How to solve algebraic fractions? Theory and practice.

This article examines operations on fractions. Rules for addition, subtraction, multiplication, division or exponentiation of fractions of the form A B will be formed and justified, where A and B can be numbers, numerical expressions or expressions with variables. In conclusion, examples of solutions with detailed descriptions will be considered.

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Rules for performing operations with general numerical fractions

General fractions have a numerator and a denominator that contain natural numbers or numerical expressions. If we consider fractions such as 3 5, 2, 8 4, 1 + 2 3 4 (5 - 2), 3 4 + 7 8 2, 3 - 0, 8, 1 2 2, π 1 - 2 3 + π, 2 0, 5 ln 3, then it is clear that the numerator and denominator can have not only numbers, but also expressions of various types.

Definition 1

There are rules by which operations with ordinary fractions are carried out. It is also suitable for general fractions:

• When subtracting fractions with like denominators, only the numerators are added, and the denominator remains the same, namely: a d ± c d = a ± c d, the values ​​a, c and d ≠ 0 are some numbers or numerical expressions.
• When adding or subtracting a fraction with different denominators, it is necessary to reduce it to a common denominator, and then add or subtract the resulting fractions with the same exponents. Literally it looks like this: a b ± c d = a · p ± c · r s, where the values ​​a, b ≠ 0, c, d ≠ 0, p ≠ 0, r ≠ 0, s ≠ 0 are real numbers, and b · p = d · r = s . When p = d and r = b, then a b ± c d = a · d ± c · d b · d.
• When multiplying fractions, the action is performed with numerators, after which with denominators, then we get a b · c d = a · c b · d, where a, b ≠ 0, c, d ≠ 0 act as real numbers.
• When dividing a fraction by a fraction, we multiply the first by the second inverse, that is, we swap the numerator and denominator: a b: c d = a b · d c.

Rationale for the rules

Definition 2

There are the following mathematical points that you should rely on when calculating:

• the slash means the division sign;
• division by a number is treated as multiplication by its reciprocal value;
• application of the property of operations with real numbers;
• application of the basic property of fractions and numerical inequalities.

With their help, you can perform transformations of the form:

a d ± c d = a · d - 1 ± c · d - 1 = a ± c · d - 1 = a ± c d ; a b ± c d = a · p b · p ± c · r d · r = a · p s ± c · e s = a · p ± c · r s ; a b · c d = a · d b · d · b · c b · d = a · d · a · d - 1 · b · c · b · d - 1 = = a · d · b · c · b · d - 1 · b · d - 1 = a · d · b · c b · d · b · d - 1 = = (a · c) · (b · d) - 1 = a · c b · d

Examples

In the previous paragraph it was said about operations with fractions. It is after this that the fraction needs to be simplified. This topic was discussed in detail in the paragraph on converting fractions.

First, let's look at an example of adding and subtracting fractions with the same denominator.

Example 1

Given the fractions 8 2, 7 and 1 2, 7, then according to the rule it is necessary to add the numerator and rewrite the denominator.

Solution

Then we get a fraction of the form 8 + 1 2, 7. After performing the addition, we obtain a fraction of the form 8 + 1 2, 7 = 9 2, 7 = 90 27 = 3 1 3. So, 8 2, 7 + 1 2, 7 = 8 + 1 2, 7 = 9 2, 7 = 90 27 = 3 1 3.

Answer: 8 2 , 7 + 1 2 , 7 = 3 1 3

There is another solution. To begin with, we switch to the form of an ordinary fraction, after which we perform a simplification. It looks like this:

8 2 , 7 + 1 2 , 7 = 80 27 + 10 27 = 90 27 = 3 1 3

Example 2

Let's subtract from 1 - 2 3 · log 2 3 · log 2 5 + 1 a fraction of the form 2 3 3 · log 2 3 · log 2 5 + 1 .

Since equal denominators are given, it means that we are calculating a fraction with the same denominator. We get that

1 - 2 3 log 2 3 log 2 5 + 1 - 2 3 3 log 2 3 log 2 5 + 1 = 1 - 2 - 2 3 3 log 2 3 log 2 5 + 1

There are examples of calculating fractions with different denominators. An important point is reduction to a common denominator. Without this, we will not be able to perform further operations with fractions.

The process is vaguely reminiscent of reduction to a common denominator. That is, the least common divisor in the denominator is searched for, after which the missing factors are added to the fractions.

If the fractions being added do not have common factors, then their product can become one.

Example 3

Let's look at the example of adding fractions 2 3 5 + 1 and 1 2.

Solution

In this case, the common denominator is the product of the denominators. Then we get that 2 · 3 5 + 1. Then, when setting additional factors, we have that for the first fraction it is equal to 2, and for the second it is 3 5 + 1. After multiplication, the fractions are reduced to the form 4 2 · 3 5 + 1. The general reduction of 1 2 will be 3 5 + 1 2 · 3 5 + 1. We add the resulting fractional expressions and get that

2 3 5 + 1 + 1 2 = 2 2 2 3 5 + 1 + 1 3 5 + 1 2 3 5 + 1 = = 4 2 3 5 + 1 + 3 5 + 1 2 3 5 + 1 = 4 + 3 5 + 1 2 3 5 + 1 = 5 + 3 5 2 3 5 + 1

Answer: 2 3 5 + 1 + 1 2 = 5 + 3 5 2 3 5 + 1

When we are dealing with general fractions, then we usually do not talk about the lowest common denominator. It is unprofitable to take the product of the numerators as the denominator. First you need to check if there is a number that is less in value than their product.

Example 4

Let's consider the example of 1 6 · 2 1 5 and 1 4 · 2 3 5, when their product is equal to 6 · 2 1 5 · 4 · 2 3 5 = 24 · 2 4 5. Then we take 12 · 2 3 5 as the common denominator.

Let's look at examples of multiplying general fractions.

Example 5

To do this, you need to multiply 2 + 1 6 and 2 · 5 3 · 2 + 1.

Solution

Following the rule, it is necessary to rewrite and write the product of the numerators as a denominator. We get that 2 + 1 6 2 5 3 2 + 1 2 + 1 2 5 6 3 2 + 1. Once a fraction has been multiplied, you can make reductions to simplify it. Then 5 · 3 3 2 + 1: 10 9 3 = 5 · 3 3 2 + 1 · 9 3 10.

Using the rule for transition from division to multiplication by a reciprocal fraction, we obtain a fraction that is the reciprocal of the given one. To do this, the numerator and denominator are swapped. Let's look at an example:

5 3 3 2 + 1: 10 9 3 = 5 3 3 2 + 1 9 3 10

Then they must multiply and simplify the resulting fraction. If necessary, get rid of irrationality in the denominator. We get that

5 3 3 2 + 1: 10 9 3 = 5 3 3 9 3 10 2 + 1 = 5 2 10 2 + 1 = 3 2 2 + 1 = 3 2 - 1 2 2 + 1 2 - 1 = 3 2 - 1 2 2 2 - 1 2 = 3 2 - 1 2

Answer: 5 3 3 2 + 1: 10 9 3 = 3 2 - 1 2

This paragraph is applicable when a number or numerical expression can be represented as a fraction with a denominator equal to 1, then the operation with such a fraction is considered a separate paragraph. For example, the expression 1 6 · 7 4 - 1 · 3 shows that the root of 3 can be replaced by another 3 1 expression. Then this entry will look like multiplying two fractions of the form 1 6 · 7 4 - 1 · 3 = 1 6 · 7 4 - 1 · 3 1.

Performing Operations on Fractions Containing Variables

The rules discussed in the first article are applicable to operations with fractions containing variables. Consider the subtraction rule when the denominators are the same.

It is necessary to prove that A, C and D (D not equal to zero) can be any expressions, and the equality A D ± C D = A ± C D is equivalent to its range of permissible values.

It is necessary to take a set of ODZ variables. Then A, C, D must take the corresponding values ​​a 0 , c 0 and d 0. Substitution of the form A D ± C D results in a difference of the form a 0 d 0 ± c 0 d 0 , where, using the addition rule, we obtain a formula of the form a 0 ± c 0 d 0 . If we substitute the expression A ± C D, then we get the same fraction of the form a 0 ± c 0 d 0. From here we conclude that the selected value that satisfies the ODZ, A ± C D and A D ± C D are considered equal.

For any value of the variables, these expressions will be equal, that is, they are called identically equal. This means that this expression is considered a provable equality of the form A D ± C D = A ± C D .

Examples of adding and subtracting fractions with variables

When you have the same denominators, you only need to add or subtract the numerators. This fraction can be simplified. Sometimes you have to work with fractions that are identically equal, but at first glance this is not noticeable, since some transformations must be performed. For example, x 2 3 x 1 3 + 1 and x 1 3 + 1 2 or 1 2 sin 2 α and sin a cos a. Most often, a simplification of the original expression is required in order to see the same denominators.

Example 6

Calculate: 1) x 2 + 1 x + x - 2 - 5 - x x + x - 2, 2) l g 2 x + 4 x · (l g x + 2) + 4 · l g x x · (l g x + 2) , x - 1 x - 1 + x x + 1 .

Solution

1. To make the calculation, you need to subtract fractions that have the same denominator. Then we get that x 2 + 1 x + x - 2 - 5 - x x + x - 2 = x 2 + 1 - 5 - x x + x - 2 . After which you can expand the brackets and add similar terms. We get that x 2 + 1 - 5 - x x + x - 2 = x 2 + 1 - 5 + x x + x - 2 = x 2 + x - 4 x + x - 2
2. Since the denominators are the same, all that remains is to add the numerators, leaving the denominator: l g 2 x + 4 x (l g x + 2) + 4 l g x x (l g x + 2) = l g 2 x + 4 + 4 x (l g x + 2)
   The addition has been completed. It can be seen that it is possible to reduce the fraction. Its numerator can be folded using the formula for the square of the sum, then we get (l g x + 2) 2 from abbreviated multiplication formulas. Then we get that
   l g 2 x + 4 + 2 l g x x (l g x + 2) = (l g x + 2) 2 x (l g x + 2) = l g x + 2 x
3. Given fractions of the form x - 1 x - 1 + x x + 1 with different denominators. After the transformation, you can move on to addition.

Let's consider a twofold solution.

The first method is that the denominator of the first fraction is factorized using squares, with its subsequent reduction. We get a fraction of the form

x - 1 x - 1 = x - 1 (x - 1) x + 1 = 1 x + 1

So x - 1 x - 1 + x x + 1 = 1 x + 1 + x x + 1 = 1 + x x + 1 .

In this case, it is necessary to get rid of irrationality in the denominator.

1 + x x + 1 = 1 + x x - 1 x + 1 x - 1 = x - 1 + x x - x x - 1

The second method is to multiply the numerator and denominator of the second fraction by the expression x - 1. Thus, we get rid of irrationality and move on to adding fractions with the same denominator. Then

x - 1 x - 1 + x x + 1 = x - 1 x - 1 + x x - 1 x + 1 x - 1 = = x - 1 x - 1 + x x - x x - 1 = x - 1 + x · x - x x - 1

Answer: 1) x 2 + 1 x + x - 2 - 5 - x x + x - 2 = x 2 + x - 4 x + x - 2, 2) l g 2 x + 4 x · (l g x + 2) + 4 · l g x x · (l g x + 2) = l g x + 2 x, 3) x - 1 x - 1 + x x + 1 = x - 1 + x · x - x x - 1 .

In the last example we found that reduction to a common denominator is inevitable. To do this, you need to simplify the fractions. When adding or subtracting, you always need to look for a common denominator, which looks like the product of the denominators with additional factors added to the numerators.

Example 7

Calculate the values ​​of the fractions: 1) x 3 + 1 x 7 + 2 2, 2) x + 1 x ln 2 (x + 1) (2 x - 4) - sin x x 5 ln (x + 1) (2 x - 4) , 3) ​​1 cos 2 x - x + 1 cos 2 x + 2 cos x x + x

Solution

1. The denominator does not require any complex calculations, so you need to choose their product of the form 3 x 7 + 2 · 2, then choose x 7 + 2 · 2 for the first fraction as an additional factor, and 3 for the second. When multiplying, we get a fraction of the form x 3 + 1 x 7 + 2 2 = x x 7 + 2 2 3 x 7 + 2 2 + 3 1 3 x 7 + 2 2 = = x x 7 + 2 2 + 3 3 x 7 + 2 2 = x x 7 + 2 2 x + 3 3 x 7 + 2 2
2. It can be seen that the denominators are presented in the form of a product, which means that additional transformations are unnecessary. The common denominator will be considered to be a product of the form x 5 · ln 2 x + 1 · 2 x - 4 . Hence x 4 is an additional factor to the first fraction, and ln(x + 1) to the second. Then we subtract and get:
   x + 1 x · ln 2 (x + 1) · 2 x - 4 - sin x x 5 · ln (x + 1) · 2 x - 4 = = x + 1 · x 4 x 5 · ln 2 (x + 1 ) · 2 x - 4 - sin x · ln x + 1 x 5 · ln 2 (x + 1) · (2 ​​x - 4) = = x + 1 · x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · (2 ​​x - 4) = x · x 4 + x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · (2 ​​x - 4 )
3. This example makes sense when working with fraction denominators. It is necessary to apply the formulas for the difference of squares and the square of the sum, since they will make it possible to move on to an expression of the form 1 cos x - x · cos x + x + 1 (cos x + x) 2. It can be seen that the fractions are reduced to a common denominator. We get that cos x - x · cos x + x 2 .

Then we get that

1 cos 2 x - x + 1 cos 2 x + 2 cos x x + x = = 1 cos x - x cos x + x + 1 cos x + x 2 = = cos x + x cos x - x cos x + x 2 + cos x - x cos x - x cos x + x 2 = = cos x + x + cos x - x cos x - x cos x + x 2 = 2 cos x cos x - x cos x + x 2

Answer:

1) x 3 + 1 x 7 + 2 2 = x x 7 + 2 2 x + 3 3 x 7 + 2 2, 2) x + 1 x ln 2 (x + 1) 2 x - 4 - sin x x 5 · ln (x + 1) · 2 x - 4 = = x · x 4 + x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · ( 2 x - 4) , 3) ​​1 cos 2 x - x + 1 cos 2 x + 2 · cos x · x + x = 2 · cos x cos x - x · cos x + x 2 .

Examples of multiplying fractions with variables

When multiplying fractions, the numerator is multiplied by the numerator and the denominator by the denominator. Then you can apply the reduction property.

Example 8

Multiply the fractions x + 2 · x x 2 · ln x 2 · ln x + 1 and 3 · x 2 1 3 · x + 1 - 2 sin 2 · x - x.

Solution

Multiplication needs to be done. We get that

x + 2 x x 2 ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x) = = x - 2 x 3 x 2 1 3 x + 1 - 2 x 2 ln x 2 ln x + 1 sin (2 x - x)

The number 3 is moved to the first place for the convenience of calculations, and you can reduce the fraction by x 2, then we get an expression of the form

3 x - 2 x x 1 3 x + 1 - 2 ln x 2 ln x + 1 sin (2 x - x)

Answer: x + 2 x x 2 ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x) = 3 x - 2 x x 1 3 x + 1 - 2 ln x 2 · ln x + 1 · sin (2 · x - x) .

Division

Division of fractions is similar to multiplication, since the first fraction is multiplied by the second reciprocal. If we take for example the fraction x + 2 x x 2 ln x 2 ln x + 1 and divide by 3 x 2 1 3 x + 1 - 2 sin 2 x - x, then it can be written as

x + 2 · x x 2 · ln x 2 · ln x + 1: 3 · x 2 1 3 · x + 1 - 2 sin (2 · x - x) , then replace with a product of the form x + 2 · x x 2 · ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x)

Exponentiation

Let's move on to considering operations with general fractions with exponentiation. If there is a power with a natural exponent, then the action is considered as multiplication of equal fractions. But it is recommended to use a general approach based on the properties of degrees. Any expressions A and C, where C is not identically equal to zero, and any real r on the ODZ for an expression of the form A C r the equality A C r = A r C r is valid. The result is a fraction raised to a power. For example, consider:

x 0, 7 - π · ln 3 x - 2 - 5 x + 1 2, 5 = = x 0, 7 - π · ln 3 x - 2 - 5 2, 5 x + 1 2, 5

Procedure for performing operations with fractions

Operations on fractions are performed according to certain rules. In practice, we notice that an expression may contain several fractions or fractional expressions. Then it is necessary to perform all actions in strict order: raise to a power, multiply, divide, then add and subtract. If there are parentheses, the first action is performed in them.

Example 9

Calculate 1 - x cos x - 1 c o s x · 1 + 1 x .

Solution

Since we have the same denominator, then 1 - x cos x and 1 c o s x, but subtractions cannot be performed according to the rule; first, the actions in parentheses are performed, then multiplication, and then addition. Then when calculating we get that

1 + 1 x = 1 1 + 1 x = x x + 1 x = x + 1 x

When substituting the expression into the original one, we get that 1 - x cos x - 1 cos x · x + 1 x. When multiplying fractions we have: 1 cos x · x + 1 x = x + 1 cos x · x. Having made all the substitutions, we get 1 - x cos x - x + 1 cos x · x. Now you need to work with fractions that have different denominators. We get:

x · 1 - x cos x · x - x + 1 cos x · x = x · 1 - x - 1 + x cos x · x = = x - x - x - 1 cos x · x = - x + 1 cos x x

Answer: 1 - x cos x - 1 c o s x · 1 + 1 x = - x + 1 cos x · x .

If you notice an error in the text, please highlight it and press Ctrl+Enter

) and denominator by denominator (we get the denominator of the product).

Formula for multiplying fractions:

For example:

Before you begin multiplying numerators and denominators, you need to check whether the fraction can be reduced. If you can reduce the fraction, it will be easier for you to make further calculations.

Dividing a common fraction by a fraction.

Dividing fractions involving natural numbers.

It's not as scary as it seems. As in the case of addition, we convert the integer into a fraction with one in the denominator. For example:

Multiplying mixed fractions.

Rules for multiplying fractions (mixed):

• convert mixed fractions to improper fractions;
• multiplying the numerators and denominators of fractions;
• reduce the fraction;
• If you get an improper fraction, then we convert the improper fraction into a mixed fraction.

Note! To multiply a mixed fraction by another mixed fraction, you first need to convert them to the form of improper fractions, and then multiply according to the rule for multiplying ordinary fractions.

The second way to multiply a fraction by a natural number.

It may be more convenient to use the second method of multiplying a common fraction by a number.

Note! To multiply a fraction by a natural number, you must divide the denominator of the fraction by this number, and leave the numerator unchanged.

From the example given above, it is clear that this option is more convenient to use when the denominator of a fraction is divided without a remainder by a natural number.

Multistory fractions.

In high school, three-story (or more) fractions are often encountered. Example:

To bring such a fraction to its usual form, use division through 2 points:

Note! When dividing fractions, the order of division is very important. Be careful, it's easy to get confused here.

Note, For example:

When dividing one by any fraction, the result will be the same fraction, only inverted:

Practical tips for multiplying and dividing fractions:

1. The most important thing when working with fractional expressions is accuracy and attentiveness. Do all calculations carefully and accurately, concentratedly and clearly. It's better to write a few extra lines in your draft than to get lost in mental calculations.

2. In tasks with different types of fractions, go to the type of ordinary fractions.

3. We reduce all fractions until it is no longer possible to reduce.

4. We transform multi-level fractional expressions into ordinary ones using division through 2 points.

5. Divide a unit by a fraction in your head, simply turning the fraction over.

Fractional expressions are difficult for a child to understand. Most people have difficulties with. When studying the topic “adding fractions with whole numbers,” the child falls into a stupor, finding it difficult to solve the problem. In many examples, before performing an action, a series of calculations must be performed. For example, convert fractions or convert an improper fraction to a proper fraction.

Let’s explain it clearly to the child. Let's take three apples, two of which will be whole, and cut the third into 4 parts. Separate one slice from the cut apple, and place the remaining three next to two whole fruits. We get ¼ of an apple on one side and 2 ¾ on the other. If we combine them, we get three apples. Let's try to reduce 2 ¾ apples by ¼, that is, remove another slice, we get 2 2/4 apples.

Let's take a closer look at operations with fractions that contain integers:

First, let's remember the calculation rule for fractional expressions with a common denominator:

At first glance, everything is easy and simple. But this only applies to expressions that do not require conversion.

How to find the value of an expression where the denominators are different

In some tasks you need to find the meaning of an expression where the denominators are different. Let's look at a specific case:
3 2/7+6 1/3

Let's find the value of this expression by finding a common denominator for two fractions.

For the numbers 7 and 3, this is 21. We leave the integer parts the same, and bring the fractional parts to 21, for this we multiply the first fraction by 3, the second by 7, we get:
6/21+7/21, do not forget that whole parts cannot be converted. As a result, we get two fractions with the same denominator and calculate their sum:
3 6/21+6 7/21=9 15/21
What if the result of addition is an improper fraction that already has an integer part:
2 1/3+3 2/3
In this case, we add up the integer parts and fractional parts, we get:
5 3/3, as you know, 3/3 is one, which means 2 1/3+3 2/3=5 3/3=5+1=6

Finding the sum is all clear, let’s look at the subtraction:

From all that has been said, the rule for operations with mixed numbers follows:

• If you need to subtract an integer from a fractional expression, you do not need to represent the second number as a fraction; it is enough to perform the operation only on the integer parts.

Let's try to calculate the meaning of the expressions ourselves:

Let’s take a closer look at the example under the letter “m”:

4 5/11-2 8/11, the numerator of the first fraction is less than the second. To do this, we borrow one integer from the first fraction, we get,
3 5/11+11/11=3 whole 16/11, subtract the second from the first fraction:
3 16/11-2 8/11=1 whole 8/11

• Be careful when completing the task, do not forget to convert improper fractions into mixed fractions, highlighting the whole part. To do this, you need to divide the value of the numerator by the value of the denominator, then what happens takes the place of the whole part, the remainder will be the numerator, for example:

19/4=4 ¾, let’s check: 4*4+3=19, the denominator 4 remains unchanged.

Summarize:

Before you begin to complete a task related to fractions, you need to analyze what kind of expression it is, what transformations need to be made on the fraction in order for the solution to be correct. Look for a more rational solution. Don't go the hard way. Plan all the actions, solve them first in draft form, then transfer them to your school notebook.

To avoid confusion when solving fractional expressions, you must follow the rule of consistency. Decide everything carefully, without rushing.

Examples with fractions are one of the basic elements of mathematics. There are many different types of equations with fractions. Below are detailed instructions for solving examples of this type.

How to solve examples with fractions - general rules

To solve examples with fractions of any type, be it addition, subtraction, multiplication or division, you need to know the basic rules:

• In order to add fractional expressions with the same denominator (the denominator is the number located at the bottom of the fraction, the numerator is at the top), you need to add their numerators, and leave the denominator the same.
• In order to subtract a second fractional expression (with the same denominator) from one fraction, you need to subtract their numerators and leave the denominator the same.
• To add or subtract fractions with different denominators, you need to find the lowest common denominator.
• In order to find a fractional product, you need to multiply the numerators and denominators, and, if possible, reduce.
• To divide a fraction by a fraction, you multiply the first fraction by the second fraction reversed.

How to solve examples with fractions - practice

Rule 1, example 1:

Calculate 3/4 +1/4.

According to Rule 1, if two (or more) fractions have the same denominator, you simply add their numerators. We get: 3/4 + 1/4 = 4/4. If a fraction has the same numerator and denominator, the fraction will equal 1.

Answer: 3/4 + 1/4 = 4/4 = 1.

Rule 2, example 1:

Calculate: 3/4 – 1/4

Using rule number 2, to solve this equation you need to subtract 1 from 3 and leave the denominator the same. We get 2/4. Since two 2 and 4 can be reduced, we reduce and get 1/2.

Answer: 3/4 – 1/4 = 2/4 = 1/2.

Rule 3, Example 1

Calculate: 3/4 + 1/6

Solution: Using the 3rd rule, we find the lowest common denominator. The least common denominator is the number that is divisible by the denominators of all fractional expressions in the example. Thus, we need to find the minimum number that will be divisible by both 4 and 6. This number is 12. We write 12 as the denominator. Divide 12 by the denominator of the first fraction, we get 3, multiply by 3, write 3 in the numerator *3 and + sign. Divide 12 by the denominator of the second fraction, we get 2, multiply 2 by 1, write 2*1 in the numerator. So, we get a new fraction with a denominator equal to 12 and a numerator equal to 3*3+2*1=11. 11/12.

Answer: 11/12

Rule 3, Example 2:

Calculate 3/4 – 1/6. This example is very similar to the previous one. We do all the same steps, but in the numerator instead of the + sign, we write a minus sign. We get: 3*3-2*1/12 = 9-2/12 = 7/12.

Answer: 7/12

Rule 4, Example 1:

Calculate: 3/4 * 1/4

Using the fourth rule, we multiply the denominator of the first fraction by the denominator of the second and the numerator of the first fraction by the numerator of the second. 3*1/4*4 = 3/16.

Answer: 3/16

Rule 4, Example 2:

Calculate 2/5 * 10/4.

This fraction can be reduced. In the case of a product, the numerator of the first fraction and the denominator of the second and the numerator of the second fraction and the denominator of the first are canceled.

2 cancels from 4. 10 cancels from 5. We get 1 * 2/2 = 1*1 = 1.

Answer: 2/5 * 10/4 = 1

Rule 5, Example 1:

Calculate: 3/4: 5/6

Using the 5th rule, we get: 3/4: 5/6 = 3/4 * 6/5. We reduce the fraction according to the principle of the previous example and get 9/10.

Answer: 9/10.

How to solve examples with fractions - fractional equations

Fractional equations are examples where the denominator contains an unknown. In order to solve such an equation, you need to use certain rules.

Let's look at an example:

Solve the equation 15/3x+5 = 3

Let us remember that you cannot divide by zero, i.e. the denominator value must not be zero. When solving such examples, this must be indicated. For this purpose, there is an OA (permissible value range).

So 3x+5 ≠ 0.
Hence: 3x ≠ 5.
x ≠ 5/3

At x = 5/3 the equation simply has no solution.

Having specified the ODZ, the best way to solve this equation is to get rid of the fractions. To do this, we first present all non-fractional values ​​as a fraction, in this case the number 3. We get: 15/(3x+5) = 3/1. To get rid of fractions you need to multiply each of them by the lowest common denominator. In this case it will be (3x+5)*1. Sequencing:

1. Multiply 15/(3x+5) by (3x+5)*1 = 15*(3x+5).
2. Open the brackets: 15*(3x+5) = 45x + 75.
3. We do the same with the right side of the equation: 3*(3x+5) = 9x + 15.
4. Equate the left and right sides: 45x + 75 = 9x +15
5. Move the X's to the left, numbers to the right: 36x = – 50
6. Find x: x = -50/36.
7. We reduce: -50/36 = -25/18

Answer: ODZ x ≠ 5/3. x = -25/18.

How to solve examples with fractions - fractional inequalities

Fractional inequalities of the type (3x-5)/(2-x)≥0 are solved using the number axis. Let's look at this example.

Sequencing:

• We equate the numerator and denominator to zero: 1. 3x-5=0 => 3x=5 => x=5/3
  2. 2-x=0 => x=2
• We draw a number axis, writing the resulting values ​​on it.
• Draw a circle under the value. There are two types of circles - filled and empty. A filled circle means that the given value is within the solution range. An empty circle indicates that this value is not included in the solution range.
• Since the denominator cannot be equal to zero, there will be an empty circle under the 2nd.

• To determine the signs, we substitute any number greater than two into the equation, for example 3. (3*3-5)/(2-3)= -4. the value is negative, which means we write a minus above the area after the two. Then substitute for X any value of the interval from 5/3 to 2, for example 1. The value is again negative. We write a minus. We repeat the same with the area located up to 5/3. We substitute any number less than 5/3, for example 1. Again, minus.

• Since we are interested in the values ​​of x at which the expression will be greater than or equal to 0, and there are no such values ​​(there are minuses everywhere), this inequality has no solution, that is, x = Ø (an empty set).

Answer: x = Ø

Fraction- a number that consists of an integer number of fractions of a unit and is represented in the form: a/b

Numerator of fraction (a)- the number located above the fraction line and showing the number of shares into which the unit was divided.

Fraction denominator (b)- a number located under the fraction line and showing how many parts the unit is divided into.

2. Reducing fractions to a common denominator

3. Arithmetic operations on ordinary fractions

3.1. Addition of ordinary fractions

3.2. Subtracting fractions

3.3. Multiplying common fractions

3.4. Dividing fractions

4. Reciprocal numbers

5. Decimals

6. Arithmetic operations on decimals

6.1. Adding Decimals

6.2. Subtracting Decimals

6.3. Multiplying Decimals

6.4. Decimal division

#1. The main property of a fraction

If the numerator and denominator of a fraction are multiplied or divided by the same number that is not equal to zero, you get a fraction equal to the given one.

3/7=3*3/7*3=9/21, that is, 3/7=9/21

a/b=a*m/b*m - this is what the main property of a fraction looks like.

In other words, we get a fraction equal to the given one by multiplying or dividing the numerator and denominator of the original fraction by the same natural number.

If ad=bc, then two fractions a/b =c /d are considered equal.

For example, the fractions 3/5 and 9/15 will be equal, since 3*15=5*9, that is, 45=45

Reducing a fraction is the process of replacing a fraction in which the new fraction is equal to the original one, but with a smaller numerator and denominator.

It is customary to reduce fractions based on the basic property of the fraction.

For example, 45/60=15/ ​ ​20 =9/12=3/4 ​ ​ ​ (the numerator and denominator are divided by the number 3, by 5 and by 15).

Irreducible fraction is a fraction of the form 3/4 ​ ​ , where the numerator and denominator are mutually prime numbers. The main purpose of reducing a fraction is to make the fraction irreducible.

2. Reducing fractions to a common denominator

To bring two fractions to a common denominator, you need to:

1) factor the denominator of each fraction into prime factors;

2) multiply the numerator and denominator of the first fraction by the missing ones

factors from the expansion of the second denominator;

3) multiply the numerator and denominator of the second fraction by the missing factors from the first expansion.

Examples: Reduce fractions to a common denominator.

Let's factor the denominators into simple factors: 18=3∙3∙2, 15=3∙5

Multiply the numerator and denominator of the fraction by the missing factor 5 from the second expansion.

numerator and denominator of the fraction into the missing factors 3 and 2 from the first expansion.

= , 90 – common denominator of fractions.

3. Arithmetic operations on ordinary fractions

3.1. Addition of ordinary fractions

a) If the denominators are the same, the numerator of the first fraction is added to the numerator of the second fraction, leaving the denominator the same. As you can see in the example:

a/b+c/b=(a+c)/b ​ ​ ;

b) For different denominators, fractions are first reduced to a common denominator, and then the numerators are added according to rule a):

7/3+1/4=7*4/12+1*3/12=(28+3)/12=31/12

3.2. Subtracting fractions

a) If the denominators are the same, subtract the numerator of the second fraction from the numerator of the first fraction, leaving the denominator the same:

a/b-c/b=(a-c)/b ​ ​ ;

b) If the denominators of the fractions are different, then first the fractions are brought to a common denominator, and then the actions are repeated as in point a).

3.3. Multiplying common fractions

Multiplying fractions obeys the following rule:

a/b*c/d=a*c/b*d,

that is, they multiply the numerators and denominators separately.

For example:

3/5*4/8=3*4/5*8=12/40.

3.4. Dividing fractions

Fractions are divided in the following way:

a/b:c/d=a*d/b*c,

that is, the fraction a/b is multiplied by the inverse fraction of the given one, that is, multiplied by d/c.

Example: 7/2:1/8=7/2*8/1=56/2=28

4. Reciprocal numbers

If a*b=1, then the number b is reciprocal number for the number a.

Example: for the number 9 the reciprocal is 1/9 ​ ​ , since 9*1/9 = 1 , for the number 5 - the inverse number 1/5 ​ ​ , because 5* ​ 1/5 ​ ​ = 1 .

5. Decimals

Decimal is a proper fraction whose denominator is equal to 10, 1000, 10 000, …, 10^n 1 0 , 1 0 0 0 , 1 0 0 0 0 , . . . , 1 0 ​ n​ ​ .

For example: 6/10 =0,6; 44/1000=0,044 ​ .

Incorrect ones with a denominator are written in the same way 10^n or mixed numbers.

For example: 51/10= 5,1; 763/100=7,63

Any ordinary fraction with a denominator that is a divisor of a certain power of 10 is represented as a decimal fraction.

a changer, which is a divisor of a certain power of the number 10.

Example: 5 is a divisor of 100, so it is a fraction 1/5=1 *20/5*20=20/100=0,2 ​ 0 = 0 , 2 .

6. Arithmetic operations on decimals

6.1. Adding Decimals

To add two decimal fractions, you need to arrange them so that there are identical digits under each other and a comma under the comma, and then add the fractions like ordinary numbers.

6.2. Subtracting Decimals

It is performed in the same way as addition.

6.3. Multiplying Decimals

When multiplying decimal numbers, it is enough to multiply the given numbers, not paying attention to commas (like natural numbers), and in the resulting answer, a comma on the right separates as many digits as there are after the decimal point in both factors in total.

Let's multiply 2.7 by 1.3. We have 27\cdot 13=351 2 7 ⋅ 1 3 = 3 5 1 . We separate two digits on the right with a comma (the first and second numbers have one digit after the decimal point; 1+1=2 1 + 1 = 2 ). As a result we get 2.7\cdot 1.3=3.51 2 , 7 ⋅ 1 , 3 = 3 , 5 1 .

If the resulting result contains fewer digits than need to be separated by a comma, then the missing zeros are written in front, for example:

To multiply by 10, 100, 1000, you need to move the decimal point 1, 2, 3 digits to the right (if necessary, a certain number of zeros are assigned to the right).

For example: 1.47\cdot 10,000 = 14,700 1 , 4 7 ⋅ 1 0 0 0 0 = 1 4 7 0 0 .

6.4. Decimal division

Dividing a decimal fraction by a natural number is done in the same way as dividing a natural number by a natural number. The comma in the quotient is placed after the division of the whole part is completed.

If the integer part of the dividend is less than the divisor, then the answer is zero integers, for example:

Let's look at dividing a decimal by a decimal. Let's say we need to divide 2.576 by 1.12. First of all, let's multiply the dividend and divisor of the fraction by 100, that is, move the decimal point to the right in the dividend and divisor by as many digits as there are in the divisor after the decimal point (in this example, two). Then you need to divide the fraction 257.6 by the natural number 112, that is, the problem is reduced to the case already considered:

It happens that the final decimal fraction is not always obtained when dividing one number by another. The result is an infinite decimal fraction. In such cases, we move on to ordinary fractions.

For example, 2.8: 0.09= 28/10: 9/100= 28*100/10*9=2800/90=280/9= ​ 31 1/9 ​ ​ .