Today in the lesson we will generalize our knowledge in solving systems of inequalities and study the solution of a set of systems of inequalities.

Definition one.

It is said that several inequalities with one variable form a system of inequalities if the task is to find all common solutions of the given inequalities.

The value of the variable, at which each of the inequalities of the system turns into a true numerical inequality, is called a particular solution of the system of inequalities.

The set of all particular solutions to a system of inequalities is a general solution to a system of inequalities (more often they simply say a solution to a system of inequalities).

To solve a system of inequalities means to find all its particular solutions, or to prove that this system has no solutions.

Remember! The solution of a system of inequalities is the intersection of the solutions of the inequalities included in the system.

The inequalities included in the system are combined with a curly bracket.

Algorithm for solving a system of inequalities with one variable:

The first is to solve each inequality separately.

The second is to find the intersection of the found solutions.

This intersection is the set of solutions to the system of inequalities

Exercise 1

Solve the system of inequalities seven x minus forty two less than or equal to zero and two x minus seven greater than zero.

The solution to the first inequality - x is less than or equal to six, the second inequality - x is greater than seven second. We mark these gaps on the coordinate line. The solution of the first inequality is marked with hatching from below, the solution of the second inequality is marked with hatching from above. The solution to the system of inequalities will be the intersection of the solutions of the inequalities, that is, the interval on which both hatchings coincide. As a result, we get a half-interval from seven second to six, including six.

Task 2

Solve the system of inequalities: x squared plus x minus six is ​​greater than zero and x squared plus x plus six is ​​greater than zero.

Solution

Let's solve the first inequality - x squared plus x minus six is ​​greater than zero.

Consider the function y equals x squared plus x minus six. Zeros of the function: the first x is equal to minus three, the second x is equal to two. Schematically depicting a parabola, we find that the solution to the first inequality is the union of open numerical rays from minus infinity to minus three and from two to plus infinity.

Let's solve the second inequality of the system x square plus x plus six greater than zero.

Consider the function y equals x squared plus x plus six. The discriminant is minus twenty-three less than zero, which means that the function has no zeros. The parabola has no common points with the x-axis. Depicting a parabola schematically, we find that the solution of the inequality is the set of all numbers.

Let us depict on the coordinate line the solutions of the inequalities of the system.

It can be seen from the figure that the solution of the system is the union of open numerical rays from minus infinity to minus three and from two to plus infinity.

Answer: the union of open numerical rays from minus infinity to minus three and from two to plus infinity.

Remember! If in a system of several inequalities one is a consequence of another (or others), then the inequality-consequence can be discarded.

Consider an example of solving an inequality by a system.

Task 3

Solve the inequality logarithm of the expression x square minus thirteen x plus forty two base two greater than or equal to one.

Solution

The ODZ inequality is given by x squared minus thirteen x plus forty two greater than zero. Let's represent the number one as the logarithm of two base two and get the inequality - the logarithm of the expression x square minus thirteen x plus forty two base two is greater than or equal to the logarithm of two base two.

We see that the base of the logarithm is equal to two more than one, then we come to the equivalent inequality x square minus thirteen x plus forty two is greater than or equal to two. Therefore, solving this logarithmic inequality reduces to solving a system of two square inequalities.

Moreover, it is easy to see that if the second inequality is satisfied, then the more the first inequality is satisfied. Therefore, the first inequality is a consequence of the second, and it can be discarded. We transform the second inequality and write it in the form: x square minus thirteen x plus forty more than zero. Its solution is the union of two numerical rays from minus infinity to five and from eight to plus infinity.

Answer: the union of two numerical rays from minus infinity to five and from eight to plus infinity.

open number beams

Definition two.

It is said that several inequalities with one variable form a set of inequalities if the task is to find all such values ​​of the variable, each of which is a solution to at least one of the given inequalities.

Each such value of a variable is called a particular solution of the set of inequalities.

The set of all particular solutions of the set of inequalities is general solution of a set of inequalities.

Remember! The solution of a set of inequalities is the union of solutions of inequalities included in the set.

The inequalities included in the set are united by a square bracket.

Algorithm for solving a set of inequalities:

The first is to solve each inequality separately.

The second is to find the union of the found solutions.

This union is the solution to the set of inequalities.

Task 4

zero point two tenths multiplied by the difference of two x and three is less than x minus two;

five x minus seven is greater than x minus six.

Solution

Let's transform each of the inequalities. We get an equivalent set

x is greater than seven thirds;

x is greater than one fourth.

For the first inequality, the set of solutions is the interval from seven thirds to plus infinity, and for the second, the interval from one fourth to plus infinity.

Draw on the coordinate line a set of numbers that satisfy the inequalities x is greater than seven thirds and x is greater than one fourth.

We find that the union of these sets, i.e. the solution to this set of inequalities is an open numerical ray from one fourth to plus infinity.

Answer: an open number beam from one-fourth to plus infinity.

Task 5

Solve a set of inequalities:

two x minus one is less than three and three x minus two is greater than or equal to ten.

Solution

Let's transform each of the inequalities. We get an equivalent set of inequalities: x is greater than two and x is greater than or equal to four.

Draw on the coordinate line the set of numbers that satisfy these inequalities.

We find that the union of these sets, i.e. the solution to this set of inequalities is an open numerical ray from two to plus infinity.

Answer: an open number beam from two to plus infinity.

1. The concept of inequality with one variable

2. Equivalent inequalities. Equivalence theorems for inequalities

3. Solving inequalities with one variable

4. Graphical solution of inequalities with one variable

5. Inequalities containing a variable under the modulus sign

6. Main findings

Inequalities with one variable

Offers 2 X + 7 > 10's, x 2 +7x< 2,(х + 2)(2х-3)> 0 are called single-variable inequalities.

In general, this concept is defined as follows:

Definition. Let f(x) and g(x) be two expressions with variable x and domain X. Then an inequality of the form f(x) > g(x) or f(x)< g(х) называется неравенством с одной переменной. Мно­жество X называется областью его определения.

Variable value x from many x, under which the inequality turns into a true numerical inequality, is called its decision. Solving an inequality means finding the set of its solutions.

Thus, by solving inequality 2 x + 7 > 10 -x, x? R is the number x= 5, since 2 5 + 7 > 10 - 5 is a true numerical inequality. And the set of its solutions is the interval (1, ∞), which is found by performing the transformation of the inequality: 2 x + 7 > 10-x => 3x >3 => x >1.

Equivalent inequalities. Equivalence theorems for inequalities

The concept of equivalence underlies the solution of inequalities with one variable.

Definition. Two inequalities are said to be equivalent if their solution sets are equal.

For example, inequalities 2 x+ 7 > 10 and 2 x> 3 are equivalent, since their solution sets are equal and represent the interval (2/3, ∞).

Theorems on the equivalence of inequalities and their consequences are similar to the corresponding theorems on the equivalence of equations. When proving them, the properties of true numerical inequalities are used.

Theorem 3. Let the inequality f(x) > g(x) set on the set X and h(x) is an expression defined on the same set. Then the inequalities f(x) > g(x) and f(x) + h(x) > g(x) + h(x) are equivalent on the set x.

Consequences follow from this theorem, which are often used in solving inequalities:

1) If both sides of the inequality f(x) > g(x) add the same number d, then we get the inequality f(x) + d > g(x) + d, equivalent to the original.

2) If any term (a numerical expression or an expression with a variable) is transferred from one part of the inequality to another, changing the sign of the term to the opposite, then we obtain an inequality equivalent to the given one.

Theorem 4. Let the inequality f(x) > g(x) set on the set X and h(X X from many X expression h(x) takes positive values. Then the inequalities f(x) > g(x) and f(x) h(x) > g(x) h(x) are equivalent on the set x.

f(x) > g(x) multiply by the same positive number d, then we get the inequality f(x) d > g(x) d, equivalent to this one.

Theorem 5. Let the inequality f(x) > g(x) set on the set X and h(X) is an expression defined on the same set, and for all X their multitude X expression h(X) takes negative values. Then the inequalities f(x) > g(x) and f(x) h(x) > g(x) h(x) are equivalent on the set X.

The corollary follows from this theorem: if both sides of the inequality f(x) > g(x) multiply by the same negative number d and reverse the inequality sign, we get the inequality f(x) d > g(x) d, equivalent to this one.

Solving inequalities with one variable

Let's solve inequality 5 X - 5 < 2х - 16, X? R, and justify all the transformations that we will perform in the solution process.

Inequality solution X < 7 является промежуток (-∞, 7) и, сле­довательно, множеством решений неравенства 5X - 5 < 2x + 16 is the interval (-∞, 7).

Exercises

1. Determine which of the following entries are single-variable inequalities:

a) -12 - 7 X< 3x+ 8; d) 12 x + 3(X- 2);

b) 15( x+ 2)>4; e) 17-12 8;

c) 17-(13 + 8)< 14-9; е) 2x 2+ 3x-4> 0.

2. Is the number 3 a solution to the inequality 6(2x + 7) < 15(X + 2), X? R? And the number 4.25?

3. Are the following pairs of inequalities equivalent on the set of real numbers:

a) -17 X< -51 и X > 3;

b) (3 x-1)/4 >0 and 3 X-1>0;

c) 6-5 x>-4 and X<2?

4. Which of the following statements are true:

a) -7 X < -28 => x>4;

b) x < 6 => x < 5;

in) X< 6 => X< 20?

5. Solve inequality 3( x - 2) - 4(X + 1) < 2(х - 3) - 2 and justify all the transformations that you will perform in this case.

6. Prove that the solution of the inequality 2(x+ 1) + 5 > 3 - (1 - 2X) is any real number.

7. Prove that there is no real number that would be a solution to the inequality 3(2 - X) - 2 > 5 - 3X.

8. One side of the triangle is 5 cm, and the other is 8 cm. What can be the length of the third side if the perimeter of the triangle is:

a) less than 22 cm;

b) more than 17 cm?

GRAPHIC SOLUTION OF INEQUALITIES WITH ONE VARIABLE. For a graphical solution of the inequality f(x) > g(x) you need to plot function graphs

y = f(x) = g(x) and choose those intervals of the abscissa axis, on which the graph of the function y = f(x) located above the graph of the function y \u003d g(x).

Example 17.8. Solve Graphically an Inequality x 2- 4 > 3X.

Y - x * - 4

Solution. Let us construct graphs of functions in one coordinate system

y \u003d x 2 - 4 and y= Zx (Fig. 17.5). It can be seen from the figure that the graphs of the functions at= x 2- 4 is located above the graph of the function y \u003d 3 X at X< -1 and x > 4, i.e. the set of solutions to the original inequality is the set

(- ¥; -1) È (4; + oo) .

Answer: x O(-oo; -1) and ( 4; +oo).

Graph of a quadratic function at= ax 2 + bx + c is a parabola with branches pointing upwards if a > 0, and down if a< 0. In this case, three cases are possible: the parabola intersects the axis Oh(i.e. the equation ah 2+ bx+ c = 0 has two different roots); the parabola touches the axis X(i.e. the equation ax 2 + bx+ c = 0 has one root); the parabola does not intersect the axis Oh(i.e. the equation ah 2+ bx+ c = 0 has no roots). Thus, there are six possible positions of the parabola, which serves as a graph of the function y \u003d ah 2+b x + c(Fig. 17.6). Using these illustrations, one can solve quadratic inequalities.

Example 17.9. Solve the inequality: a) 2 x r+ 5x - 3 > 0; b) -Zx 2 - 2x- 6 < 0.

Solution, a) The equation 2x 2 + 5x -3 \u003d 0 has two roots: x, \u003d -3, x 2 = 0.5. Parabola serving as a graph of a function at= 2x 2+ 5x -3, shown in fig. a. Inequality 2x 2+ 5x -3 > 0 is performed for those values X, for which the points of the parabola lie above the axis Oh: it will be at X< х х or when X> x r> those. at X< -3 or at x > 0.5. Hence, the set of solutions to the original inequality is the set (- ¥; -3) and (0.5; + ¥).

b) Equation -Zx 2 + 2x- 6 = 0 has no real roots. Parabola serving as a graph of a function at= - 3x 2 - 2x - 6 is shown in fig. 17.6 Inequality -3x 2 - 2x - 6 < О выполняется при тех значениях X, for which the points of the parabola lie below the axis Oh. Since the entire parabola lies below the axis Oh, then the set of solutions to the original inequality is the set R .

INEQUALITIES CONTAINING A VARIABLE UNDER THE MODULUS SIGN. When solving these inequalities, keep in mind that:

|f(x) | =

f(x), if f(x) ³ 0,

- f(x), if f(x) < 0,

In this case, the region of admissible values ​​of the inequality should be divided into intervals, on each of which the expressions under the modulus sign retain their sign. Then, expanding the modules (taking into account the signs of expressions), you need to solve the inequality on each interval and combine the resulting solutions into a set of solutions to the original inequality.

Example 17.10. Solve the inequality:

|x -1| + |2-x| > 3+x.

Solution. The points x = 1 and x = 2 divide the real axis (ODZ of inequality (17.9) into three intervals: x< 1, 1 £ х £.2, х >2. Let's solve this inequality on each of them. If x< 1, то х - 1 < 0 и 2 – х >0; so |x -1| = - (x - I), |2 - x | = 2 - x. Hence, inequality (17.9) takes the form: 1- x + 2 - x > 3 + x, i.e. X< 0. Таким образом, в этом случае решениями неравенства (17.9) являются все отрицательные числа.

If 1 £ x £.2, then x - 1 ³ 0 and 2 - x ³ 0; therefore | x-1| = x - 1, |2 - x| = 2 - x. .So, there is a system:

x - 1 + 2 - x > 3 + x,

The resulting system of inequalities has no solutions. Therefore, on the interval [ 1; 2], the set of solutions to inequality (17.9) is empty.

If x > 2, then x - 1 > 0 and 2 - x<0; поэтому | х - 1| = х- 1, |2-х| = -(2- х). Значит, имеет место система:

x -1 + x - 2 > 3 + x,

x > 6 or

Combining the solutions found on all parts of the ODZ of inequality (17.9), we obtain its solution - the set (-¥; 0) È (6; + oo).

Sometimes it is useful to use the geometric interpretation of the modulus of a real number, according to which | a | means the distance of the point a of the coordinate line from the origin O, and | a - b | means the distance between points a and b on the coordinate line. Alternatively, you can use the method of squaring both sides of the inequality.

Theorem 17.5. If expressions f(x) and g(x) for any x take only non-negative values, then the inequalities f(x) > g(x) and f (x) ² > g (x) ² are equivalent.

58. Main conclusions § 12

In this section, we have defined the following concepts:

Numeric expression;

The value of a numeric expression;

An expression that doesn't make sense;

Expression with variable(s);

Expression scope;

identically equal expressions;

Identity;

Identity transformation of an expression;

Numerical equality;

Numerical inequality;

Equation with one variable;

Root of the equation;

What does it mean to solve an equation;

Equivalent Equations;

Inequality with one variable;

Solution of inequality;

What does it mean to solve an inequality;

Equivalent inequalities.

In addition, we considered theorems on the equivalence of equations and inequalities, which are the basis for their solution.

Knowledge of the definitions of all the above concepts and theorems on the equivalence of equations and inequalities is a necessary condition for a methodically competent study of algebraic material with younger students.

Municipal budgetary educational institution

"Secondary school No. 26

with in-depth study of individual subjects "

city ​​of Nizhnekamsk, Republic of Tatarstan

Summary of the lesson in mathematics
in 8th grade

Solving inequalities with one variable

and their systems

prepared

mathematic teacher

first qualification category

Kungurova Gulnaz Rafaelovna

Nizhnekamsk 2014

Lesson outline

Teacher: Kungurova G.R.

Subject: mathematics

Topic: "Solution of linear inequalities with one variable and their systems."

Grade: 8B

Date: 04/10/2014

Lesson type: lesson of generalization and systematization of the studied material.

The purpose of the lesson: consolidation of practical skills and skills in solving inequalities with one variable and their systems, inequalities containing a variable under the module sign.

Lesson objectives:

Tutorials:

generalization and systematization of students' knowledge about how to solve inequalities with one variable;

extension of the type of inequalities: double inequalities, inequalities containing a variable under the module sign, systems of inequalities;

establishment of interdisciplinary connection between mathematics, Russian language, chemistry.

Developing:

activation of attention, mental activity, development of mathematical speech, cognitive interest among students;

mastering the methods and criteria of self-assessment and self-control.

Educational:

education of independence, accuracy, ability to work in a team

The main methods used in the lesson: communicative, explanatory-illustrative, reproductive, method of programmed control.

Equipment:

a computer

computer presentation

monoblocks (performing an individual online test)

handouts (different levels of individual tasks);

self-control sheets;

Lesson plan:

1. Organizational moment.

4. Independent work

5. Reflection

6. The results of the lesson.

During the classes:

1. Organizational moment.

(The teacher tells the students the goals and objectives of the lesson.).

Today we face a very important task. We must sum up this topic. Again, it will be necessary to work out theoretical issues very carefully, to do calculations, to consider the practical application of this topic in our daily life. And we must never forget about how we reason, analyze, build logical chains. Our speech must always be correct.

Each of you has a self-control sheet on your desk. Throughout the lesson, do not forget to mark with a "+" sign your contribution to this lesson.

The teacher assigns homework, commenting on it:

№1026(a,b), No. 1019(c,d); additionally - No. 1046 (a)

2. Actualization of knowledge, skills, skills

1) Before we begin to perform practical tasks, let's turn to the theory.

The teacher announces the beginning of the definition, and the students must complete the wording

a) An inequality with one variable is an inequality of the form ax>b, ax<в;

b) Solving an inequality means finding all its solutions or proving that there are no solutions;

c) The solution of an inequality with one variable is the value of the variable that turns it into a true inequality;

d) Inequalities are called equivalent if they have the same set of solutions. If they have no solutions, then they are also called equivalent

2) On the board, inequalities with one variable, arranged in one column. And next to it, in another column, their solutions are inscribed in the form of numerical intervals. The task of students is to establish a correspondence between inequalities and corresponding gaps.

Establish a correspondence between inequalities and numerical intervals:

1. 3x > 6 a) (-∞ ; - 0.2]

2. -5x ≥ 1 b) (- ∞ ; 15)

3. 4x > 3 c) (2; + ∞)

4. 0.2x< 3 г) (0,75; + ∞)

3) Practical work in a notebook with self-examination.

On the blackboard, students write a linear inequality with one variable. After completing which one of the students voices his decision and corrects the mistakes made)

Solve the inequality:

4 (2x - 1) - 3 (x + 6) > x;

8x - 4 - 3x - 18 > x;

8x - 3x - x\u003e 4 + 18;

4x > 22;

x > 5.5.

Answer. (5.5 ; +∞)

3. Practical application of inequalities in everyday life (chemical experiment)

Inequalities in our daily lives can be good helpers. And besides, of course, there is an inextricable link between school subjects. Mathematics goes shoulder to shoulder not only with the Russian language, but also with chemistry.

(On each desk there is a reference scale for pH, ranging from 0 to 12)

If the value is 0 ≤ pH< 7, то среда кислая;

if pH = 7, then the medium is neutral;

if indicator is 7< pH ≤ 12, то среда щелочная

The teacher pours 3 colorless solutions into different test tubes. From the chemistry course, students are asked to remember the types of solution medium (acidic, neutral, alkaline). Further, empirically, involving students, the environment of each of the three solutions is determined. To do this, a universal indicator is lowered into each solution. The following happens: each indicator is painted in the corresponding color. And according to the color scheme, thanks to the reference scale, students set the environment for each of the proposed solutions.

Conclusion:

1 indicator turns red, value 0 ≤ pH< 7, значит среда первого раствора кислая, т.е. имеем кислоту в 1пробирке

2 the indicator turned green, pH = 7, which means the medium of the second solution is neutral, i.e. we had water in test tube 2

3 indicator turned blue, indicator 7< pH ≤ 12 , значит среда третьего раствора щелочная, значит в 3 пробирке была щелочь

Knowing the limits of the pH indicator, you can determine the level of acidity of the soil, soap, and many cosmetics.

Continued updating of knowledge, skills and abilities.

1) Once again the teacher starts formulating the definitions and the students have to complete them

Continue definitions:

a) Solving a system of linear inequalities means finding all its solutions or proving that there are none

b) The solution of a system of inequalities with one variable is the value of the variable for which each of the inequalities is true

c) To solve a system of inequalities with one variable, you need to find a solution to each inequality, and find the intersection of these intervals

The teacher again reminds the students that the ability to solve linear inequalities with one variable and their systems is the basis, the basis for more complex inequalities to be studied in older grades. The foundation of knowledge is being laid, the strength of which is to be confirmed at the OGE in mathematics after grade 9.

Students write in notebooks to solve systems of linear inequalities with one variable. (2 students complete these tasks on the board, explain their solution, voice the properties of inequalities used in solving systems).

№1012(e). Solve System of Linear Inequalities

0.3 x+1< 0,4х-2;

1.5x-3 > 1.3x-1. Answer. (30; +∞).

№1028(g). Solve a double inequality and indicate all the integers that are its solution

1 < (4-2х)/3 < 2 . Ответ. Целое число: 0

2) Solving inequalities containing a variable under the module sign.

Practice shows that inequalities containing a variable under the module sign cause anxiety and self-doubt in students. And often students simply do not take up such inequalities. And the reason for this is a poorly laid foundation. The teacher encourages the students to work on themselves in a timely manner, to learn all the steps in order to successfully fulfill these inequalities.

There is oral work. (Front survey)

Solving inequalities containing a variable under the module sign:

1. The module of the number x is the distance from the origin to the point with coordinate x.

| 35 | = 35,

| - 17 | = 17,

| 0 | = 0

2. Solve inequalities:

a) | x |< 3 . Ответ. (-3 ; 3)

b) | x | > 2 . Answer. (-∞; -2) U (2; +∞)

The progress of solving these inequalities is displayed on the screen in detail and the algorithm for solving inequalities containing a variable under the module sign is spoken out.

4. Independent work

In order to control the degree of assimilation of this topic, 4 students take places at the monoblocks and undergo thematic online testing. Testing time 15 minutes. After completion, a self-test is carried out both in points and in percentage terms.

The rest of the students at their desks perform independently independent work.

Independent work (run time 13min)

Option 1

Option 2

1. Solve the inequalities:

a) 6+x< 3 - 2х;

b) 0.8(x-3) - 3.2 ≤ 0.3(2 - x).

3(x+1) - (x-2)< х,

2 > 5x - (2x-1) .

-6 < 5х - 1 < 5

four*. (Additionally)

Solve the inequality:

| 2- 2x | ≤ 1

1. Solve the inequalities:

a) 4+x< 1 - 2х;

b) 0.2 (3x - 4) - 1.6 ≥ 0.3 (4-3x).

2. Solve the system of inequalities:

2(x+3) - (x - 8)< 4,

6x > 3(x+1) -1.

3. Solve the double inequality:

-1 < 3х - 1 < 2

four*. (Additionally)

Solve the inequality:

| 6x-1 | ≤ 1

After completing independent work, students submit notebooks for verification. Students who worked on monoblocks also hand over notebooks to the teacher for verification.

5. Reflection

The teacher reminds the students about the self-control sheets, on which they had to evaluate their work with the “+” sign throughout the lesson, at its various stages.

But the students will have to make the main assessment of their activity only now, after voicing one ancient parable.

Parable.

A wise man was walking, and 3 people were walking towards him. Under the hot sun, they carried carts with stones to build the temple.

The sage stopped them and asked:

- What did you do all day?

- Carried cursed stones, - answered the first.

“I did my job conscientiously,” replied the second.

- And I took part in the construction of the temple, - proudly answered the third.

In the self-control sheets, in paragraph No. 3, students must enter a phrase that would correspond to their actions in this lesson.

Self-control sheet __________________________________________

№ P / P

Lesson stages

Evaluation of educational activities

Oral work in the lesson

Practical part:

Solving inequalities with one variable;

solution of systems of inequalities;

solution of double inequalities;

solution of inequalities with module sign

Reflection

In paragraphs 1 and 2, mark the correct answers in the lesson with a “+” sign;

in paragraph 3, evaluate your work in the lesson according to the instructions

6. The results of the lesson.

The teacher, summing up the lesson, notes successful moments and problems on which additional work is to be done.

Students are invited to evaluate their work according to self-control sheets, and students receive one more mark based on the results of independent work.

At the end of the lesson, the teacher draws the students' attention to the words of the French scientist Blaise Pascal: "The greatness of a person is in his ability to think."

Bibliography:

1 . Algebra. 8th grade. Yu.N.Makarychev, N.G. Mindyuk, K.E. Neshkov, I.E. Feoktistov.-M.:

Mnemosyne, 2012

2. Algebra.8 class. Didactic materials. Guidelines / I.E. Feoktistov.

2nd edition., Ster.-M.: Mnemosyne, 2011

3. Control and measuring materials. Algebra: Grade 8 / Compiled by L.I. Martyshova.-

M.: VAKO, 2010

Internet resources:

The program for solving linear, quadratic and fractional inequalities does not just give the answer to the problem, it gives a detailed solution with explanations, i.e. displays the process of solving in order to check the knowledge of mathematics and / or algebra.

Moreover, if in the process of solving one of the inequalities it is necessary to solve, for example, a quadratic equation, then its detailed solution is also displayed (it is included in the spoiler).

This program can be useful for high school students in preparation for tests, parents to control the solution of inequalities by their children.

This program can be useful for high school students in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.

In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.

Rules for entering inequalities

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.

Numbers can be entered as integers or fractions.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.

Rules for entering decimal fractions.
In decimal fractions, the fractional part from the integer can be separated by either a dot or a comma.
For example, you can enter decimals like this: 2.5x - 3.5x^2

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &
Input: 3&1/3 - 5&6/5y +1/7y^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) y + \frac(1)(7)y^2 \)

Parentheses can be used when entering expressions. In this case, when solving the inequality, the expressions are first simplified.
For example: 5(a+1)^2+2&3/5+a > 0.6(a-2)(a+3)

Choose the desired inequality sign and enter the polynomials in the fields below.

The first inequality of the system.

Click the button to change the type of the first inequality.

> >= < <=

Solve the system of inequalities

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A bit of theory.

Systems of inequalities with one unknown. Numeric spans

You got acquainted with the concept of a system in the 7th grade and learned how to solve systems of linear equations with two unknowns. Next, systems of linear inequalities with one unknown will be considered. The solution sets of systems of inequalities can be written using intervals (intervals, half-intervals, segments, rays). You will also learn about the notation of numerical intervals.

If in the inequalities \(4x > 2000 \) and \(5x \leq 4000 \) the unknown number x is the same, then these inequalities are considered together and they are said to form a system of inequalities: $$ \left\(\begin( array)(l) 4x > 2000 \\ 5x \leq 4000 \end(array)\right.$$

The curly brace shows that you need to find such values ​​of x for which both inequalities of the system turn into true numerical inequalities. This system is an example of a system of linear inequalities with one unknown.

The solution of a system of inequalities with one unknown is the value of the unknown at which all the inequalities of the system turn into true numerical inequalities. To solve a system of inequalities means to find all solutions of this system or to establish that there are none.

The inequalities \(x \geq -2 \) and \(x \leq 3 \) can be written as a double inequality: \(-2 \leq x \leq 3 \).

The solutions of systems of inequalities with one unknown are different numerical sets. These sets have names. So, on the real axis, the set of numbers x such that \(-2 \leq x \leq 3 \) is represented by a segment with ends at points -2 and 3.

-2

3

If \(a is a segment and is denoted by [a; b]

If \(a interval and denoted by (a; b)

The sets of numbers \(x \) satisfying the inequalities \(a \leq x by half-intervals and are denoted by [a; b) and (a; b] respectively

Segments, intervals, half-intervals and rays are called numerical intervals.

Thus, numerical intervals can be specified in the form of inequalities.

A solution to an inequality with two unknowns is a pair of numbers (x; y) that turns this inequality into a true numerical inequality. To solve an inequality means to find the set of all its solutions. So, the solutions of the inequality x > y will be, for example, pairs of numbers (5; 3), (-1; -1), since \(5 \geq 3 \) and \(-1 \geq -1\)

Solving systems of inequalities

You have already learned how to solve linear inequalities with one unknown. Know what a system of inequalities and a solution to the system are. Therefore, the process of solving systems of inequalities with one unknown will not cause you any difficulties.

And yet we recall: to solve a system of inequalities, you need to solve each inequality separately, and then find the intersection of these solutions.

For example, the original system of inequalities was reduced to the form:
$$ \left\(\begin(array)(l) x \geq -2 \\ x \leq 3 \end(array)\right. $$

To solve this system of inequalities, mark the solution of each inequality on the real axis and find their intersection:

-2

3

The intersection is the segment [-2; 3] - this is the solution of the original system of inequalities.