The system of inequalities It is customary to call any set of two or more inequalities containing an unknown quantity.
This formulation is clearly illustrated, for example, by such systems of inequalities:
Solve the system of inequalities - means to find all values of the unknown variable for which each inequality of the system is realized, or to prove that there are no such .
So, for each individual system inequalities calculate the unknown variable. Further, from the resulting values, selects only those that are true for both the first and second inequalities. Therefore, when substituting the chosen value, both inequalities of the system become correct.
Let's analyze the solution of several inequalities:
Place one under the other pair of number lines; put the value on the top x, under which the first inequality o ( x> 1) become true, and on the bottom, the value X, which are the solution of the second inequality ( X> 4).
By comparing the data on number lines, note that the solution for both inequalities will be X> 4. Answer, X> 4.
Example 2
Calculating the first inequality we get -3 X< -6, или x> 2, the second - X> -8, or X < 8. Затем делаем по аналогии с предыдущим примером. На верхнюю числовую прямую наносим все те значения X, under which the first system inequality, and on the lower number line, all those values X, under which the second inequality of the system is realized.
Comparing the data, we find that both inequalities will be implemented for all values X placed from 2 to 8. Sets of values X denote double inequality 2 < X< 8.
Example 3 Let's find
In this lesson, we will begin the study of systems of inequalities. First, we will consider systems of linear inequalities. At the beginning of the lesson, we will consider where and why systems of inequalities arise. Next, we will study what it means to solve a system, and remember the union and intersection of sets. In the end, we will solve specific examples for systems of linear inequalities.
Topic: dietreal inequalities and their systems
Lesson:Mainconcepts, solution of systems of linear inequalities
Until now, we have solved individual inequalities and applied the interval method to them, these could be linear inequalities, and square and rational. Now let's move on to solving systems of inequalities - first linear systems. Let's look at an example where the need to consider systems of inequalities comes from.
Find the scope of a function
Find the scope of a function
The function exists when both square roots exist, i.e.
How to solve such a system? It is necessary to find all x satisfying both the first and second inequalities.
Draw on the x-axis the set of solutions to the first and second inequalities.
The intersection interval of two rays is our solution.
This method of representing the solution of a system of inequalities is sometimes called the roof method.
The solution of the system is the intersection of two sets.
Let's represent this graphically. We have a set A of arbitrary nature and a set B of arbitrary nature that intersect.
Definition: The intersection of two sets A and B is a third set that consists of all the elements included in both A and B.
Consider, using specific examples of solving linear systems of inequalities, how to find intersections of the sets of solutions of individual inequalities included in the system.
Solve the system of inequalities:
Answer: (7; 10].
4. Solve the system 
Where can the second inequality of the system come from? For example, from the inequality
We graphically denote the solutions of each inequality and find the interval of their intersection.
Thus, if we have a system in which one of the inequalities satisfies any value of x, then it can be eliminated.
Answer: the system is inconsistent.
We have considered typical support problems, to which the solution of any linear system of inequalities is reduced.
Consider the following system.
7. 
Sometimes a linear system is given by a double inequality; consider this case.
8. 
We considered systems of linear inequalities, understood where they come from, considered typical systems to which all linear systems reduce, and solved some of them.
1. Mordkovich A.G. and others. Algebra 9th grade: Proc. For general education Institutions. - 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.
2. Mordkovich A.G. and others. Algebra Grade 9: Task book for students educational institutions/ A. G. Mordkovich, T. N. Mishustina and others - 4th ed. — M.: Mnemosyne, 2002.-143 p.: ill.
3. Yu. N. Makarychev, Algebra. Grade 9: textbook. for general education students. institutions / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, I. E. Feoktistov. - 7th ed., Rev. and additional - M .: Mnemosyne, 2008.
4. Alimov Sh.A., Kolyagin Yu.M., Sidorov Yu.V. Algebra. Grade 9 16th ed. - M., 2011. - 287 p.
5. Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 12th ed., erased. — M.: 2010. — 224 p.: ill.
6. Algebra. Grade 9 At 2 hours. Part 2. Task book for students of educational institutions / A. G. Mordkovich, L. A. Aleksandrova, T. N. Mishustina and others; Ed. A. G. Mordkovich. - 12th ed., Rev. — M.: 2010.-223 p.: ill.
1. Portal of Natural Sciences ().
2. Electronic educational and methodological complex for the preparation of grades 10-11 for entrance exams in computer science, mathematics, Russian language ().
4. Education Center "Technology of Education" ().
5. College.ru section on mathematics ().
1. Mordkovich A.G. et al. Algebra Grade 9: Taskbook for students of educational institutions / A. G. Mordkovich, T. N. Mishustina et al. - 4th ed. - M .: Mnemosyne, 2002.-143 p.: ill. No. 53; 54; 56; 57.
The program for solving linear, square and fractional inequalities does not just give the answer to the problem, it leads 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, 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 control work, parents to control the solution of inequalities by their children.
This program can be useful for high school students general education schools in preparation for tests and exams, when testing knowledge before the Unified State Examination, 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 it done as soon as possible? homework math or algebra? 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 as a decimal, but also as 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 so: 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: /
whole part 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.
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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)
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.
In this way, number gaps can be given 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.
LINEAR EQUATIONS AND INEQUALITIES I
§ 23 Systems of linear inequalities
A system of linear inequalities is any set of two or more linear inequalities containing the same unknown quantity.
Examples of such systems are:
To solve a system of inequalities means to find all values of the unknown quantity for which each inequality of the system is satisfied.
Let's solve the above systems.
Let us place two number lines one under the other (Fig. 31); on the top note those values X , under which the first inequality ( X > 1), and on the bottom - those values X , under which the second inequality is satisfied ( X > 4).
Comparing the results on the number lines, we note that both inequalities will simultaneously be satisfied for X > 4. Answer, X > 4.
The first inequality gives -3 X < -б, или X > 2, and the second - X > -8, or X < 8. Далее поступаем так же, как и в первом примере. На одной числовой прямой отмечаем все те значения X , under which the first inequality of the system is satisfied, and on the second real line, located under the first, all those values X , for which the second inequality of the system is satisfied (Fig. 32).
Comparison of these two results shows that both inequalities will simultaneously hold for all values X , concluded from 2 to 8. The set of such values X is written as a double inequality 2< X < 8.
Example 3. Solve a system of inequalities
The first inequality of the system gives 5 X < 10, или X < 2, второе X > 4. Thus, any number that satisfies both inequalities simultaneously must be no more than 2 and no more than 4 (Fig. 33).
But there are no such numbers. Therefore, this system of inequalities is not satisfied for any values X . Such systems of inequalities are called inconsistent.
Exercises
Solve these systems of inequalities (No. 179 -184):
Solve inequalities (No. 185, 186):
185. (2X + 3) (2 - 2X ) > 0. 186. (2 - π ) (2X - 15) (X + 4) > 0.
Find the valid values of the letters included in the equality data (No. 187, 188):
Solve inequalities (No. 189, 190):
189. 1 < 2X - 5 < 2. 190. -2 < 1 - Oh < 5.
191. What should be the temperature of 10 liters of water so that when it is mixed with 6 liters of water at a temperature of 15 °, water with a temperature of at least 30 ° and not more than 40 ° is obtained?
192. One side of a triangle is 4 cm, and the sum of the other two is 10 cm. Find these sides if they are expressed as whole numbers.
193. It is known that the system of two linear inequalities is not satisfied for any values of the unknown quantity. Is it possible to say that individual inequalities of this system are not satisfied for any values of the unknown quantity?
Definition 1 . set of points in space R n , whose coordinates satisfy the equation a 1 X 1 + a 2 X 2 +…+ a n x n = b, is called ( n - 1 )-dimensional hyperplane in n-dimensional space.
Theorem 1. The hyperplane divides all space into two half-spaces. The half-space is a convex set.
The intersection of a finite number of half-spaces is a convex set.
Theorem 2 . Solving a linear inequality with n unknown
a 1 X 1 + a 2 X 2 +…+ a n x n b
is one of the half-spaces into which the whole space is divided by the hyperplane
a 1 X 1 + a 2 X 2 +…+a n x n= b.
Consider a system from m linear inequalities with n unknown.
The solution of each inequality of the system is a certain half-space. The solution of the system will be the intersection of all half-spaces. This set will be closed and convex.
Solving systems of linear inequalities
with two variables
Let a system be given m linear inequalities in two variables.
The solution of each inequality will be one of the half-planes into which the entire plane is divided by the corresponding line. The solution of the system will be the intersection of these half-planes. This problem can be solved graphically on the plane X 1 0 X 2 .
37. Representation of a convex polyhedron
Definition 1. Closed convex limited set in R n having a finite number corner points, is called convex n-dimensional polyhedron.
Definition 2 . A closed convex unbounded set in R n , which has a finite number of corner points, is called a convex polyhedral region.
Definition 3 . Lots of BUTR n is called bounded if there is n-dimensional ball containing this set.
Definition 4. A convex linear combination of points is an expression where t i , .
Theorem (representation theorem for a convex polyhedron). Any point of a convex polyhedron can be represented as a convex linear combination of its corner points.
38. The area of admissible solutions of the system of equations and inequalities.
Let a system be given m linear equations and inequalities with n unknown.
Definition 1 . Dot R n is called a possible solution of the system if its coordinates satisfy the equations and inequalities of the system. The totality of all possible solutions is called the domain of possible solutions (ROA) of the system.
Definition 2. A possible solution whose coordinates are non-negative is called an admissible solution of the system. The set of all admissible solutions is called the area of admissible solutions (DDR) of the system.
Theorem 1 . ODE is a closed, convex, bounded (or unbounded) subset in R n.
Theorem 2. An admissible solution of the system is a reference if and only if this point is the corner point of the ODS.
Theorem 3 (theorem on the representation of the ODT). If the ODE is a bounded set, then any admissible solution can be represented as a convex linear combination of the corner points of the ODE (in the form of a convex linear combination of the system's support solutions).
Theorem 4 (theorem on the existence of a support solution of the system). If the system has at least one admissible solution (ODR), then among the admissible solutions there is at least one reference solution.
