Least common denominator (LCD) of algebraic fractions, finding it. Reducing fractions to a common denominator (Moskalenko M.V.)
Most operations with algebraic fractions, such as addition and subtraction, require first reducing these fractions to the same denominators. Such denominators are also often referred to as “common denominator.” In this topic, we will look at the definition of the concepts “common denominator of algebraic fractions” and “least common denominator of algebraic fractions (LCD)”, consider the algorithm for finding the common denominator point by point and solve several problems on the topic.
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Common denominator of algebraic fractions
If we talk about ordinary fractions, then the common denominator is a number that is divisible by any of the denominators of the original fractions. For ordinary fractions 1 2 And 5 9 the number 36 can be a common denominator, since it is divisible by 2 and 9 without a remainder.
The common denominator of algebraic fractions is determined in a similar way, only polynomials are used instead of numbers, since they are the numerators and denominators of the algebraic fraction.
Definition 1
Common denominator of an algebraic fraction is a polynomial that is divisible by the denominator of any fraction.
Due to the peculiarities of algebraic fractions, which will be discussed below, we will often deal with common denominators represented as a product rather than as a standard polynomial.
Example 1
Polynomial written as a product 3 x 2 (x + 1), corresponds to a polynomial of the standard form 3 x 3 + 3 x 2. This polynomial can be the common denominator of the algebraic fractions 2 x, - 3 x y x 2 and y + 3 x + 1, due to the fact that it is divisible by x, on x 2 and on x+1. Information on the divisibility of polynomials is available in the corresponding topic of our resource.
Least common denominator (LCD)
For given algebraic fractions, the number of common denominators can be infinite.
Example 2
Let's take as an example the fractions 1 2 x and x + 1 x 2 + 3. Their common denominator is 2 x (x 2 + 3), as well as − 2 x (x 2 + 3), as well as x (x 2 + 3), as well as 6, 4 x (x 2 + 3) (y + y 4), as well as − 31 x 5 (x 2 + 3) 3, etc.
When solving problems, you can make your work easier by using a common denominator, which has the simplest form among the entire set of denominators. This denominator is often referred to as the lowest common denominator.
Definition 2
Least common denominator of algebraic fractions is the common denominator of algebraic fractions, which has the simplest form.
By the way, the term “lowest common denominator” is not generally accepted, so it is better to limit ourselves to the term “common denominator”. And here's why.
Earlier we focused your attention on the phrase “denominator of the simplest kind.” The main meaning of this phrase is the following: the denominator of the simplest form must divide without remainder any other common denominator of the data in the condition of the algebraic fractions problem. In this case, in the product, which is the common denominator of fractions, various numerical coefficients can be used.
Example 3
Let's take the fractions 1 2 · x and x + 1 x 2 + 3 . We have already found out that it will be easiest for us to work with a common denominator of the form 2 · x · (x 2 + 3). Also, the common denominator for these two fractions can be x (x 2 + 3), which does not contain a numeric coefficient. The question is which of these two common denominators is considered the least common denominator of the fractions. There is no definite answer, therefore it is more correct to simply talk about the common denominator, and to work with the option that will be most convenient to work with. So, we can use such common denominators as x 2 (x 2 + 3) (y + y 4) or − 15 x 5 (x 2 + 3) 3, which have a more complex appearance, but it may be more difficult to carry out actions with them.
Finding the common denominator of algebraic fractions: algorithm of actions
Suppose we have several algebraic fractions for which we need to find a common denominator. To solve this problem we can use the following algorithm of actions. First we need to factor the denominators of the original fractions. Then we compose a work in which we sequentially include:
- all factors from the denominator of the first fraction along with powers;
- all factors present in the denominator of the second fraction, but which are not in the written product or their degree is insufficient;
- all missing factors from the denominator of the third fraction, and so on.
The resulting product will be the common denominator of algebraic fractions.
As factors of the product, we can take all the denominators of the fractions given in the problem statement. However, the multiplier that we will get in the end will be far from the NCD in meaning and its use will be irrational.
Example 4
Determine the common denominator of the fractions 1 x 2 y, 5 x + 1 and y - 3 x 5 y.
Solution
In this case, we do not need to factor the denominators of the original fractions. Therefore, we will begin to apply the algorithm by composing the work.
From the denominator of the first fraction we take the multiplier x 2 y, from the denominator of the second fraction the multiplier x+1. We get the product x 2 y (x + 1).
The denominator of the third fraction can give us a multiplier x 5 y, however, the product we compiled earlier already has factors x 2 And y. Therefore, we add more x 5 − 2 = x 3. We get the product x 2 y (x + 1) x 3, which can be reduced to the form x 5 y (x + 1). This will be our NOZ of algebraic fractions.
Answer: x 5 · y · (x + 1) .
Now let's look at examples of problems where the denominators of algebraic fractions contain integer numerical factors. In such cases, we also follow the algorithm, having previously decomposed the integer numerical factors into simple factors.
Example 5
Find the common denominator of the fractions 1 12 x and 1 90 x 2.
Solution
Dividing the numbers in the denominators of the fractions into prime factors, we get 1 2 2 · 3 · x and 1 2 · 3 2 · 5 · x 2 . Now we can move on to compiling a common denominator. To do this, from the denominator of the first fraction we take the product 2 2 3 x and add to it the factors 3, 5 and x from the denominator of the second fraction. We get 2 2 3 x 3 5 x = 180 x 2. This is our common denominator.
Answer: 180 x 2.
If you look closely at the results of the two analyzed examples, you will notice that the common denominators of the fractions contain all the factors present in the expansions of the denominators, and if a certain factor is present in several denominators, then it is taken with the largest exponent available. And if the denominators have integer coefficients, then the common denominator contains a numerical factor equal to the least common multiple of these numerical coefficients.
Example 6
The denominators of both algebraic fractions 1 12 x and 1 90 x 2 have a factor x. In the second case, the factor x is squared. To create a common denominator, we need to take this factor to the greatest extent, i.e. x 2. There are no other multipliers with variables. Integer numeric coefficients of original fractions 12 And 90 , and their least common multiple is 180 . It turns out that the desired common denominator has the form 180 x 2.
Now we can write down another algorithm for finding the common factor of algebraic fractions. For this we:
- factor the denominators of all fractions;
- we compose the product of all letter factors (if there is a factor in several expansions, we take the option with the largest exponent);
- we add the LCM of the numerical coefficients of the expansions to the resulting product.
The given algorithms are equivalent, so any of them can be used to solve problems. It's important to pay attention to detail.
There are cases when common factors in the denominators of fractions may be invisible behind the numerical coefficients. Here it is advisable to first put the numerical coefficients at higher powers of the variables out of brackets in each of the factors present in the denominator.
Example 7
What common denominator do the fractions 3 5 - x and 5 - x · y 2 2 · x - 10 have?
Solution
In the first case, minus one must be taken out of brackets. We get 3 - x - 5 . We multiply the numerator and denominator by - 1 in order to get rid of the minus in the denominator: - 3 x - 5.
In the second case, we put the two out of brackets. This allows us to obtain the fraction 5 - x · y 2 2 · x - 5.
It is obvious that the common denominator of these algebraic fractions - 3 x - 5 and 5 - x · y 2 2 · x - 5 is 2 (x − 5).
Answer:2 (x − 5).
The data in the fraction problem condition may have fractional coefficients. In these cases, you must first get rid of fractional coefficients by multiplying the numerator and denominator by a certain number.
Example 8
Simplify the algebraic fractions 1 2 x + 1 1 14 x 2 + 1 7 and - 2 2 3 x 2 + 1 1 3 and then determine their common denominator.
Solution
Let's get rid of fractional coefficients by multiplying the numerator and denominator in the first case by 14, in the second case by 3. We get:
1 2 x + 1 1 14 x 2 + 1 7 = 14 1 2 x + 1 14 1 14 x 2 + 1 7 = 7 x + 1 x 2 + 2 and - 2 2 3 x 2 + 1 1 3 = 3 · - 2 3 · 2 3 · x 2 + 4 3 = - 6 2 · x 2 + 4 = - 6 2 · x 2 + 2 .
After the transformations, it becomes clear that the common denominator is 2 (x 2 + 2).
Answer: 2 (x 2 + 2).
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This method makes sense if the degree of the polynomial is not lower than two. In this case, the common factor can be not only a binomial of the first degree, but also of higher degrees.
To find a common factor terms of the polynomial, it is necessary to perform a number of transformations. The simplest binomial or monomial that can be taken out of brackets will be one of the roots of the polynomial. Obviously, in the case when a polynomial does not have a free term, there will be an unknown in the first degree - the polynomial, equal to 0.
More difficult to find a common factor is the case when the free term is not equal to zero. Then methods of simple selection or grouping are applicable. For example, let all the roots of the polynomial be rational, and all the coefficients of the polynomial are integers: y^4 + 3 y³ – y² – 9 y – 18.
Write down all the integer divisors of the free term. If a polynomial has rational roots, then they are among them. As a result of the selection, roots 2 and -3 are obtained. This means that the common factors of this polynomial will be the binomials (y - 2) and (y + 3).
The common factoring method is one of the components of factorization. The method described above is applicable if the coefficient of the highest degree is 1. If this is not the case, then a series of transformations must first be performed. For example: 2y³ + 19 y² + 41 y + 15.
Make a substitution of the form t = 2³·y³. To do this, multiply all the coefficients of the polynomial by 4: 2³·y³ + 19·2²·y² + 82·2·y + 60. After replacement: t³ + 19·t² + 82·t + 60. Now, to find the common factor, we apply the above method .
In addition, an effective method for finding a common factor is the elements of a polynomial. It is especially useful when the first method does not, i.e. The polynomial has no rational roots. However, groupings are not always obvious. For example: The polynomial y^4 + 4 y³ – y² – 8 y – 2 has no integer roots.
Use grouping: y^4 + 4 y³ – y² – 8 y – 2 = y^4 + 4 y³ – 2 y² + y² – 8 y – 2 = (y^4 – 2 y²) + ( 4 y³ – 8 y) + y² – 2 = (y² - 2)*(y² + 4 y + 1). The common factor of the elements of this polynomial is (y² - 2).
Multiplication and division, just like addition and subtraction, are basic arithmetic operations. Without learning to solve examples of multiplication and division, a person will encounter many difficulties not only when studying more complex branches of mathematics, but even in the most ordinary everyday affairs. Multiplication and division are closely related, and the unknown components of examples and problems involving one of these operations are calculated using the other operation. At the same time, it is necessary to clearly understand that when solving examples, it makes absolutely no difference which objects you divide or multiply.
You will need
- - multiplication table;
- - calculator or sheet of paper and pencil.
Instructions
Write down the example you need. Label the unknown factor like x. An example might look like this: a*x=b. Instead of the factor a and the product b in the example, there can be any or numbers. Remember the basic principle of multiplication: changing the places of the factors does not change the product. So unknown factor x can be placed absolutely anywhere.
To find the unknown factor in an example where there are only two factors, you just need to divide the product by the known factor. That is, this is done as follows: x=b/a. If you find it difficult to operate with abstract quantities, try to imagine this problem in the form of concrete objects. You, you have only apples and how many of them you will eat, but you don’t know how many apples everyone will get. For example, you have 5 family members, and there are 15 apples. Designate the number of apples intended for each as x. Then the equation will look like this: 5(apples)*x=15(apples). Unknown factor is found in the same way as in the equation with letters, that is, divide 15 apples among five family members, in the end it turns out that each of them ate 3 apples.
In the same way the unknown is found factor with the number of factors. For example, the example looks like a*b*c*x*=d. In theory, find with factor it is possible in the same way as in the later example: x=d/a*b*c. But you can bring the equation to a simpler form by denoting the product of known factors with another letter - for example, m. Find what m equals by multiplying the numbers a, b and c: m=a*b*c. Then the whole example can be represented as m*x=d, and the unknown quantity will be equal to x=d/m.
If known factor and the product are fractions, the example is solved in exactly the same way as with . But in this case you need to remember the actions. When multiplying fractions, their numerators and denominators are multiplied. When dividing fractions, the numerator of the dividend is multiplied by the denominator of the divisor, and the denominator of the dividend is multiplied by the numerator of the divisor. That is, in this case the example will look like this: a/b*x=c/d. In order to find an unknown quantity, you need to divide the product by the known factor. That is, x=a/b:c/d =a*d/b*c.
Video on the topic
Please note
When solving examples with fractions, the fraction of a known factor can simply be reversed and the action performed as a multiplication of fractions.
A polynomial is the sum of monomials. A monomial is the product of several factors, which are a number or a letter. Degree unknown is the number of times it is multiplied by itself.
Instructions
Please provide it if it has not already been done. Similar monomials are monomials of the same type, that is, monomials with the same unknowns of the same degree.
Take, for example, the polynomial 2*y²*x³+4*y*x+5*x²+3-y²*x³+6*y²*y²-6*y²*y². This polynomial has two unknowns - x and y.
Connect similar monomials. Monomials with the second power of y and the third power of x will come to the form y²*x³, and monomials with the fourth power of y will cancel. It turns out y²*x³+4*y*x+5*x²+3-y²*x³.
Take y as the main unknown letter. Find the maximum degree for unknown y. This is a monomial y²*x³ and, accordingly, degree 2.
Draw a conclusion. Degree polynomial 2*y²*x³+4*y*x+5*x²+3-y²*x³+6*y²*y²-6*y²*y² in x is equal to three, and in y is equal to two.
Find the degree polynomial√x+5*y by y. It is equal to the maximum degree of y, that is, one.
Find the degree polynomial√x+5*y in x. The unknown x is located, which means its degree will be a fraction. Since the root is a square root, the power of x is 1/2.
Draw a conclusion. For polynomial√x+5*y the x power is 1/2 and the y power is 1.
Video on the topic
Simplification of algebraic expressions is required in many areas of mathematics, including solving higher-order equations, differentiation and integration. Several methods are used, including factorization. To apply this method, you need to find and make a general factor for brackets.
Criss-cross multiplication
Common Divisor Method
Task. Find the meanings of the expressions:
Task. Find the meanings of the expressions:
To appreciate how much of a difference the least common multiple method makes, try calculating these same examples using the criss-cross method.
Common denominator of fractions
Of course, without a calculator. I think after this comments will be unnecessary.
See also:
I originally wanted to include common denominator techniques in the Adding and Subtracting Fractions section. But there turned out to be so much information, and its importance is so great (after all, not only numerical fractions have common denominators), that it is better to study this issue separately.
So, let's say we have two fractions with different denominators. And we want to make sure that the denominators become the same. The basic property of a fraction comes to the rescue, which, let me remind you, sounds like this:
A fraction will not change if its numerator and denominator are multiplied by the same number other than zero.
Thus, if you choose the factors correctly, the denominators of the fractions will become equal - this process is called. And the required numbers, “evening out” the denominators, are called.
Why do we need to reduce fractions to a common denominator? Here are just a few reasons:
- Adding and subtracting fractions with different denominators. There is no other way to perform this operation;
- Comparing fractions. Sometimes reduction to a common denominator greatly simplifies this task;
- Solving problems involving fractions and percentages. Percentages are essentially ordinary expressions that contain fractions.
There are many ways to find numbers that, when multiplied by them, will make the denominators of fractions equal. We will consider only three of them - in order of increasing complexity and, in a sense, effectiveness.
Criss-cross multiplication
The simplest and most reliable method, which is guaranteed to equalize the denominators. We will act “in a headlong manner”: we multiply the first fraction by the denominator of the second fraction, and the second by the denominator of the first. As a result, the denominators of both fractions will become equal to the product of the original denominators. Take a look:
Task. Find the meanings of the expressions:
As additional factors, consider the denominators of neighboring fractions. We get:
Yes, it's that simple. If you are just starting to study fractions, it is better to work using this method - this way you will insure yourself against many mistakes and are guaranteed to get the result.
The only drawback of this method is that you have to count a lot, because the denominators are multiplied “all the way”, and the result can be very large numbers. This is the price to pay for reliability.
Common Divisor Method
This technique helps to significantly reduce calculations, but, unfortunately, it is used quite rarely. The method is as follows:
- Before you go straight ahead (i.e., using the criss-cross method), take a look at the denominators. Perhaps one of them (the one that is larger) is divided into the other.
- The number resulting from this division will be an additional factor for the fraction with a smaller denominator.
- In this case, a fraction with a large denominator does not need to be multiplied by anything at all - this is where the savings lie. At the same time, the probability of error is sharply reduced.
Task. Find the meanings of the expressions:
Note that 84: 21 = 4; 72: 12 = 6. Since in both cases one denominator is divided without a remainder by the other, we use the method of common factors. We have:
Note that the second fraction was not multiplied by anything at all. In fact, we cut the amount of computation in half!
By the way, I didn’t take the fractions in this example by chance. If you're interested, try counting them using the criss-cross method. After reduction, the answers will be the same, but there will be much more work.
This is the power of the common divisors method, but, again, it can only be used when one of the denominators is divisible by the other without a remainder. Which happens quite rarely.
Least common multiple method
When we reduce fractions to a common denominator, we are essentially trying to find a number that is divisible by each denominator. Then we bring the denominators of both fractions to this number.
There are a lot of such numbers, and the smallest of them will not necessarily be equal to the direct product of the denominators of the original fractions, as is assumed in the “criss-cross” method.
For example, for denominators 8 and 12, the number 24 is quite suitable, since 24: 8 = 3; 24: 12 = 2. This number is much less than the product 8 12 = 96.
The smallest number that is divisible by each of the denominators is called their (LCM).
Notation: The least common multiple of a and b is denoted LCM(a; b). For example, LCM(16, 24) = 48; LCM(8; 12) = 24.
If you manage to find such a number, the total amount of calculations will be minimal. Look at the examples:
How to Find the Lowest Common Denominator
Find the meanings of the expressions:
Note that 234 = 117 2; 351 = 117 · 3. Factors 2 and 3 are coprime (have no common factors other than 1), and factor 117 is common. Therefore LCM(234, 351) = 117 2 3 = 702.
Likewise, 15 = 5 3; 20 = 5 · 4. Factors 3 and 4 are coprime, and factor 5 is common. Therefore LCM(15, 20) = 5 3 4 = 60.
Now let's bring the fractions to common denominators:
Notice how useful it was to factorize the original denominators:
- Having discovered identical factors, we immediately arrived at the least common multiple, which, generally speaking, is a non-trivial problem;
- From the resulting expansion you can find out which factors are “missing” in each fraction. For example, 234 · 3 = 702, therefore, for the first fraction the additional factor is 3.
Don't think that there won't be such complex fractions in the real examples. They meet all the time, and the above tasks are not the limit!
The only problem is how to find this very NOC. Sometimes everything can be found in a few seconds, literally “by eye,” but in general this is a complex computational task that requires separate consideration. We won't touch on that here.
See also:
Reducing fractions to a common denominator
I originally wanted to include common denominator techniques in the Adding and Subtracting Fractions section. But there turned out to be so much information, and its importance is so great (after all, not only numerical fractions have common denominators), that it is better to study this issue separately.
So, let's say we have two fractions with different denominators. And we want to make sure that the denominators become the same. The basic property of a fraction comes to the rescue, which, let me remind you, sounds like this:
A fraction will not change if its numerator and denominator are multiplied by the same number other than zero.
Thus, if you choose the factors correctly, the denominators of the fractions will become equal - this process is called. And the required numbers, “evening out” the denominators, are called.
Why do we need to reduce fractions to a common denominator?
Common denominator, concept and definition.
Here are just a few reasons:
- Adding and subtracting fractions with different denominators. There is no other way to perform this operation;
- Comparing fractions. Sometimes reduction to a common denominator greatly simplifies this task;
- Solving problems involving fractions and percentages. Percentages are essentially ordinary expressions that contain fractions.
There are many ways to find numbers that, when multiplied by them, will make the denominators of fractions equal. We will consider only three of them - in order of increasing complexity and, in a sense, effectiveness.
Criss-cross multiplication
The simplest and most reliable method, which is guaranteed to equalize the denominators. We will act “in a headlong manner”: we multiply the first fraction by the denominator of the second fraction, and the second by the denominator of the first. As a result, the denominators of both fractions will become equal to the product of the original denominators. Take a look:
Task. Find the meanings of the expressions:
As additional factors, consider the denominators of neighboring fractions. We get:
Yes, it's that simple. If you are just starting to study fractions, it is better to work using this method - this way you will insure yourself against many mistakes and are guaranteed to get the result.
The only drawback of this method is that you have to count a lot, because the denominators are multiplied “all the way”, and the result can be very large numbers. This is the price to pay for reliability.
Common Divisor Method
This technique helps to significantly reduce calculations, but, unfortunately, it is used quite rarely. The method is as follows:
- Before you go straight ahead (i.e., using the criss-cross method), take a look at the denominators. Perhaps one of them (the one that is larger) is divided into the other.
- The number resulting from this division will be an additional factor for the fraction with a smaller denominator.
- In this case, a fraction with a large denominator does not need to be multiplied by anything at all - this is where the savings lie. At the same time, the probability of error is sharply reduced.
Task. Find the meanings of the expressions:
Note that 84: 21 = 4; 72: 12 = 6. Since in both cases one denominator is divided without a remainder by the other, we use the method of common factors. We have:
Note that the second fraction was not multiplied by anything at all. In fact, we cut the amount of computation in half!
By the way, I didn’t take the fractions in this example by chance. If you're interested, try counting them using the criss-cross method. After reduction, the answers will be the same, but there will be much more work.
This is the power of the common divisors method, but, again, it can only be used when one of the denominators is divisible by the other without a remainder. Which happens quite rarely.
Least common multiple method
When we reduce fractions to a common denominator, we are essentially trying to find a number that is divisible by each denominator. Then we bring the denominators of both fractions to this number.
There are a lot of such numbers, and the smallest of them will not necessarily be equal to the direct product of the denominators of the original fractions, as is assumed in the “criss-cross” method.
For example, for denominators 8 and 12, the number 24 is quite suitable, since 24: 8 = 3; 24: 12 = 2. This number is much less than the product 8 12 = 96.
The smallest number that is divisible by each of the denominators is called their (LCM).
Notation: The least common multiple of a and b is denoted LCM(a; b). For example, LCM(16, 24) = 48; LCM(8; 12) = 24.
If you manage to find such a number, the total amount of calculations will be minimal. Look at the examples:
Task. Find the meanings of the expressions:
Note that 234 = 117 2; 351 = 117 · 3. Factors 2 and 3 are coprime (have no common factors other than 1), and factor 117 is common. Therefore LCM(234, 351) = 117 2 3 = 702.
Likewise, 15 = 5 3; 20 = 5 · 4. Factors 3 and 4 are coprime, and factor 5 is common. Therefore LCM(15, 20) = 5 3 4 = 60.
Now let's bring the fractions to common denominators:
Notice how useful it was to factorize the original denominators:
- Having discovered identical factors, we immediately arrived at the least common multiple, which, generally speaking, is a non-trivial problem;
- From the resulting expansion you can find out which factors are “missing” in each fraction. For example, 234 · 3 = 702, therefore, for the first fraction the additional factor is 3.
To appreciate how much of a difference the least common multiple method makes, try calculating these same examples using the criss-cross method. Of course, without a calculator. I think after this comments will be unnecessary.
Don't think that there won't be such complex fractions in the real examples. They meet all the time, and the above tasks are not the limit!
The only problem is how to find this very NOC. Sometimes everything can be found in a few seconds, literally “by eye,” but in general this is a complex computational task that requires separate consideration. We won't touch on that here.
See also:
Reducing fractions to a common denominator
I originally wanted to include common denominator techniques in the Adding and Subtracting Fractions section. But there turned out to be so much information, and its importance is so great (after all, not only numerical fractions have common denominators), that it is better to study this issue separately.
So, let's say we have two fractions with different denominators. And we want to make sure that the denominators become the same. The basic property of a fraction comes to the rescue, which, let me remind you, sounds like this:
A fraction will not change if its numerator and denominator are multiplied by the same number other than zero.
Thus, if you choose the factors correctly, the denominators of the fractions will become equal - this process is called. And the required numbers, “evening out” the denominators, are called.
Why do we need to reduce fractions to a common denominator? Here are just a few reasons:
- Adding and subtracting fractions with different denominators. There is no other way to perform this operation;
- Comparing fractions. Sometimes reduction to a common denominator greatly simplifies this task;
- Solving problems involving fractions and percentages. Percentages are essentially ordinary expressions that contain fractions.
There are many ways to find numbers that, when multiplied by them, will make the denominators of fractions equal. We will consider only three of them - in order of increasing complexity and, in a sense, effectiveness.
Criss-cross multiplication
The simplest and most reliable method, which is guaranteed to equalize the denominators. We will act “in a headlong manner”: we multiply the first fraction by the denominator of the second fraction, and the second by the denominator of the first. As a result, the denominators of both fractions will become equal to the product of the original denominators.
Take a look:
Task. Find the meanings of the expressions:
As additional factors, consider the denominators of neighboring fractions. We get:
Yes, it's that simple. If you are just starting to study fractions, it is better to work using this method - this way you will insure yourself against many mistakes and are guaranteed to get the result.
The only drawback of this method is that you have to count a lot, because the denominators are multiplied “all the way”, and the result can be very large numbers. This is the price to pay for reliability.
Common Divisor Method
This technique helps to significantly reduce calculations, but, unfortunately, it is used quite rarely. The method is as follows:
- Before you go straight ahead (i.e., using the criss-cross method), take a look at the denominators. Perhaps one of them (the one that is larger) is divided into the other.
- The number resulting from this division will be an additional factor for the fraction with a smaller denominator.
- In this case, a fraction with a large denominator does not need to be multiplied by anything at all - this is where the savings lie. At the same time, the probability of error is sharply reduced.
Task. Find the meanings of the expressions:
Note that 84: 21 = 4; 72: 12 = 6. Since in both cases one denominator is divided without a remainder by the other, we use the method of common factors. We have:
Note that the second fraction was not multiplied by anything at all. In fact, we cut the amount of computation in half!
By the way, I didn’t take the fractions in this example by chance. If you're interested, try counting them using the criss-cross method. After reduction, the answers will be the same, but there will be much more work.
This is the power of the common divisors method, but, again, it can only be used when one of the denominators is divisible by the other without a remainder. Which happens quite rarely.
Least common multiple method
When we reduce fractions to a common denominator, we are essentially trying to find a number that is divisible by each denominator. Then we bring the denominators of both fractions to this number.
There are a lot of such numbers, and the smallest of them will not necessarily be equal to the direct product of the denominators of the original fractions, as is assumed in the “criss-cross” method.
For example, for denominators 8 and 12, the number 24 is quite suitable, since 24: 8 = 3; 24: 12 = 2. This number is much less than the product 8 12 = 96.
The smallest number that is divisible by each of the denominators is called their (LCM).
Notation: The least common multiple of a and b is denoted LCM(a; b). For example, LCM(16, 24) = 48; LCM(8; 12) = 24.
If you manage to find such a number, the total amount of calculations will be minimal. Look at the examples:
Task. Find the meanings of the expressions:
Note that 234 = 117 2; 351 = 117 · 3. Factors 2 and 3 are coprime (have no common factors other than 1), and factor 117 is common. Therefore LCM(234, 351) = 117 2 3 = 702.
Likewise, 15 = 5 3; 20 = 5 · 4. Factors 3 and 4 are coprime, and factor 5 is common. Therefore LCM(15, 20) = 5 3 4 = 60.
Now let's bring the fractions to common denominators:
Notice how useful it was to factorize the original denominators:
- Having discovered identical factors, we immediately arrived at the least common multiple, which, generally speaking, is a non-trivial problem;
- From the resulting expansion you can find out which factors are “missing” in each fraction. For example, 234 · 3 = 702, therefore, for the first fraction the additional factor is 3.
To appreciate how much of a difference the least common multiple method makes, try calculating these same examples using the criss-cross method. Of course, without a calculator. I think after this comments will be unnecessary.
Don't think that there won't be such complex fractions in the real examples. They meet all the time, and the above tasks are not the limit!
The only problem is how to find this very NOC. Sometimes everything can be found in a few seconds, literally “by eye,” but in general this is a complex computational task that requires separate consideration. We won't touch on that here.
See also:
Reducing fractions to a common denominator
I originally wanted to include common denominator techniques in the Adding and Subtracting Fractions section. But there turned out to be so much information, and its importance is so great (after all, not only numerical fractions have common denominators), that it is better to study this issue separately.
So, let's say we have two fractions with different denominators. And we want to make sure that the denominators become the same. The basic property of a fraction comes to the rescue, which, let me remind you, sounds like this:
A fraction will not change if its numerator and denominator are multiplied by the same number other than zero.
Thus, if you choose the factors correctly, the denominators of the fractions will become equal - this process is called. And the required numbers, “evening out” the denominators, are called.
Why do we need to reduce fractions to a common denominator? Here are just a few reasons:
- Adding and subtracting fractions with different denominators. There is no other way to perform this operation;
- Comparing fractions. Sometimes reduction to a common denominator greatly simplifies this task;
- Solving problems involving fractions and percentages. Percentages are essentially ordinary expressions that contain fractions.
There are many ways to find numbers that, when multiplied by them, will make the denominators of fractions equal. We will consider only three of them - in order of increasing complexity and, in a sense, effectiveness.
Criss-cross multiplication
The simplest and most reliable method, which is guaranteed to equalize the denominators. We will act “in a headlong manner”: we multiply the first fraction by the denominator of the second fraction, and the second by the denominator of the first. As a result, the denominators of both fractions will become equal to the product of the original denominators. Take a look:
Task. Find the meanings of the expressions:
As additional factors, consider the denominators of neighboring fractions. We get:
Yes, it's that simple. If you are just starting to study fractions, it is better to work using this method - this way you will insure yourself against many mistakes and are guaranteed to get the result.
The only drawback of this method is that you have to count a lot, because the denominators are multiplied “all the way”, and the result can be very large numbers.
Reducing fractions to a common denominator
This is the price to pay for reliability.
Common Divisor Method
This technique helps to significantly reduce calculations, but, unfortunately, it is used quite rarely. The method is as follows:
- Before you go straight ahead (i.e., using the criss-cross method), take a look at the denominators. Perhaps one of them (the one that is larger) is divided into the other.
- The number resulting from this division will be an additional factor for the fraction with a smaller denominator.
- In this case, a fraction with a large denominator does not need to be multiplied by anything at all - this is where the savings lie. At the same time, the probability of error is sharply reduced.
Task. Find the meanings of the expressions:
Note that 84: 21 = 4; 72: 12 = 6. Since in both cases one denominator is divided without a remainder by the other, we use the method of common factors. We have:
Note that the second fraction was not multiplied by anything at all. In fact, we cut the amount of computation in half!
By the way, I didn’t take the fractions in this example by chance. If you're interested, try counting them using the criss-cross method. After reduction, the answers will be the same, but there will be much more work.
This is the power of the common divisors method, but, again, it can only be used when one of the denominators is divisible by the other without a remainder. Which happens quite rarely.
Least common multiple method
When we reduce fractions to a common denominator, we are essentially trying to find a number that is divisible by each denominator. Then we bring the denominators of both fractions to this number.
There are a lot of such numbers, and the smallest of them will not necessarily be equal to the direct product of the denominators of the original fractions, as is assumed in the “criss-cross” method.
For example, for denominators 8 and 12, the number 24 is quite suitable, since 24: 8 = 3; 24: 12 = 2. This number is much less than the product 8 12 = 96.
The smallest number that is divisible by each of the denominators is called their (LCM).
Notation: The least common multiple of a and b is denoted LCM(a; b). For example, LCM(16, 24) = 48; LCM(8; 12) = 24.
If you manage to find such a number, the total amount of calculations will be minimal. Look at the examples:
Task. Find the meanings of the expressions:
Note that 234 = 117 2; 351 = 117 · 3. Factors 2 and 3 are coprime (have no common factors other than 1), and factor 117 is common. Therefore LCM(234, 351) = 117 2 3 = 702.
Likewise, 15 = 5 3; 20 = 5 · 4. Factors 3 and 4 are coprime, and factor 5 is common. Therefore LCM(15, 20) = 5 3 4 = 60.
Now let's bring the fractions to common denominators:
Notice how useful it was to factorize the original denominators:
- Having discovered identical factors, we immediately arrived at the least common multiple, which, generally speaking, is a non-trivial problem;
- From the resulting expansion you can find out which factors are “missing” in each fraction. For example, 234 · 3 = 702, therefore, for the first fraction the additional factor is 3.
To appreciate how much of a difference the least common multiple method makes, try calculating these same examples using the criss-cross method. Of course, without a calculator. I think after this comments will be unnecessary.
Don't think that there won't be such complex fractions in the real examples. They meet all the time, and the above tasks are not the limit!
The only problem is how to find this very NOC. Sometimes everything can be found in a few seconds, literally “by eye,” but in general this is a complex computational task that requires separate consideration. We won't touch on that here.
To reduce fractions to the lowest common denominator, you need to: 1) find the least common multiple of the denominators of the given fractions, it will be the lowest common denominator. 2) find an additional factor for each fraction by dividing 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 their 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. Find for the 1st fraction an additional factor 4 (20 : 5=4). For the 2nd fraction the additional factor 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 have reduced these fractions to the lowest common denominator ( 20 ).
The lowest common denominator of these fractions is the number 8, since 8 is divisible by 4 and itself. There will be no additional factor for the 1st fraction (or we can say that it is equal to one), for the 2nd fraction the additional factor is 2 (8 : 4=2). We multiply the numerator and denominator of the 2nd fraction by 2. We have reduced these fractions to the lowest common denominator ( 8 ).
These fractions are not irreducible.
Let's reduce the 1st fraction by 4, and reduce the 2nd fraction by 2. ( see examples on reducing ordinary fractions: Sitemap → 5.4.2. Examples of reducing common fractions). Find the LOC(16 ; 20)=2 4 · 5=16· 5=80. The additional multiplier for the 1st fraction is 5 (80 : 16=5). The additional factor 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 have reduced these fractions to the lowest common denominator ( 80 ).
We find the lowest common denominator NCD(5 ; 6 and 15)=NOK(5 ; 6 and 15)=30. The additional factor to the 1st fraction is 6 (30 : 5=6), the additional factor to the 2nd fraction is 5 (30 : 6=5), the additional factor 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 have reduced these fractions to the lowest common denominator ( 30 ).
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In this lesson we will look at reducing fractions to a common denominator and solve problems on this topic. Let's define the concept of a common denominator and an additional factor, and remember about relatively prime numbers. Let's define the concept of the lowest 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. The main property of a fraction.
If the numerator and denominator of a fraction are multiplied or divided by the same natural number, you get an equal fraction.
For example, the numerator and denominator of a fraction can be divided by 2. We get the 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. To bring a fraction to a new denominator, its numerator and denominator are multiplied by an additional factor.
1. Reduce the fraction to the denominator 35.
The number 35 is a multiple of 7, that is, 35 is divisible by 7 without a remainder. This means that this transformation is possible. Let's find an additional factor. To do this, divide 35 by 7. We get 5. Multiply the numerator and denominator of the original fraction by 5.
2. Reduce the fraction to denominator 18.
Let's find an additional factor. To do this, divide the new denominator by the original one. We get 3. Multiply the numerator and denominator of this fraction by 3.
3. Reduce the fraction to a denominator of 60.
Dividing 60 by 15 gives an additional factor. It is equal to 4. Multiply the numerator and denominator by 4.
4. Reduce the fraction to the denominator 24
In simple cases, reduction to a new denominator is performed mentally. It is only customary to indicate the additional factor behind a bracket slightly 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 also 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 their lowest common denominator. It is equal to the least common multiple of the denominators of the given fractions.
Example. Reduce to the lowest common denominator of the fraction and .
First, let's 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, divide 12 by 4 and 6. Three is an additional factor for the first fraction, and two is for the second. Let's bring the fractions to the denominator 12.
We brought the fractions to a common denominator, that is, we found equal fractions that have the same denominator.
Rule. To reduce fractions to their lowest common denominator, you must
First, find the least common multiple of the denominators of these fractions, it will be their least common denominator;
Secondly, divide the lowest common denominator by the denominators of these fractions, i.e. find an additional factor for each fraction.
Third, 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 reduce 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 gives 5 and 3, respectively. We reduce the fractions to the denominator 45.
c) Reduce the fractions and to a common denominator.
The common denominator is 24. Additional factors are 2 and 3, respectively.
Sometimes it can be difficult to verbally find the least common multiple of the denominators of given fractions. Then the common denominator and additional factors are found using prime factorization.
Reduce the fractions and to a common denominator.
Let's factor 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. Let's 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 bring the fractions to a common denominator of 840.
References
1. Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S. and others. Mathematics 6. - M.: Mnemosyne, 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., Tchaikovsky I.V. Assignments for the mathematics course for grades 5-6. - ZSh MEPhI, 2011.
5. Rurukin A.N., Sochilov S.V., Tchaikovsky K.G. Mathematics 5-6. A manual for 6th grade students at the MEPhI correspondence school. - ZSh MEPhI, 2011.
6. Shevrin L.N., Gein A.G., Koryakov I.O. and others. Mathematics: Textbook-interlocutor for 5-6 grades of secondary school. Math teacher's library. - Enlightenment, 1989.
You can download the books specified in clause 1.2. of this lesson.
Homework
Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S. and others. Mathematics 6. - M.: Mnemosyne, 2012. (link see 1.2)
Homework: No. 297, No. 298, No. 300.
Other tasks: No. 270, No. 290