How to specify the solution of a system of inequalities. Solving systems of linear inequalities graphically

There are only "X's" and only the abscissa axis, now "Ys" are added and the field of activity expands to the entire coordinate plane. Further in the text, the phrase "linear inequality" is understood in a two-dimensional sense, which will become clear in a matter of seconds.

In addition to analytical geometry, the material is relevant for a number of problems of mathematical analysis, economic and mathematical modeling, so I recommend that you study this lecture with all seriousness.

Linear inequalities

There are two types of linear inequalities:

1) Strict inequalities: .

2) Non-strict inequalities: .

What is the geometric meaning of these inequalities? If a linear equation defines a straight line, then a linear inequality defines half-plane.

To understand the information below, you need to know the types of lines on the plane and be able to build lines. If you have any difficulties in this part, read the help Graphs and properties of functions– a paragraph about a linear function.

Let's start with the simplest linear inequalities. The blue dream of any loser is a coordinate plane on which there is nothing at all:

As you know, the abscissa axis is given by the equation - “y” is always (for any value of “x”) equals zero

Let's consider the inequality. How to understand it informally? "Y" is always (for any value of "x") positive. It is obvious that this inequality determines the upper half-plane, since all points with positive "games" are located there.

In the event that the inequality is not strict, to the upper half-plane additionally axis is added.

Similarly: the inequality is satisfied by all points of the lower half-plane, the non-strict inequality corresponds to the lower half-plane + axis .

With the y-axis, the same prosaic story:

– the inequality defines the right half-plane;
– the inequality defines the right half-plane, including the y-axis;
– the inequality defines the left half-plane;
– the inequality defines the left half-plane, including the y-axis.

At the second step, we consider inequalities in which one of the variables is missing.

Missing "y":

Or missing "x":

These inequalities can be dealt with in two ways. please consider both approaches. Along the way, let's remember and consolidate school actions with inequalities already discussed in the lesson Function scope.

Example 1

Solve linear inequalities:

What does it mean to solve a linear inequality?

To solve a linear inequality means to find a half-plane, whose points satisfy the given inequality (plus the line itself, if the inequality is not strict). Solution, usually, graphic.

It is more convenient to immediately execute the drawing, and then comment everything out:

a) Solve the inequality

Method one

The method is very similar to the story with coordinate axes, which we discussed above. The idea is to transform the inequality - to leave one variable on the left side without any constants, in this case, the x variable.

rule: In the inequality, the terms are transferred from part to part with a sign change, while the sign of the inequality itself does not change(for example, if there was a “less than” sign, then it will remain “less”).

We transfer the "five" to the right side with a change of sign:

rule POSITIVE does not change.

Now draw a straight line (dashed blue line). The straight line is dashed because the inequality strict, and the points belonging to this line will certainly not be included in the solution.

What is the meaning of inequality? "X" is always (for any value of "y") less than . Obviously, this assertion is satisfied by all points of the left half-plane. This half-plane, in principle, can be shaded, but I will limit myself to small blue arrows so as not to turn the drawing into an artistic palette.

Method two

This is a universal way. READ VERY CAREFULLY!

First, draw a straight line. For clarity, by the way, it is advisable to represent the equation in the form .

Now choose any point of the plane, not belonging to a straight line. In most cases, the most delicious point, of course. Substitute the coordinates of this point into the inequality:

Received wrong inequality(in simple words, this cannot be), which means that the point does not satisfy the inequality .

The key rule of our task:
does not satisfy inequality, then ALL points of a given half-plane do not satisfy to this inequality.
– If any point of the half-plane (not belonging to the line) satisfies inequality, then ALL points of a given half-plane satisfy to this inequality.

You can test: any point to the right of the line will not satisfy the inequality .

What is the conclusion from the experiment with the dot? There is nowhere to go, the inequality is satisfied by all points of the other - the left half-plane (you can also check).

b) Solve the inequality

Method one

Let's transform the inequality:

rule: Both sides of the inequality can be multiplied (divided) by NEGATIVE number, while the inequality sign CHANGING to the opposite (for example, if there was a “greater than or equal to” sign, then it will become “less than or equal to”).

Multiply both sides of the inequality by:

Let's draw a straight line (red color), moreover, draw a solid line, since we have inequality non-strict, and the line certainly belongs to the solution.

After analyzing the resulting inequality, we come to the conclusion that its solution is the lower half-plane (+ the line itself).

A suitable half-plane is hatched or marked with arrows.

Method two

Let's draw a straight line. Let's choose an arbitrary point of the plane (not belonging to a straight line), for example, and substitute its coordinates into our inequality:

Received correct inequality, then the point satisfies the inequality , and in general, ALL points of the lower half-plane satisfy this inequality.

Here, with the experimental point, we “hit” the desired half-plane.

The solution to the problem is indicated by a red straight line and red arrows.

Personally, I like the first solution more, because the second one is more formal.

Example 2

Solve linear inequalities:

This is a do-it-yourself example. Try to solve the problem in two ways (by the way, this is a good way to check the solution). In the answer at the end of the lesson there will be only the final drawing.

I think that after all the actions done in the examples, you will have to marry them, it will not be difficult to solve the simplest inequality, like, etc.

We turn to the consideration of the third, general case, when both variables are present in the inequality:

Alternatively, the free term "ce" may be zero.

Example 3

Find half-planes corresponding to the following inequalities:

Solution: This uses the universal point substitution method.

a) Let's construct the equation of a straight line, while the line should be drawn with a dotted line, since the inequality is strict and the straight line itself will not be included in the solution.

We select an experimental point of the plane that does not belong to the given line, for example, and substitute its coordinates into our inequality:

Received wrong inequality, so the point and ALL points of this half-plane do not satisfy the inequality . The solution to the inequality will be another half-plane, we admire the blue lightning:

b) Let's solve the inequality. Let's draw a straight line first. This is easy to do, we have a canonical direct proportionality. The line is drawn solid, since the inequality is not strict.

We choose an arbitrary point of the plane that does not belong to the line. I would like to use the origin again, but, alas, now it is not suitable. Therefore, you will have to work with another girlfriend. It is more profitable to take a point with small coordinate values, for example, . Substitute its coordinates into our inequality:

Received correct inequality, so the point and all points of the given half-plane satisfy the inequality . The desired half-plane is marked with red arrows. In addition, the solution includes the line itself.

Example 4

Find half-planes corresponding to the inequalities:

This is a do-it-yourself example. A complete solution, a rough sample of finishing and an answer at the end of the lesson.

Let's look at the reverse problem:

Example 5

a) Given a straight line. Define the half-plane in which the point is located, while the line itself must be included in the solution.

b) Given a straight line. Define the half-plane in which the point is located. The line itself is not included in the solution.

Solution: there is no need for a drawing here and the solution will be analytical. Nothing hard:

a) Compose an auxiliary polynomial and calculate its value at the point :
. Thus, the desired inequality will be with the "less than" sign. By condition, the line is included in the solution, so the inequality will not be strict:

b) Compose the polynomial and calculate its value at the point :
. Thus, the desired inequality will be with a "greater than" sign. By condition, the line is not included in the solution, therefore, the inequality will be strict: .

Answer:

Creative example for self-study:

Example 6

Given points and a line. Among the listed points, find those that, together with the origin, lie on the same side of the given line.

A little hint: first you need to write an inequality that defines the half-plane in which the origin is located. Analytical solution and answer at the end of the lesson.

Systems of linear inequalities

A system of linear inequalities is, as you understand, a system composed of several inequalities. Lol, well, I gave out the definition =) A hedgehog is a hedgehog, a knife is a knife. But the truth is - it turned out simply and affordable! No, seriously, I don’t want to give some examples in a general way, so let’s immediately move on to pressing issues:

What does it mean to solve a system of linear inequalities?

Solve a system of linear inequalities- this means find the set of points in the plane that satisfy to each system inequality.

As the simplest examples, consider systems of inequalities that determine the coordinate quarters of a rectangular coordinate system (“the drawing of twos” is at the very beginning of the lesson):

The system of inequalities defines the first coordinate quarter (upper right). Coordinates of any point of the first quarter, for example, etc. satisfy to each inequality of this system.

Similarly:
– the system of inequalities defines the second coordinate quarter (upper left);
– the system of inequalities defines the third coordinate quarter (lower left);
– the system of inequalities defines the fourth coordinate quarter (lower right).

A system of linear inequalities may not have solutions, that is, to be incompatible. Again, the simplest example: . It is quite obvious that "x" cannot be more than three and less than two at the same time.

The solution to the system of inequalities can be a straight line, for example: . Swan, crayfish, without pike, pulling the cart in two different directions. Yes, things are still there - the solution to this system is a straight line.

But the most common case, when the solution of the system is some plane area. Decision area may be unlimited(for example, coordinate quarters) or limited. The restricted domain of solutions is called polygon solution system.

Example 7

Solve a system of linear inequalities

In practice, in most cases, you have to deal with non-strict inequalities, so they will dance the rest of the lesson.

Solution: the fact that there are too many inequalities should not be scary. How many inequalities can there be in a system? Yes, as much as you want. The main thing is to adhere to a rational algorithm for constructing the solution area:

1) First, we deal with the simplest inequalities. The inequalities define the first coordinate quarter, including the boundary of the coordinate axes. Already much easier, since the search area has narrowed significantly. In the drawing, we immediately mark the corresponding half-planes with arrows (red and blue arrows)

2) The second simplest inequality - there is no “y” here. Firstly, we build the line itself, and, secondly, after transforming the inequality to the form, it immediately becomes clear that all "x"s are less than 6. We mark the corresponding half-plane with green arrows. Well, the search area has become even smaller - such a rectangle that is not limited from above.

3) At the last step, we solve the “with full ammunition” inequalities: . We discussed the solution algorithm in detail in the previous section. In short: first we build a straight line, then with the help of an experimental point we find the half-plane we need.

Stand up, children, stand in a circle:


The solution area of ​​the system is a polygon, in the drawing it is circled with a crimson line and shaded. I overdid it a little =) In the notebook, it is enough to either shade the area of ​​\u200b\u200bsolutions, or outline it more boldly with a simple pencil.

Any point of this polygon satisfies EVERY inequality of the system (for interest, you can check).

Answer: the solution of the system is a polygon.

When making a clean copy, it would be nice to describe in detail at what points you built straight lines (see lesson Graphs and properties of functions), and how the half-planes were determined (see the first paragraph of this lesson). However, in practice, in most cases, you will be credited with just the correct drawing. The calculations themselves can be carried out on a draft or even verbally.

In addition to the solution polygon of the system, in practice, albeit less frequently, there is an open area. Try to parse the following example yourself. Although, for the sake of accuracy, there is no torture here - the construction algorithm is the same, it’s just that the area will turn out to be not limited.

Example 8

Solve the system

Solution and answer at the end of the lesson. You will most likely have other letter designations for the vertices of the resulting area. This is not important, the main thing is to find the vertices correctly and build the area correctly.

It is not uncommon when in tasks it is required not only to construct the domain of solutions of the system, but also to find the coordinates of the vertices of the domain. In the two previous examples, the coordinates of these points were obvious, but in practice everything is far from ice:

Example 9

Solve the system and find the coordinates of the vertices of the resulting area

Solution: we will depict the area of ​​solutions of this system in the drawing. The inequality sets the left half-plane with the y-axis, and there is no more freebies here. After calculations on a clean / draft or deep thought processes, we get the following decision area:

In the article we will consider solution of inequalities. Let's talk plainly about how to build a solution to inequalities with clear examples!

Before considering the solution of inequalities with examples, let's deal with the basic concepts.

Introduction to inequalities

inequality is called an expression in which functions are connected by relation signs >, . Inequalities can be both numerical and alphabetic.
Inequalities with two relation signs are called double, with three - triple, etc. For example:
a(x) > b(x),
a(x) a(x) b(x),
a(x) b(x).
a(x) Inequalities containing the sign > or or are not strict.
Inequality solution is any value of the variable for which this inequality is true.
"Solve the inequality" means that you need to find the set of all its solutions. There are various methods for solving inequalities. For inequality solutions use a number line that is infinite. For example, solving the inequality x > 3 is an interval from 3 to +, and the number 3 is not included in this interval, so the point on the line is denoted by an empty circle, because the inequality is strict.
+
The answer will be: x (3; +).
The value x=3 is not included in the set of solutions, so the parenthesis is round. The infinity sign is always enclosed in a parenthesis. The sign means "belonging".
Consider how to solve inequalities using another example with the sign:
x2
-+
The value x=2 is included in the set of solutions, so the square bracket and the point on the line is denoted by a filled circle.
The answer will be: x .

Let's summarize what we've learned.
Suppose we need to solve a system of inequalities: $\begin(cases)f_1 (x)>f_2 (x)\\g_1 (x)>g_2 (x)\end(cases)$.
Then, the interval ($x_1; x_2$) is the solution to the first inequality.
The interval ($y_1; y_2$) is the solution to the second inequality.
The solution of a system of inequalities is the intersection of the solutions of each inequality.

Systems of inequalities can consist of inequalities not only of the first order, but also of any other types of inequalities.

Important rules for solving systems of inequalities.
If one of the inequalities of the system has no solutions, then the whole system has no solutions.
If one of the inequalities is satisfied for any values ​​of the variable, then the solution of the system will be the solution of the other inequality.

Examples.
Solve the system of inequalities:$\begin(cases)x^2-16>0\\x^2-8x+12≤0 \end(cases)$
Solution.
Let's solve each inequality separately.
$x^2-16>0$.
$(x-4)(x+4)>0$.

Let's solve the second inequality.
$x^2-8x+12≤0$.
$(x-6)(x-2)≤0$.

The solution to the inequality is a gap.
Let's draw both intervals on one straight line and find the intersection.
The intersection of the intervals is the segment (4; 6].
Answer: (4;6].

Solve the system of inequalities.
a) $\begin(cases)3x+3>6\\2x^2+4x+4 b) $\begin(cases)3x+3>6\\2x^2+4x+4>0\end(cases )$.

Solution.
a) The first inequality has a solution x>1.
Let's find the discriminant for the second inequality.
$D=16-4 * 2 * 4=-16$. $D Recall the rule, when one of the inequalities has no solutions, then the whole system has no solutions.
Answer: There are no solutions.

B) The first inequality has a solution x>1.
The second inequality is greater than zero for all x. Then the solution of the system coincides with the solution of the first inequality.
Answer: x>1.

Problems on systems of inequalities for independent solution

Solve systems of inequalities:
a) $\begin(cases)4x-5>11\\2x-12 b) $\begin(cases)-3x+1>5\\3x-11 c) $\begin(cases)x^2-25 d) $\begin(cases)x^2-16x+55>0\\x^2-17x+60≥0 \end(cases)$
e) $\begin(cases)x^2+36

Solving an Inequality with Two Variables, and even more so systems of inequalities with two variables, appears to be quite a challenge. However, there is a simple algorithm that helps to easily and effortlessly solve seemingly very complex problems of this kind. Let's try to figure it out.

Suppose we have an inequality with two variables of one of the following types:

y > f(x); y ≥ f(x); y< f(x); y ≤ f(x).

To depict the set of solutions of such an inequality on the coordinate plane, proceed as follows:

1. We build a graph of the function y = f(x), which divides the plane into two regions.

2. We choose any of the obtained areas and consider an arbitrary point in it. We check the satisfiability of the original inequality for this point. If, as a result of the check, a correct numerical inequality is obtained, then we conclude that the original inequality is satisfied in the entire area to which the selected point belongs. Thus, the set of solutions to the inequality is the area to which the selected point belongs. If as a result of the check an incorrect numerical inequality is obtained, then the set of solutions to the inequality will be the second region, to which the selected point does not belong.

3. If the inequality is strict, then the boundaries of the region, that is, the points of the graph of the function y = f(x), are not included in the set of solutions and the boundary is shown as a dotted line. If the inequality is not strict, then the boundaries of the region, that is, the points of the graph of the function y = f(x), are included in the set of solutions to this inequality, and the boundary in this case is depicted as a solid line.
Now let's look at a few problems on this topic.

Task 1.

What set of points is given by the inequality x · y ≤ 4?

Solution.

1) We build a graph of the equation x · y = 4. To do this, we first transform it. Obviously, x does not turn to 0 in this case, since otherwise we would have 0 · y = 4, which is not true. So we can divide our equation by x. We get: y = 4/x. The graph of this function is a hyperbola. It divides the entire plane into two regions: the one between the two branches of the hyperbola and the one outside them.

2) We choose an arbitrary point from the first region, let it be the point (4; 2).
Checking the inequality: 4 2 ≤ 4 is false.

This means that the points of this region do not satisfy the original inequality. Then we can conclude that the set of solutions to the inequality will be the second region, to which the selected point does not belong.

3) Since the inequality is not strict, we draw the boundary points, that is, the points of the graph of the function y = 4/x, with a solid line.

Let's color the set of points that defines the original inequality with yellow color (Fig. 1).

Task 2.

Draw the area defined on the coordinate plane by the system
( y > x 2 + 2;
(y + x > 1;
( x 2 + y 2 ≤ 9.

Solution.

We build graphs of the following functions to begin with (Fig. 2):

y \u003d x 2 + 2 - parabola,

y + x = 1 - straight line

x 2 + y 2 \u003d 9 is a circle.

1) y > x 2 + 2.

We take the point (0; 5), which lies above the graph of the function.
Checking the inequality: 5 > 0 2 + 2 is true.

Therefore, all points lying above the given parabola y = x 2 + 2 satisfy the first inequality of the system. Let's color them yellow.

2) y + x > 1.

We take the point (0; 3), which lies above the graph of the function.
Checking the inequality: 3 + 0 > 1 is true.

Therefore, all points lying above the line y + x = 1 satisfy the second inequality of the system. Let's color them in green.

3) x2 + y2 ≤ 9.

We take a point (0; -4), which lies outside the circle x 2 + y 2 = 9.
Checking the inequality: 0 2 + (-4) 2 ≤ 9 is wrong.

Therefore, all points lying outside the circle x 2 + y 2 = 9, do not satisfy the third inequality of the system. Then we can conclude that all points lying inside the circle x 2 + y 2 = 9 satisfy the third inequality of the system. Let's paint them with purple shading.

Do not forget that if the inequality is strict, then the corresponding boundary line should be drawn with a dotted line. We get the following picture (Fig. 3).

(Fig. 4).

Task 3.

Draw the area defined on the coordinate plane by the system:
(x 2 + y 2 ≤ 16;
(x ≥ -y;
(x 2 + y 2 ≥ 4.

Solution.

To begin with, we build graphs of the following functions:

x 2 + y 2 \u003d 16 - circle,

x \u003d -y - straight

x 2 + y 2 \u003d 4 - circle (Fig. 5).

Now we deal with each inequality separately.

1) x2 + y2 ≤ 16.

We take the point (0; 0), which lies inside the circle x 2 + y 2 = 16.
Checking the inequality: 0 2 + (0) 2 ≤ 16 is correct.

Therefore, all points lying inside the circle x 2 + y 2 = 16 satisfy the first inequality of the system.
Let's color them in red.

We take the point (1; 1), which lies above the graph of the function.
We check the inequality: 1 ≥ -1 - true.

Therefore, all points lying above the line x = -y satisfy the second inequality of the system. Let's color them in blue.

3) x2 + y2 ≥ 4.

We take the point (0; 5), which lies outside the circle x 2 + y 2 = 4.
We check the inequality: 0 2 + 5 2 ≥ 4 is correct.

Therefore, all points outside the circle x 2 + y 2 = 4 satisfy the third inequality of the system. Let's color them blue.

In this problem, all inequalities are not strict, which means that we draw all the boundaries with a solid line. We get the following picture (Fig. 6).

The area of ​​interest is the area where all three colored areas intersect each other. (fig 7).

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The system of inequalities.
Example 1. Find the scope of an expression
Solution. There must be a non-negative number under the square root sign, which means that two inequalities must simultaneously hold: In such cases, the problem is said to be reduced to solving the system of inequalities

But we have not met with such a mathematical model (system of inequalities) yet. This means that we are not yet able to complete the solution of the example.

The inequalities that form a system are combined with a curly bracket (the same is the case in systems of equations). For example, the entry

means that the inequalities 2x - 1 > 3 and 3x - 2< 11 образуют систему неравенств.

Sometimes the system of inequalities is written as a double inequality. For example, the system of inequalities

can be written as a double inequality 3<2х-1<11.

In the 9th grade algebra course, we will only consider systems of two inequalities.

Consider the system of inequalities

You can pick up several of its particular solutions, for example x = 3, x = 4, x = 3.5. Indeed, for x = 3 the first inequality takes the form 5 > 3, and the second - the form 7< 11. Получились два верных числовых неравенства, значит, х = 3 - решение системы неравенств. Точно так же можно убедиться в том, что х = 4, х = 3,5 - решения системы неравенств.

At the same time, the value x = 5 is not a solution to the system of inequalities. For x = 5, the first inequality takes the form 9 > 3 - the correct numerical inequality, and the second - the form 13< 11- неверное числовое неравенство .
To solve a system of inequalities means to find all its particular solutions. It is clear that such guessing as demonstrated above is not a method for solving a system of inequalities. In the following example, we will show how one usually argues when solving a system of inequalities.

Example 3 Solve the system of inequalities:

Solution.

but) Solving the first inequality of the system, we find 2x > 4, x > 2; solving the second inequality of the system, we find Zx< 13 Отметим эти промежутки на одной координатной прямой , использовав для выделения первого промежутка верхнюю штриховку, а для второго - нижнюю штриховку (рис. 22). Решением системы неравенств будет пересечение решений неравенств системы, т.е. промежуток, на котором обе штриховки совпали. В рассматриваемом примере получаем интервал
b) Solving the first inequality of the system, we find x > 2; solving the second inequality of the system, we find We mark these gaps on one coordinate line, using the top hatching for the first gap, and the bottom hatching for the second (Fig. 23). The solution of the system of inequalities will be the intersection of the solutions of the inequalities of the system, i.e. the interval where both hatches coincide. In the example under consideration, we get a beam

in) Solving the first inequality of the system, we find x< 2; решая второе неравенство системы, находим Отметим эти промежутки на одной координатной прямой, использовав для первого промежутка верхнюю штриховку, а для второго - нижнюю штриховку (рис. 24). Решением системы неравенств будет пересечение решений неравенств системы, т.е. промежуток, на котором обе штриховки совпали. Здесь такого промежутка нет, значит, система неравенств не имеет решений.

Let us generalize the reasoning carried out in the considered example. Suppose we need to solve a system of inequalities

Let, for example, the interval (a, b) be a solution to the inequality fx 2 > g (x), and the interval (c, d) be the solution to the inequality f 2 (x) > s 2 (x). We mark these gaps on one coordinate line, using the top hatching for the first gap, and the bottom hatching for the second (Fig. 25). The solution of the system of inequalities is the intersection of the solutions of the inequalities of the system, i.e. the interval where both hatches coincide. On fig. 25 is the interval (s, b).

Now we can easily solve the system of inequalities that we got above, in example 1:

Solving the first inequality of the system, we find x > 2; solving the second inequality of the system, we find x< 8. Отметим эти промежутки (лучи) на одной координатной прямой, использовав для первого -верхнюю, а для второго - нижнюю штриховку (рис. 26). Решением системы неравенств будет пересечение решений неравенств системы, т.е. промежуток, на котором обе штриховки совпали, - отрезок . Это - область определения того выражения, о котором шла речь в примере 1.

Of course, the system of inequalities does not have to consist of linear inequalities, as has been the case so far; any rational (and not only rational) inequalities can occur. Technically, working with a system of rational non-linear inequalities is, of course, more difficult, but there is nothing fundamentally new (compared to systems of linear inequalities).

Example 4 Solve the system of inequalities

Solution.

1) Solve the inequality We have
Note the points -3 and 3 on the number line (Fig. 27). They divide the line into three intervals, and on each interval the expression p (x) = (x - 3) (x + 3) retains a constant sign - these signs are indicated in Fig. 27. We are interested in the intervals where the inequality p(x) > 0 is satisfied (they are shaded in Fig. 27), and the points where the equality p(x) = 0 is satisfied, i.e. points x \u003d -3, x \u003d 3 (they are marked in Fig. 2 7 with dark circles). Thus, in fig. 27 shows a geometric model for solving the first inequality.

2) Solve the inequality We have
Note the points 0 and 5 on the number line (Fig. 28). They divide the line into three intervals, and on each interval the expression<7(х) = х(5 - х) сохраняет постоянный знак - эти знаки указаны на рис. 28. Нас интересуют промежутки, на которых выполняется неравенство g(х) >O (shaded in Fig. 28), and the points at which the equality g (x) - O is satisfied, i.e. points x = 0, x = 5 (they are marked in Fig. 28 by dark circles). Thus, in fig. 28 shows a geometric model for solving the second inequality of the system.

3) We mark the solutions found for the first and second inequalities of the system on the same coordinate line, using the upper hatching for the solutions of the first inequality, and the lower hatching for the solutions of the second (Fig. 29). The solution of the system of inequalities will be the intersection of the solutions of the inequalities of the system, i.e. the interval where both hatches coincide. Such an interval is a segment.

Example 5 Solve the system of inequalities:

Solution:

but) From the first inequality we find x >2. Consider the second inequality. The square trinomial x 2 + x + 2 has no real roots, and its leading coefficient (the coefficient at x 2) is positive. This means that for all x the inequality x 2 + x + 2>0 is satisfied, and therefore the second inequality of the system has no solutions. What does this mean for the system of inequalities? This means that the system has no solutions.

b) From the first inequality we find x > 2, and the second inequality holds for any values ​​of x. What does this mean for the system of inequalities? This means that its solution has the form x>2, i.e. coincides with the solution of the first inequality.

Answer:

a) there are no decisions; b) x>2.

This example is an illustration for the following useful

1. If in a system of several inequalities with one variable one inequality has no solutions, then the system has no solutions.

2. If in a system of two inequalities with one variable one inequality is satisfied for any values ​​of the variable , then the solution of the system is the solution of the second inequality of the system.

Concluding this section, let us return to the problem of the conceived number given at the beginning of it and solve it, as they say, according to all the rules.

Example 2(see p. 29). Think of a natural number. It is known that if 13 is added to the square of the conceived number, then the sum will be greater than the product of the conceived number and the number 14. If 45 is added to the square of the conceived number, then the sum will be less than the product of the conceived number and the number 18. What number is conceived?

Solution.

First step. Drawing up a mathematical model.
The intended number x, as we saw above, must satisfy the system of inequalities

Second phase. Working with the compiled mathematical model. Let's transform the first inequality of the system to the form
x2- 14x+ 13 > 0.

Let's find the roots of the trinomial x 2 - 14x + 13: x 2 \u003d 1, x 2 \u003d 13. Using the parabola y \u003d x 2 - 14x + 13 (Fig. 30), we conclude that the inequality of interest to us is satisfied for x< 1 или x > 13.

Let us transform the second inequality of the system to the form x2 - 18 2 + 45< 0. Найдем корни трехчлена х 2 - 18x + 45: = 3, х 2 = 15.