Solving inequalities examples. Solving inequalities

For example, the inequality is the expression \(x>5\).

Types of inequalities:

If \(a\) and \(b\) are numbers or , then the inequality is called numerical. It's actually just comparing two numbers. Such inequalities are divided into faithful And unfaithful.

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

\(17+3\geq 115\) is an incorrect numerical inequality, since \(17+3=20\), and \(20\) is less than \(115\) (and not greater than or equal to).

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

What is the solution to an inequality?

If you substitute a number instead of a variable into an inequality, it will turn into a numeric one.

If a given value for x turns the original inequality into a true numerical one, then it is called solution to inequality. If not, then this value is not a solution. And so that solve inequality– you need to find all its solutions (or show that there are none).

For example, if we substitute the number \(7\) into the linear inequality \(x+6>10\), we get the correct numerical inequality: \(13>10\). And if we substitute \(2\), there will be an incorrect numerical inequality \(8>10\). That is, \(7\) is a solution to the original inequality, but \(2\) is not.

However, the inequality \(x+6>10\) has other solutions. Indeed, we will get the correct numerical inequalities when substituting \(5\), and \(12\), and \(138\)... And how can we find all possible solutions? For this they use For our case we have:

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

That is, any number greater than four will suit us. Now you need to write down the answer. Solutions to inequalities are usually written numerically, additionally marking them on number axis hatching. For our case we have:

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

When does the sign of an inequality change?

There is one big trap in inequalities that students really “love” to fall into:

When multiplying (or dividing) an inequality by a negative number, it is reversed (“more” by “less”, “more or equal” by “less than or equal”, and so on)

Why is this happening? To understand this, let's look at the transformations numerical inequality\(3>1\). It is correct, three is indeed greater than one. First, let's try to multiply it by any positive number, for example, two:

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

As we can see, after multiplication the inequality remains true. And no matter what positive number we multiply by, we will always get the correct inequality. Now let's try to multiply by negative number, for example, minus three:

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

The result is an incorrect inequality, because minus nine is less than minus three! That is, in order for the inequality to become true (and therefore, the transformation of multiplication by negative was “legal”), you need to reverse the comparison sign, like this: \(−9<− 3\).
With division it will work out the same way, you can check it yourself.

The rule written above applies to all types of inequalities, not just numerical ones.

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

Inequalities and disability

Inequalities, just like equations, can have restrictions on , that is, on the values ​​of x. Accordingly, those values ​​that are unacceptable according to the DZ should be excluded from the range of solutions.

Example: Solve the inequality \(\sqrt(x+1)<3\)

Solution: It is clear that in order for the left side to be less than \(3\), the radical expression must be less than \(9\) (after all, from \(9\) just \(3\)). We get:

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

All? Any value of x smaller than \(8\) will suit us? No! Because if we take, for example, the value \(-5\) that seems to fit the requirement, it will not be a solution to the original inequality, since it will lead us to calculating the root of a negative number.

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

Therefore, we must also take into account the restrictions on the value of X - it cannot be such that there is a negative number under the root. Thus, we have the second requirement for x:

\(x+1\geq0\)
\(x\geq-1\)

And for x to be the final solution, it must satisfy both requirements at once: it must be less than \(8\) (to be a solution) and greater than \(-1\) (to be admissible in principle). Plotting it on the number line, we have the final answer:

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

Now you can understand how linear inequalities a x + b are solved<0 (они могут быть записаны и с помощью любого другого знака неравенства).

The main way to solve them is to use equivalent transformations that allow one to arrive at a≠0 to elementary inequalities type x

, ≥), p - a certain number, which are the desired solution, and for a=0 - to numerical inequalities of the form a

, ≥), from which a conclusion is drawn about the solution of the original inequality. We will analyze it first.

It also doesn’t hurt to look at solving linear inequalities in one variable from other perspectives. Therefore, we will also show how linear inequality can be solved graphically and using the interval method.

Using equivalent transformations

Let us need to solve the linear inequality a x+b<0 (≤, >, ≥). Let's show how to do this using equivalent inequality transformations.

The approaches differ depending on whether the coefficient a of the variable x is equal or not equal to zero. Let's look at them one by one. Moreover, when considering, we will adhere to a three-point scheme: first we will give the essence of the process, then we will give an algorithm for solving a linear inequality, and finally, we will give solutions to typical examples.

Let's start with algorithm for solving linear inequality a x+b<0 (≤, >, ≥) for a≠0.

• Firstly, the number b is transferred to the right side of the inequality with the opposite sign. This allows us to pass to the equivalent inequality a x<−b (≤, >, ≥).
• Secondly, both sides of the resulting inequality are divided by a non-zero number a. Moreover, if a is a positive number, then the inequality sign is preserved, and if a is a negative number, then the inequality sign is reversed. The result is an elementary inequality equivalent to the original linear inequality, and this is the answer.

It remains to understand the application of the announced algorithm using examples. Let's consider how it can be used to solve linear inequalities for a≠0.

Example.

Solve the inequality 3·x+12≤0.

Solution.

For a given linear inequality we have a=3 and b=12. Obviously, the coefficient a for the variable x is different from zero. Let's use the corresponding solution algorithm given above.

First, we move the term 12 to the right side of the inequality, not forgetting to change its sign, that is, −12 will appear on the right side. As a result, we arrive at the equivalent inequality 3·x≤−12.

And, secondly, we divide both sides of the resulting inequality by 3, since 3 is a positive number, we do not change the sign of the inequality. We have (3 x):3≤(−12):3, which is the same as x≤−4.

The resulting elementary inequality x≤−4 is equivalent to the original linear inequality and is its desired solution.

So, the solution to the linear inequality 3 x + 12≤0 is any real number less than or equal to minus four. The answer can also be written in the form of a numerical interval corresponding to the inequality x≤−4, that is, as (−∞, −4] .

Having acquired skill in working with linear inequalities, their solutions can be written down briefly without explanation. In this case, first write down the original linear inequality, and below - equivalent inequalities obtained at each step of the solution:
3 x+12≤0 ;
3 x≤−12 ;
x≤−4 .

Answer:

x≤−4 or (−∞, −4] .

Example.

List all solutions to the linear inequality −2.7·z>0.

Solution.

Here the coefficient a for the variable z is equal to −2.7. And the coefficient b is absent in explicit form, that is, it is equal to zero. Therefore, the first step of the algorithm for solving a linear inequality with one variable does not need to be performed, since moving a zero from the left side to the right will not change the form of the original inequality.

It remains to divide both sides of the inequality by −2.7, not forgetting to change the sign of the inequality to the opposite one, since −2.7 is a negative number. We have (−2.7 z):(−2.7)<0:(−2,7) , and then z<0 .

And now briefly:
−2.7·z>0;
z<0 .

Answer:

z<0 или (−∞, 0) .

Example.

Solve the inequality .

Solution.

We need to solve a linear inequality with coefficient a for the variable x equal to −5, and with coefficient b, which corresponds to the fraction −15/22. We proceed according to the well-known scheme: first we transfer −15/22 to the right side with the opposite sign, after which we divide both sides of the inequality by the negative number −5, while changing the sign of the inequality:

The last transition on the right side uses , then executed .

Answer:

Now let's move on to the case when a=0. The principle of solving the linear inequality a x+b<0 (знак, естественно, может быть и другим) при a=0 , то есть, неравенства 0·x+b<0 , заключается в рассмотрении числового неравенства b<0 и выяснении, верное оно или нет.

What is this based on? Very simple: on determining the solution to the inequality. How? Yes, here’s how: no matter what value of the variable x we ​​substitute into the original linear inequality, we will get a numerical inequality of the form b<0 (так как при подстановке любого значения t вместо переменной x мы имеем 0·t+b<0 , откуда b<0 ). Если оно верное, то это означает, что любое число является решением исходного неравенства. Если же числовое неравенство b<0 оказывается неверным, то это говорит о том, что исходное линейное неравенство не имеет решений, так как не существует ни одного значения переменной, которое обращало бы его в верное числовое равенство.

Let us formulate the above arguments in the form algorithm for solving linear inequalities 0 x+b<0 (≤, >, ≥) :

• Consider the numerical inequality b<0 (≤, >, ≥) and
• if it is true, then the solution to the original inequality is any number;
• if it is false, then the original linear inequality has no solutions.

Now let's understand this with examples.

Example.

Solve the inequality 0·x+7>0.

Solution.

For any value of the variable x, the linear inequality 0 x+7>0 will turn into the numerical inequality 7>0. The last inequality is true, therefore, any number is a solution to the original inequality.

Answer:

the solution is any number or (−∞, +∞) .

Example.

Does the linear inequality 0·x−12.7≥0 have solutions?

Solution.

If we substitute any number instead of the variable x, then the original inequality turns into a numerical inequality −12.7≥0, which is incorrect. This means that not a single number is a solution to the linear inequality 0·x−12.7≥0.

Answer:

no, it doesn't.

To conclude this section, we will analyze the solutions to two linear inequalities, both of whose coefficients are equal to zero.

Example.

Which of the linear inequalities 0·x+0>0 and 0·x+0≥0 has no solutions, and which has infinitely many solutions?

Solution.

If you substitute any number instead of the variable x, then the first inequality will take the form 0>0, and the second – 0≥0. The first of them is incorrect, and the second is correct. Consequently, the linear inequality 0·x+0>0 has no solutions, and the inequality 0·x+0≥0 has infinitely many solutions, namely, its solution is any number.

Answer:

the inequality 0 x+0>0 has no solutions, and the inequality 0 x+0≥0 has infinitely many solutions.

Interval method

In general, the method of intervals is studied in a school algebra course later than the topic of solving linear inequalities in one variable. But the interval method allows you to solve a variety of inequalities, including linear ones. Therefore, let's dwell on it.

Let us immediately note that it is advisable to use the interval method to solve linear inequalities with a non-zero coefficient for the variable x. Otherwise, it is faster and more convenient to draw a conclusion about the solution of the inequality using the method discussed at the end of the previous paragraph.

The interval method implies

• introducing a function corresponding to the left side of the inequality, in our case – linear function y=a x+b ,
• finding its zeros, which divide the domain of definition into intervals,
• determination of the signs that have function values ​​on these intervals, on the basis of which a conclusion is made about the solution of a linear inequality.

Let's collect these moments in algorithm, revealing how to solve linear inequalities a x+b<0 (≤, >, ≥) for a≠0 using the interval method:

• The zeros of the function y=a·x+b are found, for which a·x+b=0 is solved. As is known, for a≠0 it has a single root, which we denote as x 0 .
• It is constructed, and a point with coordinate x 0 is depicted on it. Moreover, if it is decided strict inequality(with sign< или >), then this point is made punctuated (with an empty center), and if it is not strict (with a sign ≤ or ≥), then a regular point is placed. This point divides the coordinate line into two intervals (−∞, x 0) and (x 0, +∞).
• The signs of the function y=a·x+b on these intervals are determined. To do this, the value of this function is calculated at any point in the interval (−∞, x 0), and the sign of this value will be the desired sign on the interval (−∞, x 0). Similarly, the sign on the interval (x 0 , +∞) coincides with the sign of the value of the function y=a·x+b at any point in this interval. But you can do without these calculations, and draw conclusions about the signs based on the value of the coefficient a: if a>0, then on the intervals (−∞, x 0) and (x 0, +∞) there will be signs − and +, respectively, and if a >0, then + and −.
• If inequalities with signs > or ≥ are being solved, then a hatch is placed over the gap with a plus sign, and if inequalities with signs are being solved< или ≤, то – со знаком минус. В результате получается , которое и является искомым решением линейного неравенства.

Let's consider an example of solving a linear inequality using the interval method.

Example.

Solve the inequality −3·x+12>0.

Solution.

Since we are analyzing the interval method, we will use it. According to the algorithm, first we find the root of the equation −3·x+12=0, −3·x=−12, x=4. Next, we draw a coordinate line and mark a point on it with coordinate 4, and we make this point punctured, since we are solving a strict inequality:

Now we determine the signs on the intervals. To determine the sign on the interval (−∞, 4), you can calculate the value of the function y=−3·x+12, for example, at x=3. We have −3·3+12=3>0, which means there is a + sign on this interval. To determine the sign on another interval (4, +∞), you can calculate the value of the function y=−3 x+12, for example, at point x=5. We have −3·5+12=−3<0 , значит, на этом промежутке знак −. Эти же выводы можно было сделать на основании значения коэффициента при x : так как он равен −3 , то есть, он отрицательный, то на промежутке (−∞, 4) будет знак +, а на промежутке (4, +∞) знак −. Проставляем определенные знаки над соответствующими промежутками:

Since we are solving the inequality with the > sign, we draw shading over the gap with the + sign, the drawing takes the form

Based on the resulting image, we conclude that the desired solution is (−∞, 4) or in another notation x<4 .

Answer:

(−∞, 4) or x<4 .

Graphically

It is useful to have an understanding of the geometric interpretation of solving linear inequalities in one variable. To get it, let's consider four linear inequalities with the same left-hand side: 0.5 x−1<0 , 0,5·x−1≤0 , 0,5·x−1>0 and 0.5 x−1≥0 , their solutions are x<2 , x≤2 , x>2 and x≥2, and also draw a graph of the linear function y=0.5 x−1.

It's easy to notice that

• solution to the inequality 0.5 x−1<0 представляет собой промежуток, на котором график функции y=0,5·x−1 располагается ниже оси абсцисс (эта часть графика изображена синим цветом),
• the solution to the inequality 0.5 x−1≤0 represents the interval in which the graph of the function y=0.5 x−1 is below the Ox axis or coincides with it (in other words, not above the abscissa axis),
• similarly, the solution to the inequality 0.5 x−1>0 is the interval in which the graph of the function is above the Ox axis (this part of the graph is shown in red),
• and the solution to the inequality 0.5·x−1≥0 is the interval in which the graph of the function is higher or coincides with the abscissa axis.

Graphical method for solving inequalities, in particular linear, and implies finding intervals in which the graph of the function corresponding to the left side of the inequality is located above, below, not below or not above the graph of the function corresponding to the right side of the inequality. In our case of linear inequality, the function corresponding to the left side is y=a·x+b, and the right side is y=0, coinciding with the Ox axis.

Given the information given, it is easy to formulate algorithm for solving linear inequalities graphically:

• A graph of the function y=a x+b is constructed (schematically possible) and
• when solving the inequality a x+b<0 определяется промежуток, на котором график ниже оси Ox ,
• when solving the inequality a x+b≤0, the interval is determined in which the graph is lower or coincides with the Ox axis,
• when solving the inequality a x+b>0, the interval is determined in which the graph is above the Ox axis,
• when solving the inequality a·x+b≥0, the interval in which the graph is higher or coincides with the Ox axis is determined.

Example.

Solve the inequality graphically.

Solution.

Let's sketch a graph of a linear function . This is a straight line that is decreasing, since the coefficient of x is negative. We also need the coordinate of the point of its intersection with the x-axis, it is the root of the equation , which is equal to . For our needs, we don’t even need to depict the Oy axis. So our schematic drawing will look like this

Since we are solving an inequality with a > sign, we are interested in the interval in which the graph of the function is above the Ox axis. For clarity, let’s highlight this part of the graph in red, and in order to easily determine the interval corresponding to this part, let’s highlight in red the part of the coordinate plane in which the selected part of the graph is located, as in the figure below:

The gap we are interested in is the part of the Ox axis that is highlighted in red. Obviously this is an open number beam . This is the solution we are looking for. Note that if we were solving the inequality not with the sign >, but with the sign of the non-strict inequality ≥, then we would have to add in the answer, since at this point the graph of the function coincides with the Ox axis .y=0·x+7, which is the same as y=7, defines a straight line on the coordinate plane parallel to the Ox axis and lying above it. Therefore, the inequality 0 x+7<=0 не имеет решений, так как нет промежутков, на которых график функции y=0·x+7 ниже оси абсцисс.

And the graph of the function y=0·x+0, which is the same as y=0, is a straight line coinciding with the Ox axis. Therefore, the solution to the inequality 0·x+0≥0 is the set of all real numbers.

Answer:

second inequality, its solution is any real number.

Inequalities that reduce to linear

A huge number of inequalities can be replaced by equivalent linear inequalities using equivalent transformations, in other words, reduced to a linear inequality. Such inequalities are called inequalities that reduce to linear.

At school, almost simultaneously with solving linear inequalities, simple inequalities that reduce to linear ones are also considered. They are special cases entire inequalities, namely, in their left and right parts there are whole expressions that represent or linear binomials, or are converted to them by and . For clarity, we give several examples of such inequalities: 5−2·x>0, 7·(x−1)+3≤4·x−2+x, .

Inequalities that are similar in form to those indicated above can always be reduced to linear ones. This can be done by opening parentheses, bringing similar terms, rearranging terms, and moving terms from one side of the inequality to another with the opposite sign.

For example, to reduce the inequality 5−2 x>0 to linear, it is enough to rearrange the terms on its left side, we have −2 x+5>0. To reduce the second inequality 7·(x−1)+3≤4·x−2+x to linear, you need a little more steps: on the left side we open the brackets 7·x−7+3≤4·x−2+x , after To achieve this, we present similar terms in both sides 7 x−4≤5 x−2 , then we transfer the terms from the right side to the left side 7 x−4−5 x+2≤0 , finally, we present similar terms in the left side 2 ·x−2≤0 . Similarly, the third inequality can be reduced to a linear inequality.

Due to the fact that such inequalities can always be reduced to linear ones, some authors even call them linear as well. But we will still consider them reducible to linear.

Now it becomes clear why such inequalities are considered together with linear inequalities. And the principle of their solution is absolutely the same: by performing equivalent transformations, they can be reduced to elementary inequalities, which are the desired solutions.

To solve an inequality of this type, you can first reduce it to a linear one, and then solve this linear inequality. But it’s more rational and convenient to do this:

• after opening the brackets, collect all the terms with the variable on the left side of the inequality, and all the numbers on the right,
• then bring similar terms,
• and then divide both sides of the resulting inequality by the coefficient of x (if it is, of course, different from zero). This will give the answer.

Example.

Solve the inequality 5·(x+3)+x≤6·(x−3)+1.

Solution.

First, let's open the brackets, as a result we come to the inequality 5 x + 15 + x ≤ 6 x − 18 + 1 . Now let’s give similar terms: 6 x+15≤6 x−17 . Next we move the terms from left side, we get 6 x+15−6 x+17≤0, and again we bring similar terms (which leads us to the linear inequality 0 x+32≤0) and we have 32≤0. This is how we came to an incorrect numerical inequality, from which we conclude that the original inequality has no solutions.

Answer:

no solutions.

In conclusion, we note that there are a lot of other inequalities that can be reduced to linear inequalities, or to inequalities of the type considered above. For example, the solution exponential inequality 5 2 x−1 ≥1 reduces to solving the linear inequality 2 x−1≥0 . But we will talk about this when analyzing solutions to inequalities of the corresponding form.

References.

• Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
• Mordkovich A. G. Algebra. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
• Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., erased. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
• Mordkovich A. G. Algebra and the beginnings of mathematical analysis. 11th grade. In 2 hours. Part 1. Textbook for students of general education institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.

The main types of inequalities are presented, including Bernoulli, Cauchy - Bunyakovsky, Minkowski, Chebyshev inequalities. The properties of inequalities and actions on them are considered. The basic methods for solving inequalities are given.

Formulas for basic inequalities

Formulas for universal inequalities

Universal inequalities are satisfied for any values ​​of the quantities included in them. The main types of universal inequalities are listed below.

1) | a b | ≤ |a| + |b| ; | a 1 a 2 ... a n | ≤ |a 1 | + |a 2 | + ... + |a n |

2) |a| + |b| ≥ | a - b | ≥ | |a| - |b| |

3)
Equality occurs only when a 1 = a 2 = ... = a n.

4) Cauchy-Bunyakovsky inequality

Equality holds if and only if α a k = β b k for all k = 1, 2, ..., n and some α, β, |α| + |β| > 0 .

5) Minkowski's inequality, for p ≥ 1

Formulas of satisfiable inequalities

Satisfiable inequalities are satisfied for certain values ​​of the quantities included in them.

1) Bernoulli's inequality:
.
More generally:
,
where , numbers of the same sign and greater than -1 : .
Bernoulli's Lemma:
.
See "Proofs of inequalities and Bernoulli's lemma".

2)
for a i ≥ 0 (i = 1, 2, ..., n) .

3) Chebyshev's inequality
at 0 < a 1 ≤ a 2 ≤ ... ≤ a n And 0 < b 1 ≤ b 2 ≤ ... ≤ b n
.
At 0 < a 1 ≤ a 2 ≤ ... ≤ a n And b 1 ≥ b 2 ≥ ... ≥ b n > 0
.

4) Generalized Chebyshev inequalities
at 0 < a 1 ≤ a 2 ≤ ... ≤ a n And 0 < b 1 ≤ b 2 ≤ ... ≤ b n and k natural
.
At 0 < a 1 ≤ a 2 ≤ ... ≤ a n And b 1 ≥ b 2 ≥ ... ≥ b n > 0
.

Properties of inequalities

Properties of inequalities are a set of those rules that are satisfied when transforming them. Below are the properties of the inequalities. It is understood that the original inequalities are satisfied for values ​​of x i (i = 1, 2, 3, 4) belonging to some predetermined interval.

1) When the order of the sides changes, the inequality sign changes to the opposite.
If x 1< x 2 , то x 2 >x 1 .
If x 1 ≤ x 2, then x 2 ≥ x 1.
If x 1 ≥ x 2, then x 2 ≤ x 1.
If x 1 > x 2 then x 2< x 1 .

2) One equality is equivalent to two weak inequalities different sign.
If x 1 = x 2, then x 1 ≤ x 2 and x 1 ≥ x 2.
If x 1 ≤ x 2 and x 1 ≥ x 2, then x 1 = x 2.

3) Transitivity property
If x 1< x 2 и x 2 < x 3 , то x 1 < x 3 .
If x 1< x 2 и x 2 ≤ x 3 , то x 1 < x 3 .
If x 1 ≤ x 2 and x 2< x 3 , то x 1 < x 3 .
If x 1 ≤ x 2 and x 2 ≤ x 3, then x 1 ≤ x 3.

4) The same number can be added (subtracted) to both sides of the inequality.
If x 1< x 2 , то x 1 + A < x 2 + A .
If x 1 ≤ x 2, then x 1 + A ≤ x 2 + A.
If x 1 ≥ x 2, then x 1 + A ≥ x 2 + A.
If x 1 > x 2, then x 1 + A > x 2 + A.

5) If there are two or more inequalities with the sign of the same direction, then their left and right sides can be added.
If x 1< x 2 , x 3 < x 4 , то x 1 + x 3 < x 2 + x 4 .
If x 1< x 2 , x 3 ≤ x 4 , то x 1 + x 3 < x 2 + x 4 .
If x 1 ≤ x 2 , x 3< x 4 , то x 1 + x 3 < x 2 + x 4 .
If x 1 ≤ x 2, x 3 ≤ x 4, then x 1 + x 3 ≤ x 2 + x 4.
Similar expressions apply to the signs ≥, >.
If the original inequalities contain signs of non-strict inequalities and at least one strict inequality (but all signs have the same direction), then the addition results in a strict inequality.

6) Both sides of the inequality can be multiplied (divided) by a positive number.
If x 1< x 2 и A >0, then A x 1< A · x 2 .
If x 1 ≤ x 2 and A > 0, then A x 1 ≤ A x 2.
If x 1 ≥ x 2 and A > 0, then A x 1 ≥ A x 2.
If x 1 > x 2 and A > 0, then A · x 1 > A · x 2.

7) Both sides of the inequality can be multiplied (divided) by a negative number. In this case, the sign of inequality will change to the opposite.
If x 1< x 2 и A < 0 , то A · x 1 >A x 2.
If x 1 ≤ x 2 and A< 0 , то A · x 1 ≥ A · x 2 .
If x 1 ≥ x 2 and A< 0 , то A · x 1 ≤ A · x 2 .
If x 1 > x 2 and A< 0 , то A · x 1 < A · x 2 .

8) If there are two or more inequalities with positive terms, with the sign of the same direction, then their left and right sides can be multiplied by each other.
If x 1< x 2 , x 3 < x 4 , x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4 .
If x 1< x 2 , x 3 ≤ x 4 , x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4 .
If x 1 ≤ x 2 , x 3< x 4 , x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4 .
If x 1 ≤ x 2, x 3 ≤ x 4, x 1, x 2, x 3, x 4 > 0 then x 1 x 3 ≤ x 2 x 4.
Similar expressions apply to the signs ≥, >.
If the original inequalities contain signs of non-strict inequalities and at least one strict inequality (but all signs have the same direction), then multiplication results in a strict inequality.

9) Let f(x) be a monotonically increasing function. That is, for any x 1 > x 2, f(x 1) > f(x 2). Then this function can be applied to both sides of the inequality, which will not change the sign of the inequality.
If x 1< x 2 , то f(x 1) < f(x 2) .
If x 1 ≤ x 2 then f(x 1) ≤ f(x 2) .
If x 1 ≥ x 2 then f(x 1) ≥ f(x 2) .
If x 1 > x 2, then f(x 1) > f(x 2).

10) Let f(x) be a monotonically decreasing function, That is, for any x 1 > x 2, f(x 1)< f(x 2) . Тогда к обеим частям неравенства можно применить эту функцию, от чего знак неравенства изменится на противоположный.
If x 1< x 2 , то f(x 1) >f(x 2) .
If x 1 ≤ x 2 then f(x 1) ≥ f(x 2) .
If x 1 ≥ x 2 then f(x 1) ≤ f(x 2) .
If x 1 > x 2 then f(x 1)< f(x 2) .

Methods for solving inequalities

Solving inequalities using the interval method

The interval method is applicable if the inequality includes one variable, which we denote as x, and it has the form:
f(x) > 0
where f(x) - continuous function, having final number break points. The inequality sign can be anything: >, ≥,<, ≤ .

The interval method is as follows.

1) Find the domain of definition of the function f(x) and mark it with intervals on the number axis.

2) Find the discontinuity points of the function f(x). For example, if this is a fraction, then we find the points at which the denominator becomes zero. We mark these points on the number axis.

3) Solve the equation
f(x) = 0 .
We mark the roots of this equation on the number axis.

4) As a result, the number axis will be divided into intervals (segments) by points. Within each interval included in the domain of definition, we select any point and at this point we calculate the value of the function. If this value is greater than zero, then we place a “+” sign above the segment (interval). If this value is less than zero, then we put a “-” sign above the segment (interval).

5) If the inequality has the form: f(x) > 0, then select intervals with the “+” sign. The solution to the inequality is to combine these intervals, which do not include their boundaries.
If the inequality has the form: f(x) ≥ 0, then to the solution we add points at which f(x) = 0. That is, some intervals may have closed boundaries (the boundary belongs to the interval). the other part may have open boundaries (the boundary does not belong to the interval).
Similarly, if the inequality has the form: f(x)< 0 , то выбираем интервалы с знаком „-“ . Решением неравенства будет объединение этих интервалов, в которые не входят их границы.
If the inequality has the form: f(x) ≤ 0, then to the solution we add points at which f(x) = 0.

Solving inequalities using their properties

This method is applicable to inequalities of any complexity. It consists in applying the properties (presented above) to bring the inequalities to more simple view and get a solution. It is quite possible that this will result in not just one, but a system of inequalities. This is a universal method. It applies to any inequalities.

Used literature:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.

Inequalities and systems of inequalities are one of the topics covered in high school in algebra. In terms of difficulty level, it is not the most difficult, since it has simple rules (more on them a little later). As a rule, schoolchildren learn to solve systems of inequalities quite easily. This is also due to the fact that teachers simply “train” their students on this topic. And they cannot help but do this, because it is studied in the future using other mathematical quantities, and is also tested on the OGE and the Unified State Exam. IN school textbooks The topic of inequalities and systems of inequalities is covered in great detail, so if you are going to study it, it is best to resort to them. This article only summarizes larger material and there may be some omissions.

The concept of a system of inequalities

If you turn to scientific language, then we can define the concept of “system of inequalities”. This is a mathematical model that represents several inequalities. This model, of course, requires a solution, and this will be the general answer for all the inequalities of the system proposed in the task (usually it is written like this, for example: “Solve the system of inequalities 4 x + 1 > 2 and 30 - x > 6... "). However, before moving on to the types and methods of solutions, you need to understand something else.

Systems of inequalities and systems of equations

In the process of studying new topic very often misunderstandings arise. On the one hand, everything is clear and you want to start solving tasks as soon as possible, but on the other hand, some moments remain in the “shadow” and are not fully understood. Also, some elements of already acquired knowledge may be intertwined with new ones. As a result of this “overlapping”, errors often occur.

Therefore, before we begin to analyze our topic, we should remember the differences between equations and inequalities and their systems. To do this, we need to clarify once again what the data represents. mathematical concepts. An equation is always an equality, and it is always equal to something (in mathematics this word is denoted by the sign "="). Inequality is a model in which one quantity is either greater or less than another, or contains a statement that they are not the same. Thus, in the first case, it is appropriate to talk about equality, and in the second, no matter how obvious it may sound from the name itself, about the inequality of the initial data. Systems of equations and inequalities practically do not differ from each other and the methods for solving them are the same. The only difference is that in the first case equalities are used, and in the second inequalities are used.

Types of inequalities

There are two types of inequalities: numerical and with an unknown variable. The first type represents provided values ​​(numbers) that are unequal to each other, for example, 8 > 10. The second is inequalities containing an unknown variable (denoted by some letter Latin alphabet, most often X). This variable needs to be found. Depending on how many there are, the mathematical model distinguishes between inequalities with one (they make up a system of inequalities with one variable) or several variables (they make up a system of inequalities with several variables).

The last two types, according to the degree of their construction and the level of complexity of the solution, are divided into simple and complex. Simple ones are also called linear inequalities. They, in turn, are divided into strict and non-strict. Strict ones specifically “say” that one quantity must necessarily be either less or more, so this is pure inequality. Several examples can be given: 8 x + 9 > 2, 100 - 3 x > 5, etc. Non-strict ones also include equality. That is, one value can be greater than or equal to another value (the “≥” sign) or less than or equal to another value (the “≤” sign). Even in linear inequalities, the variable is not at the root, square, or divisible by anything, which is why they are called “simple.” Complex ones involve unknown variables that require execution to find. more mathematical operations. They are often located in a square, cube or under a root, they can be modular, logarithmic, fractional, etc. But since our task is the need to understand the solution of systems of inequalities, we will talk about a system of linear inequalities. However, before that, a few words should be said about their properties.

Properties of inequalities

The properties of inequalities include the following:

1. The inequality sign is reversed if an operation is used to change the order of the sides (for example, if t 1 ≤ t 2, then t 2 ≥ t 1).
2. Both sides of the inequality allow you to add the same number to itself (for example, if t 1 ≤ t 2, then t 1 + number ≤ t 2 + number).
3. Two or more inequalities with a sign in the same direction allow their left and right sides to be added (for example, if t 1 ≥ t 2, t 3 ≥ t 4, then t 1 + t 3 ≥ t 2 + t 4).
4. Both sides of the inequality can be multiplied or divided by the same positive number (for example, if t 1 ≤ t 2 and a number ≤ 0, then the number · t 1 ≥ number · t 2).
5. Two or more inequalities that have positive terms and a sign in the same direction allow themselves to be multiplied by each other (for example, if t 1 ≤ t 2, t 3 ≤ t 4, t 1, t 2, t 3, t 4 ≥ 0 then t 1 · t 3 ≤ t 2 · t 4).
6. Both parts of the inequality allow themselves to be multiplied or divided by the same negative number, but in this case the sign of the inequality changes (for example, if t 1 ≤ t 2 and a number ≤ 0, then the number · t 1 ≥ number · t 2).
7. All inequalities have the property of transitivity (for example, if t 1 ≤ t 2 and t 2 ≤ t 3, then t 1 ≤ t 3).

Now, after studying the basic principles of the theory related to inequalities, we can proceed directly to the consideration of the rules for solving their systems.

Solving systems of inequalities. General information. Solutions

As mentioned above, the solution is the values ​​of the variable that are suitable for all the inequalities of the given system. Solving systems of inequalities is the implementation of mathematical operations that ultimately lead to a solution to the entire system or prove that it has no solutions. In this case, the variable is said to refer to empty numerical set(written like this: letter denoting a variable∈ (sign “belongs”) ø (sign “empty set”), for example, x ∈ ø (read: “The variable “x” belongs to the empty set”). There are several ways to solve systems of inequalities: graphical, algebraic, substitution method. It is worth noting that they are among those mathematical models, which have several unknown variables. In the case where there is only one, the interval method is suitable.

Graphic method

Allows you to solve a system of inequalities with several unknown quantities (from two and above). Thanks to this method, a system of linear inequalities can be solved quite easily and quickly, so it is the most common method. This is explained by the fact that plotting a graph reduces the amount of writing mathematical operations. It becomes especially pleasant to take a little break from the pen, pick up a pencil with a ruler and start further actions with their help when a lot of work has been done and you want a little variety. However this method some people don’t like it because they have to break away from the task and switch their mental activity to drawing. However, this is a very effective method.

To solve a system of inequalities using graphic method, it is necessary to transfer all terms of each inequality to their left side. The signs will be reversed, zero should be written on the right, then each inequality needs to be written separately. As a result, functions will be obtained from inequalities. After this, you can take out a pencil and a ruler: now you need to draw a graph of each function obtained. The entire set of numbers that will be in the interval of their intersection will be a solution to the system of inequalities.

Algebraic way

Allows you to solve a system of inequalities with two unknown variables. Also, inequalities must have with the same sign inequalities (i.e. they must contain either only the “greater than” sign, or only the “less than” sign, etc.) Despite its limitations, this method is also more complex. It is applied in two stages.

The first involves actions to get rid of one of the unknown variables. First you need to select it, then check for the presence of numbers in front of this variable. If they are not there (then the variable will look like a single letter), then we do not change anything, if there are (the type of the variable will be, for example, 5y or 12y), then it is necessary to make sure that in each inequality the number in front of the selected variable is the same. To do this, you need to multiply each term of the inequalities by common multiplier, for example, if 3y is written in the first inequality, and 5y in the second, then it is necessary to multiply all terms of the first inequality by 5, and the second by 3. The result is 15y and 15y, respectively.

Second stage of solution. It is necessary to transfer the left side of each inequality to their right sides, changing the sign of each term to the opposite, and write zero on the right. Then comes the fun part: getting rid of the selected variable (otherwise known as “reduction”) while adding the inequalities. This results in an inequality with one variable that needs to be solved. After this, you should do the same thing, only with another unknown variable. The results obtained will be the solution of the system.

Substitution method

Allows you to solve a system of inequalities if it is possible to introduce a new variable. Typically, this method is used when the unknown variable in one term of the inequality is raised to the fourth power, and in the other term it is squared. Thus, this method is aimed at reducing the degree of inequalities in the system. The sample inequality x 4 - x 2 - 1 ≤ 0 is solved in this way. A new variable is introduced, for example t. They write: “Let t = x 2,” then the model is rewritten in a new form. In our case, we get t 2 - t - 1 ≤0. This inequality needs to be solved using the interval method (more on that a little later), then back to the variable X, then do the same with the other inequality. The answers received will be the solution of the system.

Interval method

This is the simplest way to solve systems of inequalities, and at the same time it is universal and widespread. It is used in secondary schools and even in higher schools. Its essence lies in the fact that the student looks for intervals of inequality on a number line, which is drawn in a notebook (this is not a graph, but just an ordinary line with numbers). Where the intervals of inequalities intersect, the solution to the system is found. To use the interval method, you need to follow these steps:

1. All terms of each inequality are transferred to the left side with the sign changing to the opposite (zero is written on the right).
2. The inequalities are written out separately, and the solution to each of them is determined.
3. The intersections of inequalities on the number line are found. All numbers located at these intersections will be a solution.

Which method should I use?

Obviously the one that seems easiest and most convenient, but there are cases when tasks require a certain method. Most often they say that you need to solve either using a graph or the interval method. Algebraic way and substitution are used extremely rarely or not at all, since they are quite complex and confusing, and besides, they are more used for solving systems of equations rather than inequalities, so you should resort to drawing graphs and intervals. They bring clarity, which cannot but contribute to the efficient and fast execution of mathematical operations.

If something doesn't work out

While studying a particular topic in algebra, naturally, problems may arise with its understanding. And this is normal, because our brain is designed in such a way that it is not able to understand complex material at one time. Often you need to reread a paragraph, take help from a teacher, or practice solving a problem. typical tasks. In our case, they look, for example, like this: “Solve the system of inequalities 3 x + 1 ≥ 0 and 2 x - 1 > 3.” Thus, personal desire, help from outsiders and practice help in understanding any complex topic.

Solver?

A solution book is also very suitable, but not for copying homework, but for self-help. In them you can find systems of inequalities with solutions, look at them (as if they were patterns), try to understand exactly how the author of the solution coped with the task, and then try to do the same on your own.

Conclusions

Algebra is one of the most difficult subjects in school. Well, what can you do? Mathematics has always been like this: for some it is easy, but for others it is difficult. But in any case, it should be remembered that general education program It is built in such a way that any student can handle it. Moreover, one must keep in mind huge amount assistants Some of them have been mentioned above.

What you need to know about inequality icons? Inequalities with icon more (> ), or less (< ) are called strict. With icons greater than or equal to (≥ ), less than or equal to (≤ ) are called not strict. Icon not equal (≠ ) stands apart, but you also have to solve examples with this icon all the time. And we will decide.)

The icon itself does not have much influence on the solution process. But at the end of the decision, when choosing the final answer, the meaning of the icon appears in full force! This is what we will see below in examples. There are some jokes there...

Inequalities, like equalities, exist faithful and unfaithful. Everything is simple here, no tricks. Let's say 5 > 2 is a true inequality. 5 < 2 - incorrect.

This preparation works for inequalities any kind and simple to the point of horror.) You just need to correctly perform two (only two!) elementary actions. These actions are familiar to everyone. But, characteristically, mistakes in these actions are the main mistake in solving inequalities, yes... Therefore, these actions must be repeated. These actions are called like this:

Identical transformations of inequalities.

Identical transformations of inequalities are very similar to identical transformations of equations. Actually, this is the main problem. The differences go over your head and... here you are.) Therefore, I will especially highlight these differences. So, the first identical transformation of inequalities:

1. The same number or expression can be added (subtracted) to both sides of the inequality. Any. This will not change the inequality sign.

In practice, this rule is used as a transfer of terms from the left side of the inequality to the right (and vice versa) with a change of sign. With a change in the sign of the term, not the inequality! The one-to-one rule is the same as the rule for equations. Here are the next ones identity transformations in inequalities differs significantly from those in equations. So I highlight them in red:

2. Both sides of the inequality can be multiplied (divided) by the same thingpositivenumber. For anypositive will not change.

3. Both sides of the inequality can be multiplied (divided) by the same thingnegative number. For anynegativenumber. The inequality sign from thiswill change to the opposite.

You remember (I hope...) that the equation can be multiplied/divided by anything. And for any number, and for an expression with an X. If only it wasn't zero. This makes him, the equation, neither hot nor cold.) It does not change. But inequalities are more sensitive to multiplication/division.

A good example for a long memory. Let us write an inequality that does not raise doubts:

5 > 2

Multiply both sides by +3, we get:

15 > 6

Any objections? There are no objections.) And if we multiply both sides of the original inequality by -3, we get:

15 > -6

And this is an outright lie.) A complete lie! Deception of the people! But as soon as you change the inequality sign to the opposite one, everything falls into place:

15 < -6

I’m not just swearing about lies and deception.) "Forgot to change the equal sign..."- This home error in solving inequalities. This is trivial and simple rule so many people were hurt! Which they forgot...) So I’m swearing. Maybe I'll remember...)

Particularly attentive people will notice that inequality cannot be multiplied by an expression with an X. Respect to those who are attentive!) Why not? The answer is simple. We don’t know the sign of this expression with an X. It can be positive, negative... Therefore, we do not know which inequality sign to put after multiplication. Should I change it or not? Unknown. Of course, this restriction (the prohibition of multiplying/dividing an inequality by an expression with an x) can be circumvented. If you really need it. But this is a topic for other lessons.

That's all the identical transformations of inequalities. Let me remind you once again that they work for any inequalities Now you can move on to specific types.

Linear inequalities. Solution, examples.

Linear inequalities are inequalities in which x is in the first power and there is no division by x. Type:

x+3 > 5x-5

How are such inequalities resolved? They are very easy to solve! Namely: with the help of we reduce the most confusing linear inequality straight to the answer. That's the solution. I will highlight the main points of the decision. To avoid stupid mistakes.)

Let's solve this inequality:

x+3 > 5x-5

We solve it in exactly the same way as a linear equation. With the only difference:

We are closely monitoring inequality sign!

The first step is the most common. With X's - to the left, without X's - to the right... This is the first identical transformation, simple and trouble-free.) Just don't forget to change the signs of the transferred terms.

The inequality sign remains:

x-5x > -5-3

Here are similar ones.

The inequality sign remains:

4x > -8

It remains to apply the last identical transformation: divide both sides by -4.

Divide by negative number.

The inequality sign will change to the opposite:

X < 2

This is the answer.

This is how all linear inequalities are solved.

Attention! Point 2 is drawn white, i.e. unpainted. Empty inside. This means that she is not included in the answer! I drew her so healthy on purpose. Such a point (empty, not healthy!)) in mathematics is called punctured point.

The remaining numbers on the axis can be marked, but not necessary. Extraneous numbers that are not related to our inequality can be confusing, yes... You just need to remember that the numbers increase in the direction of the arrow, i.e. numbers 3, 4, 5, etc. are to the right are twos, and numbers are 1, 0, -1, etc. - to the left.

Inequality x < 2 - strict. X is strictly less than two. If in doubt, checking is simple. We substitute the dubious number into the inequality and think: “Two is less than two? No, of course!” That's right. Inequality 2 < 2 incorrect. A two in return is not appropriate.

Is one okay? Certainly. Less... And zero is good, and -17, and 0.34... Yes, all numbers that are less than two are good! And even 1.9999.... At least a little bit, but less!

So let's mark all these numbers on the number axis. How? There are options here. Option one - shading. We move the mouse over the picture (or touch the picture on the tablet) and see that the area of ​​​​all x's that meet the condition x is shaded < 2 . That's it.

Let's look at the second option using the second example:

X ≥ -0,5

Draw an axis and mark the number -0.5. Like this:

Notice the difference?) Well, yes, it’s hard not to notice... This dot is black! Painted over. This means -0.5 is included in the answer. Here, by the way, the verification may confuse someone. Let's substitute:

-0,5 ≥ -0,5

How so? -0.5 is no more than -0.5! And there is more icon...

It's OK. In a non-strict inequality, everything that fits the icon is suitable. AND equals good, and more good. Therefore, -0.5 is included in the response.

So, we marked -0.5 on the axis; it remains to mark all the numbers that are greater than -0.5. This time I mark the area of ​​suitable x values bow(from the word arc), rather than shading. We hover the cursor over the drawing and see this bow.

There is no particular difference between the shading and the arms. Do as the teacher says. If there is no teacher, draw arches. In more difficult tasks shading is less obvious. You can get confused.

This is how linear inequalities are drawn on an axis. Let us move on to the next feature of the inequalities.

Writing the answer for inequalities.

The equations were good.) We found x and wrote down the answer, for example: x=3. There are two forms of writing answers in inequalities. One is in the form of final inequality. good for simple cases. For example:

X< 2.

This is a complete answer.

Sometimes you need to write down the same thing, but in a different form, using numerical intervals. Then the recording starts to look very scientific):

x ∈ (-∞; 2)

Under the icon ∈ the word is hidden "belongs"

The entry reads like this: x belongs to the interval from minus infinity to two not including. Quite logical. X can be any number out of all of them possible numbers from minus infinity to two. There cannot be a double X, which is what the word tells us "not including".

And where in the answer is it clear that "not including"? This fact is noted in the answer round parenthesis immediately after the two. If the two were included, the bracket would be square. Here it is:]. IN following example such a bracket is used.

Let's write down the answer: x ≥ -0,5 at intervals:

x ∈ [-0.5; +∞)

Reads: x belongs to the interval from minus 0.5, including, to plus infinity.

Infinity can never be turned on. It's not a number, it's a symbol. Therefore, in such records, infinity is always adjacent to parenthesis.

This form of recording is convenient for complex answers consisting of several spaces. But - just for final answers. IN intermediate results, where a further solution is expected, it is better to use the usual form, in the form simple inequality. We will deal with this in the relevant topics.

Popular tasks with inequalities.

The linear inequalities themselves are simple. Therefore, tasks often become more difficult. So it was necessary to think. This, if you’re not used to it, is not very pleasant.) But it’s useful. I will show examples of such tasks. Not for you to learn them, it's unnecessary. And in order not to be afraid when meeting with similar examples. Just think a little - and it’s simple!)

1. Find any two solutions to the inequality 3x - 3< 0

If it’s not very clear what to do, remember the main rule of mathematics:

If you don’t know what you need, do what you can!)

X < 1

And what? Nothing special. What are they asking us? We are asked to find two specific numbers that are the solution to an inequality. Those. fit the answer. Two any numbers. Actually, this is confusing.) A couple of 0 and 0.5 are suitable. A couple -3 and -8. Yes these couples infinite set! Which answer is correct?!

I answer: everything! Any pair of numbers, each of which is less than one, will be the correct answer. Write which one you want. Let's move on.

2. Solve the inequality:

4x - 3 ≠ 0

Tasks in this form are rare. But, as auxiliary inequalities, when finding ODZ, for example, or when finding the domain of definition of a function, they occur all the time. Such a linear inequality can be solved as an ordinary linear equation. Only everywhere except the "=" sign ( equals) put a sign " ≠ " (not equal). This is how you approach the answer, with an inequality sign:

X ≠ 0,75

In more complex examples, it's better to do things differently. Make inequality out of equality. Like this:

4x - 3 = 0

Calmly solve it as taught and get the answer:

x = 0.75

The main thing is, at the very end, when writing down the final answer, do not forget that we found x, which gives equality. And we need - inequality. Therefore, we don’t really need this X.) And we need to write it down with the correct symbol:

X ≠ 0,75

This approach results in fewer errors. Those who solve equations automatically. And for those who don’t solve equations, inequalities are, in fact, of no use...) Another example of a popular task:

3. Find the smallest integer solution to the inequality:

3(x - 1) < 5x + 9

First we simply solve the inequality. We open the brackets, move them, bring similar ones... We get:

X > - 6

Didn't it work out that way!? Did you follow the signs!? And behind the signs of members, and behind the sign of inequality...

Let's think again. We need to find a specific number that matches both the answer and the condition "smallest integer". If it doesn’t dawn on you right away, you can just take any number and figure it out. Two over minus six? Certainly! Is there a suitable smaller number? Of course. For example, zero is greater than -6. And even less? We need the smallest thing possible! Minus three is more than minus six! You can already catch the pattern and stop stupidly going through numbers, right?)

Let's take a number closer to -6. For example, -5. The answer is fulfilled, -5 > - 6. Is it possible to find another number less than -5 but greater than -6? You can, for example, -5.5... Stop! We are told whole solution! Doesn't roll -5.5! What about minus six? Uh-uh! The inequality is strict, minus 6 is in no way less than minus 6!

Therefore, the correct answer is -5.

Hopefully with a selection of values ​​from general solution everything is clear. Another example:

4. Solve inequality:

7 < 3x+1 < 13

Wow! This expression is called triple inequality. Strictly speaking, this is an abbreviated form of a system of inequalities. But to decide such triple inequalities It still happens in some tasks... It can be solved without any systems. According to the same identical transformations.

We need to simplify, bring this inequality to pure X. But... What should be transferred where?! This is where it’s time to remember that moving left and right is short form first identity transformation.

A full form sounds like this: Any number or expression can be added/subtracted to both sides of the equation (inequality).

There are three parts here. So we will apply identical transformations to all three parts!

So, let's get rid of the one in the middle part of the inequality. Let's subtract one from the entire middle part. So that the inequality does not change, we subtract one from the remaining two parts. Like this:

7 -1< 3x+1-1 < 13-1

6 < 3x < 12

That’s better, right?) All that remains is to divide all three parts into three:

2 < X < 4

That's it. This is the answer. X can be any number from two (not including) to four (not including). This answer is also written at intervals; such entries will be in quadratic inequalities. There they are the most common thing.

At the end of the lesson I will repeat the most important thing. Success in solving linear inequalities depends on the ability to transform and simplify linear equations. If at the same time watch for the inequality sign, there won't be any problems. That's what I wish for you. No problems.)

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