Solve the system of inequalities with a detailed solution. Systems of inequalities

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 easy to find the vertices of the region as the intersection points of the corresponding lines

Thus, 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.

In this lesson, we will start exploring 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 the system, and recall the union and intersection of sets. At the end, we will solve specific examples for systems of linear inequalities.

Theme: The dietReal inequalities and their systems

Lesson:Mainconcepts, solution of systems of linear inequalities

Until now, we have solved individual inequalities and applied the method of intervals to them, it 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 domain 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 ox axis the set of solutions to the first and second inequalities.

The interval of intersection of two rays is our solution.

This method of depicting the solution to a system of inequalities is sometimes called the roof method.

The solution to the system is the intersection of two sets.

Let's depict this graphically. We have a set A of an arbitrary nature and a set B of an arbitrary nature, which 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.

Let us consider, using specific examples, solving linear systems of inequalities, how to find the 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 does the second inequality of the system come from? For example, from the inequality

Let us graphically designate the solutions to 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.

Sometimes a linear system is given by a double inequality; consider this case.

We examined systems of linear inequalities, understood where they come from, considered the typical systems to which all linear systems are reduced, and solved some of them.

1. Mordkovich A.G. and others. Algebra 9th grade: Textbook. For general education. Institutions. - 4th ed. - M .: Mnemosina, 2002.-192 p .: ill.

2. Mordkovich A.G. and others. Algebra 9th grade: Problem book for students of educational institutions / A. G. Mordkovich, T. N. Mishustina, etc. - 4th ed. - M .: Mnemosina, 2002.-143 p .: ill.

3. Makarychev Yu. N. Algebra. Grade 9: textbook. for general education students. institutions / Yu. N. Makarychev, NG Mindyuk, KI Neshkov, IE Feoktistov. - 7th ed., Rev. and add. - M .: Mnemosina, 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 pm, Part 2. Problem 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.

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1. Mordkovich A.G. and others. Algebra 9th grade: Problem book for students of educational institutions / A. G. Mordkovich, T. N. Mishustina, etc. - 4th ed. - M.: Mnemozina, 2002.-143 p .: ill. No. 53; 54; 56; 57.

For example:

\ (\ begin (cases) 5x + 2≥0 \\ x<2x+1\\x-4>2 \ end (cases) \)

\ (\ begin (cases) x ^ 2-55x + 250<(x-14)^2\\x^2-55x+250≥0\\x-14>0 \ end (cases) \)

\ (\ begin (cases) (x ^ 2 + 1) (x ^ 2 + 3) (x ^ 2-1) ≥0 \\ x<3\end{cases}\)

Solving a system of inequalities

To solve the system of inequalities you need to find the x values that fit all inequalities in the system - this means that they are executed simultaneously.

Example. Let's solve the system \ (\ begin (cases) x> 4 \\ x \ leq7 \ end (cases) \)
Solution: The first inequality becomes true if x is greater than \ (4 \). That is, the solutions to the first inequality are all x values from \ ((4; \ infty) \), or on the number axis:

The second inequality is suitable for x values less than 7, that is, any x from the interval \ ((- \ infty; 7] \) or on the numeric axis:

And what values are suitable for both inequalities? Those that belong to both gaps, that is, where the gaps intersect.

Answer: \((4;7]\)

As you may have noticed, it is convenient to use number axes to intersect solutions to inequalities in the system.

The general principle for solving systems of inequalities: you need to find a solution to each inequality, and then intersect these solutions using the number line.

Example:(Assignment from the OGE) Solve the system \ (\ begin (cases) 7 (3x + 2) -3 (7x + 2)> 2x \\ (x-5) (x + 8)<0\end{cases}\)

Solution:

\ (\ begin (cases) 7 (3x + 2) -3 (7x + 2)> 2x \\ (x-5) (x + 8)<0\end{cases}\)	Let's solve each inequality separately from the other.


	Let's reverse the resulting inequality.
	We divide all inequality by \ (2 \).

	Let's write down the answer for the first inequality.
\ (x∈ (-∞; 4) \)	Now let's solve the second inequality.
2) \ ((x-5) (x + 8)<0\)	Inequality is already ideal for use.
	Let's write down the answer for the second inequality.
	Let's combine both solutions using number axes.
	Let us write out in response an interval on which there is a solution to both inequalities - both the first and the second.

Answer: \((-8;4)\)

Example:(Assignment from the OGE) Solve the system \ (\ begin (cases) \ frac (10-2x) (3+ (5-2x) ^ 2) ≥0 \\ 2-7x≤14-3x \ end (cases) \)

Solution:

\ (\ begin (cases) \ frac (10-2x) (3+ (5-2x) ^ 2) ≥0 \\ 2-7x≤14-3x \ end (cases) \)	Again, we will solve the inequalities separately.
1) \ (\ frac (10-2x) (3+ (5-2x) ^ 2) \) \ (≥0 \)	If you are frightened by the denominator - do not be afraid, now we will remove it. The fact is that \ (3+ (5-2x) ^ 2 \) is always a positive expression. Judge for yourself: \ ((5-2x) ^ 2 \) because of the square is either positive or equal to zero. \ ((5-2x) ^ 2 + 3 \) - exactly positive. So you can safely multiply the inequality by \ (3+ (5-2x) ^ 2 \)
	Before us is the usual - we will express \ (x \). To do this, move \ (10 \) to the right side.
	We divide the inequality by \ (- 2 \). Since the number is negative, we change the sign of inequality.
	Let us mark the solution on the number line.
	Let's write down the answer to the first inequality.
\ (x∈ (-∞; 5] \)	At this stage, the main thing is not to forget that there is a second inequality.
2) \ (2-7x≤14-3x \)	Again linear inequality - again we express \ (x \).
\ (- 7x + 3x≤14-2 \)	We present similar terms.
	We divide all inequality by \ (- 4 \), inverting the sign.
	Draw the solution on the number axis and write out the answer for this inequality.
\ (x∈ [-3; ∞) \)	Now let's combine the solutions.
	Let's write down the answer.

Answer: \([-3;5]\)

Example: Solve the system \ (\ begin (cases) x ^ 2-55x + 250<(x-14)^2\\x^2-55x+250≥0\\x-14>0 \ end (cases) \)

Solution:

\ (\ begin (cases) x ^ 2-55x + 250<(x-14)^2\\x^2-55x+250≥0\\x-14>0 \ end (cases) \)

Solving inequalities online

Before solving inequalities, it is necessary to understand well how the equations are solved.

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

Let us explain what it means to solve inequality?

After studying the equations in the student's head, the following picture develops: you need to find such values of the variable for which both sides of the equation take on the same values. In other words, find all the points where equality holds. That's right!

When we talk about inequalities, we mean finding the 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 what will be the solution to the inequality in three variables?

How to deal with inequalities?

A universal method for solving inequalities is considered the method of intervals (aka the method of intervals), which consists in determining all the intervals within the boundaries of which a given inequality will be satisfied.

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

How to write down the solution to an inequality correctly?

When you have determined the intervals of solutions to an 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 of the equation satisfies the GDV and the inequality is not strict, then the boundary of the interval is included in the solution of the inequality. Otherwise, no.

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

An important point

Do not think that only intervals, half-intervals, and line segments can be a solution to an inequality. No, the solution may include individual points.

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

And the inequality | x |

What is the inequality calculator for?

The inequality calculator gives the correct final answer. In this case, in most cases, an illustration of a numerical axis or plane is given. It can be seen whether the boundaries of the intervals are included in the solution or not - the points are displayed filled or punctured.

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

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

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 the values of the unknown variable for which each inequality of the system is realized, or to prove that there are no such .

Hence, for each individual system inequalities calculate the unknown variable. Further, from the resulting values, it selects only those that are true for both the first and second inequalities. Therefore, when the chosen value is substituted, both inequalities of the system become correct.

Let's analyze the solution to several inequalities:

Place one pair of number lines under the other; on the top we will apply the value x for which the first inequalities about ( x> 1) becomes true, and at the bottom, the value NS, which are a solution to the second inequality ( NS> 4).

Comparing the data on numeric straight lines, note that the solution for both inequalities will NS> 4. Answer, NS> 4.

Example 2.

Calculating the first inequality we get -3 NS< -6, или x> 2, the second - NS> -8, or NS < 8. Затем делаем по аналогии с предыдущим примером. На верхнюю числовую прямую наносим все те значения NS at which the first system inequality, and on the lower number line, all those values NS, at which the second inequality of the system is realized.

Comparing the data, we find that both inequalities will be realized for all values NS placed from 2 to 8. Sets of values NS denote double inequality 2 < NS< 8.

Example 3. Find

Solving a system of inequalities

To solve the system of inequalities you need to find the x values ​​that fit all inequalities in the system - this means that they are executed simultaneously.

The general principle for solving systems of inequalities: you need to find a solution to each inequality, and then intersect these solutions using the number line.

To solve the system of inequalities you need to find the x values that fit all inequalities in the system - this means that they are executed simultaneously.