Solution of inequality х 2 3. Linear inequalities, examples, solutions. Situations where irrationality appears in inequality

see also Solving a linear programming problem graphically, Canonical form of linear programming problems

The system of constraints for such a problem consists of inequalities in two variables:
and the objective function has the form F = C 1 x + C 2 y to be maximized.

Let's answer the question: what pairs of numbers ( x; y) are solutions to the system of inequalities, i.e., satisfy each of the inequalities simultaneously? In other words, what does it mean to solve the system graphically?
First, you need to understand what is the solution to one linear inequality with two unknowns.
Solving a linear inequality with two unknowns means determining all pairs of values of the unknowns for which the inequality is satisfied.
For example, the inequality 3 x – 5y≥ 42 satisfy the pairs ( x , y): (100, 2); (3, –10), etc. The problem is to find all such pairs.
Consider two inequalities: ax + by≤ c, ax + by≥ c... Straight ax + by = c divides the plane into two half-planes so that the coordinates of the points of one of them satisfy the inequality ax + by >c and the other inequality ax + +by <c.
Indeed, take a point with a coordinate x = x 0; then a point lying on a straight line and having an abscissa x 0, has ordinate

Let for definiteness a& lt 0, b>0, c> 0. All points with abscissa x 0 lying above P(for example, dot M) have y M>y 0, and all points below the point P, with abscissa x 0, have y N<y 0. Insofar as x 0 is an arbitrary point, then there will always be points on one side of the straight line for which ax+ by > c forming a half-plane, and on the other hand, points for which ax + by< c.

Picture 1

The sign of the inequality in the half-plane depends on the numbers a, b , c.
This implies the following method for graphically solving systems of linear inequalities in two variables. To solve the system, you must:

For each inequality, write down the equation corresponding to the given inequality.
Construct straight lines that are graphs of functions defined by equations.
For each straight line, determine the half-plane, which is given by the inequality. To do this, take an arbitrary point that does not lie on a straight line, substitute its coordinates into the inequality. if the inequality is true, then the half-plane containing the selected point is the solution to the original inequality. If the inequality is not true, then the half-plane on the other side of the straight line is the set of solutions to this inequality.
To solve the system of inequalities, it is necessary to find the area of intersection of all half-planes that are the solution to each inequality in the system.

This area may be empty, then the system of inequalities has no solutions, is inconsistent. Otherwise, the system is said to be compatible.
There can be a finite number and an infinite number of solutions. The area can be a closed polygon or it can be unlimited.

Let's look at three relevant examples.

Example 1. Solve the system graphically:
x + y - 1 ≤ 0;
–2x - 2y + 5 ≤ 0.

consider the equations x + y – 1 = 0 and –2x – 2y + 5 = 0 corresponding to the inequalities;
we construct the straight lines given by these equations.

Picture 2

Let us define the half-planes given by the inequalities. Take an arbitrary point, let (0; 0). Consider x+ y– 1 0, substitute the point (0; 0): 0 + 0 - 1 ≤ 0. Hence, in the half-plane where the point (0; 0) lies, x + y – 1 ≤ 0, i.e. the half-plane below the line is the solution to the first inequality. Substituting this point (0; 0) into the second, we get: –2 ∙ 0 - 2 ∙ 0 + 5 ≤ 0, i.e. in the half-plane where the point (0; 0) lies, –2 x – 2y+ 5≥ 0, and we were asked where -2 x – 2y+ 5 ≤ 0, therefore, in the other half-plane - in the one that is higher than the line.
Let's find the intersection of these two half-planes. Lines are parallel, so the planes do not intersect anywhere, which means that the system of these inequalities does not have solutions, it is incompatible.

Example 2. Find graphically solutions to the system of inequalities:

Figure 3
1. Let us write out the equations corresponding to the inequalities and construct straight lines.
x + 2y– 2 = 0

x	2	0
y	0	1

y – x – 1 = 0

x	0	2
y	1	3

y + 2 = 0;
y = –2.
2. Having chosen the point (0; 0), we define the signs of the inequalities in the half-planes:
0 + 2 ∙ 0 - 2 ≤ 0, i.e. x + 2y- 2 ≤ 0 in the half-plane below the straight line;
0 - 0 - 1 ≤ 0, i.e. y –x- 1 ≤ 0 in the half-plane below the straight line;
0 + 2 = 2 ≥ 0, i.e. y+ 2 ≥ 0 in the half-plane above the line.
3. The intersection of these three half-planes will be the area that is a triangle. It is not difficult to find the vertices of the region as the intersection points of the corresponding lines

In this way, A(–3; –2), V(0; 1), WITH(6; –2).

Let's consider one more example in which the resulting solution area of the system is not limited.

The form ax 2 + bx + 0 0, where (instead of the> sign, there can, of course, be any other inequality sign). We have at our disposal all the facts of theory necessary for solving such inequalities, of which we will now be convinced.

Example 1... Solve inequality:

a) x 2 - 2x - 3> 0; b) x 2 - 2x - 3< 0;
c) x 2 - 2x - 3> 0; d) x 2 - 2x - 3< 0.
Solution,

a) Consider the parabola y = x 2 - 2x - 3 shown in Fig. 117.

To solve the inequality x 2 - 2x - 3> 0 - this means to answer the question at what values of x the ordinates of the points of the parabola are positive.

Note that y> 0, i.e. the graph of the function is located above the x-axis, for x< -1 или при х > 3.

Hence, the solutions to the inequality are all points of the open ray(- 00, - 1), as well as all points of the open beam (3, +00).

Using the sign U (the sign of union of sets), the answer can be written like this: (-00, - 1) U (3, +00). However, the answer can be written like this: x< - 1; х > 3.

b) Inequality x 2 - 2x - 3< 0, или у < 0, где у = х 2 - 2х - 3, также можно решить с помощью рис. 117: schedule located below the x-axis if -1< х < 3. Поэтому решениями данного неравенства служат все точки интервала (- 1, 3).

c) The inequality x 2 - 2x - 3> 0 differs from the inequality x 2 - 2x - 3> 0 in that the answer must also include the roots of the equation x 2 - 2x - 3 = 0, that is, the points x = -1

and x = 3. Thus, the solutions of this nonstrict inequality are all points of the ray (-00, - 1], as well as all points of the ray.

Practical mathematicians usually say this: why do we need, when solving the inequality ax 2 + bx + c> 0, to carefully construct a parabola graph of a quadratic function

y = ax 2 + bx + c (as was done in example 1)? It is enough to make a schematic sketch of the graph, for which you just need to find roots square trinomial (the point of intersection of the parabola with the x-axis) and determine where the branches of the parabola are directed - up or down. This schematic sketch will provide a visual interpretation of the solution to inequality.

Example 2. Solve the inequality - 2x 2 + 3x + 9< 0.
Solution.

1) Find the roots of the square trinomial - 2x 2 + 3x + 9: x 1 = 3; x 2 = - 1.5.

2) The parabola, which serves as the graph of the function y = -2x 2 + 3x + 9, crosses the x-axis at points 3 and - 1.5, and the branches of the parabola are directed downward, since the senior coefficient- negative number - 2. In fig. 118 is a sketch of a graph.

3) Using fig. 118, we conclude: y< 0 на тех промежутках оси х, где график расположен ниже оси х, т.е. на открытом луче (-оо, -1,5) или на открытом луче C, +оо).
Answer: x< -1,5; х > 3.

Example 3. Solve the 4x inequality 2 - 4x + 1< 0.
Solution.

1) From the equation 4x 2 - 4x + 1 = 0 we find.

2) A square trinomial has one root; this means that the parabola, which serves as the graph of the square trinomial, does not intersect the x-axis, but touches it at a point. The branches of the parabola are directed upward (Fig. 119.)

3) Using the geometric model shown in Fig. 119, we establish that the specified inequality is satisfied only at the point, since for all other values of x the ordinates of the graph are positive.
Answer: .
You may have noticed that in fact, examples 1, 2, 3 used a very specific algorithm solution of square inequalities, let us formalize it.

Algorithm for solving the square inequality ax 2 + bx + 0 0 (ax 2 + bx + c< 0)

The first step of this algorithm is to find the roots of the square trinomial. But the roots may not exist, so what to do? Then the algorithm is inapplicable, which means that one must reason somehow differently. The following theorems provide a key to this reasoning.

In other words, if D< 0, а >0, then the inequality ax 2 + bx + c> 0 is satisfied for all x; on the contrary, the inequality ax 2 + bx + c< 0 не имеет решений.
Proof. Schedule functions y = ax 2 + bx + c is a parabola whose branches are directed upward (since a> 0) and which does not intersect the x-axis, since the square trinomial has no roots by condition. The graph is shown in Fig. 120. We see that for all x the graph is located above the x-axis, which means that for all x the inequality ax 2 + bx + c> 0 is fulfilled, which was required to prove.

In other words, if D< 0, а < 0, то неравенство ах 2 + bх + с < 0 выполняется при всех х; напротив, неравенство ах 2 + bх + с >0 has no solutions.

Proof. The graph of the function y = ax 2 + bx + c is a parabola, the branches of which are directed downward (since a< 0) и которая не пересекает ось х, так как корней у квадратного трехчлена по условию нет. График представлен на рис. 121. Видим, что при всех х график расположен ниже оси х, а это значит, что при всех х выполняется неравенство ах 2 + bх + с < 0, что и требовалось доказать.

Example 4... Solve inequality:

a) 2x 2 - x + 4> 0; b) -x 2 + 3x - 8> 0.

a) Find the discriminant of the square trinomial 2x 2 - x + 4. We have D = (-1) 2 - 4 2 4 = - 31< 0.
The leading coefficient of the trinomial (number 2) is positive.

Hence, by Theorem 1, for all x the inequality 2x 2 - x + 4> 0 holds, that is, the whole (-00, + 00) serves as a solution to the given inequality.

b) Find the discriminant of the square trinomial - x 2 + Zx - 8. We have D = Z2 - 4 (- 1) (- 8) = - 23< 0. Старший коэффициент трехчлена (число - 1) отрицателен. Следовательно, по теореме 2, при всех х выполняется неравенство - х 2 + Зx - 8 < 0. Это значит, что неравенство - х 2 + Зх - 8 0 не выполняется ни при каком значении х, т. е. заданное неравенство не имеет решений.

Answer: a) (-00, + 00); b) there are no solutions.

In the next example, we will get acquainted with another line of reasoning that is used to solve square inequalities.

Example 5. Solve the inequality Зх 2 - 10х + 3< 0.
Solution. Let us factorize the square trinomial Зx 2 - 10x + 3. The roots of the trinomial are the numbers 3 and, therefore, using ax 2 + bx + c = a (x - x 1) (x - x 2), we get 3x 2 - 10x + 3 = 3 (x - 3) (x -)
Note on the number line the roots of the trinomial: 3 and (Fig. 122).

Let x> 3; then x-3> 0 and x-> 0, and hence the product 3 (x - 3) (x -) is positive. Further, let< х < 3; тогда x-3< 0, а х- >0. Therefore, the product 3 (x-3) (x-) is negative. Finally, let x<; тогда x-3< 0 и x- < 0. Но в таком случае произведение
3 (x -3) (x -) is positive.

Summing up the reasoning, we come to the conclusion: the signs of the square trinomial Зx 2 - 10x + 3 change as shown in Fig. 122. We are interested in at what x the square trinomial takes negative values. From fig. 122 we conclude: the square trinomial Zx 2 - 10x + 3 takes negative values for any value of x from the interval (, 3)
Answer (, 3), or< х < 3.

Comment. The reasoning method we used in Example 5 is commonly referred to as the spacing method (or the spacing method). It is actively used in mathematics to solve rational inequalities. In the 9th grade, we will study the method of intervals in more detail.

Example 6... At what values of the parameter p is the quadratic equation x 2 - 5x + p 2 = 0:
a) has two different roots;

b) has one root;

c) does not have roots?

Solution. The number of roots of a quadratic equation depends on the sign of its discriminant D. In this case, we find D = 25 - 4p 2.

a) The quadratic equation has two different roots, if D> 0, then the problem is reduced to solving the inequality 25 - 4p 2> 0. We multiply both sides of this inequality by -1 (not forgetting to change the sign of the inequality). We obtain the equivalent inequality 4p 2 - 25< 0. Далее имеем 4 (р - 2,5) (р + 2,5) < 0.

The signs of the expression 4 (p - 2.5) (p + 2.5) are shown in Fig. 123.

We conclude that the inequality 4 (p - 2.5) (p + 2.5)< 0 выполняется для всех значений р из интервала (-2,5; 2,5). Именно при этих значениях параметра р данное квадратное уравнение имеет два различных корня.

b) quadratic equation has one root if D is 0.
As we stated above, D = 0 when p = 2.5 or p = -2.5.

It is for these values of the parameter p that this quadratic equation has only one root.

c) A quadratic equation has no roots if D< 0. Решим неравенство 25 - 4р 2 < 0.

We get 4p 2 - 25> 0; 4 (p-2.5) (p + 2.5)> 0, whence (see Fig. 123) p< -2,5; р >2.5. For these values of the parameter p, this quadratic equation has no roots.

Answer: a) at p (-2.5, 2.5);

b) when p = 2.5 or p = -2.5;
c) at p< - 2,5 или р > 2,5.

A. G. Mordkovich, Algebra... 8th grade: Textbook. for general education. institutions. - 3rd ed., revised. - M .: Mnemozina, 2001 .-- 223 s: ill.

Help for the student online, Mathematics for grade 8 download, calendar-thematic planning

Hello! My dear students, in this article we will learn how to solve exponential inequalities with you. .

No matter how complicated the exponential inequality may seem to you, after some transformations (we will talk about them a little later) all inequalities are reduced to solving the simplest exponential inequalities:

a x> b, a x< b and a x ≥ b, a x ≤ b.

Let's try to figure out how such inequalities are resolved.

We will consider the solution strict inequalities... The only difference in solving nonstrict inequalities is that the corresponding roots obtained are included in the answer.

Let it be necessary to solve an inequality of the form a f (x)> b, where a> 1 and b> 0.

Look at the scheme for solving such inequalities (Figure 1):

Now let's look at a specific example. Solve Inequality: 5 x - 1> 125.

Since 5> 1 and 125> 0, then
x - 1> log 5 125, that is
x - 1> 3,
x> 4.

Answer: (4; +∞) .

And what will be the solution to the same inequality a f (x)> b, if 0 and b> 0?

So, the diagram in Figure 2

Example: Solve inequality (1/2) 2x - 2 ≥ 4

Applying the rule (Figure 2), we get
2x - 2 ≤ log 1/2 4,
2x - 2 ≤ –2,
2x ≤ 0,
x ≤ 0.

Answer: (–∞; 0] .

Consider again the same inequality a f (x)> b, if a> 0 and b<0 .

So, the diagram in Figure 3:

An example of solving an inequality (1/3) x + 2> –9... As we notice, no matter what number we substitute for x, (1/3) x + 2 is always greater than zero.

Answer: (–∞; +∞) .

But how are inequalities of the form a f (x)< b , where a> 1 and b> 0?

The diagram in Figure 4:

And the next example: 3 3 - x ≥ 8.
Since 3> 1 and 8> 0, then
3 - x> log 3 8, that is
–X> log 3 8 - 3,
X< 3 – log 3 8.

Answer: (0; 3 – log 3 8) .

How to change the solution to inequality a f (x)< b , at 0 and b> 0?

The diagram in Figure 5:

And next example: Solve inequality 0.6 2x - 3< 0,36 .

Following the scheme in Figure 5, we obtain
2x - 3> log 0.6 0.36,
2x - 3> 2,
2x> 5,
x> 2.5

Answer: (2,5; +∞) .

Consider the last scheme for solving an inequality of the form a f (x)< b , at a> 0 and b<0 shown in Figure 6:

For example, let's solve the inequality:

We note that no matter what number we substitute for x, the left side of the inequality is always greater than zero, and our expression is less than -8, i.e. and zero, then there are no solutions.

Answer: no solutions.

Knowing how the simplest exponential inequalities are solved, one can proceed to solving exponential inequalities.

Example 1.

Find the largest integer value x satisfying the inequality

Since 6 x is greater than zero (for any x the denominator does not vanish), we multiply both sides of the inequality by 6 x, we get:

440 - 2 6 2x> 8, then
- 2 6 2x> 8 - 440,
- 2 6 2x> - 332,
6 2x< 216,
2x< 3,

x< 1,5. Наибольшее целое число из помежутка (–∞; 1,5) это число 1.

Answer: 1.

Example 2.

Solve inequality 2 2 x - 3 2 x + 2 ≤ 0

We denote 2 x through y, we obtain the inequality y 2 - 3y + 2 ≤ 0, we solve this square inequality.

y 2 - 3y +2 = 0,
y 1 = 1 and y 2 = 2.

The branches of the parabola are directed upwards, we will depict the graph:

Then the solution to the inequality is the inequality 1< у < 2, вернемся к нашей переменной х и получим неравенство 1< 2 х < 2, решая которое и найдем ответ 0 < x < 1.

Answer: (0; 1) .

Example 3... Solve inequality 5 x +1 - 3 x +2< 2·5 x – 2·3 x –1
Let's collect expressions with the same bases in one part of the inequality

5 x +1 - 2.5 x< 3 x +2 – 2·3 x –1

We take out 5 x on the left side of the inequality, and 3 х on the right side of the inequality, and we obtain the inequality

5 x (5 - 2)< 3 х (9 – 2/3),
3 5 x< (25/3)·3 х

We divide both sides of the inequality by the expression 3 3 x, the inequality sign does not change, since 3 3 x is a positive number, we get the inequality:

X< 2 (так как 5/3 > 1).

Answer: (–∞; 2) .

If you have any questions about solving exponential inequalities or want to practice solving similar examples, sign up for my lessons. Tutor Valentina Galinevskaya.

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Attention!
There are additional
materials in Special Section 555.
For those who are very "not very ..."
And for those who are "very even ...")

What's happened "square inequality"? No question!) If you take any quadratic equation and replace the sign in it "=" (equal) to any inequality icon ( > ≥ < ≤ ≠ ), we get a square inequality. For instance:

1. x 2 -8x + 12 ≥ 0

2. -x 2 + 3x > 0

3. x 2 ≤ 4

Well, you get the idea ...)

It is not for nothing that I connected equations and inequalities here. The point is that the first step in solving any square inequality - solve the equation from which this inequality is made. For this reason, the inability to solve quadratic equations automatically leads to a complete failure in inequalities. Is the hint clear?) If anything, see how to solve any quadratic equations. Everything is detailed there. And in this lesson we will deal specifically with inequalities.

The inequality ready for solution has the form: on the left - a square trinomial ax 2 + bx + c, on the right - zero. The inequality sign can be absolutely any. The first two examples are here are already ready for a solution. The third example still needs to be prepared.

If you like this site ...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Instant validation testing. Learning - with interest!)

you can get acquainted with functions and derivatives.

After receiving the initial information about inequalities with variables, we turn to the question of their solution. We will analyze the solution of linear inequalities with one variable and all methods for their resolution with algorithms and examples. Only linear equations with one variable will be considered.

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 inequalities do not have a fundamental difference, therefore it is necessary to use several definitions.
Definition 1
Linear inequality with one variable x is an inequality of the form ax + b> 0, when instead of> any inequality sign is used< , ≤ , ≥ , 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 some numbers, is called linear inequalities in 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:

the notation a · x + b> 0 in the first, and a · x> c - in the second;

the admissibility of the equality to zero of the coefficient a, 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 considered to be 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 an ordinary 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 converging to linear.

How to solve linear inequality

The main way to solve such inequalities is reduced to equivalent transformations in order to find elementary inequalities x< p (≤ , >, ≥), p which is some number, for a ≠ 0, and of the form a< p (≤ , >, ≥) for a = 0.

To solve an inequality with one variable, you can apply the method of intervals or plot it graphically. Any of them can be applied in isolation.

Using equivalent transformations

To solve a linear inequality of the form ax + 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, it is necessary to adhere to a scheme consisting of 3 points: the essence of the process, the algorithm, the solution itself.
Definition 4
Algorithm for solving linear inequality a x + b< 0 (≤ , >, ≥) for a ≠ 0

the number b will be transferred to the right-hand side of the inequality with the opposite sign, which will allow us to arrive at an equivalent a x< − b (≤ , > , ≥) ;

both sides of the inequality will be divided by a number not equal to 0. Moreover, when a is positive, then the sign remains, when a is negative, it changes to the opposite.

Let's consider the application of this algorithm to the solution of examples.
Example 1
Solve an inequality of the form 3 x + 12 ≤ 0.

Solution

This linear inequality has a = 3 and b = 12. Hence, the coefficient a at x is not equal to zero. Let's apply the above algorithms and decide.

It is necessary to transfer the term 12 to the other part of the inequality with a change in the sign in front of it. Then we obtain an inequality of the form 3 x ≤ - 12. It is necessary to divide both parts by 3. The sign does 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 as follows:

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 at z is equal to - 2, 7, and b is explicitly absent or equal to zero. The first step of the algorithm can be skipped over 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 sign of inequality. That is, we get that (- 2, 7 z): (- 2, 7)< 0: (− 2 , 7) , и дальше z < 0 .

We write the entire algorithm in a short form:

- 2, 7 · z> 0; z< 0 .

Answer: z< 0 или (− ∞ , 0) .
Example 3
Solve the inequality - 5 x - 15 22 ≤ 0.

Solution

By condition, we see that it is necessary to solve the inequality with the coefficient a at the variable x, which equals - 5, with the coefficient b, which corresponds to the fraction - 15 22. It is necessary to solve the inequality following the algorithm, that is: transfer - 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

At the last transition, for the right side, the rule for dividing numbers with different signs is used 15 22: - 5 = - 15 22: 5, after which we perform division of an ordinary fraction by a natural number - 15 22: 5 = - 15 22 1 5 = - 15 1 22 5 = - 3 22.

Answer: x ≥ - 3 22 and [- 3 22 + ∞).

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 the definition of the solution to 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 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 if 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 of x. Then we obtain an inequality of the form 7> 0. The last inequality is considered true, which means that 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 get that the inequality will take the form - 12, 7 ≥ 0. It is wrong. That is, 0 x - 12, 7 ≥ 0 has no solutions.

Answer: no solutions.

Consider the solution to linear inequalities where both coefficients are equal to zero.
Example 6
Determine an inequality that has no solution from 0 x + 0> 0 and 0 x + 0 ≥ 0.

Solution

Substituting any number instead of x, we obtain two inequalities of the form 0> 0 and 0 ≥ 0. The first is not true. Hence, 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, and 0 x + 0 ≥ 0 has solutions.

This method is considered in the school mathematics course. The interval method is able to resolve various types of inequalities, including linear ones.

The method of intervals is used for linear inequalities when the value of the coefficient x is not equal to 0. Otherwise, you will have to calculate using another method.
Definition 6
The spacing method is:

introduction of the function y = a x + b;

searching for zeros to split the domain into intervals;

definition of signs for understanding them at intervals.

Let's put together an algorithm for solving linear equations a x + b< 0 (≤ , >, ≥) for a ≠ 0 using the method of intervals:

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 is the only root that takes the notation x 0;

construction of a coordinate line with an image of a point with a coordinate x 0, with strict inequality, the point is denoted by a punctured one, with a non-strict one - filled;

determination of the signs of the function y = a · x + b at intervals, for this it is necessary to find the values of the function at points in the interval;

the solution of the inequality with signs> or ≥ on the coordinate line is added shading over the positive interval,< или ≤ над отрицательным промежутком.

Let's consider several examples of solving a linear inequality using the interval method.
Example 6
Solve the inequality - 3 x + 12> 0.

Solution

From the algorithm it follows 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 since the inequality is severe. Consider the drawing below.

It is necessary to identify the signs in between. 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 in between is positive.

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.

It can be seen from the drawing that the sought solution has the form (- ∞, 4) or x< 4 .

Answer: (- ∞, 4) or x< 4 .

To understand how to represent 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 x< 2 , x ≤ 2 , x >2 and x ≥ 2. To do this, we will depict the graph of the linear function y = 0.5 · x - 1, shown below.

It's clear that
Definition 7
solution of 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 the interval, the function is located above O x;

the solution 0, 5 · x - 1 ≥ 0 is considered to be the interval where the graph is higher than O x or coincides.

The meaning of the graphical solution of inequalities is to find the intervals that must 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 8
The plotting of the function y = a x + b is performed:

while solving the inequality a x + b< 0 определяется промежуток, где график изображен ниже О х;

during the solution of the inequality a · x + b ≤ 0, the interval is determined, where the graph is displayed below the O x axis or coincides;

during the solution of the inequality a x + b> 0, the interval is determined, where the graph is depicted above O x;

while 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 the graph.

Solution

It is necessary to plot a linear function - 5 x - 3> 0. This line is decreasing because the coefficient at x is negative. To determine the coordinates of the point of its intersection with O x - 5 · x - 3> 0, we get the value - 3 5. Let's draw it graphically.

The solution to the inequality with the> sign, then it is necessary to pay attention to the interval above O x. Highlight the necessary part of the plane in red and get that

The required spacing is part of the red O x. Hence, an open number ray - ∞, - 3 5 will be a solution to the inequality. If by condition they had a non-strict inequality, then the value of the point - 3 5 would also be a solution to the inequality. And it would be the same as Oh.

Answer: - ∞, - 3 5 or x< - 3 5 .

The graphical solution is used when the left side will correspond to the function y = 0 x + b, that is, y = b. Then the line will be parallel to O x or coinciding at b = 0. These cases show that the inequality may not have solutions, or any number may be the solution.
Example 8
Determine from the inequalities 0 x + 7< = 0 , 0 · x + 0 ≥ 0 то, которое имеет хотя бы одно решение.

Solution

The representation y = 0 x + 7 is y = 7, then a coordinate plane with a straight line parallel to O x and located above O x will be given. Hence, 0 x + 7< = 0 решений не имеет, потому как нет промежутков.

The graph of the function y = 0 x + 0, y = 0 is considered, that is, the straight line coincides with O x. Hence, the inequality 0 x + 0 ≥ 0 has a set of solutions.

Answer: the second inequality has a solution for any value of x.

Inequalities Reducing to Linear

Solving inequalities can be reduced to solving 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 above inequalities are always reduced to the form of a linear equation. Then the brackets are opened and similar terms are given, transferred from different parts, changing the sign to the opposite.

Reducing the inequality 5 - 2 x> 0 to a linear one, 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 and bring similar terms. It looks like this:

7x - 7 + 3 ≤ 4x - 2 + x 7x - 4 ≤ 5x - 2 7x - 4 - 5x + 2 ≤ 0 2x - 2 ≤ 0

This brings the solution to a linear inequality.

These inequalities are considered as linear, since they have the same principle of solution, after which it is possible to reduce them to elementary inequalities.

To solve this kind of inequality of this kind, it is necessary to reduce it to a linear one. This should be done in this way:
Definition 9
expand brackets;

collect variables on the left, and numbers on the right;

bring 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 expand the parentheses, 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 transferring the terms from the left to the right, we get that 6 x + 15 - 6 x + 17 ≤ 0. Hence, it has an inequality of the form 32 ≤ 0 from the one obtained in the calculation of 0 x + 32 ≤ 0. It can be seen that the inequality is incorrect, which means that the inequality given by the condition has no solutions.

Answer: no solutions.

It is worth noting that there are many inequalities of another kind, which can be reduced to a linear one or an inequality of the kind shown above. For example, 5 2 x - 1 ≥ 1 is an exponential equation that reduces to a linear solution 2 x - 1 ≥ 0. These cases will be considered when solving inequalities of this type.

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