How to find entire solutions to a system of inequalities. Solving integer and fractional rational inequalities. Respecting your privacy at the company level
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:
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\(2x+1\geq4(5-x)\) |
Variable only to the first power |
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\(3x^2-x+5>0\) |
There is a variable in the second power (square), but there are no higher powers (third, fourth, etc.) |
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\(\log_(4)((x+1))<3\) |
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\(2^(x)\leq8^(5x-2)\) |
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 to 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 the number axis with shading. 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 of the 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 a 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.
Example: Solve the inequality \(2(x+1)-1<7+8x\)Solution:
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\(2x+2-1<7+8x\) |
Let's move \(8x\) to the left, and \(2\) and \(-1\) to the right, not forgetting to change the signs |
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\(2x-8x<7-2+1\) |
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\(-6x<6\) \(|:(-6)\) |
Let's divide both sides of the inequality by \(-6\), not forgetting to change from “less” to “more” |
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Let's mark a numerical interval on the axis. Inequality, therefore we “prick out” the value \(-1\) itself and do not take it as an answer |
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Let's write the answer as an interval |
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: \(\left[-1;8\right)\)
After obtaining initial information about inequalities with variables, we move on to the question of solving them. We will analyze the solution of linear inequalities with one variable and all the methods for solving them with algorithms and examples. Only linear equations with one variable will be considered.
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What is linear inequality?
First, you need to define a linear equation and find out its standard form and how it will differ from others. From the school course we have that there is no fundamental difference between inequalities, so it is necessary to use several definitions.
Definition 1
Linear inequality with one variable x is an inequality of the form a · x + b > 0, when any inequality sign is used instead of >< , ≤ , ≥ , а и b являются действительными числами, где a ≠ 0 .
Definition 2
Inequalities a x< c или a · x >c, with x being a variable and a and c being some numbers, is called linear inequalities with one variable.
Since nothing is said about whether the coefficient can be equal to 0, then a strict inequality of the form 0 x > c and 0 x< c может быть записано в виде нестрогого, а именно, a · x ≤ c , a · x ≥ c . Такое уравнение считается линейным.
Their differences are:
- notation form a · x + b > 0 in the first, and a · x > c – in the second;
- admissibility of coefficient a being equal to zero, a ≠ 0 - in the first, and a = 0 - in the second.
It is believed that the inequalities a · x + b > 0 and a · x > c are equivalent, because they are obtained by transferring a term from one part to another. Solving the inequality 0 x + 5 > 0 will lead to the fact that it will need to be solved, and the case a = 0 will not work.
Definition 3
It is believed that linear inequalities in one variable x are inequalities of the form a x + b< 0 , a · x + b >0, a x + b ≤ 0 And a x + b ≥ 0, where a and b are real numbers. Instead of x there can be a regular number.
Based on the rule, we have that 4 x − 1 > 0, 0 z + 2, 3 ≤ 0, - 2 3 x - 2< 0 являются примерами линейных неравенств. А неравенства такого плана, как 5 · x >7 , − 0 , 5 · y ≤ − 1 , 2 are called reducible to linear.
How to solve linear inequality
The main way to solve such inequalities is to use equivalent transformations in order to find the elementary inequalities x< p (≤ , >, ≥) , p which is a certain number, for a ≠ 0, and of the form a< p (≤ , >, ≥) for a = 0.
To solve inequalities in one variable, you can use the interval method or represent it graphically. Any of them can be used separately.
Using equivalent transformations
To solve a linear inequality of the form a x + b< 0 (≤ , >, ≥), it is necessary to apply equivalent inequality transformations. The coefficient may or may not be zero. Let's consider both cases. To find out, you need to adhere to a scheme consisting of 3 points: the essence of the process, the algorithm, and the solution itself.
Definition 4
Algorithm for solving linear inequality a x + b< 0 (≤ , >, ≥) for a ≠ 0
- the number b will be moved to the right side of the inequality with the opposite sign, which will allow us to arrive at the equivalent a x< − b (≤ , > , ≥) ;
- Both sides of the inequality will be divided by a number not equal to 0. Moreover, when a is positive, the sign remains; when a is negative, it changes to the opposite.
Let's consider the application of this algorithm to solve examples.
Example 1
Solve the inequality of the form 3 x + 12 ≤ 0.
Solution
This linear inequality has a = 3 and b = 12. This means that the coefficient a of x is not equal to zero. Let's apply the above algorithms and solve it.
It is necessary to move term 12 to another part of the inequality and change the sign in front of it. Then we get an inequality of the form 3 x ≤ − 12. It is necessary to divide both parts by 3. The sign will not change since 3 is a positive number. We get that (3 x) : 3 ≤ (− 12) : 3, which gives the result x ≤ − 4.
An inequality of the form x ≤ − 4 is equivalent. That is, the solution for 3 x + 12 ≤ 0 is any real number that is less than or equal to 4. The answer is written as an inequality x ≤ − 4, or a numerical interval of the form (− ∞, − 4].
The entire algorithm described above is written like this:
3 x + 12 ≤ 0 ; 3 x ≤ − 12 ; x ≤ − 4 .
Answer: x ≤ − 4 or (− ∞ , − 4 ] .
Example 2
Indicate all available solutions to the inequality − 2, 7 · z > 0.
Solution
From the condition we see that the coefficient a for z is equal to - 2.7, and b is explicitly absent or equal to zero. You can not use the first step of the algorithm, but immediately move on to the second.
We divide both sides of the equation by the number - 2, 7. Since the number is negative, it is necessary to reverse the inequality sign. That is, we get that (− 2, 7 z) : (− 2, 7)< 0: (− 2 , 7) , и дальше z < 0 .
Let us write the entire algorithm in brief form:
− 2, 7 z > 0; z< 0 .
Answer: z< 0 или (− ∞ , 0) .
Example 3
Solve the inequality - 5 x - 15 22 ≤ 0.
Solution
According to the condition, we see that it is necessary to solve the inequality with coefficient a for the variable x, which is equal to - 5, with coefficient b, which corresponds to the fraction - 15 22. It is necessary to solve the inequality by following the algorithm, that is: move - 15 22 to another part with the opposite sign, divide both parts by - 5, change the sign of the inequality:
5 x ≤ 15 22 ; - 5 x: - 5 ≥ 15 22: - 5 x ≥ - 3 22
During the last transition for the right side, the rule for dividing the number with different signs is used 15 22: - 5 = - 15 22: 5, after which we divide the ordinary fraction by the natural number - 15 22: 5 = - 15 22 · 1 5 = - 15 · 1 22 · 5 = - 3 22 .
Answer: x ≥ - 3 22 and [ - 3 22 + ∞) .
Let's consider the case when a = 0. Linear expression of the form a x + b< 0 является неравенством 0 · x + b < 0 , где на рассмотрение берется неравенство вида b < 0 , после чего выясняется, оно верное или нет.
Everything is based on determining the solution to the inequality. For any value of x we obtain a numerical inequality of the form b< 0 , потому что при подстановке любого t вместо переменной x , тогда получаем 0 · t + b < 0 , где b < 0 . В случае, если оно верно, то для его решения подходит любое значение. Когда b < 0 неверно, тогда линейное уравнение не имеет решений, потому как не имеется ни одного значения переменной, которое привело бы верному числовому равенству.
We will consider all judgments in the form of an algorithm for solving linear inequalities 0 x + b< 0 (≤ , > , ≥) :
Definition 5
Numerical inequality of the form b< 0 (≤ , >, ≥) is true, then the original inequality has a solution for any value, and it is false when the original inequality has no solutions.
Example 4
Solve the inequality 0 x + 7 > 0.
Solution
This linear inequality 0 x + 7 > 0 can take any value x. Then we get an inequality of the form 7 > 0. The last inequality is considered true, which means any number can be its solution.
Answer: interval (− ∞ , + ∞) .
Example 5
Find a solution to the inequality 0 x − 12, 7 ≥ 0.
Solution
When substituting the variable x of any number, we obtain that the inequality takes the form − 12, 7 ≥ 0. It is incorrect. That is, 0 x − 12, 7 ≥ 0 has no solutions.
Answer: there are no solutions.
Let's consider solving linear inequalities where both coefficients are equal to zero.
Example 6
Determine the unsolvable inequality from 0 x + 0 > 0 and 0 x + 0 ≥ 0.
Solution
When substituting any number instead of x, we obtain two inequalities of the form 0 > 0 and 0 ≥ 0. The first is incorrect. This means that 0 x + 0 > 0 has no solutions, and 0 x + 0 ≥ 0 has an infinite number of solutions, that is, any number.
Answer: the inequality 0 x + 0 > 0 has no solutions, but 0 x + 0 ≥ 0 has solutions.
This method is discussed in the school mathematics course. The interval method is capable of resolving various types of inequalities, including linear ones.
The interval method is used for linear inequalities when the value of the coefficient x is not equal to 0. Otherwise you will have to calculate using a different method.
Definition 6
The interval method is:
- introducing the function y = a · x + b ;
- searching for zeros to split the domain of definition into intervals;
- definition of signs for their concepts on intervals.
Let's assemble an algorithm for solving linear equations a x + b< 0 (≤ , >, ≥) for a ≠ 0 using the interval method:
- finding the zeros of the function y = a · x + b to solve an equation of the form a · x + b = 0 . If a ≠ 0, then the solution will be a single root, which will take the designation x 0;
- construction of a coordinate line with an image of a point with coordinate x 0, with a strict inequality the point is denoted by a punctured one, with a non-strict inequality – by a shaded one;
- determining the signs of the function y = a · x + b on intervals; for this it is necessary to find the values of the function at points on the interval;
- solving an inequality with signs > or ≥ on the coordinate line, adding shading over the positive interval,< или ≤ над отрицательным промежутком.
Let's look at several examples of solving linear inequalities using the interval method.
Example 6
Solve the inequality − 3 x + 12 > 0.
Solution
It follows from the algorithm that first you need to find the root of the equation − 3 x + 12 = 0. We get that − 3 · x = − 12 , x = 4 . It is necessary to draw a coordinate line where we mark point 4. It will be punctured because the inequality is strict. Consider the drawing below.
It is necessary to determine the signs at the intervals. To determine it on the interval (− ∞, 4), it is necessary to calculate the function y = − 3 x + 12 at x = 3. From here we get that − 3 3 + 12 = 3 > 0. The sign on the interval is positive.
We determine the sign from the interval (4, + ∞), then substitute the value x = 5. We have that − 3 5 + 12 = − 3< 0 . Знак на промежутке является отрицательным. Изобразим на числовой прямой, приведенной ниже.
We solve the inequality with the > sign, and the shading is performed over the positive interval. Consider the drawing below.
From the drawing it is clear that the desired solution has the form (− ∞ , 4) or x< 4 .
Answer: (− ∞ , 4) or x< 4 .
To understand how to depict graphically, it is necessary to consider 4 linear inequalities as an example: 0, 5 x − 1< 0 , 0 , 5 · x − 1 ≤ 0 , 0 , 5 · x − 1 >0 and 0, 5 x − 1 ≥ 0. Their solutions will be the values of x< 2 , x ≤ 2 , x >2 and x ≥ 2. To do this, let's plot the linear function y = 0, 5 x − 1 shown below.
It's clear that
Definition 7
- solving the inequality 0, 5 x − 1< 0 считается промежуток, где график функции y = 0 , 5 · x − 1 располагается ниже О х;
- the solution 0, 5 x − 1 ≤ 0 is considered to be the interval where the function y = 0, 5 x − 1 is lower than O x or coincides;
- the solution 0, 5 · x − 1 > 0 is considered to be an interval, the function is located above O x;
- the solution 0, 5 · x − 1 ≥ 0 is considered to be the interval where the graph above O x or coincides.
The point of graphically solving inequalities is to find the intervals that need to be depicted on the graph. In this case, we find that the left side has y = a · x + b, and the right side has y = 0, and coincides with O x.
Definition 8The graph of the function y = a x + b is plotted:
- while solving the inequality a x + b< 0 определяется промежуток, где график изображен ниже О х;
- when solving the inequality a · x + b ≤ 0, the interval is determined where the graph is depicted below the O x axis or coincides;
- when solving the inequality a · x + b > 0, the interval is determined where the graph is depicted above O x;
- When solving the inequality a · x + b ≥ 0, the interval is determined where the graph is above O x or coincides.
Example 7
Solve the inequality - 5 · x - 3 > 0 using a graph.
Solution
It is necessary to construct a graph of the linear function - 5 · x - 3 > 0. This line is decreasing because the coefficient of x is negative. To determine the coordinates of the point of its intersection with O x - 5 · x - 3 > 0, we obtain the value - 3 5. Let's depict it graphically.
Solving the inequality with the > sign, then you need to pay attention to the interval above O x. Let us highlight the required part of the plane in red and get that
The required gap is part O x red. This means that the open number ray - ∞ , - 3 5 will be a solution to the inequality. If, according to the condition, we had a non-strict inequality, then the value of the point - 3 5 would also be a solution to the inequality. And it would coincide with O x.
Answer: - ∞ , - 3 5 or x< - 3 5 .
The graphical solution is used when the left side corresponds to the function y = 0 x + b, that is, y = b. Then the straight line will be parallel to O x or coinciding at b = 0. These cases show that the inequality may have no solutions, or the solution may be any number.
Example 8
Determine from the inequalities 0 x + 7< = 0 , 0 · x + 0 ≥ 0 то, которое имеет хотя бы одно решение.
Solution
The representation of y = 0 x + 7 is y = 7, then a coordinate plane will be given with a line parallel to O x and located above O x. So 0 x + 7< = 0 решений не имеет, потому как нет промежутков.
The graph of the function y = 0 x + 0 is considered to be y = 0, that is, the straight line coincides with O x. This means that the inequality 0 x + 0 ≥ 0 has many solutions.
Answer: The second inequality has a solution for any value of x.
Inequalities that reduce to linear
The solution of inequalities can be reduced to the solution of a linear equation, which are called inequalities that reduce to linear.
These inequalities were considered in the school course, since they were a special case of solving inequalities, which led to the opening of parentheses and the reduction of similar terms. For example, consider that 5 − 2 x > 0, 7 (x − 1) + 3 ≤ 4 x − 2 + x, x - 3 5 - 2 x + 1 > 2 7 x.
The inequalities given above are always reduced to the form of a linear equation. After that, the brackets are opened and similar terms are given, transferred from different parts, changing the sign to the opposite.
When reducing the inequality 5 − 2 x > 0 to linear, we represent it in such a way that it has the form − 2 x + 5 > 0, and to reduce the second we obtain that 7 (x − 1) + 3 ≤ 4 x − 2 + x . It is necessary to open the brackets, bring similar terms, move all terms to the left side and bring similar terms. It looks like this:
7 x − 7 + 3 ≤ 4 x − 2 + x 7 x − 4 ≤ 5 x − 2 7 x − 4 − 5 x + 2 ≤ 0 2 x − 2 ≤ 0
This leads the solution to a linear inequality.
These inequalities are considered linear, since they have the same solution principle, after which it is possible to reduce them to elementary inequalities.
To solve this type of inequality, it is necessary to reduce it to a linear one. It should be done this way:
Definition 9
- open parentheses;
- collect variables on the left and numbers on the right;
- give similar terms;
- divide both sides by the coefficient of x.
Example 9
Solve the inequality 5 · (x + 3) + x ≤ 6 · (x − 3) + 1.
Solution
We open the brackets, then we get an inequality of the form 5 x + 15 + x ≤ 6 x − 18 + 1. After reducing similar terms, we have that 6 x + 15 ≤ 6 x − 17. After moving the terms from the left to the right, we find that 6 x + 15 − 6 x + 17 ≤ 0. Hence there is an inequality of the form 32 ≤ 0 from that obtained by calculating 0 x + 32 ≤ 0. It can be seen that the inequality is false, which means that the inequality given by condition has no solutions.
Answer: no solutions.
It is worth noting that there are many other types of inequalities that can be reduced to linear or inequalities of the type shown above. For example, 5 2 x − 1 ≥ 1 is an exponential equation that reduces to a solution of the linear form 2 x − 1 ≥ 0. These cases will be considered when solving inequalities of this type.
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We continue to delve into the topic of “solving inequalities with one variable.” We are already familiar with linear inequalities and quadratic inequalities. They are special cases rational inequalities, which we will now study. Let's start by finding out what type of inequalities are called rational. Next we will look at their division into whole rational and fractional rational inequalities. And after this we will study how to solve rational inequalities with one variable, write down the corresponding algorithms and consider solutions to typical examples with detailed explanations.
Page navigation.
What are rational inequalities?
In algebra classes at school, as soon as the conversation starts about solving inequalities, we immediately encounter rational inequalities. However, at first they are not called by their name, since at this stage the types of inequalities are of little interest, and the main goal is to gain initial skills in working with inequalities. The term “rational inequality” itself is introduced later in the 9th grade, when a detailed study of inequalities of this particular type begins.
Let's find out what rational inequalities are. Here's the definition:
The stated definition does not say anything about the number of variables, which means that any number of them is allowed. Depending on this, rational inequalities with one, two, etc. are distinguished. variables. By the way, the textbook gives a similar definition, but for rational inequalities with one variable. This is understandable, since the school focuses on solving inequalities with one variable (below we will also talk only about solving rational inequalities with one variable). Inequalities with two variables are considered little, and inequalities with three or more variables are practically not given any attention.
So, a rational inequality can be recognized by its notation; to do this, just look at the expressions on its left and right sides and make sure that they are rational expressions. These considerations allow us to give examples of rational inequalities. For example, x>4 , x 3 +2 y≤5 (y−1) (x 2 +1), are rational inequalities. And inequality 
is not rational, since its left side contains a variable under the root sign, and, therefore, is not a rational expression. Inequality is also not rational, since both its parts are not rational expressions.
For the convenience of further description, we introduce the division of rational inequalities into integer and fractional ones.
Definition.
We will call the rational inequality whole, if both its parts are whole rational expressions.
Definition.
Fractional rational inequality is a rational inequality, at least one part of which is a fractional expression.
So 0.5 x≤3 (2−5 y) , 
are integer inequalities, and 1:x+3>0 and 
- fractionally rational.
Now we have a clear understanding of what rational inequalities are, and we can safely begin to understand the principles of solving integer and fractional rational inequalities with one variable.
Solving entire inequalities
Let’s set ourselves a task: let’s say we need to solve a whole rational inequality with one variable x of the form r(x)
Let us move the expression from the right side to the left, which will lead us to an equivalent inequality of the form r(x)−s(x)<0 (≤, >, ≥) with a zero on the right. Obviously, the expression r(x)−s(x) formed on the left side is also an integer, and it is known that any . Having transformed the expression r(x)−s(x) into the identically equal polynomial h(x) (here we note that the expressions r(x)−s(x) and h(x) have the same variable x ), we move on to the equivalent inequality h(x)<0 (≤, >, ≥).
In the simplest cases, the transformations performed will be enough to obtain the desired solution, since they will lead us from the original whole rational inequality to an inequality that we know how to solve, for example, to a linear or quadratic one. Let's look at examples.
Example.
Find the solution to the whole rational inequality x·(x+3)+2·x≤(x+1) 2 +1.
Solution.
First we move the expression from the right side to the left: x·(x+3)+2·x−(x+1) 2 −1≤0. Having completed everything on the left side, we arrive at the linear inequality 3 x−2≤0, which is equivalent to the original integer inequality. The solution is not difficult:
3 x≤2 ,
x≤2/3.
Answer:
x≤2/3.
Example.
Solve the inequality (x 2 +1) 2 −3 x 2 >(x 2 −x) (x 2 +x).
Solution.
We start as usual by transferring the expression from the right side, and then perform transformations on the left side using:
(x 2 +1) 2 −3 x 2 −(x 2 −x) (x 2 +x)>0,
x 4 +2 x 2 +1−3 x 2 −x 4 +x 2 >0,
1>0
.
Thus, by performing equivalent transformations, we arrived at the inequality 1>0, which is true for any value of the variable x. This means that the solution to the original integer inequality is any real number.
Answer:
x - any.
Example.
Solve the inequality x+6+2 x 3 −2 x (x 2 +x−5)>0.
Solution.
There is a zero on the right side, so there is no need to move anything from it. Let's transform the whole expression on the left side into a polynomial:
x+6+2 x 3 −2 x 3 −2 x 2 +10 x>0,
−2 x 2 +11 x+6>0 .
We obtained a quadratic inequality, which is equivalent to the original inequality. We solve it using any method known to us. Let's solve the quadratic inequality graphically.
Find the roots of the quadratic trinomial −2 x 2 +11 x+6: 
We make a schematic drawing on which we mark the found zeros, and take into account that the branches of the parabola are directed downward, since the leading coefficient is negative: 
Since we are solving an inequality with a > sign, we are interested in the intervals in which the parabola is located above the x-axis. This occurs on the interval (−0.5, 6), which is the desired solution.
Answer:
(−0,5, 6) .
In more complex cases, on the left side of the resulting inequality h(x)<0 (≤, >, ≥) will be a polynomial of the third or higher degree. To solve such inequalities, the interval method is suitable, in the first step of which you will need to find all the roots of the polynomial h(x), which is often done through .
Example.
Find the solution to the whole rational inequality (x 2 +2)·(x+4)<14−9·x .
Solution.
Let's move everything to the left side, after which there is:
(x 2 +2)·(x+4)−14+9·x<0
,
x 3 +4 x 2 +2 x+8−14+9 x<0
,
x 3 +4 x 2 +11 x−6<0
.
The manipulations performed lead us to an inequality that is equivalent to the original one. On its left side is a polynomial of the third degree. It can be solved using the interval method. To do this, first of all, you need to find the roots of the polynomial that rests on x 3 +4 x 2 +11 x−6=0. Let's find out whether it has rational roots, which can only be among the divisors of the free term, that is, among the numbers ±1, ±2, ±3, ±6. Substituting these numbers in turn instead of the variable x into the equation x 3 +4 x 2 +11 x−6=0, we find out that the roots of the equation are the numbers 1, 2 and 3. This allows us to represent the polynomial x 3 +4 x 2 +11 x−6 as a product (x−1) (x−2) (x−3) , and the inequality x 3 +4 x 2 +11 x−6<0 переписать как (x−1)·(x−2)·(x−3)<0 . Такой вид неравенства в дальнейшем позволит с меньшими усилиями определить знаки на промежутках.
And then all that remains is to carry out the standard steps of the interval method: mark on the number line the points with coordinates 1, 2 and 3, which divide this line into four intervals, determine and place the signs, draw shading over the intervals with a minus sign (since we are solving an inequality with a minus sign<) и записать ответ.
Whence we have (−∞, 1)∪(2, 3) .
Answer:
(−∞, 1)∪(2, 3) .
It should be noted that sometimes it is inappropriate from the inequality r(x)−s(x)<0 (≤, >, ≥) go to the inequality h(x)<0 (≤, >, ≥), where h(x) is a polynomial of degree higher than two. This applies to cases where it is more difficult to factor the polynomial h(x) than to represent the expression r(x)−s(x) as a product of linear binomials and quadratic trinomials, for example, by factoring out the common factor. Let's explain this with an example.
Example.
Solve the inequality (x 2 −2·x−1)·(x 2 −19)≥2·x·(x 2 −2·x−1).
Solution.
This is a whole inequality. If we move the expression from its right side to the left, then open the brackets and add similar terms, we get the inequality x 4 −4 x 3 −16 x 2 +40 x+19≥0. Solving it is very difficult, since it involves finding the roots of a fourth-degree polynomial. It is easy to verify that it does not have rational roots (they could be the numbers 1, −1, 19 or −19), but it is problematic to look for its other roots. Therefore this path is a dead end.
Let's look for other possible solutions. It is easy to see that after transferring the expression from the right side of the original integer inequality to the left, we can take out the common factor x 2 −2 x−1 out of brackets:
(x 2 −2·x−1)·(x 2 −19)−2·x·(x 2 −2·x−1)≥0,
(x 2 −2·x−1)·(x 2 −2·x−19)≥0.
The transformation performed is equivalent, therefore the solution to the resulting inequality will also be a solution to the original inequality.
And now we can find the zeros of the expression located on the left side of the resulting inequality, for this we need x 2 −2·x−1=0 and x 2 −2·x−19=0. Their roots are numbers 
. This allows us to go to the equivalent inequality, and we can solve it using the interval method: 
We write down the answer according to the drawing.
Answer:
To conclude this point, I would just like to add that it is not always possible to find all the roots of the polynomial h(x) and, as a consequence, expand it into a product of linear binomials and square trinomials. In these cases there is no way to solve the inequality h(x)<0 (≤, >, ≥), which means there is no way to find a solution to the original integer rational equation.
Solving fractional rational inequalities
Now let’s solve the following problem: let’s say we need to solve a fractional rational inequality with one variable x of the form r(x)
Algorithm for solving fractional rational inequalities with one variable r(x)
- First you need to find the range of acceptable values (APV) of the variable x for the original inequality.
- Next, you need to move the expression from the right side of the inequality to the left, and convert the expression r(x)−s(x) formed there to the form of a fraction p(x)/q(x) , where p(x) and q(x) are integers expressions that are products of linear binomials, indecomposable quadratic trinomials and their powers with a natural exponent.
- Next, we need to solve the resulting inequality using the interval method.
- Finally, from the solution obtained in the previous step, it is necessary to exclude points that are not included in the ODZ of the variable x for the original inequality, which was found in the first step.
This way the desired solution to the fractional rational inequality will be obtained.
The second step of the algorithm requires explanation. Transferring the expression from the right side of the inequality to the left gives the inequality r(x)−s(x)<0 (≤, >, ≥), which is equivalent to the original one. Everything is clear here. But questions are raised by its further transformation to the form p(x)/q(x)<0 (≤, >, ≥).
The first question is: “Is it always possible to carry it out”? Theoretically, yes. We know that anything is possible. The numerator and denominator of a rational fraction contain polynomials. And from the fundamental theorem of algebra and Bezout’s theorem it follows that any polynomial of degree n with one variable can be represented as a product of linear binomials. This explains the possibility of carrying out this transformation.
In practice, it is quite difficult to factor polynomials, and if their degree is higher than four, then it is not always possible. If factorization is impossible, then there will be no way to find a solution to the original inequality, but such cases usually do not occur in school.
Second question: “Will the inequality p(x)/q(x)<0
(≤, >, ≥) is equivalent to the inequality r(x)−s(x)<0
(≤, >, ≥), and therefore to the original”? It can be either equivalent or unequal. It is equivalent when the ODZ for the expression p(x)/q(x) coincides with the ODZ for the expression r(x)−s(x) . In this case, the last step of the algorithm will be redundant. But the ODZ for the expression p(x)/q(x) may be wider than the ODZ for the expression r(x)−s(x) . Expansion of ODZ can occur when fractions are reduced, as, for example, when moving from 
To . Also, the expansion of ODZ can be facilitated by bringing similar terms, as, for example, when moving from 
To . The last step of the algorithm is intended for this case, at which extraneous decisions arising due to the expansion of the ODZ are excluded. Let's follow this when we look at the solutions to the examples below.
Inequality is an expression with, ≤, or ≥. For example, 3x - 5 Solving an inequality means finding all the values of the variables for which the inequality is true. Each of these numbers is a solution to the inequality, and the set of all such solutions is its many solutions. Inequalities that have the same set of solutions are called equivalent inequalities.
Linear inequalities
The principles for solving inequalities are similar to the principles for solving equations.Principles for solving inequalities
For any real numbers a, b, and c:
The principle of adding inequalities: If a Multiplication principle for inequalities: If a 0 is true then ac If a bc is also true.
Similar statements also apply for a ≤ b.
When both sides of an inequality are multiplied by a negative number, the sign of the inequality must be reversed.
First-level inequalities, as in example 1 (below), are called linear inequalities.
Example 1 Solve each of the following inequalities. Then draw a set of solutions.
a) 3x - 5 b) 13 - 7x ≥ 10x - 4
Solution
Any number less than 11/5 is a solution.
The set of solutions is (x|x
To check, we can draw a graph of y 1 = 3x - 5 and y 2 = 6 - 2x. Then it is clear that for x 
The solution set is (x|x ≤ 1), or (-∞, 1]. The graph of the solution set is shown below. 
Double inequalities
When two inequalities are connected by a word And, or, then it is formed double inequality. Double inequality like
-3
And 2x + 5 ≤ 7
called connected, because it uses And. Entry -3 Double inequalities can be solved using the principles of addition and multiplication of inequalities.
Example 2 Solve -3 Solution We have
Set of solutions (x|x ≤ -1 or x > 3). We can also write the solution using interval notation and the symbol for associations or including both sets: (-∞ -1] (3, ∞). The graph of the solution set is shown below. 
To check, let's plot y 1 = 2x - 5, y 2 = -7, and y 3 = 1. Note that for (x|x ≤ -1 or x > 3), y 1 ≤ y 2 or y 1 > y 3 . 
Inequalities with absolute value (modulus)
Inequalities sometimes contain moduli. The following properties are used to solve them.
For a > 0 and algebraic expression x:
|x| |x| > a is equivalent to x or x > a.
Similar statements for |x| ≤ a and |x| ≥ a.
For example,
|x| |y| ≥ 1 is equivalent to y ≤ -1 or y ≥ 1;
and |2x + 3| ≤ 4 is equivalent to -4 ≤ 2x + 3 ≤ 4.
Example 4 Solve each of the following inequalities. Graph the set of solutions.
a) |3x + 2| b) |5 - 2x| ≥ 1
Solution
a) |3x + 2|
b) |5 - 2x| ≥ 1
The solution set is (x|x ≤ 2 or x ≥ 3), or (-∞, 2] )
