Lesson “Systems of inequalities with two variables. Inequalities with two variables and their systems Examples of solving inequalities with two variables

It is often necessary to depict on the coordinate plane the set of solutions to an inequality with two variables. A solution to an inequality with two variables is a pair of values ​​of these variables that turns the given inequality into a true numerical inequality.

2y+ Zx< 6.

Let's draw a straight line first. To do this, we write the inequality as an equation 2y+ Zx = 6 and express y. Thus, we get: y=(6-3x)/2.

This line divides the set of all points of the coordinate plane into points above it and points below it.

Take a meme from each area checkpoint, for example A (1; 1) and B (1; 3)

The coordinates of point A satisfy the given inequality 2y + 3x< 6, т. е. 2 . 1 + 3 . 1 < 6.

Point B coordinates Not satisfy this inequality 2∙3 + 3∙1< 6.

Since this inequality can change the sign on the line 2y + Zx = 6, then the inequality satisfies the set of points of the area where the point A is located. Let's shade this area.

Thus, we have depicted the set of solutions to the inequality 2y + Zx< 6.

Example

We depict the set of solutions to the inequality x 2 + 2x + y 2 - 4y + 1 > 0 on the coordinate plane.

First, we construct a graph of the equation x 2 + 2x + y 2 - 4y + 1 \u003d 0. We divide the circle equation in this equation: (x 2 + 2x + 1) + (y 2 - 4y + 4) \u003d 4, or (x + 1) 2 + (y - 2) 2 = 2 2 .

This is the equation of a circle centered at point 0 (-1; 2) and radius R = 2. Let's construct this circle.

Since this inequality is strict and the points lying on the circle itself do not satisfy the inequality, we construct the circle with a dotted line.

It is easy to check that the coordinates of the center O of the circle do not satisfy this inequality. The expression x 2 + 2x + y 2 - 4y + 1 changes its sign on the constructed circle. Then the inequality is satisfied by points located outside the circle. These points are shaded.

Example

Let us depict on the coordinate plane the set of solutions of the inequality

(y - x 2) (y - x - 3)< 0.

First, we construct a graph of the equation (y - x 2) (y - x - 3) \u003d 0. It is a parabola y \u003d x 2 and a straight line y \u003d x + 3. We build these lines and note that the change in the sign of the expression (y - x 2) (y - x - 3) occurs only on these lines. For the point A (0; 5), we determine the sign of this expression: (5-3) > 0 (i.e., this inequality is not satisfied). Now it is easy to mark the set of points for which this inequality is satisfied (these areas are shaded).

Algorithm for Solving Inequalities with Two Variables

1. We reduce the inequality to the form f (x; y)< 0 (f (х; у) >0; f (x; y) ≤ 0; f (x; y) ≥ 0;)

2. We write the equality f (x; y) = 0

3. Recognize the graphs recorded on the left side.

4. We build these graphs. If the inequality is strict (f (x; y)< 0 или f (х; у) >0), then - with strokes, if the inequality is not strict (f (x; y) ≤ 0 or f (x; y) ≥ 0), then - with a solid line.

5. Determine how many parts of the graphics are divided into the coordinate plane

6. Select a control point in one of these parts. Determine the sign of the expression f (x; y)

7. We arrange signs in other parts of the plane, taking into account the alternation (as by the method of intervals)

8. We select the parts we need in accordance with the sign of the inequality that we are solving, and apply hatching

Let f(x,y) And g(x, y)- two expressions with variables X And at and domain of definition X. Then inequalities of the form f(x, y) > g(x, y) or f(x, y) < g(x, y) called inequality with two variables .

Meaning of variables x, y from many X, under which the inequality turns into a true numerical inequality, is called its decision and denoted (x, y). Solve the inequality is to find a set of such pairs.

If each pair of numbers (x, y) from the set of solutions to the inequality, put in correspondence a point M(x, y), we obtain the set of points on the plane given by this inequality. He is called graph of this inequality . An inequality plot is usually an area on a plane.

To depict the set of solutions to the inequality f(x, y) > g(x, y), proceed as follows. First, replace the inequality sign with an equals sign and find a line that has the equation f(x,y) = g(x,y). This line divides the plane into several parts. After that, it suffices to take one point in each part and check whether the inequality holds at this point f(x, y) > g(x, y). If it is executed at this point, then it will also be executed in the entire part where this point lies. Combining such parts, we obtain a set of solutions.

Task. y > x.

Solution. First, we replace the inequality sign with an equals sign and construct a line in a rectangular coordinate system that has the equation y = x.

This line divides the plane into two parts. After that, we take one point in each part and check whether the inequality holds at this point y > x.

Task. Solve graphically inequality
X 2 + at 2 £25.

Rice. 18.

Solution. First, replace the inequality sign with an equals sign and draw a line X 2 + at 2 = 25. This is a circle with a center at the origin and a radius of 5. The resulting circle divides the plane into two parts. Checking the validity of the inequality X 2 + at 2 £ 25 in each part, we get that the graph is the set of points of the circle and part of the plane inside the circle.

Let two inequalities be given f 1(x, y) > g 1(x, y) And f 2(x, y) > g 2(x, y).

Systems of sets of inequalities with two variables

System of inequalities is yourself conjunction of these inequalities. System solution is any value (x, y), which turns each of the inequalities into a true numerical inequality. Lots of solutions systems inequalities is the intersection of the solution sets of inequalities that form the given system.

Set of inequalities is yourself disjunction of these inequalities. Set decision is any value (x, y), which turns into a true numerical inequality at least one of the inequalities in the set. Lots of solutions aggregates is the union of sets of solutions to inequalities forming a set.

Task. Solve graphically a system of inequalities

Solution. y = x And X 2 + at 2 = 25. We solve each inequality of the system.

The graph of the system will be a set of points in the plane, which are the intersection (double shading) of the solution sets of the first and second inequalities.

Task. Solve graphically a set of inequalities

Solution. First, we replace the inequality sign with an equals sign and draw lines in the same coordinate system y = x+ 4 and X 2 + at 2 = 16. Solve each population inequality. The aggregate graph will be a set of points in the plane, which are the union of the sets of solutions of the first and second inequalities.

Exercises for independent work

1. Solve graphically inequalities: a) at> 2x; b) at< 2x + 3;

V) x 2+y 2 > 9; G) x 2+y 2 £4.

2. Solve graphically systems of inequalities:

a) c)

, and even more so systems of inequalities with two variablesseems quite a 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 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 \u003d 4 / x, with a solid line.

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

Task 2.

Draw the area defined on the coordinate plane by the system

Solution.

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

y \u003d x 2 + 2 - parabola,

y + x = 1 - straight line

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

Now we deal with each inequality separately.

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 correct.

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. Check the inequality: 0 2 + (-4) 2 ≤ 9 is incorrect.

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 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).

The desired area is the area where all three colored areas intersect with each other (Fig. 4).

Questions for abstracts

Write an inequality whose solution is a circle and points inside the circle:

Find the points that are the solution of the inequality:
1) (6;10)
2) (-12;0)
3) (8;9)
4) (9;7)
5) (-12;12)

1. Inequalities with two variables. Methods for solving a system of two inequalities with two variables: an analytical method and a graphical method.

2. Systems of two inequalities with two variables: recording the result of the solution.

3. Sets of inequalities with two variables.

INEQUALITIES AND SYSTEMS OF INEQUALITIES WITH TWO VARIABLES. Predicate of the form f₁(x, y)>< f 2 (х, у), хÎХ, уÎ У, где f₁(х, у) и f 2 (х, у) - expressions with variables x and y defined on the set XxY is called inequality with two variables (with two unknowns) x and y. It is clear that any two-variable inequality can be written as f(x, y) > 0, хОХ, уО U. Inequality solution with two variables is a pair of values ​​of variables that turns the inequality into a true numerical inequality. It is known that a pair of real numbers (x, y) uniquely defines a point in the coordinate plane. This makes it possible to depict the solutions of an inequality or a system of inequalities with two variables geometrically, in the form of a certain set of points on the coordinate plane. If equation.

f(x, y)= 0 defines some line on the coordinate plane, then the set of points of the plane that do not lie on this line consists of a finite number of regions С₁, C 2 ,..., C p(Fig. 17.8). In each of the regions C, the function f(x, y) is different from zero, because the points where f(x, y)= 0 belong to the boundaries of these regions.

Solution. Let us transform the inequality to the form x > y 2 + 2y - 3. Construct a parabola on the coordinate plane X= y 2 + 2y - 3. It will split the plane into two regions G₁ and G 2 (Fig. 17.9). Since the abscissa of any point lying to the right of the parabola X= y 2 + 2y- 3, greater than the abscissa of a point that has the same ordinate but lies on a parabola, etc. inequality x>y z + 2y -3 is not strict, then the geometric representation of the solutions of this inequality will be the set of points of the plane lying on the parabola X= at 2+ 2y - 3 and to the right of it (Fig. 17.9).

Rice. 17.9

Rice. 17.10

Example 17.15. Draw on the coordinate plane the set of solutions to the system of inequalities

y > 0,

xy > 5,

x + y<6.

Solution. Geometric representation of the solution of the system of inequalities x > 0, y > 0 is the set of points of the first coordinate angle. Geometric representation of the solutions of the inequality x + y< 6 or at< 6 - X is the set of points below the line and on the line itself, which serves as the graph of the function y= 6 - X. Geometric representation of the solutions of the inequality xy > 5 or because X> 0 inequalities y > 5/x is the set of points lying above the branch of the hyperbola serving as the graph of the function y = 5/x. As a result, we obtain a set of points of the coordinate plane lying in the first coordinate angle below the straight line serving as the graph of the function y \u003d 6 - x, and above the branch of the hyperbola serving as the graph of the function y = 5x(Fig. 17.10).

Chapter III. NATURAL NUMBERS AND ZERO

Subject: Equations and inequalities. Systems of equations and inequalities

Lesson:Equations and inequalities with two variables

Consider in general terms an equation and an inequality with two variables.

An equation with two variables;

Inequality with two variables, the sign of the 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 to such an equation or inequality if, when substituting this pair into the expression, we obtain the correct equation or inequality, respectively.

The problem is to find or represent on the plane the set of all solutions. You can rephrase this problem - find the locus of points (GMT), plot an equation or inequality.

Example 1 - solve equation and inequality:

In other words, the task involves finding the GMT.

Consider the solution of the equation. In this case, the value of the variable x can be any, in connection with this we have:

Obviously, the solution to the equation is the set of points that form a 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 the equality . If we take a point in the half-plane above the line, we have . If we take a point in a half-plane under a straight line, then it will not satisfy our inequality: .

Now consider a 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 centered at the origin and with radius 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, not taking into account 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 suffices 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 a square.

Let's consider the so-called area method using a specific example.

Example 4 - depict the set of solutions to the inequality:

According to the method of regions, first of all we consider the function on the left side, if the right side is zero. This is a function of two variables:

Similarly to the method of intervals, we temporarily depart from the inequality and study the features and properties of the composed function.

ODZ: which means that the x-axis is punctured.

Now we indicate that the function is zero when the numerator of the fraction is zero, we have:

We build a graph of the function.

Rice. 4. Graph of the function, given the ODZ

Now consider the areas of constancy of the function, they are formed by a straight line and a broken line. inside the broken line there is an area D 1 . Between a segment of a polyline and a straight line - area D 2, below a straight line - area D 3, between a segment of a polyline and a straight line - area D 4

In each of the selected areas, the function retains its sign, which means that it is enough to check an arbitrary test point in each area.

Let's take a point (0;1) in the area. We have:

Let's take a point (10;1) in the area. We have:

Thus, the entire region is negative and does not satisfy the given inequality.

Take a point (0;-5) in the area. We have:

Thus, the entire region is positive and satisfies the given inequality.