The lowest common denominator is used to simplify given 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).
Find the lowest common denominator of the fractions (or least common multiple). NOZ is smallest number, which is evenly divisible by each denominator.
- Sometimes NPD is an obvious number. For example, if given the equation: x/3 + 1/2 = (3x +1)/6, then it is obvious that the least common multiple of the numbers 3, 2 and 6 is 6.
- If the NCD is not obvious, write down the multiples of the largest denominator and find among them one that will be a multiple of the other denominators. Often the NOD can be found by simply multiplying two denominators. For example, if the equation is given x/8 + 2/6 = (x - 3)/9, then NOS = 8*9 = 72.
- If one or more denominators contain a variable, the process becomes somewhat more complicated (but not impossible). In this case, the NOC is an expression (containing a variable) that is divided by each denominator. For example, in the equation 5/(x-1) = 1/x + 2/(3x) NOZ = 3x(x-1), because this expression is divided by each denominator: 3x(x-1)/(x-1 ) = 3x; 3x(x-1)/3x = (x-1); 3x(x-1)/x = 3(x-1).
Multiply both the numerator and denominator of each fraction by a number equal to the result of dividing the NOC by the corresponding denominator of each fraction. Since you are multiplying both the numerator and denominator by the same number, you are effectively multiplying the fraction by 1 (for example, 2/2 = 1 or 3/3 = 1).
- So in our example, multiply x/3 by 2/2 to get 2x/6, and 1/2 multiply by 3/3 to get 3/6 (the fraction 3x +1/6 does not need to be multiplied because it the denominator is 6).
- Proceed similarly when the variable is in the denominator. In our second example, NOZ = 3x(x-1), so multiply 5/(x-1) by (3x)/(3x) to get 5(3x)/(3x)(x-1); 1/x multiplied by 3(x-1)/3(x-1) and you get 3(x-1)/3x(x-1); 2/(3x) multiplied by (x-1)/(x-1) and you get 2(x-1)/3x(x-1).
Find x. Now that you have reduced the fractions to common denominator, you can get rid of the denominator. To do this, multiply each side of the equation by the common denominator. Then solve the resulting equation, that is, find “x”. To do this, isolate the variable on one side of the equation.
- In our example: 2x/6 + 3/6 = (3x +1)/6. You can add 2 fractions with the same denominator, so write the equation as: (2x+3)/6=(3x+1)/6. Multiply both sides of the equation by 6 and get rid of the denominators: 2x+3 = 3x +1. Solve and get x = 2.
- In our second example (with a variable in the denominator), the equation looks like (after reduction to a common denominator): 5(3x)/(3x)(x-1) = 3(x-1)/3x(x-1) + 2 (x-1)/3x(x-1). By multiplying both sides of the equation by N3, you get rid of the denominator and get: 5(3x) = 3(x-1) + 2(x-1), or 15x = 3x - 3 + 2x -2, or 15x = x - 5 Solve and get: x = -5/14.
Solution fractional rational equations
Rational equations are equations in which both the left and right sides are rational expressions.
(Recall: rational expressions are integers and fractional expressions without radicals, involving addition, subtraction, multiplication or division operations - 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 roots found.
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 in the following way:
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
Because on the left and right sides 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 get 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 out of 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.
\(\bullet\) 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.
ODZ (region acceptable values) of a rational equation are all values of \(x\) for 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 ODZ equation– these 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 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 terms to left side and bring it 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 profile level exam. 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 based on the results of passing the Unified State Exam.
How to prepare for the exam using the Shkolkovo educational portal?
Sometimes you can find a source that fully presents the basic theory for solving mathematical problems turns out to be quite difficult. The textbook may simply not be at hand. And find necessary formulas sometimes it can be quite difficult even on the Internet.
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All necessary theory on the topic “Rational Equations” our specialists prepared and presented to the maximum accessible form. After studying the information presented, students will be able to fill gaps in knowledge.
For successful preparation To Unified State Examination for graduates it is necessary not only to brush up on the basic theoretical material on the topic “Rational Equations”, but to practice completing tasks on specific examples. A large selection of tasks is presented in the “Catalog” section.
For each exercise on the site, our experts have written a solution algorithm and indicated the correct answer. Students can practice solving problems varying degrees difficulties depending on the level of preparation. The list of tasks in the corresponding section is constantly supplemented and updated.
Study theoretical material and hone problem-solving skills on the topic “Rational Equations”, similar to those included in Unified State Exam tests, can be done online. If necessary, any of the presented tasks can be added to the “Favorites” section. Repeating again basic theory on the topic “Rational Equations”, a 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.
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