How to solve inequality in two variables. Inequalities with two variables. Municipal educational institution "Upshinskaya basic secondary school"

Solving an inequality in two variables, and even more so systems of inequalities with two variables, seems to be quite a difficult task. However, there is a simple algorithm that helps you easily and effortlessly solve the seemingly very complex tasks of such kind. Let's try to figure it out.

Let us 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 to such an inequality on 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 select any of the resulting areas and consider an arbitrary point in it. We check the feasibility of the original inequality for this point. If the test results in a correct numerical inequality, then we conclude that the original inequality is satisfied in the entire region to which the selected point belongs. Thus, the set of solutions to the inequality is the region to which the selected point belongs. If the check results in an incorrect numerical inequality, 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 depicted with 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 several 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 in this case does not turn to 0, since otherwise we would have 0 · y = 4, which is incorrect. This means 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) Let’s select an arbitrary point from the first region, let it be point (4; 2).
Let's check the inequality: 4 · 2 ≤ 4 – 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 us color the set of points that defines the original inequality, yellow (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.

Let's start by building graphics following functions (Fig. 2):

y = x 2 + 2 – parabola,

y + x = 1 – straight line

x 2 + y 2 = 9 – circle.

1) y > x 2 + 2.

We take the point (0; 5), which lies above the graph of the function.
Let's check the inequality: 5 > 0 2 + 2 – true.

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

2) y + x > 1.

We take the point (0; 3), which lies above the graph of the function.
Let's check the inequality: 3 + 0 > 1 – true.

Consequently, all points lying above the straight line y + x = 1 satisfy the second inequality of the system. Let's paint them with green shading.

3) x 2 + y 2 ≤ 9.

Take the point (0; -4), which lies outside the circle x 2 + y 2 = 9.
Let's check the inequality: 0 2 + (-4) 2 ≤ 9 – incorrect.

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 = 16 – circle,

x = -y – straight line

x 2 + y 2 = 4 – circle (Fig. 5).

Now let's look at each inequality separately.

1) x 2 + y 2 ≤ 16.

Take the point (0; 0), which lies inside the circle x 2 + y 2 = 16.
Let's check the inequality: 0 2 + (0) 2 ≤ 16 – true.

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

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

Consequently, all points lying above the line x = -y satisfy the second inequality of the system. Let's paint them with blue shading.

3) x 2 + y 2 ≥ 4.

Take the point (0; 5), which lies outside the circle x 2 + y 2 = 4.
Let's check the inequality: 0 2 + 5 2 ≥ 4 – true.

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

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

The search area is the area where all three colored areas intersect with each other (Figure 7).

Still have questions? Don't know how to solve a system of inequalities with two variables?
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The first lesson is free!

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The video lesson “Inequalities with two variables” is intended for teaching algebra on this topic in the 9th grade of a secondary school. Video tutorial contains description theoretical foundations solving inequalities, describes in detail the process of solving inequalities graphically, its features, and demonstrates examples of solving tasks on the topic. The purpose of this video lesson is to facilitate the understanding of the material using a visual presentation of information, to promote the formation of skills in solving problems using the studied mathematical methods.

The main tools of the video lesson are the use of animation in the presentation of graphs and theoretical information, highlighting concepts, features important for understanding and memorizing the material, color and others graphically, voice explanation of the explanation for the purpose of easier memorization of information and the formation of the ability to use mathematical language.

The video lesson begins by introducing the topic and an example demonstrating the concept of solving an inequality. To form an understanding of the meaning of the concept of a solution, the inequality 3x 2 -y is presented<10, в которое подставляется пара значений х=2 и у=6. Демонстрируется, как после подстановки данных значений неравенство становится верным. Понятие решения данного неравенства как пары значений (2;6) выведено на экран, подчеркивая его важность. Затем представляется определение рассмотренного понятия для запоминания его учениками или записи в тетрадь.

An important part of the ability to solve inequalities is the ability to depict the set of its solutions on a coordinate plane. The formation of such a skill in this lesson begins with a demonstration of finding a set of solutions to linear inequalities ax+by c. The peculiarities of defining the inequality are noted - x and y are variables, a, b, c are some numbers, among which a and b are not equal to zero.

An example of such an inequality is x+3y>6. To transform the inequality into an equivalent inequality reflecting the dependence of the values ​​of y on the values ​​of x, both sides of the inequality are divided by 3, y remains on one side of the equation, and x is moved to the other. The value x=3 is arbitrarily selected for substitution into the inequality. It is noted that if you substitute this x value into the inequality and replace the inequality sign with an equal sign, you can find the corresponding value y=1. The pair (3;1) will be a solution to the equation y=-(1/3)x+2. If we substitute any values ​​of y greater than 1, then the inequality with a given value of x will be true: (3;2), (3;8), etc. Similar to this process of finding a solution, the general case for finding a set of solutions to a given inequality is considered. The search for a set of solutions to the inequality begins with the substitution of a certain value x 0. On the right side of the inequality we get the expression -(1/3)x 0 +2. A certain pair of numbers (x 0;y 0) is a solution to the equation y=-(1/3)x+2. Accordingly, the solutions to the inequality y>-(1/3)x 0 +2 will be the corresponding pairs of values ​​with x 0, where y is greater than the values ​​of y 0. That is, the solutions to this inequality will be pairs of values ​​(x 0 ; y).

To find the set of solutions to the inequality x+3y>6 on the coordinate plane, the construction of a straight line corresponding to the equation y=-(1/3)x+2 is demonstrated on it. On this line, point M is marked with coordinates (x 0; y 0). It is noted that all points K(x 0 ;y) with ordinates y>y 0, that is, located above this line, will satisfy the conditions of inequality y>-(1/3)x+2. From the analysis it is concluded that this inequality is given by a set of points that are located above the straight line y=-(1/3)x+2. This set of points constitutes a half-plane over a given line. Since the inequality is strict, the straight line itself is not among the solutions. In the figure, this fact is marked with a dotted designation.

Summarizing the data obtained as a result of describing the solution to the inequality x+3y>6, we can say that the straight line x+3y=6 divides the plane into two half-planes, while the half-plane located above reflects the set of values ​​satisfying the inequality x+3y>6, and located below the line - solution to the inequality x+3y<6. Данный вывод является важным для понимания, каким образом решаются неравенства, поэтому выведен на экран отдельно в рамке.

Next, we consider an example of solving a non-strict inequality of the second degree y>=(x-3) 2. To determine the set of solutions, a parabola y = (x-3) 2 is constructed nearby in the figure. The point M(x 0 ; y 0) is marked on the parabola, the values ​​of which will be solutions to the equation y = (x-3) 2. At this point, a perpendicular is constructed, on which a point K(x 0 ;y) is marked above the parabola, which will be the solution to the inequality y>(x-3) 2. We can conclude that the original inequality is satisfied by the coordinates of points located on a given parabola y = (x-3) 2 and above it. In the figure, this solution area is marked by shading.

The next example demonstrating the position on the plane of points that are a solution to an inequality of the second degree is a description of the solution to the inequality x 2 + y 2<=9. На координатной плоскости строится окружность радиусом 3 с центром в начале координат. Отмечается, что решениями уравнения будут точки, сумма квадратов координат которых не превышает квадрата радиуса. Также отмечается, что окружность х 2 +у 2 =9 разбивает плоскость на области внутри окружности и вне круга. Очевидно, что множество точек внутренней части круга удовлетворяют неравенству х 2 +у 2 <9, а внешняя часть - неравенству х 2 +у 2 >9. Accordingly, the solution to the original inequality will be the set of points on the circle and the region inside it.

Next, we consider the solution to the equation xy>8. On the coordinate plane next to the task, a hyperbola is constructed that satisfies the equation xy=8. Mark the point M(x 0;y 0) belonging to the hyperbola and K(x 0;y) above it parallel to the y-axis. It is obvious that the coordinates of point K correspond to the inequality xy>8, since the product of the coordinates of this point exceeds 8. It is pointed out that in the same way one can prove the correspondence of points belonging to area B to the inequality xy<8. Следовательно, решением неравенства ху>8 there will be a set of points lying in areas A and C.

The video lesson “Inequalities with two variables” can serve as a visual aid for the teacher in the classroom. The material will also help students who are learning the material independently. It is useful to use a video lesson during distance learning.

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Slide captions:

Inequalities with two variables and their systems Lesson 1

Inequalities with two variables Inequalities 3х – 4у  0; and are inequalities with two variables x and y. The solution to an inequality in two variables is a pair of values ​​of the variables that turns it into a true numerical inequality. For x = 5 and y = 3, the inequality 3x - 4y  0 turns into the correct numerical inequality 3  0. The pair of numbers (5;3) is a solution to this inequality. The pair of numbers (3;5) is not its solution.

Is the pair of numbers (-2; 3) a solution to the inequality: No. 482 (b, c) Is not Is

The solution to the inequality is the ordered pair real numbers, which turns this inequality into a true numerical inequality. Graphically, this corresponds to specifying a point on the coordinate plane. Solving an inequality means finding many solutions to it.

Inequalities with two variables have the form: The set of solutions to an inequality is the set of all points of the coordinate plane that satisfy a given inequality.

Solution sets for the inequality F(x,y) ≥ 0 x y F(x,y)≤0 x y

F(x, y)>0 F(x, y)

Trial point rule Construct F(x ; y)=0 Taking a trial point from any area, determine whether its coordinates are a solution to the inequality Draw a conclusion about the solution to the inequality x y 1 1 2 A(1;2) F(x ; y) =0

Linear inequalities with two variables A linear inequality with two variables is called an inequality of the form ax + bx +c  0 or ax + bx +c

Find the error! No. 484 (b) -4 2 x 2 -6 y 6 -2 0 4 -2 - 4

Solve graphically the inequality: -1 -1 0 x 1 -2 y -2 2 2 1 We draw graphs with solid lines:

Let's determine the inequality sign in each of the areas -1 -1 0 x 1 -2 y -2 2 2 1 3 4 - + 1 + 2 - 7 + 6 - 5 +

The solution to the inequality is a set of points from the areas containing the plus sign and solutions to the equation -1 -1 0 x 1 -2 y -2 2 2 1 3 4 - + 1 + 2 - 7 + 6 - 5 +

Let's solve together No. 485 (b) No. 486 (b, d) No. 1. Set the inequality and draw on the coordinate plane the set of points for which: a) the abscissa is greater than the ordinate; b) the sum of the abscissa and ordinate is greater than their double difference.

Let's solve together No. 2. Define by inequality an open half-plane located above the straight line AB passing through points A(1;4) and B(3;5). Answer: y  0.5x +3.5 No. 3. For what values ​​of b does the set of solutions to the inequality 3x – b y + 7  0 represent an open half-plane located above the straight line 3x – b y + 7 = 0. Answer: b  0.

Homework P. 21, No. 483; No. 484(c,d); No. 485(a); No. 486(c).

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Slide captions:

Inequalities with two variables and their systems Lesson 2

Systems of inequalities with two variables

The solution to a system of inequalities with two variables is a pair of values ​​of variables that turns each of the inequalities of the system into a true numerical inequality. No. 1. Draw the set of solutions to systems of inequalities. No. 496 (oral)

a) x y 2 2 x y 2 2 b)

Let's solve together No. 1. At what values ​​of k does the system of inequalities define a triangle on the coordinate plane? Answer: 0

We solve together x y 2 2 2 2 No. 2. The figure shows a triangle with vertices A(0;5), B(4;0), C(1;-2), D(-4;2). Define this quadrilateral with a system of inequalities. A B C D

Let's solve together No. 3. For what k and b is the set of points of the coordinate plane defined by the system of inequalities: a) strip; b) angle; c) empty set. Answer: a) k= 2,b  3; b) k ≠ 2, b – any number; c) k = 2; b

Let's solve number 4 together. What figure is given by the equation? (orally) 1) 2) 3) No. 5. Draw on the coordinate plane the set of solutions of points specified by the inequality.

Let's solve together No. 497 (c, d), 498 (c)

Homework P.22 No. 496, No. 497 (a, b), No. 498 (a, b), No. 504.

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Slide captions:

Inequalities with two variables and their systems Lesson 3

Find the error! -4 2 x 2 -6 y 6 -2 0 4 -2 - 4

Find the error! | | | | | | | | | | | | | | | | | | 1 x y 2

Determine the inequality 0 - 6 - 1 5 3 1 2 y x - 3 - 2 1 -3 4

0 - 6 - 1 5 3 1 2 y x - 3 - 2 1 -3 4 Determine the inequality

0 - 3 - 1 5 3 1 2 y x - 3 - 2 1 Determine the inequality sign ≤

Solve graphically the system of inequalities -1 -1 0 x 1 -2 y -2 2 2 1

Inequalities and systems of inequalities of higher degrees with two variables No. 1. Draw on the coordinate plane the set of points specified by the system of inequalities

Inequalities and systems of inequalities of higher degrees with two variables No. 2. Draw on the coordinate plane the set of points specified by the system of inequalities

Inequalities and systems of inequalities of higher degrees with two variables No. 3. Draw on the coordinate plane the set of points specified by the system of inequalities. Let us transform the first inequality of the system:

Inequalities and systems of inequalities of higher degrees with two variables We obtain an equivalent system

Inequalities and systems of inequalities of higher degrees with two variables No. 4. Draw on the coordinate plane the set of points specified by the system of inequalities

Let's decide together No. 502 Collection of Galitsky. No. 9.66 b) y ≤ |3x -2| 0 - 6 - 1 5 3 1 2 y x - 3 - 2 1 -3 4

. No. 9.66(c) Solve together 0 - 6 - 1 5 3 1 2 y x - 3 - 2 1 -3 4 |y| ≥ 3x - 2

We solve together No. 9.66(g) 0 - 6 - 1 5 3 1 2 y x - 3 - 2 1 -3 4 |y|

Solve the inequality: x y -1 -1 0 1 -2 -2 2 2 1

0 - 6 - 1 5 3 1 2 y x - 3 - 2 1 -3 4 Write down the system of inequalities

11:11 3) What figure is determined by the set of solutions to the system of inequalities? Find the area of ​​each figure. 6) How many pairs natural numbers are solutions to the system of inequalities? Calculate the sum of all such numbers. Solution of training exercises 2) Write down a system of inequalities with two variables, the set of solutions of which is shown in Figure 0 2 x y 2 1) Draw the set of solutions of the system on the coordinate plane: 4) Define the ring shown in the figure as a system of inequalities. 5) Solve the system of inequalities y x 0 5 10 5 10

Solution of training exercises 7) Calculate the area of ​​the figure given by the set of solutions to the system of inequalities and find the greatest distance between the points of this figure 8) At what value of m does the system of inequalities have only one solution? 9) Indicate some values ​​of k and b at which the system of inequalities defines on the coordinate plane: a) a strip; b) angle.

This is interesting. The English mathematician Thomas Harriot (Harriot T., 1560-1621) introduced the familiar inequality sign, arguing it as follows: “If two parallel segments serve as a symbol of equality, then intersecting segments must be a symbol of inequality.” In 1585, young Harriot was sent by the Queen of England on an exploratory expedition to North America. There he saw a tattoo popular among Indians in the form. This is probably why Harriot proposed the inequality sign in two of its forms: “>” is greater than... and “

This is interesting. The symbols ≤ and ≥ for non-strict comparison were proposed by Wallis in 1670. Originally, the line was above the comparison sign, and not below it, as it is now. These symbols became widespread after the support of the French mathematician Pierre Bouguer (1734), from whom they acquired their modern form.

Solving an inequality in two variables, and even more so systems of inequalities with two variables, seems to be quite a difficult task. However, there is a simple algorithm that helps solve seemingly very complex problems of this kind easily and without much effort. Let's try to figure it out.

Let us 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 to 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 select any of the resulting areas and consider an arbitrary point in it. We check the feasibility of the original inequality for this point. If the test results in a correct numerical inequality, then we conclude that the original inequality is satisfied in the entire region to which the selected point belongs. Thus, the set of solutions to the inequality is the region to which the selected point belongs. If the check results in an incorrect numerical inequality, 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 depicted with 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 several 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 in this case does not turn to 0, since otherwise we would have 0 · y = 4, which is incorrect. This means 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) Let’s select an arbitrary point from the first region, let it be point (4; 2).
Let's check the inequality: 4 · 2 ≤ 4 – 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 paint the set of points that defines the original inequality in yellow (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.

To begin with, we build graphs of the following functions (Fig. 2):

y = x 2 + 2 – parabola,

y + x = 1 – straight line

x 2 + y 2 = 9 – circle.

1) y > x 2 + 2.

We take the point (0; 5), which lies above the graph of the function.
Let's check the inequality: 5 > 0 2 + 2 – true.

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

2) y + x > 1.

We take the point (0; 3), which lies above the graph of the function.
Let's check the inequality: 3 + 0 > 1 – true.

Consequently, all points lying above the straight line y + x = 1 satisfy the second inequality of the system. Let's paint them with green shading.

3) x 2 + y 2 ≤ 9.

Take the point (0; -4), which lies outside the circle x 2 + y 2 = 9.
Let's check the inequality: 0 2 + (-4) 2 ≤ 9 – incorrect.

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 = 16 – circle,

x = -y – straight line

x 2 + y 2 = 4 – circle (Fig. 5).

Now let's look at each inequality separately.

1) x 2 + y 2 ≤ 16.

Take the point (0; 0), which lies inside the circle x 2 + y 2 = 16.
Let's check the inequality: 0 2 + (0) 2 ≤ 16 – true.

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

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

Consequently, all points lying above the line x = -y satisfy the second inequality of the system. Let's paint them with blue shading.

3) x 2 + y 2 ≥ 4.

Take the point (0; 5), which lies outside the circle x 2 + y 2 = 4.
Let's check the inequality: 0 2 + 5 2 ≥ 4 – true.

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

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

The search area is the area where all three colored areas intersect with each other (Figure 7).

Still have questions? Don't know how to solve a system of inequalities with two variables?
To get help from a tutor, register.
The first lesson is free!

website, when copying material in full or in part, a link to the source is required.

Systems of inequalities

with two variables

To the textbook by Yu.N. Makarychev

Algebra, 9th grade, Chapter III §

mathematics teacher of the highest category

Municipal educational institution "Upshinskaya main" comprehensive school»

Orsha district of the Republic of Mari El

Solving the system of inequalities

with two variables

A solution to a system of inequalities with two variables is a pair of values ​​of these variables that is both a solution to the first inequality and the second inequality of the system.

(1; 2) – solution?

(2; 1) – solution?

(1; 2) – solution

(2; 1) is not a solution

Representation of the set of solutions to an inequality in two variables on the coordinate plane

A parabola divides the coordinate plane into two regions. The solution to the inequality is the region with point A.

Representation of the set of solutions to a system of inequalities with two variables on the coordinate plane

The set of solutions to a system of inequalities with two variables is the intersection of the sets of solutions to inequalities included in the system. On the coordinate plane, the set of solutions to a system of inequalities is depicted by a set of points that are a common part of the sets that represent solutions to each inequality of the system.

X = 2

• Let's build a straight line X = 2.

• Let's build a straight line at = -3.
• It divides the plane into two areas, select the area we need and apply shading

at = -3

The solutions of this system are the coordinates of the intersection points of the sets of solutions to the inequalities of the system (right angle)

• Let's build a straight line 2y + 3x = 6
• It divides the plane into two areas, select the area we need and apply shading

• Let's build a straight line at- 2x = -3
• It divides the plane into two areas, select the area we need and apply shading

The solutions of a given system are the coordinates of the intersection points of the sets of solutions to the system’s inequalities (angle)

• Let's construct a straight line y = 2 x + 1
• It divides the plane into two areas, select the area we need and apply shading

• Let's construct a straight line y = 2 x - 1
• It divides the plane into two areas, select the area we need and apply shading

The solutions of a given system are the coordinates of the intersection points of the sets of solutions to the inequalities of the system (strip)

• Let's build a circle X 2 + y 2 = 1
• It divides the plane into two areas, select the area we need and apply shading

• Let's construct a straight line 2x + y = 0
• It divides the plane into two areas, select the area we need and apply shading

The solutions of this system are the points of the semicircle

• Let's construct a parabola y = (x - 1) 2 -2
• It divides the plane into two areas, select the area we need and apply shading

• Let's construct a circle (x-1) 2 + (y+2) 2 =1
• It divides the plane into two areas, select the area we need and apply shading

The solutions of a given system are the intersection points of the sets of solutions to the inequalities of the system

Draw a set of points that are solutions to the system and calculate the area of ​​the resulting figure