Addition and subtraction of fractions with the same denominators. Everything is in place. Adding fractions with the same denominators

Solving problems from the problem book Vilenkin, Zhokhov, Chesnokov, Schwarzburd for grade 5 on the topic:

§ 5. Ordinary fractions:
26. Adding and subtracting fractions with the same denominators

1005 A salad was made from tomatoes weighing 5/16 kg and cucumbers weighing 9/16 kg. What is the mass of lettuce?
SOLUTION

1006 The weight of the machine is 73/100 t and the weight of its package is 23/100 t. Find the weight of the machine including the package.
SOLUTION

1007 On the first day potatoes were planted on 2/7 of the plot, and on the second day on 3/7 of the plot. What part of the plot was planted with potatoes during these two days?
SOLUTION

1008 One brigade received 7/10 tons of nails, and the second 3/10 tons less. How many nails did the second brigade get?
SOLUTION

1009 10/11 fields were sown in two days. On the first day, 4/11 fields were sown. What part of the field was sown on the second day?
SOLUTION

1010 The tank is 3/5 filled with gasoline, 1/5 of the tank was poured into a barrel. What part of the tank was left filled with gasoline?
SOLUTION

1012 Find the value of the expression
SOLUTION

1013 Of the 11 greenhouses of the vegetable farm, 4 are planted with tomatoes and 2 with cucumbers. What part of the greenhouses is occupied by cucumbers and tomatoes? Solve the problem in two ways.
SOLUTION

1014 An area of ​​300 hectares was allocated for planting forests. Spruce was planted on 3/10 of the plot, and pine on 4/10 of the plot. How many hectares are occupied by spruce and pine together?
SOLUTION

1015 The team decided to make 175 products over the plan. On the first day she made 9/25 of that amount, on the second day 13/25 of that amount. How many products did the brigade produce in these two days? How many items does she have left to make?
SOLUTION

1016 11/17 fields of the vegetable farm were planted with potatoes. Cucumbers are sown on 1/17 of the field more than carrots, and 8/17 of the field less than potatoes. What part of the field is sown with cucumbers and what with carrots? What part of the field is occupied by potatoes, cucumbers and carrots together?
SOLUTION

1019 There were 2 q 70 kg of fruit in the tent. Apples made up 5/9 of all fruits and pears 1/9 of all fruits. By how much is the weight of the apples greater than the weight of the pears? Solve the problem in two ways.
SOLUTION

1020 On the first day, the tourist walked 5/14 of the entire path, and on the second day, 7/14. It is known that during these two days the tourist walked 36 km. How many kilometers is the entire journey of the tourist?
SOLUTION

1021 The first story occupied 5/13 of the book, and the second story 2/13 of the book. It is known that the first story was 12 pages longer than the second. How many pages are in the whole book?
SOLUTION

1022 Using the equation 4/25 + 12/25= 16/25 find the values ​​of the expression and solve the equations
SOLUTION

1024 260 people go on the tour. How many buses should be ordered if each bus should have no more than 30 passengers?
SOLUTION

1025 Draw a line. Then draw a line segment whose length is
SOLUTION

1026 Find the coordinates of points A, B, C, D, E, M, K (Fig. 128) and compare these coordinates with 1.
SOLUTION

1027 Calculate the perimeter and area of ​​triangle ABC (Fig. 129)
SOLUTION

1030 Find all values ​​of x such that x/15 is a proper fraction and 8/x is an improper fraction.
SOLUTION

1031 Name 3 proper fractions whose numerator is greater than 100. Name 3 improper fractions whose denominator is greater than 200.
SOLUTION

1033 The length of a rectangular cuboid is 8 m, the width is 6 m and the height is 12 m. Find the sum of the areas of the largest and smallest faces of this cuboid.
SOLUTION

1034 For the manufacture of 750 m of viscose fabric, 10 kg of cellulose is required. From 1 m3 of wood, 200 kg of pulp can be obtained. How many meters of viscose fabric can be obtained from 20 m3 of wood?
SOLUTION

1035 The combination lock has six buttons. To open it, you need to press the buttons in a certain sequence to dial the code. How many code options are there for this lock?
SOLUTION

1036 Solve the equation: a) (x - 111) 59 = 11 918; b) 975(x - 615) = 12 675; c) (30 901 - a): 605 = 51; d) 39 765: (b - 893) = 1205.
SOLUTION

1037 Solve the problem: 1) Of the 30 seeds planted, 23 sprouted. What part of the planted seeds sprouted? 2) 40 swans swam on the pond. Of these, 30 were white. What fraction of all swans were white swans?
SOLUTION

1038 Find the value of the expression: 1) 76 (3569 + 2795) - (24 078 + 30 785); 2) (43 512-43 006) 805 - (48 987 + 297 305)
SOLUTION

1039 In the first hour, 5/17 of the entire road was cleared of snow, and in the second hour, 9/17 of the entire road. What part of the road was cleared of snow during these two hours? What part of the road was less cleared in the first hour than in the second?
SOLUTION

1040 6/25 m of fabric was used for the dress for the first doll, and 9/25 m of fabric for the dress for the second doll. How much fabric was used for both dresses? How much more fabric was used for the dress of the second doll than for the dress of the first doll?

Fractions are regular numbers, they can also be added and subtracted. But due to the fact that they have a denominator, more complex rules are required here than for integers.

Consider the simplest case, when there are two fractions with the same denominators. Then:

To add fractions with the same denominators, add their numerators and leave the denominator unchanged.

To subtract fractions with the same denominators, it is necessary to subtract the numerator of the second from the numerator of the first fraction, and again leave the denominator unchanged.

Within each expression, the denominators of the fractions are equal. By definition of addition and subtraction of fractions, we get:

As you can see, nothing complicated: just add or subtract the numerators - and that's it.

But even in such simple actions people manage to make mistakes. Most often they forget that the denominator does not change. For example, when adding them, they also begin to add up, and this is fundamentally wrong.

Get rid of bad habit Adding the denominators is easy enough. Try to do the same when subtracting. As a result, the denominator will be zero, and the fraction (suddenly!) will lose its meaning.

Therefore, remember once and for all: when adding and subtracting, the denominator does not change!

Also, many people make mistakes when adding several negative fractions. There is confusion with the signs: where to put a minus, and where - a plus.

This problem is also very easy to solve. It is enough to remember that the minus before the fraction sign can always be transferred to the numerator - and vice versa. And of course, do not forget two simple rules:

1. Plus times minus gives minus;
2. Two negatives make an affirmative.

Let's analyze all this with specific examples:

A task. Find the value of the expression:

In the first case, everything is simple, and in the second, we will add minuses to the numerators of fractions:

What if the denominators are different

You cannot directly add fractions with different denominators. At least, this method is unknown to me. However, the original fractions can always be rewritten so that the denominators become the same.

There are many ways to convert fractions. Three of them are discussed in the lesson " Bringing fractions to a common denominator", so we will not dwell on them here. Let's take a look at some examples:

A task. Find the value of the expression:

In the first case, we bring the fractions to a common denominator using the "cross-wise" method. In the second, we will look for the LCM. Note that 6 = 2 3; 9 = 3 · 3. The last factors in these expansions are equal, and the first ones are coprime. Therefore, LCM(6; 9) = 2 3 3 = 18.

What if the fraction has an integer part

I can make you happy different denominators with fractions, this is not the greatest evil. Much more errors occur when whole part.

Of course, for such fractions there are own addition and subtraction algorithms, but they are rather complicated and require a long study. Better use a simple circuit below:

1. Convert all fractions containing an integer part to improper. We get normal terms (even if with different denominators), which are calculated according to the rules discussed above;
2. Actually, calculate the sum or difference of the resulting fractions. As a result, we will practically find the answer;
3. If this is all that was required in the task, we perform the inverse transformation, i.e. get rid of not proper fraction, separating the whole part in it.

The rules for switching to improper fractions and highlighting the integer part are described in detail in the lesson "What is a numerical fraction". If you don't remember, be sure to repeat. Examples:

A task. Find the value of the expression:

Everything is simple here. The denominators inside each expression are equal, so it remains to convert all fractions to improper ones and count. We have:

To simplify the calculations, I skipped some obvious steps in the last examples.

A small note to the last two examples, where fractions with a highlighted integer part are subtracted. The minus before the second fraction means that it is the whole fraction that is subtracted, and not just its whole part.

Reread this sentence again, look at the examples, and think about it. This is where beginners allow great amount errors. They love to give such tasks to control work. You will also meet them repeatedly in the tests for this lesson, which will be published shortly.

Summary: General Scheme of Computing

In conclusion, I will give a general algorithm that will help you find the sum or difference of two or more fractions:

1. If an integer part is highlighted in one or more fractions, convert these fractions to improper ones;
2. Bring all the fractions to a common denominator in any way convenient for you (unless, of course, the compilers of the problems did this);
3. Add or subtract the resulting numbers according to the rules for adding and subtracting fractions with the same denominators;
4. Reduce the result if possible. If the fraction turned out to be incorrect, select the whole part.

Remember that it is better to highlight the whole part at the very end of the task, just before writing the answer.

§ 87. Addition of fractions.

Adding fractions has many similarities to adding integers. Addition of fractions is an action consisting in the fact that several given numbers (terms) are combined into one number (sum), which contains all units and fractions of units of terms.

We will consider three cases in turn:

1. Addition of fractions with the same denominators.
2. Addition of fractions with different denominators.
3. Addition mixed numbers.

1. Addition of fractions with the same denominators.

Consider an example: 1 / 5 + 2 / 5 .

Take the segment AB (Fig. 17), take it as a unit and divide by 5 equal parts, then the part AC of this segment will be equal to 1/5 of the segment AB, and the part of the same segment CD will be equal to 2/5 AB.

It can be seen from the drawing that if we take the segment AD, then it will be equal to 3/5 AB; but segment AD is precisely the sum of segments AC and CD. So, we can write:

1 / 5 + 2 / 5 = 3 / 5

Considering these terms and the resulting amount, we see that the numerator of the sum was obtained by adding the numerators of the terms, and the denominator remained unchanged.

From this we get the following rule: To add fractions with the same denominators, you must add their numerators and leave the same denominator.

Consider an example:

2. Addition of fractions with different denominators.

Let's add fractions: 3/4 + 3/8 First they need to be reduced to the lowest common denominator:

The intermediate link 6/8 + 3/8 could not have been written; we have written it here for greater clarity.

Thus, to add fractions with different denominators, you must first bring them to the lowest common denominator, add their numerators and sign common denominator.

Consider an example (we will write additional factors over the corresponding fractions):

3. Addition of mixed numbers.

Let's add the numbers: 2 3 / 8 + 3 5 / 6.

Let us first bring the fractional parts of our numbers to a common denominator and rewrite them again:

Now add the integer and fractional parts in sequence:

§ 88. Subtraction of fractions.

Subtraction of fractions is defined in the same way as subtraction of whole numbers. This is an action by which, given the sum of two terms and one of them, another term is found. Let's consider three cases in turn:

1. Subtraction of fractions with the same denominators.
2. Subtraction of fractions with different denominators.
3. Subtraction of mixed numbers.

1. Subtraction of fractions with the same denominators.

Consider an example:

13 / 15 - 4 / 15

Let's take the segment AB (Fig. 18), take it as a unit and divide it into 15 equal parts; then the AC part of this segment will be 1/15 of AB, and the AD part of the same segment will correspond to 13/15 AB. Let's set aside another segment ED, equal to 4/15 AB.

We need to subtract 4/15 from 13/15. In the drawing, this means that the segment ED must be subtracted from the segment AD. As a result, segment AE will remain, which is 9/15 of segment AB. So we can write:

The example we made shows that the numerator of the difference was obtained by subtracting the numerators, and the denominator remained the same.

Therefore, in order to subtract fractions with the same denominators, you need to subtract the numerator of the subtrahend from the numerator of the minuend and leave the same denominator.

2. Subtraction of fractions with different denominators.

Example. 3/4 - 5/8

First, let's reduce these fractions to the smallest common denominator:

The intermediate link 6 / 8 - 5 / 8 is written here for clarity, but it can be skipped in the future.

Thus, in order to subtract a fraction from a fraction, you must first bring them to the smallest common denominator, then subtract the numerator of the subtrahend from the numerator of the minuend and sign the common denominator under their difference.

Consider an example:

3. Subtraction of mixed numbers.

Example. 10 3 / 4 - 7 2 / 3 .

Let's bring the fractional parts of the minuend and the subtrahend to the lowest common denominator:

We subtracted a whole from a whole and a fraction from a fraction. But there are cases when the fractional part of the subtrahend is greater than the fractional part of the minuend. In such cases, you need to take one unit from the integer part of the reduced, split it into those parts in which the fractional part is expressed, and add to the fractional part of the reduced. And then the subtraction will be performed in the same way as in the previous example:

§ 89. Multiplication of fractions.

When studying the multiplication of fractions, we will consider the following questions:

1. Multiplying a fraction by an integer.
2. Finding a fraction of a given number.
3. Multiplication of a whole number by a fraction.
4. Multiplying a fraction by a fraction.
5. Multiplication of mixed numbers.
6. The concept of interest.
7. Finding percentages of a given number. Let's consider them sequentially.

1. Multiplying a fraction by an integer.

Multiplying a fraction by an integer has the same meaning as multiplying an integer by an integer. Multiplying a fraction (multiplicand) by an integer (multiplier) means composing the sum of identical terms, in which each term is equal to the multiplicand, and the number of terms is equal to the multiplier.

So, if you need to multiply 1/9 by 7, then this can be done like this:

We easily got the result, since the action was reduced to adding fractions with the same denominators. Consequently,

Consideration of this action shows that multiplying a fraction by an integer is equivalent to increasing this fraction as many times as there are units in the integer. And since the increase in the fraction is achieved either by increasing its numerator

or by decreasing its denominator , then we can either multiply the numerator by the integer, or divide the denominator by it, if such a division is possible.

From here we get the rule:

To multiply a fraction by an integer, you need to multiply the numerator by this integer and leave the denominator the same, or, if possible, divide the denominator by this number, leaving the numerator unchanged.

When multiplying, abbreviations are possible, for example:

2. Finding a fraction of a given number. There are many problems in which you have to find, or calculate, a part of a given number. The difference between these tasks and others is that they give the number of some objects or units of measurement and you need to find a part of this number, which is also indicated here by a certain fraction. To facilitate understanding, we will first give examples of such problems, and then introduce the method of solving them.

Task 1. I had 60 rubles; 1 / 3 of this money I spent on the purchase of books. How much did the books cost?

Task 2. The train must cover the distance between cities A and B, equal to 300 km. He has already covered 2/3 of that distance. How many kilometers is this?

Task 3. There are 400 houses in the village, 3/4 of them are brick, the rest are wooden. How many brick houses?

Here are some of those numerous tasks to find a part of a given number that we have to meet. They are usually called problems for finding a fraction of a given number.

Solution of problem 1. From 60 rubles. I spent 1 / 3 on books; So, to find the cost of books, you need to divide the number 60 by 3:

Problem 2 solution. The meaning of the problem is that you need to find 2 / 3 of 300 km. Calculate first 1/3 of 300; this is achieved by dividing 300 km by 3:

300: 3 = 100 (that's 1/3 of 300).

To find two-thirds of 300, you need to double the resulting quotient, that is, multiply by 2:

100 x 2 = 200 (that's 2/3 of 300).

Solution of problem 3. Here you need to determine the number of brick houses, which are 3/4 of 400. Let's first find 1/4 of 400,

400: 4 = 100 (that's 1/4 of 400).

To calculate three quarters of 400, the resulting quotient must be tripled, that is, multiplied by 3:

100 x 3 = 300 (that's 3/4 of 400).

Based on the solution of these problems, we can derive the following rule:

To find the value of a fraction of a given number, you need to divide this number by the denominator of the fraction and multiply the resulting quotient by its numerator.

3. Multiplication of a whole number by a fraction.

Earlier (§ 26) it was established that the multiplication of integers should be understood as the addition of identical terms (5 x 4 \u003d 5 + 5 + 5 + 5 \u003d 20). In this paragraph (paragraph 1) it was established that multiplying a fraction by an integer means finding the sum of identical terms equal to this fraction.

In both cases, the multiplication consisted in finding the sum of identical terms.

Now we move on to multiplying a whole number by a fraction. Here we will meet with such, for example, multiplication: 9 2 / 3. It is quite obvious that the previous definition of multiplication does not apply to this case. This is evident from the fact that we cannot replace such multiplication by adding equal numbers.

Because of this, we will have to give a new definition of multiplication, i.e., in other words, to answer the question of what should be understood by multiplication by a fraction, how this action should be understood.

The meaning of multiplying a whole number by a fraction is clarified from next definition: to multiply an integer (multiplier) by a fraction (multiplier) means to find this fraction of the multiplier.

Namely, multiplying 9 by 2/3 means finding 2/3 of nine units. In the previous paragraph, such problems were solved; so it's easy to figure out that we end up with 6.

But now an interesting and important question arises: why such seemingly different actions as finding the sum of equal numbers and finding the fraction of a number are called the same word “multiplication” in arithmetic?

This happens because the previous action (repeating the number with terms several times) and the new action (finding the fraction of a number) give an answer to homogeneous questions. This means that we proceed here from the considerations that homogeneous questions or tasks are solved by one and the same action.

To understand this, consider the following problem: “1 m of cloth costs 50 rubles. How much will 4 m of such cloth cost?

This problem is solved by multiplying the number of rubles (50) by the number of meters (4), i.e. 50 x 4 = 200 (rubles).

Let's take the same problem, but in it the amount of cloth will be expressed as a fractional number: “1 m of cloth costs 50 rubles. How much will 3/4 m of such cloth cost?

This problem also needs to be solved by multiplying the number of rubles (50) by the number of meters (3/4).

You can also change the numbers in it several times without changing the meaning of the problem, for example, take 9/10 m or 2 3/10 m, etc.

Since these problems have the same content and differ only in numbers, we call the actions used in solving them the same word - multiplication.

How is a whole number multiplied by a fraction?

Let's take the numbers encountered in the last problem:

According to the definition, we must find 3 / 4 of 50. First we find 1 / 4 of 50, and then 3 / 4.

1/4 of 50 is 50/4;

3/4 of 50 is .

Consequently.

Consider another example: 12 5 / 8 = ?

1/8 of 12 is 12/8,

5/8 of the number 12 is .

Consequently,

From here we get the rule:

To multiply an integer by a fraction, you need to multiply the integer by the numerator of the fraction and make this product the numerator, and sign the denominator of the given fraction as the denominator.

We write this rule using letters:

To make this rule perfectly clear, it should be remembered that a fraction can be considered as a quotient. Therefore, it is useful to compare the found rule with the rule for multiplying a number by a quotient, which was set out in § 38

It must be remembered that before performing multiplication, you should do (if possible) cuts, for example:

4. Multiplying a fraction by a fraction. Multiplying a fraction by a fraction has the same meaning as multiplying an integer by a fraction, that is, when multiplying a fraction by a fraction, you need to find the fraction in the multiplier from the first fraction (multiplier).

Namely, multiplying 3/4 by 1/2 (half) means finding half of 3/4.

How do you multiply a fraction by a fraction?

Let's take an example: 3/4 times 5/7. This means that you need to find 5 / 7 from 3 / 4 . Find first 1/7 of 3/4 and then 5/7

1/7 of 3/4 would be expressed like this:

5 / 7 numbers 3 / 4 will be expressed as follows:

In this way,

Another example: 5/8 times 4/9.

1/9 of 5/8 is ,

4/9 numbers 5/8 are .

In this way,

From these examples, the following rule can be deduced:

To multiply a fraction by a fraction, you need to multiply the numerator by the numerator, and the denominator by the denominator and make the first product the numerator and the second product the denominator of the product.

This is the rule in general view can be written like this:

When multiplying, it is necessary to make (if possible) reductions. Consider examples:

5. Multiplication of mixed numbers. Since mixed numbers can easily be replaced by improper fractions, this circumstance is usually used when multiplying mixed numbers. This means that in those cases where the multiplicand, or the multiplier, or both factors are expressed as mixed numbers, then they are replaced by improper fractions. Multiply, for example, mixed numbers: 2 1/2 and 3 1/5. We turn each of them into an improper fraction and then we will multiply the resulting fractions according to the rule of multiplying a fraction by a fraction:

Rule. To multiply mixed numbers, you must first convert them to improper fractions and then multiply according to the rule of multiplying a fraction by a fraction.

Note. If one of the factors is an integer, then the multiplication can be performed based on the distribution law as follows:

6. The concept of interest. When solving problems and when performing various practical calculations, we use all kinds of fractions. But one must keep in mind that many quantities admit not any, but natural subdivisions for them. For example, you can take one hundredth (1/100) of a ruble, it will be a penny, two hundredths is 2 kopecks, three hundredths is 3 kopecks. You can take 1/10 of the ruble, it will be "10 kopecks, or a dime. You can take a quarter of the ruble, i.e. 25 kopecks, half a ruble, i.e. 50 kopecks (fifty kopecks). But they practically don’t take, for example , 2/7 rubles because the ruble is not divided into sevenths.

The unit of measurement for weight, i.e., the kilogram, allows, first of all, decimal subdivisions, for example, 1/10 kg, or 100 g. And such fractions of a kilogram as 1/6, 1/11, 1/13 are uncommon.

In general our (metric) measures are decimal and allow decimal subdivisions.

However, it should be noted that it is extremely useful and convenient in a wide variety of cases to use the same (uniform) method of subdividing quantities. Many years of experience have shown that such a well-justified division is the "hundredths" division. Let's consider a few examples related to the most diverse areas of human practice.

1. The price of books has decreased by 12/100 of the previous price.

Example. The previous price of the book is 10 rubles. She went down by 1 ruble. 20 kop.

2. Savings banks pay out during the year to depositors 2/100 of the amount that is put into savings.

Example. 500 rubles are put into the cash desk, the income from this amount for the year is 10 rubles.

3. The number of graduates of one school was 5/100 of the total number of students.

EXAMPLE Only 1,200 students studied at the school, 60 of them graduated from school.

The hundredth of a number is called a percentage..

The word "percentage" is borrowed from Latin and its root "cent" means one hundred. Together with the preposition (pro centum), this word means "for a hundred." The meaning of this expression follows from the fact that initially in ancient Rome interest was the money that the debtor paid to the lender “for every hundred”. The word "cent" is heard in such familiar words: centner (one hundred kilograms), centimeter (they say centimeter).

For example, instead of saying that the plant produced 1/100 of all the products produced by it during the past month, we will say this: the plant produced one percent of the rejects during the past month. Instead of saying: the plant produced 4/100 more products than the established plan, we will say: the plant exceeded the plan by 4 percent.

The above examples can be expressed differently:

1. The price of books has decreased by 12 percent of the previous price.

2. Savings banks pay depositors 2 percent per year of the amount put into savings.

3. The number of graduates of one school was 5 percent of the number of all students in the school.

To shorten the letter, it is customary to write the% sign instead of the word "percentage".

However, it must be remembered that the % sign is usually not written in calculations, it can be written in the problem statement and in the final result. When performing calculations, you need to write a fraction with a denominator of 100 instead of an integer with this icon.

You need to be able to replace an integer with the specified icon with a fraction with a denominator of 100:

Conversely, you need to get used to writing an integer with the indicated icon instead of a fraction with a denominator of 100:

7. Finding percentages of a given number.

Task 1. The school received 200 cubic meters. m of firewood, with birch firewood accounting for 30%. How much birch wood was there?

The meaning of this problem is that birch firewood was only a part of the firewood that was delivered to the school, and this part is expressed as a fraction of 30 / 100. So, we are faced with the task of finding a fraction of a number. To solve it, we must multiply 200 by 30 / 100 (tasks for finding the fraction of a number are solved by multiplying a number by a fraction.).

So 30% of 200 equals 60.

The fraction 30 / 100 encountered in this problem can be reduced by 10. It would be possible to perform this reduction from the very beginning; the solution to the problem would not change.

Task 2. There were 300 children of various ages in the camp. Children aged 11 were 21%, children aged 12 were 61% and finally 13 year olds were 18%. How many children of each age were in the camp?

In this problem, you need to perform three calculations, that is, successively find the number of children 11 years old, then 12 years old, and finally 13 years old.

So, here it will be necessary to find a fraction of a number three times. Let's do it:

1) How many children were 11 years old?

2) How many children were 12 years old?

3) How many children were 13 years old?

After solving the problem, it is useful to add the numbers found; their sum should be 300:

63 + 183 + 54 = 300

You should also pay attention to the fact that the sum of the percentages given in the condition of the problem is 100:

21% + 61% + 18% = 100%

This suggests that total number children who were in the camp was taken as 100%.

3 a da cha 3. The worker received 1,200 rubles per month. Of these, he spent 65% on food, 6% on an apartment and heating, 4% on gas, electricity and radio, 10% on cultural needs and 15% he saved. How much money was spent on the needs indicated in the task?

To solve this problem, you need to find a fraction of the number 1,200 5 times. Let's do it.

1) How much money is spent on food? The task says that this expense is 65% of all earnings, i.e. 65/100 of the number 1,200. Let's do the calculation:

2) How much money was paid for an apartment with heating? Arguing like the previous one, we arrive at the following calculation:

3) How much money did you pay for gas, electricity and radio?

4) How much money is spent on cultural needs?

5) How much money did the worker save?

For verification, it is useful to add the numbers found in these 5 questions. The amount should be 1,200 rubles. All earnings are taken as 100%, which is easy to check by adding up the percentages given in the problem statement.

We have solved three problems. Despite the fact that these tasks were about different things (delivery of firewood for the school, the number of children of different ages, the expenses of the worker), they were solved in the same way. This happened because in all tasks it was necessary to find a few percent of the given numbers.

§ 90. Division of fractions.

When studying the division of fractions, we will consider the following questions:

1. Divide an integer by an integer.
2. Division of a fraction by an integer
3. Division of an integer by a fraction.
4. Division of a fraction by a fraction.
5. Division of mixed numbers.
6. Finding a number given its fraction.
7. Finding a number by its percentage.

Let's consider them sequentially.

1. Divide an integer by an integer.

As it was indicated in the section of integers, division is the action consisting in the fact that, given the product of two factors (the dividend) and one of these factors (the divisor), another factor is found.

The division of an integer by an integer we considered in the department of integers. We met there two cases of division: division without a remainder, or "entirely" (150: 10 = 15), and division with a remainder (100: 9 = 11 and 1 in the remainder). We can therefore say that in the realm of integers, exact division is not always possible, because the dividend is not always the product of the divisor and the integer. After the introduction of multiplication by a fraction, we can consider any case of division of integers as possible (only division by zero is excluded).

For example, dividing 7 by 12 means finding a number whose product times 12 would be 7. This number is the fraction 7/12 because 7/12 12 = 7. Another example: 14: 25 = 14/25 because 14/25 25 = 14.

Thus, to divide an integer by an integer, you need to make a fraction, the numerator of which is equal to the dividend, and the denominator is the divisor.

2. Division of a fraction by an integer.

Divide the fraction 6 / 7 by 3. According to the definition of division given above, we have here the product (6 / 7) and one of the factors (3); it is required to find such a second factor that, when multiplied by 3, would give the given product 6 / 7. Obviously, it should be three times smaller than this product. This means that the task set before us was to reduce the fraction 6 / 7 by 3 times.

We already know that the reduction of a fraction can be done either by decreasing its numerator or by increasing its denominator. Therefore, you can write:

In this case, the numerator 6 is divisible by 3, so the numerator should be reduced by 3 times.

Let's take another example: 5 / 8 divided by 2. Here the numerator 5 is not divisible by 2, which means that the denominator will have to be multiplied by this number:

Based on this, we can state the rule: To divide a fraction by an integer, you need to divide the numerator of the fraction by that integer(if possible), leaving the same denominator, or multiply the denominator of the fraction by this number, leaving the same numerator.

3. Division of an integer by a fraction.

Let it be required to divide 5 by 1 / 2, i.e. find a number that, after multiplying by 1 / 2, will give the product 5. Obviously, this number must be greater than 5, since 1 / 2 is a proper fraction, and when multiplying a number by a proper fraction, the product must be less than the multiplicand. To make it clearer, let's write our actions as follows: 5: 1 / 2 = X , so x 1 / 2 \u003d 5.

We must find such a number X , which, when multiplied by 1/2, would give 5. Since multiplying a certain number by 1/2 means finding 1/2 of this number, then, therefore, 1/2 unknown number X is 5, and the whole number X twice as much, i.e. 5 2 \u003d 10.

So 5: 1 / 2 = 5 2 = 10

Let's check:

Let's consider one more example. Let it be required to divide 6 by 2 / 3 . Let's first try to find the desired result using the drawing (Fig. 19).

Fig.19

Draw a segment AB, equal to 6 of some units, and divide each unit into 3 equal parts. In each unit, three-thirds (3 / 3) in the entire segment AB is 6 times larger, i.e. e. 18/3. We connect with the help of small brackets 18 obtained segments of 2; There will be only 9 segments. This means that the fraction 2/3 is contained in b units 9 times, or, in other words, the fraction 2/3 is 9 times less than 6 integer units. Consequently,

How to get this result without a drawing using only calculations? We will argue as follows: it is required to divide 6 by 2 / 3, i.e., it is required to answer the question, how many times 2 / 3 is contained in 6. Let's find out first: how many times is 1 / 3 contained in 6? In a whole unit - 3 thirds, and in 6 units - 6 times more, i.e. 18 thirds; to find this number, we must multiply 6 by 3. This means that 1/3 is contained in b units 18 times, and 2/3 is contained in b not 18 times, but twice less times, i.e. 18: 2 = 9. Therefore, when dividing 6 by 2/3, we did the following:

From here we get the rule for dividing an integer by a fraction. To divide an integer by a fraction, you need to multiply this integer by the denominator of the given fraction and, making this product the numerator, divide it by the numerator of the given fraction.

We write the rule using letters:

To make this rule perfectly clear, it should be remembered that a fraction can be considered as a quotient. Therefore, it is useful to compare the found rule with the rule for dividing a number by a quotient, which was set out in § 38. Note that the same formula was obtained there.

When dividing, abbreviations are possible, for example:

4. Division of a fraction by a fraction.

Let it be required to divide 3/4 by 3/8. What will denote the number that will be obtained as a result of division? It will answer the question how many times the fraction 3/8 is contained in the fraction 3/4. To understand this issue, let's make a drawing (Fig. 20).

Take the segment AB, take it as a unit, divide it into 4 equal parts and mark 3 such parts. Segment AC will be equal to 3/4 of segment AB. Let us now divide each of the four initial segments in half, then the segment AB will be divided into 8 equal parts and each such part will be equal to 1/8 of the segment AB. We connect 3 such segments with arcs, then each of the segments AD and DC will be equal to 3/8 of the segment AB. The drawing shows that the segment equal to 3/8 is contained in the segment equal to 3/4 exactly 2 times; So the result of the division can be written like this:

3 / 4: 3 / 8 = 2

Let's consider one more example. Let it be required to divide 15/16 by 3/32:

We can reason like this: we need to find a number that, after being multiplied by 3 / 32, will give a product equal to 15 / 16. Let's write the calculations like this:

15 / 16: 3 / 32 = X

3 / 32 X = 15 / 16

3/32 unknown number X make up 15 / 16

1/32 unknown number X is ,

32 / 32 numbers X make up .

Consequently,

Thus, to divide a fraction by a fraction, you need to multiply the numerator of the first fraction by the denominator of the second, and multiply the denominator of the first fraction by the numerator of the second and make the first product the numerator and the second the denominator.

Let's write the rule using letters:

When dividing, abbreviations are possible, for example:

5. Division of mixed numbers.

When dividing mixed numbers, they must first be converted to improper fractions, then divide the resulting fractions according to the division rules fractional numbers. Consider an example:

Convert mixed numbers to improper fractions:

Now let's split:

Thus, to divide mixed numbers, you need to convert them to improper fractions and then divide according to the rule for dividing fractions.

6. Finding a number given its fraction.

Among the various tasks on fractions, there are sometimes those in which the value of some fraction of an unknown number is given and it is required to find this number. This type of problem will be inverse to the problem of finding a fraction of a given number; there a number was given and it was required to find some fraction of this number, here a fraction of a number is given and it is required to find this number itself. This idea will become even clearer if we turn to the solution of this type of problem.

Task 1. On the first day, glaziers glazed 50 windows, which is 1 / 3 of all windows of the built house. How many windows are in this house?

Solution. The problem says that 50 glazed windows make up 1/3 of all the windows of the house, which means that there are 3 times more windows in total, i.e.

The house had 150 windows.

Task 2. The shop sold 1,500 kg of flour, which is 3/8 of the total stock of flour in the shop. What was the store's initial supply of flour?

Solution. It can be seen from the condition of the problem that the sold 1,500 kg of flour make up 3/8 of the total stock; this means that 1/8 of this stock will be 3 times less, i.e., to calculate it, you need to reduce 1500 by 3 times:

1,500: 3 = 500 (that's 1/8 of the stock).

Obviously, the entire stock will be 8 times larger. Consequently,

500 8 \u003d 4,000 (kg).

The initial supply of flour in the store was 4,000 kg.

From the consideration of this problem, the following rule can be deduced.

To find a number by a given value of its fraction, it is enough to divide this value by the numerator of the fraction and multiply the result by the denominator of the fraction.

We solved two problems on finding a number given its fraction. Such problems, as it is especially well seen from the last one, are solved by two actions: division (when one part is found) and multiplication (when the whole number is found).

However, after we have studied the division of fractions, the above problems can be solved in one action, namely: division by a fraction.

For example, the last task can be solved in one action like this:

In the future, we will solve the problem of finding a number by its fraction in one action - division.

7. Finding a number by its percentage.

In these tasks, you will need to find a number, knowing a few percent of this number.

Task 1. At the beginning of this year, I received 60 rubles from the savings bank. income from the amount I put into savings a year ago. How much money did I put in the savings bank? (Cash offices give depositors 2% of income per year.)

The meaning of the problem is that a certain amount of money was put by me in a savings bank and lay there for a year. After a year, I received 60 rubles from her. income, which is 2/100 of the money I put in. How much money did I deposit?

Therefore, knowing the part of this money, expressed in two ways (in rubles and in fractions), we must find the entire, as yet unknown, amount. This is an ordinary problem of finding a number given its fraction. The following tasks are solved by division:

So, 3,000 rubles were put into the savings bank.

Task 2. In two weeks, fishermen fulfilled the monthly plan by 64%, having prepared 512 tons of fish. What was their plan?

From the condition of the problem, it is known that the fishermen completed part of the plan. This part is equal to 512 tons, which is 64% of the plan. How many tons of fish need to be harvested according to the plan, we do not know. The solution of the problem will consist in finding this number.

Such tasks are solved by dividing:

So, according to the plan, you need to prepare 800 tons of fish.

Task 3. The train went from Riga to Moscow. When he passed the 276th kilometer, one of the passengers asked the passing conductor how much of the journey they had already traveled. To this the conductor replied: “We have already covered 30% of the entire journey.” What is the distance from Riga to Moscow?

It can be seen from the condition of the problem that 30% of the journey from Riga to Moscow is 276 km. We need to find the entire distance between these cities, i.e., for this part, find the whole:

§ 91. Reciprocal numbers. Replacing division with multiplication.

Take the fraction 2/3 and rearrange the numerator to the place of the denominator, we get 3/2. We got a fraction, the reciprocal of this one.

In order to get a fraction reciprocal of a given one, you need to put its numerator in the place of the denominator, and the denominator in the place of the numerator. In this way, we can get a fraction that is the reciprocal of any fraction. For example:

3 / 4 , reverse 4 / 3 ; 5 / 6 , reverse 6 / 5

Two fractions that have the property that the numerator of the first is the denominator of the second and the denominator of the first is the numerator of the second are called mutually inverse.

Now let's think about what fraction will be the reciprocal of 1/2. Obviously, it will be 2 / 1, or just 2. Looking for the reciprocal of this, we got an integer. And this case is not isolated; on the contrary, for all fractions with a numerator of 1 (one), the reciprocals will be integers, for example:

1 / 3, inverse 3; 1 / 5, reverse 5

Since when finding reciprocals we also met with integers, in the future we will not talk about reciprocals, but about reciprocals.

Let's figure out how to write the reciprocal of a whole number. For fractions, this is solved simply: you need to put the denominator in the place of the numerator. In the same way, you can get the reciprocal of an integer, since any integer can have a denominator of 1. Therefore, the reciprocal of 7 will be 1 / 7, because 7 \u003d 7 / 1; for the number 10 the reverse is 1 / 10 since 10 = 10 / 1

This idea can be expressed in another way: the reciprocal of a given number is obtained by dividing one by given number . This statement is true not only for integers, but also for fractions. Indeed, if you want to write a number that is the reciprocal of the fraction 5 / 9, then we can take 1 and divide it by 5 / 9, i.e.

Now let's point out one property mutually reciprocal numbers, which will be useful to us: the product of mutually reciprocal numbers is equal to one. Indeed:

Using this property, we can find reciprocals in the following way. Let's find the reciprocal of 8.

Let's denote it with the letter X , then 8 X = 1, hence X = 1 / 8 . Let's find another number, the inverse of 7/12, denote it by a letter X , then 7 / 12 X = 1, hence X = 1:7 / 12 or X = 12 / 7 .

We introduced here the concept of reciprocal numbers in order to slightly supplement information about the division of fractions.

When we divide the number 6 by 3 / 5, then we do the following:

Pay Special attention to the expression and compare it with the given one: .

If we take the expression separately, without connection with the previous one, then it is impossible to solve the question of where it came from: from dividing 6 by 3/5 or from multiplying 6 by 5/3. In both cases the result is the same. So we can say that dividing one number by another can be replaced by multiplying the dividend by the reciprocal of the divisor.

The examples that we give below fully confirm this conclusion.

Municipal educational institution

« Medium comprehensive school No. 13"

Republic of Komi, city of Vorkuta

Math lesson in grade 5 on the topic

"Addition and subtraction of fractions

with the same denominators"

The lesson was developed by a math teacher

Babenko N.E.

Vorkuta

Routing math lesson in 5th grade

Lesson topic: Addition and subtraction ordinary fractions with the same denominators.

Grade: 5

Didactic purpose: create conditions for the formation of a new educational information.

Content Goals:

-training: teach how to add and subtract fractions with the same denominators; repeat the concepts of "Correct, improper fraction", generalize and consolidate students' knowledge of comparing fractions.

-developing: develop attention, the ability to analyze, compare, generalize and draw conclusions. - educational: cultivate accuracy when writing examples and problems with ordinary fractions; promote understanding of the need for intellectual effort for successful learning.

Tasks: gain new knowledge on the topic of addition and subtraction of fractions with the same denominators; learn to work independently, draw conclusions.

Lesson type: a lesson in mastering new material

Forms of work: individual, frontal, in groups.

Forms of control: control by the teacher, self-control, mutual control.

Teaching methods:

Sources of knowledge: verbal, visual;

According to the degree of teacher-student interaction: conversation;

Regarding didactic tasks: preparation for perception;

Regarding the nature of cognitive activity: reproductive, partially exploratory, practical.

Educational and methodological support: textbook"Maths. Grade 5 "author Vilenkina N. Ya ., presentation.

Equipment: computer, multimedia projector, blackboard, chalk.

Lesson stages	Stage tasks	Time	Teacher activity	Student activities
1. Organizational moment	Create a favorable psychological attitude to work. Provide motivation for learning by children, their acceptance of the objectives of the lesson.		Greeting, checking readiness for the lesson, organizing the attention of children. (slide number 3) Do you remember what you learned in the previous lessons? Came to our lesson today Dunno and asked to help him deal with the concept of ordinary fractions and learn tasks using fractions. And as you may have guessed, in this lesson we will continue to work with ordinary fractions. The topic of today's lesson (slide number 1)"Addition and subtraction of common fractions with like denominators". What are our goals for this lesson? (slide number 4-7) Goals have been set, but, as you know, in order to achieve them, you need to remember what you learned earlier.	Included in the business rhythm of the lesson. with common fractions. Learn to distinguish between proper and improper fractions and compare them. Students write the date and topic of the lesson in their notebooks. Lesson Objectives: Identify the rule and learn how to add and subtract fractions with the same denominators. develop attention, logical thinking, competent mathematical speech. Cultivate accuracy when writing examples and problems with ordinary fractions.	Personal: self-determination. Regulatory: goal setting. Communicative: planning educational cooperation with the teacher and peers.
2. Actualization of knowledge and skills	Updating of basic knowledge and methods of action; repetition of the ability to translate text into a record in the form of a fraction, restoration of the definition of the correct and improper fraction, fixing individual difficulties		And here are the first questions from Dunno; How are natural numbers different from fractional numbers? What does the denominator show and where is it written? What does the numerator show and where is it written? Working with drawings .(slide number 8-11) Read the answers you received, but how else are these fractions read? (slide number 12) oral work. (slide number 13) Help Dunno collect pears with improper fractions written on them. What is a proper fraction? What is an improper fraction? (slide number 14) Self works. (slide number 15).	Integers represent whole units and fractional numbers represent parts of units. The denominator shows how many shares are divided and write it under the line. The numerator shows how many shares were taken and write it above the line. Students write their answers to the questions on the slides in their notebooks. 1/2 - Half 1/3 - third 1/4 - quarter 𝟖/𝟖; 𝟏𝟕/𝟏𝟑; 𝟏𝟏/𝟗. A fraction in which the numerator is less than the denominator is called a proper fraction. A fraction in which the numerator is greater than the denominator is called an improper fraction. Work in pairs. Students change notebooks and perform a check by evaluating each other.	Personal: evaluation of learning material. Communicative: the ability to use speech to regulate one's actions, to build statements that are understandable to others. Regulatory: control and evaluation of the process and results of activities. Cognitive: structuring your own knowledge.
3. Goal setting and motivation.	Ensuring children are motivated to learn, accepting the objectives of the lesson		Dunno guys are very surprised that fractions can be compared so easily. Let's show him what else he can do with ordinary fractions. I propose to build a broken line from three segments of 2 cm each and calculate its length in cm. (slide number 16) problem situation; Try to calculate the length of the polyline in dm. Clue: Find what part is 2 cm from a decimeter. (slide №17-18) How did you calculate the length in dm? And now let's try together to formulate the rule for adding fractions with the same denominators. (slide number 19) Let's write the addition rule using letters. Dunno asked me to help him solve the problem. (slide number 20) Friends came to visit him, he decided to treat them with apples put 10 (shares) on a plate, 4 shares ate how many shares are left? What action did you use to solve the problem? Formulate a rule for subtracting fractions with the same denominators. Let's write this rule in letters. (slide number 21)	In a notebook, a drawing is made and calculated; 2+2+2=6cm. Students face a problem 2cm from dm., 2/10dm. Mark in the figure and again calculate the length of the polyline. 2/10+2/10+2/10=2+2+2/10=6/10 Perform addition of fractions. When adding fractions with the same denominators, add the numerators and keep the denominator the same. 10/10-4/10=10-4/10=6/10 In a notebook, write down the rule using letters	Cognitive: the ability to consciously and voluntarily build a speech statement in oral form. Personal: self-determination. Regulatory: goal setting. Communicative: manifestation of activity in interaction for solving cognitive problems; the ability to use speech to regulate one's actions, the structure of statements that are understandable to others.
4. Application of knowledge and skills in a new situation	Ensuring the perception, comprehension and primary memorization by children of the studied topic: "Addition and subtraction of ordinary fractions with the same denominators."		So you have completed one of the learning objectives of our lesson, you have identified the rules for adding and subtracting fractions with the same denominators, it remains to learn how to apply these rules in practice. To do this, we will work with the textbook; (slide number 22) 1.Page 156, No. 1005. What is the mass of tomatoes? What is the mass of cucumbers? How to find mass of lettuce? - Read the answer. 2. Page 156, No. 1006. What is the mass of the machine? What is the mass of the package? How to find the mass of the machine with packaging? - Read the answer. 3. Page 156, No. 1008. What mass of nails did the first brigade receive? How many tons less did the second brigade receive? How many tons of nails did the second brigade receive?	Problem solving by new topic (kg) lettuce Answer: (kg). (t) machine weight and packing together. Answer: (t). (t) the second brigade received nails. Answer: (t).	Cognitive: generating interest in the topic. Personal: formation of readiness for self-education. Communicative: the ability to formulate their thoughts orally; listen to and understand the speech of others. Regulatory: planning their activities to solve the task and control the result.
5. Physical education	Change of activity.		Change activities, provide emotional relief for students. (slide number 23) Physical education minute	Students have changed activities and are ready to continue working.
6. Primary fastening	Establishing the correctness and awareness of the study of the topic. Identification of gaps in the primary comprehension of the studied material, correction of the identified gaps, ensuring the consolidation in the memory of children of the knowledge and methods of action that they need for independent work on new material.		The first five students who complete the work receive grades. And so that we can quickly check the correctness of the solution. I invite 4 students to the board. At the board, each performs one column. Dunno turned to us for help, he asks you to check the work he did. (slide number 24)	Solution with commenting; b) ; G) and); h) . Independent work: (slide number 25)	Regulatory: implementation of ascertaining and predictive control by the result and by the method of action. Cognitive:- ability to navigate in the system of knowledge, Communicative:, control, correction, evaluation.
7. Control of assimilation, discussion of the mistakes made and their correction.	Give a qualitative assessment of the work of the class and individual students.		What did you learn in class today? Who wants to formulate a rule for finding the addition of fractions with the same denominators. Who wants to formulate a rule for finding the subtraction of fractions with the same denominators.	Students formulate the rules for adding and subtracting fractions with the same denominators.	Personal: the formation of positive self-esteem Communicative:; the ability to express one's thoughts with sufficient completeness and accuracy; Regulatory: ability to be independent but analyze the correctness of the actions and make the necessary my corrections.
8. Reflection (summarizing the lesson)			It was difficult … It was interesting … I learned … I was surprised... Am I……….mood? (slide number 26)	Students answer questions. They express their opinions.	Regulatory: evaluating one's own performance in the classroom. Communicative: ability to analyze own successes, failures, to determine the ways of correction. Cognitive: reflection.
9. Information about homework	Ensuring children understand the purpose, content and methods of doing homework		Reports homework: Execute in writing №1017, №1019, №1020. (slide number 27)	Open diaries, write down homework, ask questions.

Literature:

1. Vilenkin N.Ya., "Mathematics 5", "Mnemosyne", 2007

2. Chesnokov A.S., "Didactic materials in mathematics, grade 5", M, 2006

3. Super physical education http://videouroki.net/diski.php

View presentation content
"ADDITION AND SUBTRACTION OF FRACTIONS WITH THE SAME DENOMINATORS"

ADDITION AND SUBTRACTION OF FRACTIONS WITH THE SAME DENOMINATORS

Organizing time

Control buttons

"Go back" (return to the previous slide)

"To the beginning" (return 1 slide)

"To get out"

Well, check it out buddy

Are you ready to start the lesson?

Everything is in place

Is it all right

Pen, book and notebook?

Is everyone seated correctly?

Is everyone watching closely?

Everyone wants to receive

Only a rating of "5".

Lesson Objectives:

Tutorial:

Developing:

Educational:

Tutorial:

Developing:

Educational:

- repeat the concepts of "Correct, improper fraction",

- generalize and consolidate knowledge about comparing fractions,

Learn how to add and subtract fractions with the same denominators.

Developing:

Educational:

Tutorial:

- develop attention

- develop logical thinking,

- to develop competent mathematical speech.

Educational:

Developing:

Tutorial:

- cultivate accuracy when writing examples and problems with ordinary fractions.

What part of the figure is:

a) triangle ABO from quadrilateral ABCO;

b) triangle АOL from polygon CВALK;

c) what part of the figure is painted in red;

What part of the figure is:

a) triangle ABO from quadrilateral ABCO;

b) triangle АOL from polygon CВAL;

c) quadrilateral ABCO from the whole figure.

What part of the figure is:

a) triangle ABO from quadrilateral ABCO;

b) triangle АOL from polygon CВALK;

c) quadrilateral ABCO from the whole figure.

Additional names for some fractions

Half (One of two equal parts that together make up a whole).

Third (One of three equal parts into which something is divided).

Quarter (One of four equal parts into which something is divided).

Harvest

Help Dunno to collect pears on which wrong fractions are written.

A fraction in which the numerator is less than the denominator is called correct shot.

A fraction in which the numerator is greater than the denominator is called wrong shot.

Compare fractions

When adding fractions with the same denominators, add the numerators and keep the denominator the same.

Using letters, the addition rule can be written as follows:

When subtracting fractions with the same denominators, the numerator of the subtrahend is subtracted from the numerator of the minuend, and the denominator is left the same.

Using letters, the subtraction rule can be written as follows:

Working with the textbook

Page 156

№ 1005

№ 1006

№ 1008

• It was difficult …
• It was interesting …
• I learned …
• I was surprised...
• I have……….mood

Babenko Natalia Emanoilovna

Mathematic teacher

MOU "Secondary School No. 13"

Vorkuta Komi.

You can perform various actions with fractions, for example, adding fractions. Addition of fractions can be divided into several types. Each type of addition of fractions has its own rules and algorithm of actions. Let's take a closer look at each type of addition.

Adding fractions with the same denominators.

For example, let's see how to add fractions with a common denominator.

The hikers went on a hike from point A to point E. On the first day, they walked from point A to B, or \(\frac(1)(5)\) all the way. On the second day they went from point B to D or \(\frac(2)(5)\) the whole way. How far did they travel from the beginning of the journey to point D?

To find the distance from point A to point D, add the fractions \(\frac(1)(5) + \frac(2)(5)\).

Adding fractions with the same denominators is that you need to add the numerators of these fractions, and the denominator will remain the same.

\(\frac(1)(5) + \frac(2)(5) = \frac(1 + 2)(5) = \frac(3)(5)\)

In literal form, the sum of fractions with the same denominators will look like this:

\(\bf \frac(a)(c) + \frac(b)(c) = \frac(a + b)(c)\)

Answer: the tourists traveled \(\frac(3)(5)\) all the way.

Adding fractions with different denominators.

Consider an example:

Add two fractions \(\frac(3)(4)\) and \(\frac(2)(7)\).

To add fractions with different denominators, you must first find, and then use the rule for adding fractions with the same denominators.

For denominators 4 and 7, the common denominator is 28. The first fraction \(\frac(3)(4)\) must be multiplied by 7. The second fraction \(\frac(2)(7)\) must be multiplied by 4.

\(\frac(3)(4) + \frac(2)(7) = \frac(3 \times \color(red) (7) + 2 \times \color(red) (4))(4 \ times \color(red) (7)) = \frac(21 + 8)(28) = \frac(29)(28) = 1\frac(1)(28)\)

In literal form, we get the following formula:

\(\bf \frac(a)(b) + \frac(c)(d) = \frac(a \times d + c \times b)(b \times d)\)

Addition of mixed numbers or mixed fractions.

Addition occurs according to the law of addition.

For mixed fractions, add the integer parts to the integer parts and the fractional parts to the fractional parts.

If the fractional parts of mixed numbers have the same denominators, then add the numerators, and the denominator remains the same.

Add mixed numbers \(3\frac(6)(11)\) and \(1\frac(3)(11)\).

\(3\frac(6)(11) + 1\frac(3)(11) = (\color(red) (3) + \color(blue) (\frac(6)(11))) + ( \color(red) (1) + \color(blue) (\frac(3)(11))) = (\color(red) (3) + \color(red) (1)) + (\color( blue) (\frac(6)(11)) + \color(blue) (\frac(3)(11))) = \color(red)(4) + (\color(blue) (\frac(6 + 3)(11))) = \color(red)(4) + \color(blue) (\frac(9)(11)) = \color(red)(4) \color(blue) (\frac (9)(11))\)

If the fractional parts of mixed numbers have different denominators, then we find a common denominator.

Let's add mixed numbers \(7\frac(1)(8)\) and \(2\frac(1)(6)\).

The denominator is different, so you need to find a common denominator, it is equal to 24. Multiply the first fraction \(7\frac(1)(8)\) by an additional factor of 3, and the second fraction \(2\frac(1)(6)\) on 4.

\(7\frac(1)(8) + 2\frac(1)(6) = 7\frac(1 \times \color(red) (3))(8 \times \color(red) (3) ) = 2\frac(1 \times \color(red) (4))(6 \times \color(red) (4)) =7\frac(3)(24) + 2\frac(4)(24 ) = 9\frac(7)(24)\)

Related questions:
How to add fractions?
Answer: first you need to decide what type the expression belongs to: fractions have the same denominators, different denominators or mixed fractions. Depending on the type of expression, we proceed to the solution algorithm.

How to solve fractions with different denominators?
Answer: you need to find a common denominator, and then follow the rule of adding fractions with the same denominators.

How to solve mixed fractions?
Answer: Add integer parts to integer parts and fractional parts to fractional parts.

Example #1:
Can the sum of two result in a proper fraction? Wrong fraction? Give examples.

\(\frac(2)(7) + \frac(3)(7) = \frac(2 + 3)(7) = \frac(5)(7)\)

The fraction \(\frac(5)(7)\) is a proper fraction, it is the result of the sum of two proper fractions \(\frac(2)(7)\) and \(\frac(3)(7)\).

\(\frac(2)(5) + \frac(8)(9) = \frac(2 \times 9 + 8 \times 5)(5 \times 9) =\frac(18 + 40)(45) = \frac(58)(45)\)

The fraction \(\frac(58)(45)\) is an improper fraction, it is the result of the sum of the proper fractions \(\frac(2)(5)\) and \(\frac(8)(9)\).

Answer: The answer is yes to both questions.

Example #2:
Add fractions: a) \(\frac(3)(11) + \frac(5)(11)\) b) \(\frac(1)(3) + \frac(2)(9)\).

a) \(\frac(3)(11) + \frac(5)(11) = \frac(3 + 5)(11) = \frac(8)(11)\)

b) \(\frac(1)(3) + \frac(2)(9) = \frac(1 \times \color(red) (3))(3 \times \color(red) (3)) + \frac(2)(9) = \frac(3)(9) + \frac(2)(9) = \frac(5)(9)\)

Example #3:
write down mixed fraction as a sum natural number and a proper fraction: a) \(1\frac(9)(47)\) b) \(5\frac(1)(3)\)

a) \(1\frac(9)(47) = 1 + \frac(9)(47)\)

b) \(5\frac(1)(3) = 5 + \frac(1)(3)\)

Example #4:
Calculate the sum: a) \(8\frac(5)(7) + 2\frac(1)(7)\) b) \(2\frac(9)(13) + \frac(2)(13) \) c) \(7\frac(2)(5) + 3\frac(4)(15)\)

a) \(8\frac(5)(7) + 2\frac(1)(7) = (8 + 2) + (\frac(5)(7) + \frac(1)(7)) = 10 + \frac(6)(7) = 10\frac(6)(7)\)

b) \(2\frac(9)(13) + \frac(2)(13) = 2 + (\frac(9)(13) + \frac(2)(13)) = 2\frac(11 )(13) \)

c) \(7\frac(2)(5) + 3\frac(4)(15) = 7\frac(2 \times 3)(5 \times 3) + 3\frac(4)(15) = 7\frac(6)(15) + 3\frac(4)(15) = (7 + 3)+(\frac(6)(15) + \frac(4)(15)) = 10 + \frac (10)(15) = 10\frac(10)(15) = 10\frac(2)(3)\)

Task #1:
At dinner they ate \(\frac(8)(11)\) of the cake, and in the evening at dinner they ate \(\frac(3)(11)\). Do you think the cake was completely eaten or not?

Solution:
The denominator of the fraction is 11, it indicates how many parts the cake was divided into. At lunch, we ate 8 pieces of cake out of 11. At dinner, we ate 3 pieces of cake out of 11. Let's add 8 + 3 = 11, we ate pieces of cake out of 11, that is, the whole cake.

\(\frac(8)(11) + \frac(3)(11) = \frac(11)(11) = 1\)

Answer: They ate the whole cake.