The video lesson "Systems of inequalities with two variables" contains visual educational material on this topic. The lesson includes a consideration of the concept of solving a system of inequalities with two variables, examples of solving such systems in a graphical way. The task of this video lesson is to form the ability of students to solve systems of inequalities with two variables in a graphical way, to facilitate the understanding of the process of finding solutions to such systems and memorizing the solution method.
Each description of the solution is accompanied by figures that represent the solution to the problem on the coordinate plane. These figures clearly show the features of plotting and the location of the points corresponding to the solution. All important details and concepts are highlighted with color. Thus, the video lesson is a convenient tool for solving the teacher's problems in the classroom, frees the teacher from submitting a standard block of material for conducting individual work with students.
The video tutorial begins with a presentation of the topic and an example of finding solutions to a system consisting of inequalities x<=y 2 и у<х+3. Примером точки, координаты которой удовлетворяют условиям обеих неравенств, является (1;3). Отмечается, что, так как данная пара значений является решением обоих неравенств, то она является одним из множества решений. А все множество решений будет охватывать пересечение множеств, которые являются решениями каждого из неравенств. Данный вывод выделен в рамку для запоминания и указания на его важность. Далее указывается, что множество решений на координатной плоскости представляет собой множество точек, которые являются общими для множеств, представляющих решения каждого из неравенств.
Understanding of the conclusions made about the solution of the system of inequalities is consolidated by considering examples. The first solution to the system of inequalities x 2 + y 2<=9 и x+y>= 2. Obviously, solutions to the first inequality on the coordinate plane include the circle x 2 + y 2 = 9 and the region inside it. This area in the figure is filled with horizontal hatching. The set of solutions to the inequality x + y> = 2 includes the line x + y = 2 and the half-plane located above. This area is also indicated on the plane by strokes in a different direction. You can now determine the intersection of the two decision sets in the figure. It is enclosed in a segment of a circle x 2 + y 2<=9, который покрыт штриховкой полуплоскости x+y>=2.
Next, we analyze the solution of the system of linear inequalities y> = x-3 and y> = - 2x + 4. In the figure, a coordinate plane is drawn next to the task condition. A straight line is built on it, corresponding to the solutions of the equation y = x-3. The area for solving the inequality y> = x-3 will be the area located above this straight line. It is shaded. The set of solutions to the second inequality is located above the line y = -2x + 4. This line is also built on the same coordinate plane and the solution region is hatched. The intersection of two sets is an angle constructed by two straight lines, together with its inner region. The solution area of the system of inequalities is filled with double shading.
When considering the third example, the case is described when the graphs of the equations corresponding to the inequalities of the system are parallel straight lines. It is necessary to solve the system of inequalities y<=3x+1 и y>= 3x-2. A straight line is constructed on the coordinate plane corresponding to the equation y = 3x + 1. The range of values corresponding to solutions of the inequality y<=3x+1, лежит ниже данной прямой. Множество решений второго неравенства лежит выше прямой y=3x-2. При построении отмечается, что данные прямые параллельны. Область, являющаяся пересечением двух множеств решений, представляет собой полосу между данными прямыми.
The video lesson "Systems of inequalities with two variables" can be used as a visual aid in a lesson at school or replace the teacher's explanation in independent study of the material. A detailed understandable explanation of the solution of systems of inequalities on the coordinate plane can help to submit material for distance learning.
Theme: Equations and inequalities. Systems of equations and inequalities
Lesson:Equations and inequalities in two variables
Consider in general terms the equation and inequality in two variables.
Equation in two variables;
Inequality with two variables, the sign of inequality can be any;
Here x and y are variables, p is an expression that depends on them.
A pair of numbers () is called a particular solution of such an equation or inequality if, after substituting this pair into the expression, we obtain the correct equation or inequality, respectively.
The task is to find or represent on the plane the set of all solutions. You can rephrase this problem - find the locus of points (GMT), build a graph of an equation or inequality.
Example 1 - Solve Equation and Inequality:
In other words, the task involves finding the GMT.
Consider the solution to the equation. In this case, the value of the variable x can be any, in this regard, we have:
Obviously, the solution to the equation is the set of points forming the straight line
Rice. 1. Equation graph, example 1
The solutions of the given equation are, in particular, the points (-1; 0), (0; 1), (x 0, x 0 +1)
The solution to the given inequality is the half-plane located above the line, including the line itself (see Figure 1). Indeed, if we take any point x 0 on the line, then we have equality. If we take a point in a half-plane over a straight line, we have. If we take a point in the half-plane under the straight line, then it will not satisfy our inequality:.
Now let's consider the problem with a circle and a circle.
Example 2 - Solve Equation and Inequality:
We know that the given equation is the equation of a circle with a center at the origin and a radius of 1.
Rice. 2. Illustration for example 2
At an arbitrary point x 0, the equation has two solutions: (x 0; y 0) and (x 0; -y 0).
The solution to the given inequality is the set of points located inside the circle, excluding the circle itself (see Figure 2).
Consider an equation with modules.
Example 3 - Solve the equation:
In this case, it would be possible to expand the modules, but we will consider the specifics of the equation. It is easy to see that the graph of this equation is symmetrical about both axes. Then if the point (x 0; y 0) is a solution, then the point (x 0; -y 0) is also a solution, the points (-x 0; y 0) and (-x 0; -y 0) are also a solution ...
Thus, it is enough to find a solution where both variables are non-negative, and take symmetry about the axes:
Rice. 3. Illustration for example 3
So, as we can see, the solution to the equation is the square.
Let's consider the so-called method of areas with a specific example.
Example 4 - depict a set of solutions to an inequality:
According to the method of areas, first of all we consider the function on the left side if it is zero on the right. It is a function of two variables:
Similarly to the method of intervals, we temporarily move away from inequality and study the features and properties of the composed function.
ODZ: so the x-axis is gouged out.
Now we indicate that the function is equal to zero, when the numerator of the fraction is equal to zero, we have:
We build a graph of the function.
Rice. 4. Schedule of the function, taking into account the DHS
Now we will consider the areas of constant sign of the function, they are formed by a straight line and a broken line. inside the polyline there is an area D 1. Between a segment of a broken line and a straight line - an area D 2, below a straight line - an area D 3, between a segment of a broken line and a straight line - an area D 4
In each of the selected areas, the function preserves the sign, so it is enough to check an arbitrary sample point in each area.
In the area, take the point (0; 1). We have:
Take point (10; 1) in the area. We have:
So, the whole area is negative and does not satisfy the given inequality.
In the area, take the point (0; -5). We have:
So, the whole region is positive and satisfies the given inequality.
Inequality with two variablesx and y an inequality of the form is called:
(or sign)
where is some expression with these variables.
Decision two-variable inequality is an ordered pair of numbers for which this inequality turns into a true numerical inequality.
Solve inequality- means to find the set of all its solutions. A solution to an inequality with two variables is a set of points in the coordinate plane.
The main method for solving these inequalities is graphic method. It consists in the construction of border lines (if the inequality is strict, the line is drawn with a dotted line). We obtain the boundary equation if, in the given inequality, we replace the inequality sign with the equal sign. All the lines together break the coordinate plane into parts. The desired set of points, which corresponds to a given inequality or system of inequalities, can be determined by taking a control point inside each region of the region.
The set of inequalities in two variables has the form
The solution to the set is the union of all solutions to the inequalities.
Example 1. Solve system 
Solution. Let's build in the system Ooh corresponding lines (fig. 19):
The equation defines a circle centered at a point O¢ (0; 1) and R = 2.
The equation defines a parabola with apex at a point O(0; 0).
Let us find solutions to each of the inequalities included in the system. The first inequality corresponds to the area inside the circle and the circle itself (we can verify this by substituting the coordinates of any point from this area into the inequality). The second inequality corresponds to the region under the parabola.
The solution to the system is the intersection of the two indicated regions (shown in Fig. 19 by the superposition of two shadings).
Tasks
Level I
1.1. Solve graphically:
3) ; 4) 
;
5) ; 6) 
;
7) 
;
II level
2.1. Solve graphically:
1) 
2)
2.2. Find the number of integer solutions of the system:
1) 2) 3) 
2.3. Find all integer solutions of the system:
1) 
2)
3) 
2.4. Solve inequality. In the answer, indicate the number of solutions with two integer coordinates
Solving an inequality in two variables and even more so systems of inequalities in two variables seems to be a rather difficult task. 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 display 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 areas.
2. 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 region 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 depicted by 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 of this inequality, and the boundary in this case is depicted as a solid line.
Now let's look at a few tasks on this topic.
Objective 1.
What set of points is given by the inequality x · y ≤ 4?
Solution.
1) Build the graph of the equation x · y = 4. To do this, first transform it. Obviously, x in this case does not vanish, 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 splits the entire plane into two areas: the one between the two branches of the hyperbola and the one outside them.
2) Choose an arbitrary point from the first area, let it be point (4; 2).
We check the inequality: 4 · 2 ≤ 4 - wrong.
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, the boundary points, that is, the points of the graph of the function y = 4 / x, are drawn with a solid line.
Let's paint over the set of points that defines the original inequality with yellow color (fig. 1).
Objective 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 the 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.
We 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 over them with yellow color.
2) y + x> 1.
We take the point (0; 3), which lies above the graph of the function.
We check the inequality: 3 + 0> 1 - true.
Consequently, all points lying above the line y + x = 1 satisfy the second inequality of the system. Let's fill 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.
We check the inequality: 0 2 + (-4) 2 ≤ 9 - wrong.
Therefore, all points 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 fill 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).
Objective 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 the graphs of the following functions: 
x 2 + y 2 = 16 - circle,
x = -y - straight
x 2 + y 2 = 4 - circle (fig. 5).
Now let's deal with each inequality separately.
1) x 2 + y 2 ≤ 16.
Take the point (0; 0), which lies inside the circle x 2 + y 2 = 16.
We 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 fill them with red shading. 
We take the point (1; 1), which lies above the graph of the function.
We check the inequality: 1 ≥ -1 - true.
Consequently, all points lying above the line x = -y satisfy the second inequality of the system. Let's fill 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.
We check the inequality: 0 2 + 5 2 ≥ 4 - true.
Consequently, all points outside the circle x 2 + y 2 = 4 satisfy the third inequality of the system. Color them in 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 desired area is the area where all three colored areas intersect with each other (Figure 7).
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Systems of inequalities
with two variables
To the textbook by Yu.N. Makarychev
Algebra, Grade 9, Chapter III §
mathematics teacher of the highest category
MOU "Upshinskaya basic secondary school"
Orsha district of the Republic of Mari El
Solving a system of inequalities
with two variables
A solution to a system of inequalities with two variables is a pair of values of these variables, which 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
Image of the set of solutions to an inequality in two variables on the coordinate plane
The parabola splits the coordinate plane into two areas. The solution to the inequality is the domain with point A.
Image of the set of solutions of a system of inequalities in two variables on the coordinate plane
The set of solutions to a system of inequalities in two variables is the intersection of the sets of solutions to the 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 in the system.
NS = 2
- Let's build a straight line NS = 2.
- Let's build a straight line at = -3.
- It splits 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 splits the plane into two areas, select the area we need and apply shading
- Let's build a straight line at- 2x = -3
- It splits the plane into two areas, select the area we need and apply shading
The solutions of this system are the coordinates of the intersection points of the sets of solutions to the inequalities of the system (angle)
- Construct a straight line y = 2 x + 1
- It splits the plane into two areas, select the area we need and apply shading
- Construct a straight line y = 2 x - 1
- It splits the plane into two areas, select the area we need and apply shading
The solutions of this 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 NS 2 + at 2 = 1
- It splits the plane into two areas, select the area we need and apply shading
- Construct a straight line 2x + y = 0
- It splits 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 us construct a parabola y = (x - 1) 2 -2
- It splits the plane into two areas, select the area we need and apply shading
- Construct a circle (x-1) 2 + (y + 2) 2 = 1
- It splits the plane into two areas, select the area we need and apply shading
The solutions of this 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
