Number 1 by solving the inequality x 2 1. Interval method: solving the simplest strict inequalities

Diagram in Figure 5:

And the following example: Solve inequality 0.6 2x – 3< 0,36 .

Following the diagram in Figure 5, we get
2x – 3 > log 0.6 0.36,
2х – 3 > 2,
2x > 5,
x > 2.5

Answer: (2,5; +∞) .

Let us consider the last scheme for solving an inequality of the form and f(x)< b , at a>0 And b<0 , presented 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 in our case this expression is less than -8, i.e. and zero, which means there are no solutions.

Answer: no solutions.

Knowing how to solve the simplest exponential inequalities, you can proceed to solving exponential inequalities.

Example 1.

Find the largest integer value of x that satisfies the inequality

Since 6 x is greater than zero (at no x does the denominator go to zero), multiplying both sides of the inequality by 6 x, we get:

440 – 2 6 2x > 8, then
– 2 6 2x > 8 – 440,
– 2 6 2х > – 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

Let us denote 2 x by y, obtain the inequality y 2 – 3y + 2 ≤ 0, and solve this quadratic inequality.

y 2 – 3y +2 = 0,
y 1 = 1 and y 2 = 2.

The branches of the parabola are directed upward, let's draw a graph:

Then the solution to the inequality will be 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 the 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

Let us take 5 x out of brackets on the left side of the inequality, and 3 x on the right side of the inequality and we get the inequality

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

Divide both sides of the inequality by the expression 3 3 x, the sign of the inequality 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 questions about solving exponential inequalities or would like to practice solving similar examples, sign up for my lessons. Tutor Valentina Galinevskaya.

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Inequalities are called linear the left and right sides of which are linear functions with respect to the unknown quantity. These include, for example, inequalities:

2x-1-x+3; 7x0;

5 >4 - 6x 9- x< x + 5 .

1) Strict inequalities: ax +b>0 or ax+b<0

2) Non-strict inequalities: ax +b≤0 or ax+b≫ 0

Let's analyze this task. One of the sides of the parallelogram is 7cm. What must be the length of the other side so that the perimeter of the parallelogram is greater than 44 cm?

Let the required side be X cm. In this case, the perimeter of the parallelogram will be represented by (14 + 2x) cm. The inequality 14 + 2x > 44 is a mathematical model of the problem of the perimeter of a parallelogram. If we replace the variable in this inequality X on, for example, the number 16, then we obtain the correct numerical inequality 14 + 32 > 44. In this case, they say that the number 16 is a solution to the inequality 14 + 2x > 44.

Solving the inequality name the value of a variable that turns it into a true numerical inequality.

Therefore, each of the numbers is 15.1; 20;73 act as a solution to the inequality 14 + 2x > 44, but the number 10, for example, is not its solution.

Solve inequality means to establish all its solutions or to prove that there are no solutions.

The formulation of the solution to the inequality is similar to the formulation of the root of the equation. And yet it is not customary to designate the “root of inequality.”

The properties of numerical equalities helped us solve equations. Similarly, the properties of numerical inequalities will help solve inequalities.

When solving an equation, we change it to another, simpler equation, but equivalent to the given one. The answer to inequalities is found in a similar way. When changing an equation to an equivalent equation, they use the theorem about transferring terms from one side of the equation to the opposite and about multiplying both sides of the equation by the same non-zero number. When solving an inequality, there is a significant difference between it and an equation, which lies in the fact that any solution to an equation can be verified simply by substitution into the original equation. In inequalities, this method is absent, since it is not possible to substitute countless solutions into the original inequality. Therefore, there is an important concept, these arrows<=>is a sign of equivalent, or equivalent, transformations. The transformation is called equivalent, or equivalent, if they do not change the set of solutions.

Similar rules for solving inequalities.

If we move any term from one part of the inequality to another, replacing its sign with the opposite one, we obtain an inequality equivalent to this one.

If both sides of the inequality are multiplied (divided) by the same positive number, we obtain an inequality equivalent to this one.

If both sides of the inequality are multiplied (divided) by the same negative number, replacing the inequality sign with the opposite one, we obtain an inequality equivalent to the given one.

Using these rules Let us calculate the following inequalities.

1) Let's analyze the inequality 2x - 5 > 9.

This linear inequality, we will find its solution and discuss the basic concepts.

2x - 5 > 9<=>2x>14(5 was moved to the left side with the opposite sign), then we divided everything by 2 and we have x > 7. Let us plot the set of solutions on the axis x

We have obtained a positively directed beam. We note the set of solutions either in the form of inequality x > 7, or in the form of the interval x(7; ∞). What is a particular solution to this inequality? For example, x = 10 is a particular solution to this inequality, x = 12- this is also a particular solution to this inequality.

There are many partial solutions, but our task is to find all the solutions. And there are usually countless solutions.

Let's sort it out example 2:

2) Solve inequality 4a - 11 > a + 13.

Let's solve it: A move it to one side 11 move it to the other side, we get 3a< 24, и в результате после деления обеих частей на 3 the inequality has the form a<8 .

4a - 11 > a + 13<=>3a< 24 <=>a< 8 .

Let's also display the set a< 8 , but already on the axis A.

We either write the answer in the form of inequality a< 8, либо A(-∞;8), 8 does not turn on.

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.

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; in case of strict inequality, the point is denoted by a punctured one; in case of a non-strict inequality, the point is marked;
• 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.

The 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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Solving inequalities online

Before solving inequalities, you need to have a good understanding of how equations are solved.

It doesn’t matter whether the inequality is strict () or non-strict (≤, ≥), the first step is to solve the equation by replacing the inequality sign with equality (=).

Let us explain what it means to solve an inequality?

After studying the equations, the student gets the following picture in his head: he needs to find values ​​of the variable such that both sides of the equation take on the same values. In other words, find all points at which equality holds. Everything is correct!

When we talk about inequalities, we mean finding intervals (segments) on which the inequality holds. If there are two variables in the inequality, then the solution will no longer be intervals, but some areas on the plane. Guess for yourself what will be the solution to an inequality in three variables?

How to solve inequalities?

A universal way to solve inequalities is considered to be the method of intervals (also known as the method of intervals), which consists in determining all intervals within the boundaries of which a given inequality will be satisfied.

Without going into the type of inequality, in this case this is not the point, you need to solve the corresponding equation and determine its roots, followed by the designation of these solutions on the number axis.

How to correctly write the solution to an inequality?

Once you have determined the solution intervals for the inequality, you need to correctly write out the solution itself. There is an important nuance - are the boundaries of the intervals included in the solution?

Everything is simple here. If the solution to the equation satisfies the ODZ and the inequality is not strict, then the boundary of the interval is included in the solution to the inequality. Otherwise, no.

Considering each interval, the solution to the inequality may be the interval itself, or a half-interval (when one of its boundaries satisfies the inequality), or a segment - the interval together with its boundaries.

Important point

Do not think that only intervals, half-intervals and segments can solve the inequality. No, the solution may also include individual points.

For example, the inequality |x|≤0 has only one solution - this is point 0.

And the inequality |x|

What is an inequality calculator for?

The inequalities calculator gives the correct final answer. In most cases, an illustration of a number axis or plane is provided. It is visible whether the boundaries of the intervals are included in the solution or not - the points are displayed as shaded or punctured.

Thanks to the online inequalities calculator, you can check whether you correctly found the roots of the equation, marked them on the number axis and checked the fulfillment of the inequality condition on the intervals (and boundaries)?

If your answer differs from the calculator’s answer, then you definitely need to double-check your solution and identify the mistake.