To solve examples with fractions, you need to be able to find the smallest common denominator. Below is a detailed instruction.

How to find the lowest common denominator - concept

The least common denominator (LCD) in simple words is the minimum number that is divisible by the denominators of all the fractions of a given example. In other words, it is called the Least Common Multiple (LCM). NOZ is used only if the denominators of the fractions are different.

How to find the lowest common denominator - examples

Let's consider examples of finding NOZ.

Calculate: 3/5 + 2/15.

Solution (Sequence of actions):

• We look at the denominators of fractions, make sure that they are different and the expressions are reduced as much as possible.
• We find the smallest number that is divisible by both 5 and 15. This number will be 15. Thus, 3/5 + 2/15 = ?/15.
• We figured out the denominator. What will be in the numerator? An additional multiplier will help us figure this out. An additional factor is the number obtained by dividing the NOZ by the denominator of a particular fraction. For 3/5, the additional factor is 3, since 15/5 = 3. For the second fraction, the additional factor is 1, since 15/15 = 1.
• Having found out the additional factor, we multiply it by the numerators of the fractions and add the resulting values. 3/5 + 2/15 = (3*3+2*1)/15 = (9+2)/15 = 11/15.

Answer: 3/5 + 2/15 = 11/15.

If in the example not 2, but 3 or more fractions are added or subtracted, then the NOZ must be searched for as many fractions as given.

Calculate: 1/2 - 5/12 + 3/6

Solution (sequence of actions):

• Finding the lowest common denominator. The minimum number divisible by 2, 12 and 6 is 12.
• We get: 1/2 - 5/12 + 3/6 = ?/12.
• We are looking for additional multipliers. For 1/2 - 6; for 5/12 - 1; for 3/6 - 2.
• We multiply by the numerators and assign the corresponding signs: 1/2 - 5/12 + 3/6 = (1 * 6 - 5 * 1 + 2 * 3) / 12 = 7/12.

Answer: 1/2 - 5/12 + 3/6 = 7/12.

When adding and subtracting algebraic fractions with different denominators, the fractions first lead to common denominator. This means that they find such a single denominator, which is divided by the original denominator of each algebraic fraction that is part of this expression.

As you know, if the numerator and denominator of a fraction are multiplied (or divided) by the same number other than zero, then the value of the fraction will not change. This is the main property of a fraction. Therefore, when fractions lead to a common denominator, in fact, the original denominator of each fraction is multiplied by the missing factor to a common denominator. In this case, it is necessary to multiply by this factor and the numerator of the fraction (it is different for each fraction).

For example, given the following sum of algebraic fractions:

It is required to simplify the expression, i.e., add two algebraic fractions. To do this, first of all, it is necessary to reduce the terms-fractions to a common denominator. The first step is to find a monomial that is divisible by both 3x and 2y. In this case, it is desirable that it be the smallest, i.e., find the least common multiple (LCM) for 3x and 2y.

For numerical coefficients and variables, the LCM is searched separately. LCM(3, 2) = 6 and LCM(x, y) = xy. Further, the found values ​​are multiplied: 6xy.

Now we need to determine by what factor we need to multiply 3x to get 6xy:
6xy ÷ 3x = 2y

This means that when reducing the first algebraic fraction to a common denominator, its numerator must be multiplied by 2y (the denominator has already been multiplied when reduced to a common denominator). The factor for the numerator of the second fraction is similarly searched for. It will be equal to 3x.

Thus, we get:

Further, it is already possible to act as with fractions with the same denominators: the numerators are added, and one common is written in the denominator:

After transformations, a simplified expression is obtained, which is one algebraic fraction, which is the sum of two original ones:

Algebraic fractions in the original expression may contain denominators that are polynomials rather than monomials (as in the above example). In this case, before finding a common denominator, factor the denominators (if possible). Further, the common denominator is collected from different factors. If the factor is in several initial denominators, then it is taken once. If the factor has different degrees in the original denominators, then it is taken with a larger one. For example:

Here the polynomial a 2 - b 2 can be represented as a product (a - b)(a + b). The factor 2a – 2b is expanded as 2(a – b). Thus, the common denominator will be equal to 2(a - b)(a + b).

To bring fractions to the lowest common denominator, you must: 1) find the least common multiple of the denominators of these fractions, it will be the least common denominator. 2) find an additional factor for each of the fractions, for which we divide the new denominator by the denominator of each fraction. 3) multiply the numerator and denominator of each fraction by its additional factor.

Examples. Reduce the following fractions to the lowest common denominator.

We find the least common multiple of the denominators: LCM(5; 4) = 20, since 20 is the smallest number that is divisible by both 5 and 4. We find for the 1st fraction an additional factor 4 (20 : 5=4). For the 2nd fraction, the additional multiplier is 5 (20 : 4=5). We multiply the numerator and denominator of the 1st fraction by 4, and the numerator and denominator of the 2nd fraction by 5. We reduced these fractions to the lowest common denominator ( 20 ).

The lowest common denominator of these fractions is 8, since 8 is divisible by 4 and itself. There will be no additional multiplier to the 1st fraction (or we can say that it is equal to one), to the 2nd fraction the additional multiplier is 2 (8 : 4=2). We multiply the numerator and denominator of the 2nd fraction by 2. We reduced these fractions to the lowest common denominator ( 8 ).

These fractions are not irreducible.

We reduce the 1st fraction by 4, and we reduce the 2nd fraction by 2. ( see examples on the reduction of ordinary fractions: Sitemap → 5.4.2. Examples of reduction of ordinary fractions). Find LCM(16 ; 20)=2 4 · 5=16· 5=80. The additional multiplier for the 1st fraction is 5 (80 : 16=5). The additional multiplier for the 2nd fraction is 4 (80 : 20=4). We multiply the numerator and denominator of the 1st fraction by 5, and the numerator and denominator of the 2nd fraction by 4. We reduced these fractions to the lowest common denominator ( 80 ).

Find the least common denominator of the NOC(5 ; 6 and 15) = LCM(5 ; 6 and 15)=30. The additional multiplier to the 1st fraction is 6 (30 : 5=6), the additional multiplier to the 2nd fraction is 5 (30 : 6=5), the additional multiplier to the 3rd fraction is 2 (30 : 15=2). We multiply the numerator and denominator of the 1st fraction by 6, the numerator and denominator of the 2nd fraction by 5, the numerator and denominator of the 3rd fraction by 2. We reduced these fractions to the lowest common denominator ( 30 ).

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The denominator of an arithmetic fraction a / b is the number b, showing the size of the fractions of a unit that make up the fraction. The denominator of an algebraic fraction A / B is an algebraic expression B. To perform arithmetic operations with fractions, they must be reduced to the smallest common denominator.

You will need

• To work with algebraic fractions when finding the least common denominator, you need to know the methods of factoring polynomials.

Instruction

Consider reduction to the least common denominator of two arithmetic fractions n/m and s/t, where n, m, s, t are integers. It is clear that these two fractions can be reduced to any denominator divisible by m and t. But they try to bring to the lowest common denominator. It is equal to the least common multiple of the denominators m and t of the given fractions. The least multiple (LCM) of numbers is the smallest that is divisible by all given numbers at the same time. Those. in our case, it is necessary to find the least common multiple of the numbers m and t. Denoted as LCM (m, t). Further, the fractions are multiplied by the corresponding ones: (n/m) * (LCM (m, t) / m), (s/t) * (LCM (m, t) / t).

Let's find the least common denominator of three fractions: 4/5, 7/8, 11/14. First, we expand the denominators 5, 8, 14: 5 = 1 * 5, 8 = 2 * 2 * 2 = 2^3, 14 = 2 * 7. Next, we calculate the LCM (5, 8, 14), multiplying all the numbers included in at least one of the expansions. LCM (5, 8, 14) = 5 * 2^3 * 7 = 280. Note that if the factor occurs in the expansion of several numbers (factor 2 in the expansion of denominators 8 and 14), then we take the factor to a greater degree (2^3 in our case).

So, the general is received. It is equal to 280 = 5 * 56 = 8 * 35 = 14 * 20. Here we get the numbers by which fractions with the corresponding denominators must be multiplied in order to bring them to the lowest common denominator. We get 4/5 = 56 * (4/5) = 224 / 280, 7/8 = 35 * (7/8) = 245/280, 11/14 = 20 * (11/14) = 220/280.

Reduction to the least common denominator of algebraic fractions is performed by analogy with arithmetic. For clarity, consider the problem on an example. Let two fractions (2 * x) / (9 * y^2 + 6 * y + 1) and (x^2 + 1) / (3 * y^2 + 4 * y + 1) be given. Let's factorize both denominators. Note that the denominator of the first fraction is a perfect square: 9 * y^2 + 6 * y + 1 = (3 * y + 1)^2. For

In this lesson, we will look at reducing fractions to a common denominator and solve problems on this topic. Let's give a definition of the concept of a common denominator and an additional factor, remember about coprime numbers. Let's define the concept of the least common denominator (LCD) and solve a number of problems to find it.

Topic: Adding and subtracting fractions with different denominators

Lesson: Reducing fractions to a common denominator

Repetition. Basic property of a fraction.

If the numerator and denominator of a fraction are multiplied or divided by the same natural number, then a fraction equal to it will be obtained.

For example, the numerator and denominator of a fraction can be divided by 2. We get a fraction. This operation is called fraction reduction. You can also perform the reverse transformation by multiplying the numerator and denominator of the fraction by 2. In this case, we say that we have reduced the fraction to a new denominator. The number 2 is called an additional factor.

Conclusion. A fraction can be reduced to any denominator that is a multiple of the denominator of the given fraction. In order to bring a fraction to a new denominator, its numerator and denominator are multiplied by an additional factor.

1. Bring the fraction to the denominator 35.

The number 35 is a multiple of 7, that is, 35 is divisible by 7 without a remainder. So this transformation is possible. Let's find an additional factor. To do this, we divide 35 by 7. We get 5. We multiply the numerator and denominator of the original fraction by 5.

2. Bring the fraction to the denominator 18.

Let's find an additional factor. To do this, we divide the new denominator by the original one. We get 3. We multiply the numerator and denominator of this fraction by 3.

3. Bring the fraction to the denominator 60.

By dividing 60 by 15, we get an additional multiplier. It is equal to 4. Let's multiply the numerator and denominator by 4.

4. Bring the fraction to the denominator 24

In simple cases, reduction to a new denominator is performed in the mind. It is customary to only indicate an additional factor behind the bracket a little to the right and above the original fraction.

A fraction can be reduced to a denominator of 15 and a fraction can be reduced to a denominator of 15. Fractions have a common denominator of 15.

The common denominator of fractions can be any common multiple of their denominators. For simplicity, fractions are reduced to the lowest common denominator. It is equal to the least common multiple of the denominators of the given fractions.

Example. Reduce to the least common denominator of the fraction and .

First, find the least common multiple of the denominators of these fractions. This number is 12. Let's find an additional factor for the first and second fractions. To do this, we divide 12 by 4 and by 6. Three is an additional factor for the first fraction, and two for the second. We bring the fractions to the denominator 12.

We reduced the fractions to a common denominator, that is, we found fractions that are equal to them and have the same denominator.

Rule. To bring fractions to the lowest common denominator,

First, find the least common multiple of the denominators of these fractions, which will be their least common denominator;

Secondly, divide the least common denominator by the denominators of these fractions, that is, find an additional factor for each fraction.

Thirdly, multiply the numerator and denominator of each fraction by its additional factor.

a) Reduce the fractions and to a common denominator.

The lowest common denominator is 12. The additional factor for the first fraction is 4, for the second - 3. We bring the fractions to the denominator 24.

b) Reduce the fractions and to a common denominator.

The lowest common denominator is 45. Dividing 45 by 9 by 15, we get 5 and 3, respectively. We bring the fractions to the denominator 45.

c) Reduce the fractions and to a common denominator.

The common denominator is 24. The additional factors are 2 and 3, respectively.

Sometimes it is difficult to verbally find the least common multiple for the denominators of given fractions. Then the common denominator and additional factors are found by factoring into prime factors.

Reduce to a common denominator of the fraction and .

Let's decompose the numbers 60 and 168 into prime factors. Let's write out the expansion of the number 60 and add the missing factors 2 and 7 from the second expansion. Multiply 60 by 14 and get a common denominator of 840. The additional factor for the first fraction is 14. The additional factor for the second fraction is 5. Let's reduce the fractions to a common denominator of 840.

Bibliography

1. Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S. and others. Mathematics 6. - M.: Mnemozina, 2012.

2. Merzlyak A.G., Polonsky V.V., Yakir M.S. Mathematics 6th grade. - Gymnasium, 2006.

3. Depman I.Ya., Vilenkin N.Ya. Behind the pages of a mathematics textbook. - Enlightenment, 1989.

4. Rurukin A.N., Chaikovsky I.V. Tasks for the course of mathematics grade 5-6. - ZSH MEPhI, 2011.

5. Rurukin A.N., Sochilov S.V., Chaikovsky K.G. Mathematics 5-6. A manual for students of the 6th grade of the MEPhI correspondence school. - ZSH MEPhI, 2011.

6. Shevrin L.N., Gein A.G., Koryakov I.O. and others. Mathematics: A textbook-interlocutor for grades 5-6 of high school. Library of the teacher of mathematics. - Enlightenment, 1989.

You can download the books specified in clause 1.2. this lesson.

Homework

Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S. and others. Mathematics 6. - M .: Mnemozina, 2012. (see link 1.2)

Homework: No. 297, No. 298, No. 300.

Other tasks: #270, #290