prepare for the exam using the Shkolkovo educational portal?

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Our specialists have 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.

To successfully prepare 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 using specific examples. A large selection of tasks is presented in the “Catalogue” section.

For each exercise on the site, our experts have written a solution algorithm and indicated the correct answer. Students can practice solving problems of varying degrees of difficulty depending on their skill level. The list of tasks in the corresponding section is constantly supplemented and updated.

You can study theoretical material and hone your skills in solving problems on the topic “Rational Equations”, similar to those included in the Unified State Exam tests, online. If necessary, any of the presented tasks can be added to the “Favorites” section. Having once again repeated the basic theory on the topic “Rational Equations,” the high school student will be able to return to the problem in the future to discuss the progress of its solution with the teacher in an algebra lesson.

The lowest common denominator is used to simplify this equation. This method is used when you cannot write a given equation with one rational expression on each side of the equation (and use the crisscross method of multiplication). This method is used when you are given a rational equation with 3 or more fractions (in the case of two fractions, it is better to use criss-cross multiplication).

An integer expression is a mathematical expression made up of numbers and literal variables using the operations of addition, subtraction and multiplication. Integers also include expressions that involve division by any number other than zero.

The concept of a fractional rational expression

A fractional expression is a mathematical expression that, in addition to the operations of addition, subtraction and multiplication performed with numbers and letter variables, as well as division by a number not equal to zero, also contains division into expressions with letter variables.

Rational expressions are all whole and fractional expressions. Rational equations are equations in which the left and right sides are rational expressions. If in a rational equation the left and right sides are integer expressions, then such a rational equation is called an integer.

If in a rational equation the left or right sides are fractional expressions, then such a rational equation is called fractional.

Examples of fractional rational expressions

1. x-3/x = -6*x+19

2. (x-4)/(2*x+5) = (x+7)/(x-2)

3. (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5))

Scheme for solving a fractional rational equation

1. Find the common denominator of all fractions that are included in the equation.

2. Multiply both sides of the equation by a common denominator.

3. Solve the resulting whole equation.

4. Check the roots and exclude those that make the common denominator vanish.

Since we are solving fractional rational equations, there will be variables in the denominators of the fractions. This means that they will be a common denominator. And in the second point of the algorithm we multiply by a common denominator, then extraneous roots may appear. At which the common denominator will be equal to zero, which means multiplying by it will be meaningless. Therefore, at the end it is necessary to check the obtained roots.

Let's look at an example:

Solve the fractional rational equation: (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5)).

We will adhere to the general scheme: first find the common denominator of all fractions. We get x*(x-5).

Multiply each fraction by a common denominator and write the resulting whole equation.

(x-3)/(x-5) * (x*(x-5))= x*(x+3);
1/x * (x*(x-5)) = (x-5);
(x+5)/(x*(x-5)) * (x*(x-5)) = (x+5);
x*(x+3) + (x-5) = (x+5);

Let us simplify the resulting equation. We get:

x^2+3*x + x-5 - x - 5 =0;
x^2+3*x-10=0;

We get a simple reduced quadratic equation. We solve it by any of the known methods, we get the roots x=-2 and x=5.

Now we check the obtained solutions:

Substitute the numbers -2 and 5 into the common denominator. At x=-2 the common denominator x*(x-5) does not vanish, -2*(-2-5)=14. This means that the number -2 will be the root of the original fractional rational equation.

At x=5 the common denominator x*(x-5) becomes zero. Therefore, this number is not the root of the original fractional rational equation, since there will be a division by zero.

Simply put, these are equations in which there is at least one 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 entire decision will be considered incorrect.

Algorithm for solving a fractional rational equation:

Write down and “solve” the ODZ.

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

Write the equation without opening the parentheses.

Solve the resulting equation.

Check the found roots with ODZ.

Write down in your answer the roots that passed the test in step 7.

Don’t memorize the algorithm, 3-5 solved equations and it will be remembered by itself.

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

Solution:

Answer: \(3\).

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

Solution:

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

Solving fractional rational equations

Reference Guide

Rational equations are equations in which both the left and right sides are rational expressions.

(Remember: rational expressions are integer and fractional expressions without radicals, including the operations of addition, subtraction, multiplication or division - for example: 6x; (m – n)2; x/3y, etc.)

Fractional rational equations are usually reduced to the form:

Where P(x) And Q(x) are polynomials.

To solve such equations, multiply both sides of the equation by Q(x), which can lead to the appearance of extraneous roots. Therefore, when solving fractional rational equations, it is necessary to check the found roots.

A rational equation is called whole, or algebraic, if it does not divide by an expression containing a variable.

Examples of a whole rational equation:

5x – 10 = 3(10 – x)

3x
- = 2x – 10
4

If in a rational equation there is a division by an expression containing a variable (x), then the equation is called fractional rational.

Example of a fractional rational equation:

15
x + - = 5x – 17
x

Fractional rational equations are usually solved as follows:

1) find the common denominator of the fractions and multiply both sides of the equation by it;

2) solve the resulting whole equation;

3) exclude from its roots those that reduce the common denominator of the fractions to zero.

Examples of solving integer and fractional rational equations.

Example 1. Let's solve the whole equation

x – 1 2x 5x
-- + -- = --.
2 3 6

Solution:

Finding the lowest common denominator. This is 6. Divide 6 by the denominator and multiply the resulting result by the numerator of each fraction. We obtain an equation equivalent to this:

3(x – 1) + 4x 5x
------ = --
6 6

Since the left and right sides have the same denominator, it can be omitted. Then we get a simpler equation:

3(x – 1) + 4x = 5x.

We solve it by opening the brackets and combining similar terms:

3x – 3 + 4x = 5x

3x + 4x – 5x = 3

The example is solved.

Example 2. Solve a fractional rational equation

x – 3 1 x + 5
-- + - = ---.
x – 5 x x(x – 5)

Finding a common denominator. This is x(x – 5). So:

x 2 – 3x x – 5 x + 5
--- + --- = ---
x(x – 5) x(x – 5) x(x – 5)

Now we get rid of the denominator again, since it is the same for all expressions. We reduce similar terms, equate the equation to zero and obtain a quadratic equation:

x 2 – 3x + x – 5 = x + 5

x 2 – 3x + x – 5 – x – 5 = 0

x 2 – 3x – 10 = 0.

Having solved the quadratic equation, we find its roots: –2 and 5.

Let's check whether these numbers are the roots of the original equation.

At x = –2, the common denominator x(x – 5) does not vanish. This means –2 is the root of the original equation.

At x = 5, the common denominator goes to zero, and two of the three expressions become meaningless. This means that the number 5 is not the root of the original equation.

Answer: x = –2

More examples

Example 1.

x 1 =6, x 2 = - 2.2.

Answer: -2,2;6.

Example 2.