How fractions are counted. How to solve examples with fractions. How to find the difference of fractions with the same denominator
Pupils get acquainted with fractions in the 5th grade. Previously, people who knew how to perform actions with fractions were considered very smart. The first fraction was 1/2, that is, half, then 1/3 appeared, etc. For several centuries, examples were considered too complex. Now, detailed rules have been developed for converting fractions, addition, multiplication and other actions. It is enough to understand the material a little, and the solution will be easy.
An ordinary fraction, called a simple fraction, is written as a division of two numbers: m and n.
M is the dividend, that is, the numerator of the fraction, and the divisor n is called the denominator.
Allocate correct fractions (m< n) а также неправильные (m > n).
A regular fraction is less than one (for example, 5/6 - this means that 5 parts are taken from one; 2/8 - 2 parts are taken from one). The irregular fraction is equal to or greater than 1 (8/7 - 1 is 7/7 and one more part is taken as a plus).
So, a unit is when the numerator and denominator coincide (3/3, 12/12, 100/100 and others).
Actions with ordinary fractions grade 6
With simple fractions, you can do the following:
- Expand fraction. If you multiply the upper and lower parts of the fraction by any of the same number (but not zero), then the value of the fraction will not change (3/5 \u003d 6/10 (just multiplied by 2).
- Reduction of fractions is similar to expansion, but here it is divided by some number.
- Compare. If two fractions have the same numerators, then the larger fraction will be the fraction with the lower denominator. If the denominators are the same, then the fraction with the largest numerator will be larger.
- Perform addition and subtraction. With the same denominators, this is easy to do (we sum up the upper parts, and the lower does not change). For different, you will have to find a common denominator and additional factors.
- Multiply and divide fractions.
We will consider examples of actions with fractions below.
Reduced fractions grade 6
To abbreviate means to divide the upper and lower parts of the fraction by any of the same number.
The figure shows simple examples of abbreviation. In the first option, you can immediately guess that the numerator and denominator are divisible by 2.
On a note! If the number is even, then it is in any way divisible by 2. Even numbers are 2, 4, 6 ... 32 8 (ends with even), etc.
In the second case, when dividing 6 by 18, it is immediately clear that the numbers are divisible by 2. Dividing, we get 3/9. This fraction is divisible by 3. Then the answer is 1/3. If you multiply both factors: 2 by 3, then you get 6. It turns out that the fraction was divided by six. This gradual division is called successive reduction of fractions by common factors.
Someone will immediately divide by 6, someone will need division by parts. The main thing is that at the end there is a fraction that cannot be reduced in any way.
Note that if a number consists of digits, adding up to a number divisible by 3, then the original can also be reduced by 3. Example: number 341. Add the numbers: 3 + 4 + 1 \u003d 8 (8 cannot be divided by 3, hence, the number 341 cannot be reduced by 3 without a remainder). Another example: 264. Add: 2 + 6 + 4 \u003d 12 (divisible by 3). We get: 264: 3 \u003d 88. This will simplify the reduction of large numbers.
In addition to the method of successive reduction of fractions by common factors, there are other methods.
GCD is the largest divisor for a number. Having found the GCD for the denominator and numerator, you can immediately reduce the fraction by the desired number. The search is carried out by gradually dividing each number. Next, they look at which divisors coincide, if there are several of them (as in the picture below), then you need to multiply.
Mixed fractions grade 6
All irregular fractions can be turned into mixed ones by selecting the whole part in them. The whole number is written to the left.
Often you have to make a mixed number out of an improper fraction. The transformation process in the example below: 22/4 \u003d 22 we divide by 4, we get 5 integers (5 * 4 \u003d 20). 22 - 20 \u003d 2. We get 5 integers and 2/4 (the denominator does not change). Since the fraction can be canceled, we divide the top and bottom by 2.
It is easy to turn a mixed number into an improper fraction (this is necessary when dividing and multiplying fractions). To do this: multiply the whole number by the lower part of the fraction and add the numerator to this. Done. The denominator does not change.
Calculations with fractions grade 6
Mixed numbers can be added. If the denominators are the same, then this is easy to do: add the whole parts and numerators, the denominator remains in place.
When adding numbers with different denominators, the process is more complicated. First, we bring the numbers to the one smallest denominator (NOZ).
In the example below, for the numbers 9 and 6, the denominator is 18. After that, additional factors are needed. To find them, 18 should be divided by 9, so the additional number is found - 2. We multiply it by the numerator 4 to get the fraction 8/18). The same is done with the second fraction. We are already adding up the converted fractions (integers and numerators are separate, the denominator is not changed). In the example, the answer had to be converted into a regular fraction (initially, the numerator was larger than the denominator).
Please note that the procedure is the same for the difference in fractions.
When multiplying fractions, it is important to place both under the same line. If the number is mixed, then we turn it into a simple fraction. Next, we multiply the top and bottom and write down the answer. If you can see that the fractions can be reduced, then we reduce them immediately.
In the above example, we didn't have to cut anything, we just wrote down the answer and selected the whole part.
In this example, I had to abbreviate the numbers below one line. Although you can shorten a ready-made answer.
For division, the algorithm is almost the same. First, we turn the mixed fraction into an irregular one, then write the numbers under one line, replacing division with multiplication. Do not forget to swap the upper and lower parts of the second fraction (this is the rule for dividing fractions).
If necessary, we reduce the numbers (in the example below we have reduced them by five and two). We transform the irregular fraction by selecting the whole part.
Basic problems for fractions grade 6
The video shows a few more tasks. For clarity, graphic images of solutions have been used to help visualize fractions.
Examples of multiplication of a fraction grade 6 with explanations
Multiplying fractions are written under one line. After that, they are reduced by dividing by the same numbers (for example, 15 in the denominator and 5 in the numerator can be divided by five).
Comparison of fractions grade 6
To compare fractions, you need to remember two simple rules.
Rule 1. If the denominators are different
Rule 2. When the denominators are the same
For example, let's compare the fractions 7/12 and 2/3.
- We look at the denominators, they do not coincide. So you need to find a common one.
- For fractions, the common denominator is 12.
- We divide 12 first by the lower part of the first fraction: 12: 12 \u003d 1 (this is an additional factor for the 1st fraction).
- Now we divide 12 by 3, we get 4 - add. multiplier of the 2nd fraction.
- We multiply the resulting numbers by the numerators to convert the fractions: 1 x 7 \u003d 7 (first fraction: 7/12); 4 x 2 \u003d 8 (second fraction: 8/12).
- Now we can compare: 7/12 and 8/12. Happened: 7/12< 8/12.
To represent fractions better, you can use drawings for clarity, where the object is divided into parts (for example, a cake). If you want to compare 4/7 and 2/3, then in the first case, the cake is divided into 7 parts and 4 of them are selected. In the second, they divide it into 3 parts and take 2. It will be clear to the naked eye that 2/3 will be more than 4/7.
Examples with fractions grade 6 for training
As a workout, you can do the following tasks.
- Compare fractions
- perform multiplication
Tip: if it is difficult to find the lowest common denominator for fractions (especially if their values \u200b\u200bare small), then you can multiply the denominator of the first and second fractions. Example: 2/8 and 5/9. Finding their denominator is simple: multiply 8 by 9, we get 72.
Solving equations with fractions grade 6
In solving equations, you need to remember the actions with fractions: multiplication, division, subtraction and addition. If one of the factors is unknown, then the product (total) is divided by a known factor, that is, the fractions are multiplied (the second is turned over).
If the dividend is unknown, then the denominator is multiplied by the divisor, and to find the divisor, the dividend must be divided by the quotient.
Let's present simple examples of solving equations:
Here it is only required to produce the difference of fractions without leading to a common denominator.
The answer came out as an incorrect fraction. It can be converted to 1 integer and 3/5.
In the second method, the numerator and denominator were multiplied by 4 to cancel out the bottom, rather than flip the denominator.
Instructions
It is customary to separate ordinary and decimal fractions, acquaintance with which begins in high school. There is currently no area of \u200b\u200bexpertise that does not apply this. Even in we say the first 17th century, and all at once, which means 1600-1625. You also often have to deal with elementary operations on fractions, as well as their transformation from one type to another.
Bringing fractions to a common denominator is perhaps the most important action on common fractions. This is the basis for absolutely all calculations. So, let's say there are two fractions a / b and c / d. Then, in order to bring them to a common denominator, you need to find the least common multiple (M) of the numbers b and d, and then multiply the numerator of the first fraction by (M / b), and the numerator of the second by (M / d).
Comparison of fractions is another important task. In order to do this, bring the given simple fractions to a common denominator and then compare the numerators, whose numerator is greater, that fraction and more.
In order to perform addition or subtraction of ordinary fractions, you need to bring them to a common denominator, and then perform the desired mathematical action with the numerators of these fractions. The denominator remains unchanged. Let's say you need to subtract c / d from a / b. To do this, you need to find the least common multiple M of the numbers b and d, and then subtract the other from one numerator without changing the denominator: (a * (M / b) - (c * (M / d)) / M
It is enough just to multiply one fraction by another, for this you just need to multiply their numerators and denominators:
(a / b) * (c / d) \u003d (a * c) / (b * d) To divide one fraction by another, you need to multiply the fraction of the dividend by the inverse fraction. (a / b) / (c / d) \u003d (a * d) / (b * c)
It is worth recalling that in order to get the reciprocal fraction, the numerator and denominator must be reversed.
To add 2 fractions with the same denominators, it is necessary to add their numerators, and the denominatorsleave unchanged.Adding fractions, examples:
The general formula for adding ordinary fractions and subtracting fractions with the same denominator is:
Note! Check if you can reduce the fraction that you received by writing down the answer.
Adding fractions with different denominators.
The rules for adding fractions with different denominators:
- reduce fractions to the lowest common denominator (LCN). For this we find the smallest common multiple (LCM) of denominators;
- add up the numerators of fractions, and leave the denominators unchanged;
- we reduce the fraction that we received;
- if you get an incorrect fraction, convert the incorrect fraction to a mixed fraction.
Examples of additions fractions with different denominators:
Addition of mixed numbers (mixed fractions).
Rules for adding mixed fractions:
- we bring the fractional parts of these numbers to the lowest common denominator (LCN);
- separately add whole parts and separately fractional parts, add up the results;
- if, when adding the fractional parts, we received an incorrect fraction, select the whole part from this fraction and add it to the resulting whole part;
- we reduce the resulting fraction.
Example additions mixed fraction:
Adding decimal fractions.
When adding decimal fractions, the process is written in "column" (as usual column multiplication),so that the discharges of the same name are under each other without displacement. Commas are requiredwe align clearly under each other.
The rules for adding decimal fractions:
1. If necessary, equalize the number of decimal places. To do this, add zeros tothe required fraction.
2. We write down fractions so that the commas are below each other.
3. Add fractions without paying attention to the comma.
4. We put a comma in the sum under the commas, the fractions that we add.
Note! When the given decimal fractions have a different number of digits after the decimal point,then we assign the required number of zeros to the fraction with fewer decimal places, for the equation infractions are the number of decimal places.
Let's understand example... Find the sum of decimal fractions:
0,678 + 13,7 =
We equalize the number of decimal places in decimal fractions. Add 2 zeros to the right to the decimalfractions 13,7 .
0,678 + 13,700 =
We write down answer:
0,678 + 13,7 = 14,378
If a addition of decimal fractions you have mastered it well enough, then the missing zeros can be addedin the mind.
This article begins the study of actions with algebraic fractions: we will consider in detail such actions as addition and subtraction of algebraic fractions. Let us analyze the scheme of addition and subtraction of algebraic fractions with both the same denominators and different ones. We will learn how to add an algebraic fraction with a polynomial and how to subtract them. Let us explain each step of the search for a solution to problems with specific examples.
Addition and Subtraction Actions with the Same Denominators
The scheme for adding ordinary fractions is applicable to algebraic ones. We know that when adding or subtracting ordinary fractions with the same denominators, it is necessary to add or subtract their numerators, and the denominator remains the original.
For example: 3 7 + 2 7 \u003d 3 + 2 7 \u003d 5 7 and 5 11 - 4 11 \u003d 5 - 4 11 \u003d 1 11.
Accordingly, the rule for addition and subtraction of algebraic fractions with the same denominators is written in a similar way:
Definition 1
To add or subtract algebraic fractions with the same denominators, you need to add or subtract the numerators of the original fractions, respectively, and write the denominator unchanged.
This rule makes it possible to conclude that the result of addition or subtraction of algebraic fractions is a new algebraic fraction (in a particular case: a polynomial, monomial or number).
Let us give an example of applying the formulated rule.
Example 1
Algebraic fractions are given: x 2 + 2 x y - 5 x 2 y - 2 and 3 - x y x 2 y - 2. It is necessary to add them.
Decision
The original fractions contain the same denominators. According to the rule, let's add the numerators of the given fractions, and leave the denominator unchanged.
Adding the polynomials that are the numerators of the original fractions, we get: x 2 + 2 x y - 5 + 3 - x y \u003d x 2 + (2 x y - x y) - 5 + 3 \u003d x 2 + x y - 2.
Then the required sum will be written as: x 2 + x · y - 2 x 2 · y - 2.
In practice, as in many cases, the solution is given by a chain of equalities, clearly showing all the stages of the solution:
x 2 + 2 x y - 5 x 2 y - 2 + 3 - x yx 2 y - 2 \u003d x 2 + 2 x y - 5 + 3 - x yx 2 y - 2 \u003d x 2 + x y - 2 x 2 y - 2
Answer: x 2 + 2 x y - 5 x 2 y - 2 + 3 - x y x 2 y - 2 \u003d x 2 + x y - 2 x 2 y - 2.
The result of addition or subtraction can be a cancellable fraction, in this case it is optimal to reduce it.
Example 2
It is necessary to subtract from the algebraic fraction x x 2 - 4 · y 2 the fraction 2 · y x 2 - 4 · y 2.
Decision
The denominators of the original fractions are equal. Let's perform actions with the numerators, namely: subtract the numerator of the second from the numerator of the first fraction, and then write down the result, leaving the denominator unchanged:
x x 2 - 4 y 2 - 2 y x 2 - 4 y 2 \u003d x - 2 y x 2 - 4 y 2
We see that the resulting fraction is a cancellable one. Let us reduce it by transforming the denominator using the difference of squares formula:
x - 2 y x 2 - 4 y 2 \u003d x - 2 y (x - 2 y) (x + 2 y) \u003d 1 x + 2 y
Answer: x x 2 - 4 y 2 - 2 y x 2 - 4 y 2 \u003d 1 x + 2 y.
According to the same principle, three or more algebraic fractions are added or subtracted with the same denominators. For instance:
1 x 5 + 2 x 3 - 1 + 3 x - x 4 x 5 + 2 x 3 - 1 - x 2 x 5 + 2 x 3 - 1 - 2 x 3 x 5 + 2 x 3 - 1 \u003d 1 + 3 x - x 4 - x 2 - 2 x 3 x 5 + 2 x 3 - 1
Addition and Subtraction Actions for Different Denominators
Let us again turn to the scheme of actions with ordinary fractions: in order to add or subtract ordinary fractions with different denominators, it is necessary to bring them to a common denominator, and then add the resulting fractions with the same denominators.
For example, 2 5 + 1 3 \u003d 6 15 + 5 15 \u003d 11 15 or 1 2 - 3 7 \u003d 7 14 - 6 14 \u003d 1 14.
Similarly, we will formulate the rule of addition and subtraction of algebraic fractions with different denominators:
Definition 2
To carry out addition or subtraction of algebraic fractions with different denominators, you must:
- reduce the original fractions to a common denominator;
- perform addition or subtraction of the resulting fractions with the same denominators.
Obviously, the key here will be the skill of bringing algebraic fractions to a common denominator. Let's take a closer look.
Common denominator of algebraic fractions
To bring algebraic fractions to a common denominator, it is necessary to carry out an identical transformation of the given fractions, as a result of which the denominators of the original fractions become the same. Here it is optimal to act according to the following algorithm for bringing algebraic fractions to a common denominator:
- first we determine the common denominator of algebraic fractions;
- then we find additional factors for each of the fractions by dividing the common denominator by the denominators of the original fractions;
- by the last action, the numerators and denominators of the given algebraic fractions are multiplied by the corresponding additional factors.
Algebraic fractions are given: a + 2 2 a 3 - 4 a 2, a + 3 3 a 2 - 6 a and a + 1 4 a 5 - 16 a 3. It is necessary to bring them to a common denominator.
Decision
We act according to the above algorithm. Let's determine the common denominator of the original fractions. For this purpose, we factor out the denominators of the given fractions: 2 a 3 - 4 a 2 \u003d 2 a 2 (a - 2), 3 a 2 - 6 a \u003d 3 a (a - 2) and 4 a 5 - 16 a 3 \u003d 4 a 3 (a - 2) (a + 2)... From here we can write down the common denominator: 12 a 3 (a - 2) (a + 2).
Now we have to find additional factors. Let us divide, according to the algorithm, the found common denominator into the denominators of the original fractions:
- for the first fraction: 12 a 3 (a - 2) (a + 2): (2 a 2 (a - 2)) \u003d 6 a (a + 2);
- for the second fraction: 12 a 3 (a - 2) (a + 2): (3 a (a - 2)) \u003d 4 a 2 (a + 2);
- for the third fraction: 12 a 3 (a - 2) (a + 2): (4 a 3 (a - 2) (a + 2)) \u003d 3 .
The next step is to multiply the numerators and denominators of the given fractions by the additional factors found:
a + 2 2 a 3 - 4 a 2 \u003d (a + 2) 6 a (a + 2) (2 a 3 - 4 a 2) 6 a (a + 2) \u003d 6 a (a + 2) 2 12 a 3 (a - 2) (a + 2) a + 3 3 a 2 - 6 a \u003d (a + 3) 4 a 2 ( a + 2) 3 a 2 - 6 a 4 a 2 (a + 2) \u003d 4 a 2 (a + 3) (a + 2) 12 a 3 (a - 2) (A + 2) a + 1 4 a 5 - 16 a 3 \u003d (a + 1) 3 (4 a 5 - 16 a 3) 3 \u003d 3 (a + 1) 12 a 3 (a - 2) (a + 2)
Answer: a + 2 2 a 3 - 4 a 2 \u003d 6 a (a + 2) 2 12 a 3 (a - 2) (a + 2); a + 3 3 a 2 - 6 a \u003d 4 a 2 (a + 3) (a + 2) 12 a 3 (a - 2) (a + 2); a + 1 4 a 5 - 16 a 3 \u003d 3 (a + 1) 12 a 3 (a - 2) (a + 2).
So, we brought the original fractions to a common denominator. If necessary, you can further transform the result into the form of algebraic fractions by multiplying polynomials and monomials in the numerators and denominators.
Let us also clarify the following point: it is optimal to leave the found common denominator in the form of a product in case it is necessary to cancel the finite fraction.
We examined in detail the scheme for reducing the original algebraic fractions to a common denominator, now we can begin to analyze examples for addition and subtraction of fractions with different denominators.
Example 4
Algebraic fractions are given: 1 - 2 x x 2 + x and 2 x + 5 x 2 + 3 x + 2. It is necessary to carry out the action of their addition.
Decision
The original fractions have different denominators, so the first step is to bring them to a common denominator. Factor the denominators: x 2 + x \u003d x (x + 1), and x 2 + 3 x + 2 \u003d (x + 1) (x + 2),since square trinomial roots x 2 + 3 x + 2 these are numbers: - 1 and - 2. Determine the common denominator: x (x + 1) (x + 2), then the additional factors will be: x + 2and - xfor the first and second fractions, respectively.
Thus: 1 - 2 xx 2 + x \u003d 1 - 2 xx (x + 1) \u003d (1 - 2 x) (x + 2) x (x + 1) (x + 2) \u003d x + 2 - 2 x 2 - 4 xx (x + 1) x + 2 \u003d 2 - 2 x 2 - 3 xx (x + 1) (x + 2) and 2 x + 5 x 2 + 3 x + 2 \u003d 2 x + 5 (x + 1) (x + 2) \u003d 2 x + 5 x (x + 1) (x + 2) x \u003d 2 X 2 + 5 xx (x + 1) (x + 2)
Now let's add the fractions that we brought to a common denominator:
2 - 2 x 2 - 3 xx (x + 1) (x + 2) + 2 x 2 + 5 xx (x + 1) (x + 2) \u003d \u003d 2 - 2 x 2 - 3 x + 2 x 2 + 5 xx (x + 1) (x + 2) \u003d 2 2 xx (x + 1) (x + 2)
The resulting fraction can be reduced by a common factor x + 1:
2 + 2 x x (x + 1) (x + 2) \u003d 2 (x + 1) x (x + 1) (x + 2) \u003d 2 x (x + 2)
And, finally, we write the result obtained in the form of an algebraic fraction, replacing the product in the denominator with a polynomial:
2 x (x + 2) \u003d 2 x 2 + 2 x
Let us write down the course of the solution briefly as a chain of equalities:
1 - 2 xx 2 + x + 2 x + 5 x 2 + 3 x + 2 \u003d 1 - 2 xx (x + 1) + 2 x + 5 (x + 1) (x + 2 ) \u003d \u003d 1 - 2 x (x + 2) x x + 1 x + 2 + 2 x + 5 x (x + 1) (x + 2) x \u003d 2 - 2 x 2 - 3 xx (x + 1) (x + 2) + 2 x 2 + 5 xx (x + 1) (x + 2) \u003d \u003d 2 - 2 x 2 - 3 x + 2 x 2 + 5 xx (x + 1) (x + 2) \u003d 2 x + 1 x (x + 1) (x + 2) \u003d 2 x (x + 2) \u003d 2 x 2 + 2 x
Answer: 1 - 2 x x 2 + x + 2 x + 5 x 2 + 3 x + 2 \u003d 2 x 2 + 2 x
Pay attention to this detail: before adding or subtracting algebraic fractions, if possible, it is desirable to transform them in order to simplify.
Example 5
It is necessary to subtract fractions: 2 1 1 3 · x - 2 21 and 3 · x - 1 1 7 - 2 · x.
Decision
We transform the original algebraic fractions to simplify the further solution. Let's take out the numeric coefficients of the variables in the denominator outside the brackets:
2 1 1 3 x - 2 21 \u003d 2 4 3 x - 2 21 \u003d 2 4 3 x - 1 14 and 3 x - 1 1 7 - 2 x \u003d 3 x - 1 - 2 x - 1 14
This transformation definitely gave us a benefit: we clearly see the presence of a common factor.
Let's get rid of the numerical coefficients in the denominators altogether. To do this, we use the main property of algebraic fractions: we multiply the numerator and denominator of the first fraction by 3 4, and the second by - 1 2, then we get:
2 4 3 x - 1 14 \u003d 3 4 2 3 4 4 3 x - 1 14 \u003d 3 2 x - 1 14 and 3 x - 1 - 2 x - 1 14 \u003d - 1 2 3 x - 1 - 1 2 - 2 x - 1 14 \u003d - 3 2 x + 1 2 x - 1 14.
Let's take an action that will allow us to get rid of fractional coefficients: multiply the resulting fractions by 14:
3 2 x - 1 14 \u003d 14 3 2 14 x - 1 14 \u003d 21 14 x - 1 and - 3 2 x + 1 2 x - 1 14 \u003d 14 - 3 2 x + 1 2 x - 1 14 \u003d - 21 x + 7 14 x - 1.
Finally, we perform the required action in the problem statement - subtraction:
2 1 1 3 x - 2 21 - 3 x - 1 1 7 - 2 x \u003d 21 14 x - 1 - - 21 x + 7 14 x - 1 \u003d 21 - - 21 x + 7 14 X - 1 \u003d 21 x + 14 14 x - 1
Answer: 2 1 1 3 x - 2 21 - 3 x - 1 1 7 - 2 x \u003d 21 x + 14 14 x - 1.
Addition and subtraction of an algebraic fraction and a polynomial
This action is also reduced to the addition or subtraction of algebraic fractions: it is necessary to represent the original polynomial as a fraction with denominator 1.
Example 6
It is necessary to add the polynomial x 2 - 3 with an algebraic fraction 3 x x + 2.
Decision
We write the polynomial as an algebraic fraction with denominator 1: x 2 - 3 1
Now we can perform addition according to the rule of adding fractions with different denominators:
x 2 - 3 + 3 xx + 2 \u003d x 2 - 3 1 + 3 xx + 2 \u003d x 2 - 3 (x + 2) 1 x + 2 + 3 xx + 2 \u003d \u003d x 3 + 2 X 2 - 3 x - 6 x + 2 + 3 xx + 2 \u003d x 3 + 2 x 2 - 3 x - 6 + 3 xx + 2 \u003d \u003d x 3 + 2 x 2 - 6 x + 2
Answer: x 2 - 3 + 3 x x + 2 \u003d x 3 + 2 x 2 - 6 x + 2.
If you notice an error in the text, please select it and press Ctrl + Enter
The next action you can do with fractions is subtraction. Within the framework of this material, we will consider how to correctly calculate the difference of fractions with the same and different denominators, how to subtract a fraction from a natural number and vice versa. All examples will be illustrated with tasks. Let us clarify in advance that we will analyze only the cases when the difference of the fractions results in a positive number.
How to find the difference of fractions with the same denominator
Let's start right away with an illustrative example: let's say we have an apple that has been divided into eight parts. Let's leave five pieces on the plate and take two of them. This action can be written like this:
As a result, we have 3 eighths left, since 5 - 2 \u003d 3. It turns out that 5 8 - 2 8 \u003d 3 8.
Thanks to this simple example, we saw exactly how the subtraction rule works for fractions with the same denominators. Let's formulate it.
Definition 1
To find the difference of fractions with the same denominator, you need to subtract the numerator of the other from the numerator of one, and leave the denominator the same. This rule can be written as a b - c b \u003d a - c b.
We will use this formula in the future.
Let's take specific examples.
Example 1
Subtract the fraction 17 15 from 24 15.
Decision
We see that these fractions have the same denominators. So all we have to do is subtract 17 from 24. We get 7 and add the denominator to it, we get 7 15.
Our calculations can be written as follows: 24 15 - 17 15 \u003d 24 - 17 15 \u003d 7 15
If necessary, you can reduce the complex fraction or select the whole part from the wrong one to make it easier to count.
Example 2
Find the difference 37 12 - 15 12.
Decision
Let's use the formula described above and calculate: 37 12 - 15 12 \u003d 37 - 15 12 \u003d 22 12
It is easy to see that the numerator and denominator can be divided by 2 (we talked about this earlier when we examined the divisibility criteria). Reducing the answer, we get 11 6. This is an improper fraction, from which we will select the whole part: 11 6 \u003d 1 5 6.
How to find the difference of fractions with different denominators
Such a mathematical action can be reduced to what we have already described above. To do this, we simply bring the required fractions to one denominator. Let's formulate the definition:
Definition 2
To find the difference between fractions with different denominators, you need to bring them to the same denominator and find the difference in the numerators.
Let's look at an example how this is done.
Example 3
Subtract 1 15 from 2 9.
Decision
The denominators are different, and you need to bring them to the lowest common value. In this case, the LCM is 45. For the first fraction, an additional factor of 5 is required, and for the second, an additional factor of 3.
Let's calculate: 2 9 \u003d 2 5 9 5 \u003d 10 45 1 15 \u003d 1 3 15 3 \u003d 3 45
We got two fractions with the same denominator, and now we can easily find their difference according to the previously described algorithm: 10 45 - 3 45 \u003d 10 - 3 45 \u003d 7 45
A short record of the solution looks like this: 2 9 - 1 15 \u003d 10 45 - 3 45 \u003d 10 - 3 45 \u003d 7 45.
You should not neglect to reduce the result or extract a whole part from it, if necessary. In this example, we don't need to do this.
Example 4
Find the difference 19 9 - 7 36.
Decision
Let us bring the fractions indicated in the condition to the lowest common denominator 36 and get respectively 76 9 and 7 36.
We calculate the answer: 76 36 - 7 36 \u003d 76 - 7 36 \u003d 69 36
The result can be reduced by 3 and get 23 12. The numerator is larger than the denominator, which means we can select the whole part. The final answer is 1 11 12.
A summary of the entire solution is 19 9 - 7 36 \u003d 1 11 12.
How to subtract a natural number from an ordinary fraction
This action can also be easily reduced to a simple subtraction of ordinary fractions. This can be done by representing a natural number as a fraction. Let's show with an example.
Example 5
Find the difference 83 21 - 3.
Decision
3 is the same as 3 1. Then it can be calculated like this: 83 21 - 3 \u003d 20 21.
If it is necessary to subtract an integer from an improper fraction in a condition, it is more convenient to first select an integer from it by writing it as a mixed number. Then the previous example can be solved differently.
From the fraction 83 21, when the whole part is selected, we get 83 21 \u003d 3 20 21.
Now let's just subtract 3 from it: 3 20 21 - 3 \u003d 20 21.
How to subtract a fraction from a natural number
This action is done similarly to the previous one: we rewrite the natural number as a fraction, bring both to a single denominator and find the difference. Let us illustrate this with an example.
Example 6
Find the difference: 7 - 5 3.
Decision
Make 7 as 7 1. We subtract and transform the final result, extracting the whole part from it: 7 - 5 3 \u003d 5 1 3.
There is another way to make calculations. It has some advantages that can be used in cases where the numerators and denominators of the fractions in the problem are large numbers.
Definition 3
If the fraction to be subtracted is correct, then the natural number from which we subtract must be represented as the sum of two numbers, one of which is equal to 1. After that, you need to subtract the desired fraction from one and get the answer.
Example 7
Calculate the difference 1 065 - 13 62.
Decision
The fraction to be subtracted is correct, because its numerator is less than the denominator. Therefore, we need to subtract one from 1065 and subtract the desired fraction from it: 1065 - 13 62 \u003d (1064 + 1) - 13 62
Now we need to find the answer. Using the properties of subtraction, the resulting expression can be written as 1064 + 1 - 13 62. Let's calculate the difference in parentheses. For this, we represent the unit as a fraction 1 1.
It turns out that 1 - 13 62 \u003d 1 1 - 13 62 \u003d 62 62 - 13 62 \u003d 49 62.
Now let's remember about 1064 and formulate the answer: 1064 49 62.
We use the old method to prove that it is less convenient. These are the calculations we would get:
1065 - 13 62 \u003d 1065 1 - 13 62 \u003d 1065 62 1 62 - 13 62 \u003d 66030 62 - 13 62 \u003d \u003d 66030 - 13 62 \u003d 66017 62 \u003d 1064 4 6
The answer is the same, but the calculations are obviously more cumbersome.
We have considered the case when you need to subtract the correct fraction. If it is not correct, we replace it with a mixed number and subtract according to familiar rules.
Example 8
Calculate the difference 644 - 73 5.
Decision
The second fraction is incorrect, and the whole part must be separated from it.
Now we calculate similarly to the previous example: 630 - 3 5 \u003d (629 + 1) - 3 5 \u003d 629 + 1 - 3 5 \u003d 629 + 2 5 \u003d 629 2 5
Subtraction properties for fractions
The properties that the subtraction of natural numbers possesses also apply to the cases of subtraction of ordinary fractions. Let's see how to use them when solving examples.
Example 9
Find the difference 24 4 - 3 2 - 5 6.
Decision
We have already solved similar examples when we analyzed the subtraction of a sum from a number, so we are acting according to an already known algorithm. First, let's calculate the difference 25 4 - 3 2, and then subtract the last fraction from it:
25 4 - 3 2 = 24 4 - 6 4 = 19 4 19 4 - 5 6 = 57 12 - 10 12 = 47 12
Let's transform the answer by extracting the whole part from it. The total is 3 11 12.
Summary of the entire solution:
25 4 - 3 2 - 5 6 = 25 4 - 3 2 - 5 6 = 25 4 - 6 4 - 5 6 = = 19 4 - 5 6 = 57 12 - 10 12 = 47 12 = 3 11 12
If the expression contains both fractions and natural numbers, then it is recommended to group them by type when calculating.
Example 10
Find the difference 98 + 17 20 - 5 + 3 5.
Decision
Knowing the basic properties of subtraction and addition, we can group the numbers as follows: 98 + 17 20 - 5 + 3 5 \u003d 98 + 17 20 - 5 - 3 5 \u003d 98 - 5 + 17 20 - 3 5
Let's complete the calculations: 98 - 5 + 17 20 - 3 5 \u003d 93 + 17 20 - 12 20 \u003d 93 + 5 20 \u003d 93 + 1 4 \u003d 93 1 4
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