Record an example in the form of inequality. Linear inequalities

That is, first record a variable included in the inequality, then using a sign of an accessory ∈ indicate which numerical gap belongs to the values \u200b\u200bof this variable. In this case, the expression x. ∈ [2; 8] indicates that the variable x inequality 2 ≤ x.≤ 8, takes all values \u200b\u200bin the interval from 2 to 8 inclusive. With these values, the inequality will be correct.

Note that the answer is recorded with square brackets, since the boundaries of inequality 2 ≤ x.≤ 8, namely, numbers 2 and 8 belong to many solutions of this inequality.

Many solutions of inequality 2 ≤ x.≤ 8 can also be depicted using a coordinate direct:

Here, the boundaries of the numerical gap 2 and 8 correspond to the boundaries of inequality 2 ≤ x. x. 2 ≤ x.≤ 8 .

In some border sources that do not belong to the numerical gap call open .

They are called them for the reason that the numerical gap remains open due to the fact that its boundaries do not belong to this numerical gap. An empty circle on the coordinate direct mathematics is called purifying point . It means to eliminate the point to exclude it from a numerical interval or from a variety of inequality solutions.

And in the case when the borders belong to the numerical gap, they are called closed (or closed), since such boundaries are closed (closed) a numerical gap. The creamy circle on the coordinate direct also indicates the closedness of the borders.

There are varieties of numerical gaps. Consider each of them.

Numerical beam

Numeric beam x ≥ A.where a. x - Solution of inequality.

Let be a.\u003d 3. Then inequality x ≥ A. Take view x.≥ 3. Decisions of this inequality are all numbers that are more than 3, including the number 3.

Show a numeric ray given by inequality x.≥ 3, on the coordinate direct. To do this, we note on it the point with the coordinate 3, and the rest to the right of her area We highlight strokes. It is precisely the right side, since inequality solutions x.≥ 3 are numbers, more than 3. And more than the coordinate direct are right

x.≥ 3, and the area selected by strokes corresponds to a set of values x. that are inequality solutions x.≥ 3 .

Point 3, which is the boundary of the numerical beam, is depicted in the form of a painted mug, because the border of inequality x.≥ 3 belongs to the set of its solutions.

On a letter of a numeric ray given by inequality x ≥ a

[ a.; +∞)

It can be seen that on the one hand, the boundary is framed by a square bracket, and on the other round. This is due to the fact that one border of the numeric beam belongs to him, and the other is not, because the infinity of the borders itself does not and is meant that on the other side there is no number that closes this numeric ray.

Considering that one of the boundaries of the numeric beam is closed, this gap is often called closed numeric beam.

We write the answer to inequality x.≥ 3 With the designation of the numerical beam. We have a variable a. equal to 3.

x. ∈ [ 3 ; +∞)

In this expression it is said that the variable x. inequality x.≥ 3, takes all values \u200b\u200bfrom 3 to plus infinity.

In other words, all numbers from 3 to plus infinity are inequality solutions x.≥ 3. Border 3 belongs to many solutions, because inequality x.≥ 3 is non-strict.

The closed numeric beam is also called a numerical gap that is defined in inequality. x ≤ a.Solutions inequality x ≤ A. aincluding the number a.

For example, if a. x.≤ 2. On the coordinate direct boundary 2 will be depicted with a circle, and the whole area located leftwill be highlighted with strokes. This time the left part stands out, since inequality solutions x.≤ 2 are numbers smaller than 2. And fewer numbers on the coordinate direct left

x.≤ 2, and the area selected by strokes corresponds to a set of values x. that are inequality solutions x.≤ 2 .

Point 2, which is the boundary of the numerical beam, is depicted in the form of a painted mug, because the border of inequality x.≤ 2 belongs to the set of its solutions.

We write the answer to inequality x.≤ 2 using the numerical beam designation:

x. ∈ (−∞ ; 2 ]

x.≤ 2. Border 2 belongs to set solutions, since inequality x.≤ 2 is non-strict.

Outdoor numeric beam

Open numeric beam refer to the numerical gap that is defined inequality x\u003e A. where a. - the border of this inequality, x. - Decision of inequality.

An open numeric beam is largely similar to a closed numeric beam. The difference is that the border a. does not belong a gap, like the border of inequality x\u003e A. does not belong to the set of its solutions.

Let be a.\u003d 3. Then the inequality will take a view x.\u003e 3. Decisions of this inequality are all numbers that are more than 3, with the exception of the number 3

On the coordinate direct boundary of the open numeric beam given by inequality x.\u003e 3 will be depicted in the form of an empty mug. The whole area located on the right will be highlighted by strokes:

Here, point 3 corresponds to the border of inequality x\u003e3, and the area selected by strokes corresponds to a set of values x. that are inequality solutions x\u003e 3. Point 3, which is the boundary of the open numerical beam, is depicted in the form of an empty mug, because the border of inequality x\u003e 3 does not belong to the set of its solutions.

x\u003e A, indicated as follows:

(a.; +∞)

Round brackets indicate that the boundaries of the open numeric beam do not belong to him.

We write the answer to inequality x. \u003e 3 with the designation of the open numeric beam:

x. ∈ (3 ; +∞)

In this expression it is said that all numbers from 3 to plus infinity are inequality solutions x. \u003e 3. The border 3 does not belong to the set of solutions, since inequality x. \u003e 3 is strict.

An open numeric beam is also called a numerical interval that inequality x.< a where a. - the border of this inequality, x. - Solution of inequality . Solutions inequality x.< a are all numbers that are less aexcluding Number a.

For example, if a.\u003d 2, the inequality will take a view x.< 2. On the coordinate direct border 2 will be depicted with an empty circle, and the whole area located on the left will be highlighted by strokes:

Here, point 2 corresponds to the border of inequality x.< 2, and the area selected by strokes corresponds to a set of values x. that are inequality solutions x.< 2. Point 2, which is the boundary of the open numerical beam, is depicted in the form of an empty mug, because the border of inequality x.< 2 does not belong to the set of its solutions.

On the letter an open numeric ray given by inequality x.< a , indicated as follows:

(−∞ ; a.)

We write the answer to inequality x.< 2 With the designation of the open numeric beam:

x. ∈ (−∞ ; 2)

In this expression it is said that all numbers from minus infinity to 2 are solutions inequality x.< 2. The border 2 does not belong to the set of solutions, since inequality x.< 2 is strict.

Section

Cut a ≤ x ≤ B where a. and b. x. - Decision of inequality.

Let be a. = 2 , b. \u003d 8. Then inequality a ≤ x ≤ B will take a form 2 ≤ x.≤ 8. Solutions inequality 2 ≤ x.≤ 8 are all numbers that are greater than 2 and less than 8. At the same time, the boundaries of inequality 2 and 8 belong to the set of its solutions, since inequality 2 ≤ x.≤ 8 is non-strict.

Show a segment given by double inequality 2 ≤ x.≤ 8 on the coordinate direct. To do this, we note on it the points with coordinates 2 and 8, and the area of \u200b\u200bthe area is located between them, distribute strokes:

≤ x.≤ 8, and the area selected by strokes corresponds to a set of values x. x.≤ 8. Points 2 and 8, which are the boundaries of the segment, are depicted in the form of painted circles, since the boundaries of inequality 2 ≤ x.≤ 8 belongs to the set of its solutions.

On the letter of the segment given by inequality a ≤ x ≤ B indicated as follows:

[ a; B. ]

Square brackets on both sides indicate that the borders of the segment owned his. We write the answer to inequality 2 ≤ x.

x. ∈ [ 2 ; 8 ]

In this expression it is said that all numbers from 2 to 8 inclusive are solutions of inequality 2 ≤ x.≤ 8 .

Interval

Interval call a numerical gap that is set by double inequality a.< x < b where a. and b. - borders of this inequality, x. - Decision of inequality.

Let be a \u003d 2., b \u003d 8. . Then inequality a.< x < b Type 2< x.< 8 . Решениями этого двойного неравенства являются все числа, которые больше 2 и меньше 8, исключая числа 2 и 8.

I will depict an interval on the coordinate direct:

Here, points 2 and 8 correspond to the boundaries of inequality 2< x.< 8 , а выделенная штрихами область соответствует множеству значений x. < x.< 8 . Точки 2 и 8, являющиеся границами интервала, изображены в виде пустых кружков, поскольку границы неравенства 2 < x.< 8 не принадлежат множеству его решений.

In the letter the interval defined in inequality a.< x < b, indicated as follows:

(a; B.)

Round brackets on both sides indicate that the interval boundaries do not belong his. We write the answer to inequality 2< x.< 8 с помощью этого обозначения:

x. ∈ (2 ; 8)

This expression states that all numbers from 2 to 8, excluding numbers 2 and 8, are solutions of inequality 2< x.< 8 .

Semi-interval

Half-interval refer to the numerical gap that is defined inequality a ≤ X.< b where a. and b. - borders of this inequality, x. - Decision of inequality.

The semi-interval also refers to a numerical gap that is set in inequality a.< x ≤ b .

One of the boundaries of the half-interval belongs to him. Hence the name of this numerical gap.

In a situation with a semi-interval a ≤ X.< b He (semi-interval) belongs to the left border.

And in a situation with a semi-interval a.< x ≤ b He owns the right border.

Let be a.= 2 , b.\u003d 8. Then inequality a ≤ X.< b will take a form 2 ≤ x. < 8 . Решениями этого двойного неравенства являются все числа, которые больше 2 и меньше 8, включая число 2, но исключая число 8.

Picture half-interval 2 ≤ x. < 8 на координатной прямой:

x. < 8 , а выделенная штрихами область соответствует множеству значений x. which are inequality solutions 2 ≤ x. < 8 .

Point 2, which is left border half-interval, depicted in the form of a painted mug, as the left limit of inequality 2 ≤ x. < 8 belongsthe set of its solutions.

And point 8, which is right border half-interval, depicted in the form of an empty mug, as the right limit of inequality 2 ≤ x. < 8 not belongs The set of its solutions.

a ≤ X.< b, indicated as follows:

[ a; B.)

It can be seen that on the one hand, the boundary is framed by a square bracket, and on the other round. This is due to the fact that one border of the semi-interval belongs to him, and the other is not. We write the answer to inequality 2 ≤ x. < 8 с помощью этого обозначения:

x. ∈ [ 2 ; 8)

In this expression it is said that all numbers from 2 to 8, including the number 2, but excluding the number 8, are solutions of inequality 2 ≤ x. < 8 .

Similarly, the coordinate direct can be depicted by a semi-interval defined in inequality a.< x ≤ b . Let be a.= 2 , b.\u003d 8. Then inequality a.< x ≤ b Type 2< x.≤ 8. Decisions of this double inequality are all numbers that are greater than 2 and less than 8, excluding the number 2, but including the number 8.

I will show half-interval 2.< x.≤ 8 on the coordinate direct:

Here, points 2 and 8 correspond to the boundaries of inequality 2< x.≤ 8, and the area selected by strokes corresponds to a set of values x. which are decisions of inequality 2< x.≤ 8 .

Point 2, which is left border half-interval, depicted in the form of an empty mug, as the left limit of inequality 2< x.≤ 8 not belongthe set of its solutions.

And point 8, which is right border half-interval, depicted in the form of a painted mug, as the right limit of inequality 2< x.≤ 8 belongsthe set of its solutions.

In the letter of the semi-interval defined inequality a.< x ≤ b, denotes like this: ( a; B. ]. We write the answer to inequality 2< x.≤ 8 With this designation:

x. ∈ (2 ; 8 ]

In this expression it is said that all numbers from 2 to 8, excluding number 2, but including the number 8, are solutions of inequality 2< x.≤ 8 .

An image of numerical intervals on the coordinate direct

Numerical gap can be specified using inequality or with the designation (round or square brackets). In both cases, you need to be able to depict this numerical gap on the coordinate direct. Consider several examples.

Example 1.. Pictulate a numerical gap given by inequality x.> 5

Remember that inequality of the form x.> a. Set an open numeric beam. In this case, the variable a. equal to 5. Inequality x.\u003e 5 strict, therefore the border 5 will be depicted as an empty circle. We are interested in all values. x which are more than 5, so the whole area on the right will be highlighted by strokes:

Example 2.. Portray the numerical gap (5; + ∞) on the coordinate direct

This is the same numerical gap that we depicted in the previous example. But this time it is defined not by the help of inequality, but by the designation of the numerical gap.

The border 5 is framed by a round bracket, which means it does not belong to the gap. Accordingly, the circle remains empty.

The + ∞ symbol indicates that we are interested in all the numbers that are more 5. Accordingly, the whole area to the right of the border 5 is highlighted with strokes:

Example 3.. Pictitate a numerical interval (-5; 1) on the coordinate direct.

Round brackets on both sides are the intervals. The interval boundaries do not belong to him, therefore the borders of -5 and 1 will be depicted on the coordinate line in the form of empty circles. The whole area between them will be highlighted by strokes:

Example 4.. Pictulate a numeric gap specified inequality -5< x.< 1

This is the same numerical gap that we depicted in the previous example. But this time it is defined not by the designation of the gap, but with the help of double inequality.

Inequality of type a.< x < b The interval is set. In this case, the variable a. equal to -5, and variable b. equal to one. Inequality -5< x.< 1 strict, therefore borders -5 and 1 will be depicted in the form of an empty mug. We are interested in all values. x which are more -5, but less than one, therefore the entire area between points -5 and 1 will be highlighted by strokes:

Example 5.. Portray on the coordinate direct numeric intervals [-1; 2] I.

This time, we will be shown on the coordinate direct immediately two gaps.

Square brackets are designated on both sides. The borders of the segment belong to him, therefore the boundaries of segments [-1; 2] and will be depicted on the coordinate line in the form of painted circles. The whole area between them will be highlighted by strokes.

To see the intervals [-1; 2] and, the first can be depicted on the upper area, and the second on the bottom. So do:

Example 6.. Portray on the coordinate direct numeric intervals [-1; 2) and (2; 5]

A square bracket on one side and round with the other are intervals. One of the boundaries of the half-interval belongs to him, and the other is not.

In the case of a semi-interval [-1; 2) The left border will belong to him, and the rightmost. So the left border will be depicted in the form of a painted mug. The right border will be depicted as an empty mug.

And in the case of the semi-interval (2; 5], it will only have the right limit, and the left one will not. So the left border will be depicted in the form of a painted mug. The right border will be depicted as an empty mug.

Depict the interval [-1; 2) on the upper area of \u200b\u200bthe coordinate direct, and the gap (2; 5] - on the bottom:

Examples of solutions of inequality

Inequality, which by identical transformations can be brought to mind aX\u003e B. (or to mind aX.< b ) let's call linear inequality with one variable.

In linear inequality aX\u003e B. , x. - this is a variable, the values \u200b\u200bof which you need to find, but - the coefficient of this variable, b. - The border of inequality, which, depending on the sign of inequality, can belong to the set of its solutions or not belong to it.

For example, inequality 2 x.\u003e 4 is the inequality of the type aX\u003e B. . In it the role of variable a. Plays number 2, role variable b. (The boundaries of inequality) plays the number 4.

Inequality 2. x.\u003e 4 can be done even easier. If we split both parts by 2, we will get inequality x.> 2

Received inequality x.\u003e 2 is also inequality of type aX\u003e B. , that is, linear inequality with one variable. In this inequality, the role of variable a. Plays a unit. Previously, we said that the coefficient 1 is not written. The role of the variable b. Plays number 2.

Stripping from this information, let's try to solve several simple inequalities. During the solution, we will perform elementary identical transformations in order to obtain the inequality of the form aX\u003e B.

Example 1.. Solve inequality x.− 7 < 0

Add to both parts of inequality number 7

x.− 7 + 7 < 0 + 7

In the left part will remain x. and the right side will become equal to 7

x.< 7

By the elementary transformations, we led inequality x.− 7 < 0 к равносильному неравенству x.< 7 . Решениями неравенства x.< 7 являются все числа, которые меньше 7. Граница 7 не принадлежит множеству решений, поскольку неравенство строгое.

When inequality is given to mind x.< a (or x\u003e A. ), it can be considered already solved. Our inequality x.− 7 < 0 тоже приведено к такому виду, а именно к виду x.< 7 . Но в большинстве школ требуют, чтобы ответ был записан с помощью числового промежутка и проиллюстрирован на координатной прямой.

We write the answer using a numeric gap. In this case, the answer will be an open numeric beam (you remember that the numeric ray is given inequality x.< a and denotes how (-∞; a.)

x. ∈ (−∞ ; 7)

On the coordinate direct boundary 7 will be depicted in the form of an empty mug, and the whole area located to the left of the border will be highlighted by strokes:

For checking, take any number from the interval (-∞; 7) and substitute it to inequality x.< 7 вместо переменной x. . Take, for example, number 2

2 < 7

It turned out the right numerical inequality, it means that the solution is correct. Take some other number, for example, number 4

4 < 7

It turned out true numerical inequality. So the decision is correct.

And since inequality x.< 7 равносильно исходному неравенству x -7 < 0 , то решения неравенства x.< 7 будут совпадать с решениями неравенства x -7 < 0 . Подставим те же тестовые значения 2 и 4 в неравенство x -7 < 0

2 − 7 < 0

−5 < 0 — Верное неравенство

4 − 7 < 0

−3 < 0 Верное неравенство

Example 2.. Solve inequality -4. x. < −16

We split both parts of inequality on -4. Do not forget that when dividing both parts of inequality on a negative number, sign of inequality changes to the opposite:

We led inequality -4 x. < −16 к равносильному неравенству x.\u003e 4. Solutions inequality x.\u003e 4 There will be all numbers that are more than 4. The border 4 does not belong to the set of solutions, since the inequality is strict.

x.\u003e 4 on the coordinate direct and write the answer in the form of a numerical gap:

Example 3.. Solve inequality 3y +. 1 > 1 + 6y.

Purchase 6. y. From the right side to the left, changing the sign. And 1 from the left side by transferring to the right side, again changing the sign:

3y.− 6y.> 1 − 1

We give similar terms:

−3y. > 0

We split both parts on -3. Do not forget that when dividing both parts inequality to a negative number, the sign of inequality changes to the opposite:

Solutions inequality y.< 0 являются все числа, меньшие нуля. Изобразим множество решений неравенства y.< 0 на координатной прямой и запишем ответ в виде числового промежутка:

Example 4.. Solve inequality 5(x.− 1) + 7 ≤ 1 − 3(x.+ 2)

Recall brackets in both parts of inequality:

We suffer -3 x. From the right side to the left, changing the sign. Members -5 and 7 from the left side by transferring to the right side, again changing the signs:

We give similar terms:

We divide both parts of the received inequality on 8

Decisions of inequalities are all numbers that are less. The border belongs to many decisions, since the inequality is incredible.

Example 5.. Solve inequality

Multiply both parts of inequality on 2. This will allow you to get rid of the fraraty in the left side:

Now we move 5 from the left side to the right side by changing the sign:

After bringing similar terms, we get inequality 6 x.\u003e 1. We divide both parts of this inequality by 6. Then we get:

Inequality solutions are all numbers that are more. The border does not belong to the set of solutions, since the inequality is strict.

We will show many solutions of inequality on the coordinate direct and write the answer in the form of a numerical gap:

Example 6.. Solve inequality

Multiply both parts on 6

After bringing similar terms, we get inequality 5 x.< 30 . Разделим обе части этого неравенства на 5

Solutions inequality x.< 6 являются все числа, которые меньше 6. Граница 6 не принадлежит множеству решений, поскольку неравенство является x.< 6 строгим.

I will show many solutions inequality x.< 6 на координатной прямой и запишем ответ в виде числового промежутка:

Example 7.. Solve inequality

Multiply both parts of inequality for 10

In the resulting inequality, we will reveal brackets in the left side:

We transfer members without x. On the right side

We give similar terms in both parts:

We divide both parts of the resulting inequality for 10

Solutions inequality x.≤ 3.5 are all numbers that are less than 3.5. The border 3.5 belongs to many solutions, since inequality is x.≤ 3.5 incredible.

I will show many solutions inequality x.≤ 3.5 on the coordinate direct and write the answer in the form of a numerical interval:

Example 8.. Solve inequality 4.< 4x.< 20

To solve such inequality, you need a variable x. Free from the coefficient 4. Then we can say in which interval is the solution of this inequality.

To free the variable x. from the coefficient, you can divide member 4 x. By 4. But the rule in inequalities is such that if we divide a member of inequality to some number, then the same should be done with the rest of the members belonging to this inequality. In our case, 4 need to divide all three members of inequality 4< 4x.< 20

Solutions of inequality 1.< x.< 5 являются все числа, которые больше 1 и меньше 5. Границы 1 и 5 не принадлежат множеству решений, поскольку неравенство 1 < x.< 5 является строгим.

I will show many solutions of inequality 1< x.< 5 на координатной прямой и запишем ответ в виде числового промежутка:

Example 9.. Solve inequality -1 ≤ -2 x.≤ 0

We divide all members of inequality on -2

Received inequality 0.5 ≥ x.≥ 0. Double inequality is preferably recorded so that a smaller dick is located on the left, and more on the right. Therefore, we rewrite our inequality as follows:

0 ≤ x.≤ 0,5

Decisions of inequality 0 ≤ x.≤ 0.5 are all numbers that are greater than 0 and less than 0.5. The boundaries 0 and 0.5 belong to the set of solutions, since inequality 0 ≤ x.≤ 0.5 is non-strict.

Pictures a lot of solutions of inequality 0 ≤ x.≤ 0.5 on the coordinate direct and write the answer in the form of a numerical gap:

Example 10.. Solve inequality

Multiply both inequalities at 12

We will reveal brackets in the resulting inequality and give similar terms:

We divide both parts of the received inequalities for 2

Solutions inequality x.≤ -0.5 are all numbers that are less -0.5. The border -0.5 belongs to many solutions, since inequality x.≤ -0.5 is incredible.

I will show many solutions inequality x.≤ -0.5 on the coordinate direct and write the answer in the form of a numerical gap:

Example 11.. Solve inequality

Multiply all parts of inequality on 3

Now from each part of the resulting inequality will subtract 6

Every part of the received inequality is separated by -1. Do not forget that when dividing all parts of inequality to a negative number, the sign of inequality changes to the opposite:

Solutions of inequality 3 ≤ a ≤9 are all numbers that are greater than 3 and less than 9. Borders 3 and 9 belong to multiple solutions, since inequality 3 ≤ a ≤9 is incredible.

I will show many solutions of inequality 3 ≤ a ≤9 On the coordinate direct and write the answer in the form of a numerical gap:

When there are no solutions

There are inequalities that do not have solutions. Such, for example, is inequality 6 x.> 2(3x.+ 1). In the process of solving this inequality, we will come to the fact that the sign of inequality\u003e will not justify its location. Let's see how it looks.

We will reveal brackets in the right part of this inequality, we get 6 x.> 6x.+ 2. We transfer 6. x. from the right side to the left side by changing the sign, we get 6 x.− 6x.\u003e 2. We give such components and we obtain inequality 0\u003e 2, which is not true.

For the best understanding, rewrite the creation of similar terms in the left side as follows:

Received inequality 0. x.\u003e 2. On the left side there is a work that will be zero at any x. . And zero can not be greater than the number 2. So inequality 0 x.\u003e 2 has no solutions.

x.\u003e 2, it does not have solutions and initial inequality 6 x.> 2(3x.+ 1) .

Example 2.. Solve inequality

Multiply both parts of inequality for 3

In the resulting inequality, we postpone a member 12 x. From the right side to the left, changing the sign. Then we give similar terms:

Right part of the inequality in anyone x. It will be zero. And no less zero than -8. So inequality 0 x.< −8 не имеет решений.

And if it does not have solutions given equivalent inequality 0 x.< −8 , то не имеет решений и исходное неравенство .

Answer: no solutions.

When solutions are infinitely a lot

There are inequalities that have countless solutions. Such inequalities become faithful at any x. .

Example 1.. Solve inequality 5(3x.− 9) < 15x.

We will reveal brackets in the right part of inequality:

Purchase 15. x. From the right side to the left, changing the sign:

Let's give similar terms in the left side:

Received inequality 0. x.< 45. On the left side there is a work that will be zero at any x. . And zero less than 45. means the decision of inequality 0 x.< 45 is any number.

x.< 45 has countless solutions, then the initial inequality 5(3x.− 9) < 15x. has the same solutions.

The answer can be written as a numerical interval:

x. ∈ (−∞; +∞)

This expression states that solutions inequality 5(3x.− 9) < 15x. there are all numbers from minus infinity to plus infinity.

Example 2.. Solve inequality: 31(2x.+ 1) − 12x.> 50x.

We will reveal brackets in the left part of inequality:

We suffer 50. x. From the right side to the left, changing the sign. And a member of 31 from the left side by posting to the right side, again changing the sign:

We give similar terms:

Received inequality 0. x\u003e -31. On the left side there is a work that will be zero at any x. . And zero more than -31. It means the decision of inequality 0 x.< -31 is any number.

And if the reduced equivalent inequality 0 x\u003e -31 has countless solutions, then the initial inequality 31(2x.+ 1) − 12x.> 50x. has the same solutions.

We write the answer in the form of a numerical gap:

x. ∈ (−∞; +∞)

Tasks for self-decisions

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Inequality - the reverse side of equality. The material of this article gives the definition of inequality and initial information about it in the context of mathematics.

The concept of inequality, as well as the concept of equality, is associated with the moment of comparing two objects. While equality means "the same", then inequality, on the contrary, indicates the differences in objects that are compared. For example, and - identical objects or equal. And - objects that differ from each other or unequal.

The inequality of objects is determined by the semantic load with such words as above - below (inequality on the basis of height); Thicker - thinner (inequality on the basis of thickness); Long - shorter (inequality based on length) and so on.

It is possible to argue both on equality-inequality of objects in general and the comparison of their individual characteristics. Suppose, two objects are specified: and. Without a doubt, these objects are not the same, i.e. In general, they are not equal: on the basis of size and color. But at the same time, we can argue that they are equal to their forms - both objects are circles.

In the context of mathematics, the sense load of inequality is preserved. However, in this case, we are talking about the inequality of mathematical objects: numbers, values \u200b\u200bof expressions, values \u200b\u200bof quantities (length, area, etc.), vectors, figures, etc.

Not equal, more, less

Depending on the purpose of the task of the task, we can already be just the fact of clarifying the inequality of objects, but usually following the establishment of the fact of inequality, it is explaining how much the value is greater, and what is less.

The meaning of the words "more" and "less" to us intuitively familiar from the very beginning of our life. Obvious is the skill to determine the superiority of the object in size, quantity, etc. But ultimately, any comparison leads us to comparing numbers that define some of the characteristics of the compared objects. In fact, we find out what number is more, and what - less.

Simple example:

Example 1.

In the morning the air temperature amounted to 10 degrees Celsius; At two in the afternoon, this figure was 15 degrees. Based on the comparison of natural numbers, we can argue that the temperature value in the morning was less than its value at two o'clock in the afternoon (or at two o'clock in the day the temperature increased, became greater than the temperature in the morning).

Recording inequalities with signs

There are generally accepted designations for recording inequalities:

Definition 1.

• the sign is "not equal", which is a crossed sign "equal": ≠. This sign is located between unequal objects. For example: 5 ≠ 10 five is not ten;
• the "More" sign:\u003e and the "less" sign:< . Первый записывается между большим и меньшим объектами; второй между меньшим и большим. Например, запись о сравнении отрезков вида | A B | > | C D | suggests that the cut a b is more segment with D;
• the sign "greater than or equal": ≥ and the "less or equal" sign: ≤.

More Their meaning will be described below. We give the definition of inequality according to their records.

Definition 2.

Inequalities - Algebraic expressions that have meaning and recorded with the signs of ≠,\u003e,< , ≤ , ≥ .

Strict and incredible inequalities

Signs of strict inequalities - These are the signs of "more" and "less":\u003e and< Неравенства, составленные с их помощью – strict inequalities.

Signs of non-strategic inequalities - These are "greater than or equal" signs and "less or equal": ≥ and ≤. Inequalities compiled with their help - non-fat inequalities.

How strict inequality apply, we disassembled above. Why are the incredible inequalities? In practice, such inequalities may possibly ask cases described by the words "not more" and "no less." The phrase "no more" means less or as much - this level of comparison corresponds to the "less or equal" sign ≤. In turn, "no less" means - as much or more, and this is a sign "greater than or equal" ≥. Thus, non-strict inequalities, unlike strict, give the possibility of equality of objects.

Faithful and incorrect inequalities

Faithful inequality - then inequality, which corresponds to the meaning of inequality mentioned above. Otherwise it is invalid.

We give simple examples for visibility:

Example 2.

Inequality 5 ≠ 5 is incorrect, since the number 5 and 5 is actually equal.

Or such a comparison:

Example 3.

Suppose S - the area of \u200b\u200bsome kind of figure, in this case s< - 4 является верным неравенством, поскольку площадь всегда выражена неотрицательным числом.

Similar in meaning, the term "true inequality" are phrases "Fair inequality", "there is inequality", etc.

Properties of inequalities

We describe the properties of inequalities. The obvious fact that the object can not be unequal to itself, and this is the first property of inequality. The second property sounds like this: if the first object is not equal to the second, then the second is not equal to the first.

We describe the properties corresponding to the signs "more" or "less":

Definition 5.

• antireflectivity. This property can be expressed like this: for any object K inequality k\u003e k and k< k неверны;
• antisymmetry. This property suggests that if the first object is greater or less than the second, then the second object, respectively, less or more of the first one. We write: if M\u003e n, then n< m . Или: если m < n , то n > m;
• transitivity. In an alphabone, the specified property will look like this: if it is specified that a< b и b < с, то a < c . Наоборот: a > B and B\u003e C, which means a\u003e c. This property is intuitive and natural: if the first object is greater than the second, and the second is more than the third, it becomes clear that the first object is the more than the third.

Signs of incredible inequalities are also inherent in some properties:

Definition 6.

• reflexiveness: A ≥ A and A ≤ A (this also includes a case when a \u003d a);
• antisymmetry: If a ≤ b, then b ≥ a. If a ≥ b, then b ≤ a;
• transitivity: If a ≤ b and b ≤ c, then it is obvious that a ≤ c. And also: if a ≥ b, a b ≥ s, then and ≥ s.

Double, triple, etc. inequalities

The property of transitivity makes it possible to record double, triple and so on inequalities, which are essentially chain of inequalities. For example: Double inequality - E\u003e F\u003e G or Triple inequality K 1 ≤ k 2 ≤ k 3 ≤ k 4.

Note that it is convenient to record inequality as a chain, including different signs: equally, not equal to signs of strict and incredible inequalities. For example, x \u003d 2< y ≤ z < 15 .

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For example, an inequality is an expression \\ (x\u003e 5 \\).

Types of inequalities:

If \\ (a \\) and \\ (b \\) is numbers or, then inequality is called numeric. In fact, it is just a comparison of two numbers. Such inequalities are divided into loyal and incorrect.

For example:
\(-5<2\) - верное числовое неравенство, ведь \(-5\) действительно меньше \(2\);

\\ (17 + 3 \\ GEQ 115 \\) is an incorrect numerical inequality, since \\ (17 + 3 \u003d 20 \\), and \\ (20 \\) less \\ (115 \\) (and not more than or equal).

If \\ (a \\) and \\ (b \\) are expressions containing a variable, then we have inequality with a variable. Such inequalities are divided by types depending on the contents:

What is the solution of inequality?

If in the inequality instead of a variable to substitute any number, then it will turn into a numeric.

If this value for the IKS transforms the original inequality is the correct numeric, then it is called by the decision of inequality. If not - this value is not the solution. And to solve inequality - It is necessary to find all its solutions (or show that they are not).

For example, If we are in linear inequality \\ (x + 6\u003e 10 \\), we substitute instead of the number \\ (7 \\) - the correct numerical inequality: \\ (13\u003e 10 \\). And if we substitute \\ (2 \\), there will be an incorrect numerical inequality \\ (8\u003e 10 \\). That is, \\ (7 \\) is the solution of the initial inequality, and \\ (2 \\) is not.

However, the inequality \\ (x + 6\u003e 10 \\) has other solutions. Indeed, we will get faithful numeric inequalities at substitution and \\ (5 \\), and \\ (12 \\), and \\ (138 \\) ... And how do we find all possible solutions? To do this, use for our case we have:

\\ (x + 6\u003e 10 \\) \\ (| -6 \\)
\\ (x\u003e 4 \\)

That is, we will suit any number more than four. Now you need to record the answer. Solutions of inequalities, as a rule, record numeric, additionally noting them on the numerical axis of the hatching. For our case, we have:

Answer: \\ (x \\ in (4; + \\ infty) \\)

When does the sign change in inequality?

In inequalities there is one big trap, in which students love "to come across:

When multiplying (or division) inequalities for a negative number, changes to the opposite ("more" to "less", "more or equal" to "less or equal" and so on)

Why is this happening? To understand this, let's see the transformation of numerical inequality \\ (3\u003e 1 \\). It is true, Troika is really more united. First, try to multiply it to any positive number, for example, a two:

\\ (3\u003e 1 \\) \\ (| \\ cdot2 \\)
\(6>2\)

As you can see, after multiplying, the inequality remains true. And for whatever positive number we are multiplied - we will always receive true inequality. Now let's try to multiply on a negative number, for example, minus the top:

\\ (3\u003e 1 \\) \\ (| \\ cdot (-3) \\)
\(-9>-3\)

It turned out incorrect inequality, because minus nine less than minus three! That is, in order for the inequality to be faithful (and therefore, the transformation of multiplication on a negative was "legal"), you need to turn the comparison sign, like this: \\ (- 9<− 3\).
With division, it turns out similarly, you can check yourself.

The rule recorded above applies to all types of inequalities, and not just numeric.

Answer: \\ (x \\ in (-1; \\ infty) \\)

Inequalities and ...

Inequalities, as well as equations may have restrictions on, that is, the values \u200b\u200bof the ICA. Accordingly, those values \u200b\u200bthat are unacceptable by OTZ should be excluded from the solutions.

Example: Solve inequality \\ (\\ SQRT (X + 1)<3\)

Decision: It is clear that in order for the left part there is less \\ (3 \\), the feeding expression should be less \\ (9 \\) (because of \\ (9 \\) just \\ (3 \\)). We get:

\\ (x + 1<9\) \(|-1\)
\\ (X.<8\)

Everything? We will suit any meaning of ICA less \\ (8 \\)? Not! Because if we take, for example, it seems that the value is suitable for the requirement \\ (- 5 \\) - it will not be a solution to the initial inequality, as it will lead us to the calculation of the root from a negative number.

\\ (\\ SQRT (-5 + 1)<3\)
\\ (\\ SQRT (-4)<3\)

Therefore, we must still take into account the limitations on the values \u200b\u200bof the ICA - it cannot be such that under the root there was a negative number. Thus, we have a second requirement for X:

\\ (X + 1 \\ GEQ0 \\)
\\ (X \\ GEQ-1 \\)

And so that the X is the final decision, it should satisfy immediately with both requirements: it must be less \\ (8 \\) (to be a solution) and more \\ (- 1 \\) (to be permissible in principle). Applying to a numeric axis, we have a final answer:

Answer: \\ (\\ left [-1; 8 \\ right) \\)