Solving systems of linear inequalities graphically. Solving equations, inequalities, systems using function graphs. Visual Guide (2019)


One of the most convenient methods for solving quadratic inequalities is the graphical method. In this article, we will analyze how quadratic inequalities are solved graphically. First, let's discuss what the essence of this method is. And then we give the algorithm and consider examples of solving quadratic inequalities graphically.

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The essence of the graphic method

Generally graphical way to solve inequalities with one variable is used not only to solve square inequalities, but also inequalities of other types. The essence of the graphical method for solving inequalities next: consider the functions y=f(x) and y=g(x) that correspond to the left and right parts of the inequality, build their graphs in the same rectangular coordinate system and find out at what intervals the graph of one of them is located below or above the other. Those intervals where

  • the graph of the function f above the graph of the function g are solutions to the inequality f(x)>g(x) ;
  • the graph of the function f not lower than the graph of the function g are solutions to the inequality f(x)≥g(x) ;
  • the graph of the function f below the graph of the function g are solutions to the inequality f(x)
  • the graph of the function f not above the graph of the function g are solutions to the inequality f(x)≤g(x) .

Let's also say that the abscissas of the intersection points of the graphs of functions f and g are solutions to the equation f(x)=g(x) .

Let us transfer these results to our case – to solve the quadratic inequality a x 2 +b x+c<0 (≤, >, ≥).

We introduce two functions: the first y=a x 2 +b x+c (in this case f(x)=a x 2 +b x+c) corresponds to the left side of the quadratic inequality, the second y=0 (in this case g (x)=0 ) corresponds to the right side of the inequality. schedule quadratic function f is a parabola and the graph permanent function g is a straight line coinciding with the abscissa axis Ox .

Further, according to the graphical method for solving inequalities, it is necessary to analyze at what intervals the graph of one function is located above or below the other, which will allow us to write the desired solution to the quadratic inequality. In our case, we need to analyze the position of the parabola relative to the axis Ox.

Depending on the values ​​of the coefficients a, b and c, the following six options are possible (a schematic representation is sufficient for our needs, and it is possible not to depict the Oy axis, since its position does not affect the solution of the inequality):

    In this drawing, we see a parabola whose branches are directed upwards and which intersects the axis Ox at two points, the abscissas of which are x 1 and x 2 . This drawing corresponds to the variant when the coefficient a is positive (it is responsible for the upward direction of the branches of the parabola), and when the value is positive discriminant of a square trinomial a x 2 +b x + c (in this case, the trinomial has two roots, which we denoted as x 1 and x 2, and we assumed that x 1 0 , D=b 2 −4 a c=(−1) 2 −4 1 (−6)=25>0, x 1 =−2 , x 2 =3 .

    For clarity, let's draw in red the parts of the parabola located above the abscissa axis, and in blue - located below the abscissa axis.

    Now let's find out what gaps correspond to these parts. The following drawing will help determine them (in the future, we will mentally make such selections in the form of rectangles):

    So on the abscissa axis, two intervals (−∞, x 1) and (x 2, +∞) were highlighted in red, on them the parabola is higher than the axis Ox, they make up the solution of the quadratic inequality a x 2 +b x+c>0 , and the interval (x 1 , x 2) is highlighted in blue, on it the parabola is below the axis Ox , it is a solution to the inequality a x 2 + b x + c<0 . Решениями нестрогих квадратных неравенств a·x 2 +b·x+c≥0 и a·x 2 +b·x+c≤0 будут те же промежутки, но в них следует включить числа x 1 и x 2 , отвечающие равенству a·x 2 +b·x+c=0 .

    And now briefly: for a>0 and D=b 2 −4 a c>0 (or D"=D/4>0 for an even coefficient b)

    • the solution to the quadratic inequality a x 2 +b x+c>0 is (−∞, x 1)∪(x 2 , +∞) or, in another way, x x2;
    • the solution to the quadratic inequality a x 2 +b x+c≥0 is (−∞, x 1 ]∪ or in other notation x 1 ≤x≤x 2 ,

    where x 1 and x 2 are the roots of the square trinomial a x 2 + b x + c, and x 1


    Here we see a parabola, the branches of which are directed upwards, and which touches the abscissa axis, that is, it has one common point with it, let's denote the abscissa of this point as x 0. The presented case corresponds to a>0 (the branches are directed upwards) and D=0 ( square trinomial has one root x 0 ). For example, we can take the quadratic function y=x 2 −4 x+4 , here a=1>0 , D=(−4) 2 −4 1 4=0 and x 0 =2 .

    The drawing clearly shows that the parabola is located above the Ox axis everywhere, except for the point of contact, that is, at the intervals (−∞, x 0) , (x 0 , ∞) . For clarity, we select areas in the drawing by analogy with the previous paragraph.

    We draw conclusions: for a>0 and D=0

    • the solution to the quadratic inequality a x 2 +b x+c>0 is (−∞, x 0)∪(x 0 , +∞) or in other notation x≠x 0 ;
    • the solution to the quadratic inequality a x 2 +b x+c≥0 is (−∞, +∞) or, in another notation, x∈R ;
    • quadratic inequality a x 2 +b x+c<0 не имеет решений (нет интервалов, на которых парабола расположена ниже оси Ox );
    • the quadratic inequality a x 2 +b x+c≤0 has a unique solution x=x 0 (it is given by the tangent point),

    where x 0 is the root of the square trinomial a x 2 + b x + c.


    In this case, the branches of the parabola are directed upwards, and it has no common points with the abscissa axis. Here we have the conditions a>0 (the branches are directed upwards) and D<0 (квадратный трехчлен не имеет действительных корней). Для примера можно построить график функции y=2·x 2 +1 , здесь a=2>0 , D=0 2 −4 2 1=−8<0 .

    Obviously, the parabola is located above the Ox axis throughout its entire length (there are no intervals where it is below the Ox axis, there is no point of contact).

    Thus, for a>0 and D<0 решением квадратных неравенств a·x 2 +b·x+c>0 and a x 2 +b x+c≥0 is the set of all real numbers, and the inequalities a x 2 +b x+c<0 и a·x 2 +b·x+c≤0 не имеют решений.

And there are three options for the location of the parabola with branches directed downwards, and not upwards, relative to the axis Ox. In principle, they may not be considered, since multiplying both parts of the inequality by −1 allows us to pass to an equivalent inequality with a positive coefficient at x 2 . However, it does not hurt to get an idea about these cases. The reasoning here is similar, so we write down only the main results.

Solution algorithm

The result of all previous calculations is algorithm for solving square inequalities graphically:

    A schematic drawing is performed on the coordinate plane, which depicts the Ox axis (it is not necessary to depict the Oy axis) and a sketch of a parabola corresponding to a quadratic function y=a x 2 + b x + c. To construct a sketch of a parabola, it is enough to find out two points:

    • First, by the value of the coefficient a, it is found out where its branches are directed (for a>0 - upwards, for a<0 – вниз).
    • And secondly, by the value of the discriminant of the square trinomial a x 2 + b x + c, it turns out whether the parabola intersects the x-axis at two points (for D> 0), touches it at one point (for D=0), or has no common points with the Ox axis (for D<0 ). Для удобства на чертеже указываются координаты точек пересечения или координата точки касания (при наличии этих точек), а сами точки изображаются выколотыми при решении строгих неравенств, или обычными при решении нестрогих неравенств.

  • When the drawing is ready, on it at the second step of the algorithm

    • when solving the quadratic inequality a·x 2 +b·x+c>0, the intervals at which the parabola is located above the abscissa axis are determined;
    • when solving the inequality a x 2 +b x+c≥0, the intervals are determined at which the parabola is located above the abscissa axis and the abscissas of the intersection points (or the abscissa of the tangent point) are added to them;
    • when solving the inequality a x 2 +b x+c<0 находятся промежутки, на которых парабола ниже оси Ox ;
    • finally, when solving a quadratic inequality of the form a x 2 +b x + c≤0, there are intervals where the parabola is below the Ox axis and the abscissas of the intersection points (or the abscissa of the tangency point) are added to them;

    they constitute the desired solution of the quadratic inequality, and if there are no such intervals and no points of contact, then the original quadratic inequality has no solutions.

It remains only to solve a few quadratic inequalities using this algorithm.

Examples with Solutions

Example.

Solve the inequality .

Solution.

We need to solve a quadratic inequality, we will use the algorithm from the previous paragraph. In the first step, we need to draw a sketch of the graph of the quadratic function . The coefficient at x 2 is 2, it is positive, therefore, the branches of the parabola are directed upwards. Let us also find out whether the parabola with the abscissa axis has common points, for this we calculate the discriminant of the square trinomial . We have . The discriminant turned out to be greater than zero, therefore, the trinomial has two real roots: and , that is, x 1 =−3 and x 2 =1/3.

From this it is clear that the parabola intersects the axis Ox at two points with abscissas −3 and 1/3. We will depict these points in the drawing as ordinary points, since we are solving a non-strict inequality. According to the clarified data, we obtain the following drawing (it fits the first template from the first paragraph of the article):

We pass to the second step of the algorithm. Since we are solving a non-strict quadratic inequality with the ≤ sign, we need to determine the intervals at which the parabola is located below the abscissa axis and add the abscissas of the intersection points to them.

It can be seen from the drawing that the parabola is below the abscissa in the interval (−3, 1/3) and we add the abscissas of the intersection points to it, that is, the numbers −3 and 1/3. As a result, we arrive at the numerical segment [−3, 1/3] . This is the desired solution. It can be written as a double inequality −3≤x≤1/3 .

Answer:

[−3, 1/3] or −3≤x≤1/3 .

Example.

Find a solution to the quadratic inequality −x 2 +16 x−63<0 .

Solution.

As usual, we start with a drawing. The numerical coefficient for the square of the variable is negative, −1, therefore, the branches of the parabola are directed downwards. Let's calculate the discriminant, or better, its fourth part: D"=8 2 −(−1)(−63)=64−63=1. Its value is positive, we calculate the roots of the square trinomial: and , x 1 =7 and x 2 =9. So the parabola intersects the Ox axis at two points with abscissas 7 and 9 (the initial inequality is strict, so we will depict these points with an empty center). Now we can make a schematic drawing:

Since we are solving a strict signed quadratic inequality<, то нас интересуют промежутки, на которых парабола расположена ниже оси абсцисс:

The drawing shows that the solutions to the original quadratic inequality are two intervals (−∞, 7) , (9, +∞) .

Answer:

(−∞, 7)∪(9, +∞) or in another notation x<7 , x>9 .

When solving square inequalities, when the discriminant of a square trinomial on its left side is equal to zero, you need to be careful with the inclusion or exclusion of the abscissa of the tangent point from the answer. It depends on the sign of the inequality: if the inequality is strict, then it is not a solution to the inequality, and if it is non-strict, then it is.

Example.

Does the quadratic inequality 10 x 2 −14 x+4.9≤0 have at least one solution?

Solution.

Let's plot the function y=10 x 2 −14 x+4.9 . Its branches are directed upwards, since the coefficient at x 2 is positive, and it touches the abscissa at the point with the abscissa 0.7, since D "=(−7) 2 −10 4.9=0, whence or 0.7 as a decimal.Schematically, it looks like this:

Since we are solving a quadratic inequality with the sign ≤, then its solution will be the intervals on which the parabola is below the Ox axis, as well as the abscissa of the tangent point. It can be seen from the drawing that there is not a single gap where the parabola would be below the axis Ox, therefore, its solution will be only the abscissa of the point of contact, that is, 0.7.

Answer:

this inequality has a unique solution 0.7 .

Example.

Solve the quadratic inequality –x 2 +8 x−16<0 .

Solution.

We act according to the algorithm for solving quadratic inequalities and start by plotting. The branches of the parabola are directed downwards, since the coefficient at x 2 is negative, −1. Find the discriminant of the square trinomial –x 2 +8 x−16 , we have D'=4 2 −(−1)(−16)=16−16=0 and further x 0 =−4/(−1) , x 0 =4 . So, the parabola touches the Ox axis at the point with the abscissa 4 . Let's make a drawing:

We look at the sign of the original inequality, it is<. Согласно алгоритму, решение неравенства в этом случае составляют все промежутки, на которых парабола расположена строго ниже оси абсцисс.

In our case, these are open rays (−∞, 4) , (4, +∞) . Separately, we note that 4 - the abscissa of the tangent point - is not a solution, since at the tangent point the parabola is not lower than the Ox axis.

Answer:

(−∞, 4)∪(4, +∞) or in other notation x≠4 .

Pay special attention to cases where the discriminant of the square trinomial on the left side of the square inequality is less than zero. There is no need to rush here and say that the inequality has no solutions (we are used to making such a conclusion for quadratic equations with a negative discriminant). The point is that the quadratic inequality for D<0 может иметь решение, которым является множество всех действительных чисел.

Example.

Find the solution to the quadratic inequality 3 x 2 +1>0 .

Solution.

As usual, we start with a drawing. The coefficient a is 3, it is positive, therefore, the branches of the parabola are directed upwards. Calculate the discriminant: D=0 2 −4 3 1=−12 . Since the discriminant is negative, the parabola has no common points with the x-axis. The information obtained is sufficient for a schematic diagram:

We are solving a strict quadratic inequality with > sign. Its solution will be all the intervals where the parabola is above the Ox axis. In our case, the parabola is above the x-axis along its entire length, so the desired solution will be the set of all real numbers.

Ox , and also you need to add the abscissa of the intersection points or the abscissa of the touch point to them. But the drawing clearly shows that there are no such gaps (since the parabola is everywhere below the abscissa axis), as well as there are no intersection points, just as there are no points of contact. Therefore, the original quadratic inequality has no solutions.

Answer:

there are no solutions or in another notation ∅.

Bibliography.

  • Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
  • Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
  • Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemozina, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
  • Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., Sr. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
  • Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. At 2 pm Part 1. Textbook for students of educational institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.

Goals:

1. Repeat knowledge about the quadratic function.

2. Get acquainted with the method of solving a quadratic inequality based on the properties of a quadratic function.

Equipment: multimedia, presentation “Solving square inequalities”, cards for independent work, table “Algorithm for solving square inequalities”, control sheets with carbon paper.

DURING THE CLASSES

I. Organizational moment (1 min).

II. Updating of basic knowledge(10 min).

1. Plotting a quadratic function y \u003d x 2 -6x + 8<Рисунок 1. Приложение >

  • determination of the direction of the branches of the parabola;
  • determining the coordinates of the parabola vertex;
  • determination of the axis of symmetry;
  • determination of intersection points with coordinate axes;
  • finding additional points.

2. Determine from the drawing the sign of the coefficient a and the number of roots of the equation ax 2 +in+c=0.<Рисунок 2. Приложение >

3. According to the graph of the function y \u003d x 2 -4x + 3, determine:

  • What are the zeros of the function;
  • Find the intervals on which the function takes positive values;
  • Find the intervals on which the function takes negative values;
  • At what values ​​of x does the function increase, and at what values ​​does it decrease?<Рисунок 3>

4. Learning new knowledge (12 min.)

Task 1: Solve the inequality: x 2 +4x-5 > 0.

The inequality is satisfied by the x values ​​at which the values ​​of the function y=x 2 +4x-5 are equal to zero or positive, that is, those x values ​​at which the points of the parabola lie on the x-axis or above this axis.

Let's build a graph of the function y \u003d x 2 + 4x-5.

With the x-axis: X 2 + 4x-5 \u003d 0. According to the Vieta theorem: x 1 \u003d 1, x 2 \u003d -5. Points(1;0),(-5;0).

With the y-axis: y(0)=-5. Point (0;-5).

Additional points: y(-1)=-8, y(2)=7.<Рисунок 4>

Bottom line: The values ​​of the function are positive and equal to zero (non-negative) when

  • Is it necessary to plot a quadratic function in detail every time to solve an inequality?
  • Do I need to find the coordinates of the vertex of the parabola?
  • What is important? (a, x 1, x 2)

Conclusion: To solve a quadratic inequality, it is enough to determine the zeros of the function, the direction of the branches of the parabola and build a sketch of the graph.

Task 2: Solve the inequality: x 2 -6x + 8 < 0.

Solution: Let's determine the roots of the equation x 2 -6x+8=0.

According to the Vieta theorem: x 1 \u003d 2, x 2 \u003d 4.

a>0 - the branches of the parabola are directed upwards.

Let's build a sketch of the graph.<Рисунок 5>

We mark with signs “+” and “–” the intervals on which the function takes positive and negative values. Let's choose the interval we need.

Answer: X€.

5. Consolidation of new material (7 min).

No. 660 (3). The student decides on the board.

Solve inequality-x 2 -3x-2<0.

X 2 -3x-2=0; x 2 +3x+2=0;

the roots of the equation: x 1 \u003d -1, x 2 \u003d -2.

a<0 – ветви вниз. <Рисунок 6>

No. 660 (1) - Working with a hidden board.

Solve the inequality x 2 -3x + 2 < 0.

Solution: x 2 -3x+2=0.

Let's find the roots: ; x 1 =1, x 2 =2.

a>0 - branches up. We build a sketch of the graph of the function.<Рисунок 7>

Algorithm:

  1. Find the roots of the equation ax 2 + in + c \u003d 0.
  2. Mark them on the coordinate plane.
  3. Determine the direction of the branches of the parabola.
  4. Sketch a chart.
  5. Mark with signs “+” and “-”, the intervals on which the function takes positive and negative values.
  6. Select the desired interval.

6. Independent work (10 min.).

(Reception - carbon paper).

The control sheet is signed and handed over to the teacher for verification and correction determination.

Board self-check.

Additional task:

№ 670. Find the values ​​of x at which the function takes values ​​not greater than zero: y=x 2 +6x-9.

7. Homework (2 min).

№ 660 (2, 4), № 661 (2, 4).

Fill in the table:

D Inequality a Drawing Solution
D>0 ax 2 + in + s > 0 a>0
D>0 ax 2 + in + s > 0 a<0
D>0 ax 2 + in + s < 0 a>0
D>0 ax 2 + in + s < 0 a<0

8. Summary of the lesson (3 min).

  1. Reproduce the algorithm for solving inequalities.
  2. Who did a great job?
  3. What seemed difficult?

During the lesson, you will be able to independently study the topic "Graphical solution of equations, inequalities." The teacher in the lesson will analyze the graphical methods for solving equations and inequalities. It will teach you how to build graphs, analyze them and get solutions to equations and inequalities. The lesson will also deal with specific examples on this topic.

Topic: Numeric functions

Lesson: Graphical solution of equations, inequalities

1. Lesson topic, introduction

We have considered graphs of elementary functions, including graphs of power functions with different exponents. We also considered the rules for shifting and transforming function graphs. All these skills must be applied when required. graphicsolution equations or graphic solutioninequalities.

2. Solving equations and inequalities graphically

Example 1. Graphically solve the equation:

Let's build graphs of functions (Fig. 1).

The graph of the function is a parabola passing through the points

The graph of the function is a straight line, we will build it according to the table.

Graphs intersect at a point There are no other intersection points, since the function is monotonically increasing, the function is monotonically decreasing, and, therefore, their intersection point is unique.

Example 2. Solve the inequality

a. For the inequality to hold, the graph of the function must be located above the straight line (Fig. 1). This is done when

b. In this case, on the contrary, the parabola should be under the line. This is done when

Example 3. Solve the inequality

Let's build graphs of functions (Fig. 2).

Find the root of the equation When there are no solutions. There is one solution for .

For the inequality to hold, the hyperbola must be located above the line. This is true for .

Example 4. Solve graphically the inequality:

Domain:

Let's build graphs of functions for (Fig. 3).

a. The graph of the function should be located under the graph; this is done when

b. The graph of the function is located above the graph at But since we have a non-strict sign in the condition, it is important not to lose the isolated root

3. Conclusion

We have considered a graphical method for solving equations and inequalities; considered specific examples, in the solution of which we used such properties of functions as monotonicity and evenness.

1. Mordkovich A. G. et al. Algebra 9th grade: Proc. For general education Institutions. - 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.

2. Mordkovich A. G. et al. Algebra 9th grade: Task book for students of educational institutions / A. G. Mordkovich, T. N. Mishustina et al. - 4th ed. — M.: Mnemosyne, 2002.-143 p.: ill.

3. Yu. N. Makarychev, Algebra. Grade 9: textbook. for general education students. institutions / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, I. E. Feoktistov. - 7th ed., Rev. and additional - M .: Mnemosyne, 2008.

4. Sh. A. Alimov, Yu. M. Kolyagin, and Yu. V. Sidorov, Algebra. Grade 9 16th ed. - M., 2011. - 287 p.

5. Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 12th ed., erased. — M.: 2010. — 224 p.: ill.

6. Algebra. Grade 9 At 2 hours. Part 2. Task book for students of educational institutions / A. G. Mordkovich, L. A. Aleksandrova, T. N. Mishustina and others; Ed. A. G. Mordkovich. - 12th ed., Rev. — M.: 2010.-223 p.: ill.

1. College section. ru in mathematics.

2. Internet project "Tasks".

3. Educational portal "SOLVE USE".

1. Mordkovich A. G. et al. Algebra 9th grade: Task book for students of educational institutions / A. G. Mordkovich, T. N. Mishustina et al. - 4th ed. - M .: Mnemosyne, 2002.-143 p.: ill. No. 355, 356, 364.

The graphical method consists in constructing a set of feasible LLP solutions, and finding in this set a point corresponding to the max/min objective function.

Due to the limited possibilities of a visual graphical representation, this method is used only for systems linear inequalities with two unknowns and systems that can be reduced to a given form.

In order to visually demonstrate the graphical method, we will solve the following problem:

1. At the first stage, it is necessary to construct the area of ​​feasible solutions. For this example, it is most convenient to choose X2 for the abscissa, and X1 for the ordinate, and write the inequalities in the following form:

Since both the graphs and the area of ​​​​admissible solutions are in the first quarter. In order to find the boundary points, we solve equations (1)=(2), (1)=(3) and (2)=(3).

As can be seen from the illustration, the polyhedron ABCDE forms an area of ​​feasible solutions.

If the domain of admissible solutions is not closed, then either max(f)=+ ? or min(f)= -?.

2. Now we can proceed to directly finding the maximum of the function f.

Alternately substituting the coordinates of the vertices of the polyhedron into the function f and comparing the values, we find that f(C)=f (4; 1)=19 - the maximum of the function.

This approach is quite beneficial for a small number of vertices. But this procedure can be delayed if there are quite a lot of vertices.

In this case, it is more convenient to consider a level line of the form f=a. With a monotonous increase in the number a from -? to +? straight lines f=a are displaced along the normal vector. If, with such a displacement of the level line, there exists some point X - the first common point of the area of ​​feasible solutions (polyhedron ABCDE) and the level line, then f(X) is the minimum of f on the set ABCDE. If X is the last point of intersection of the level line and the set ABCDE, then f(X) is the maximum on the set of feasible solutions. If for a>-? the line f=a intersects the set of admissible solutions, then min(f)= -?. If this happens when a>+?, then max(f)=+?.

see also Solving a linear programming problem graphically, Canonical form of linear programming problems

The system of constraints for such a problem consists of inequalities in two variables:
and the objective function has the form F = C 1 x + C 2 y, which is to be maximized.

Let's answer the question: what pairs of numbers ( x; y) are solutions to the system of inequalities, i.e., do they satisfy each of the inequalities simultaneously? In other words, what does it mean to solve a system graphically?
First you need to understand what is the solution of one linear inequality with two unknowns.
To solve a linear inequality with two unknowns means to determine all pairs of values ​​of the unknowns for which the inequality is satisfied.
For example, inequality 3 x – 5y≥ 42 satisfy the pairs ( x , y) : (100, 2); (3, –10), etc. The problem is to find all such pairs.
Consider two inequalities: ax + by≤ c, ax + by≥ c. Straight ax + by = c divides the plane into two half-planes so that the coordinates of the points of one of them satisfy the inequality ax + by >c, and the other inequality ax + +by <c.
Indeed, take a point with coordinate x = x 0; then a point lying on a straight line and having an abscissa x 0 , has an ordinate

Let for definiteness a<0, b>0, c>0. All points with abscissa x 0 above P(e.g. dot M), have y M>y 0 , and all points below the point P, with abscissa x 0 , have yN<y 0 . Because the x 0 is an arbitrary point, then there will always be points on one side of the line for which ax+ by > c, forming a half-plane, and on the other hand, points for which ax + by< c.

Picture 1

The inequality sign in the half-plane depends on the numbers a, b , c.
This leads to the following method graphic solution systems of linear inequalities in two variables. To solve the system, you need:

  1. For each inequality, write down the equation corresponding to the given inequality.
  2. Construct lines that are graphs of functions given by equations.
  3. For each straight line, determine the half-plane, which is given by the inequality. To do this, take an arbitrary point that does not lie on a straight line, substitute its coordinates into the inequality. if the inequality is true, then the half-plane containing the chosen point is the solution to the original inequality. If the inequality is false, then the half-plane on the other side of the line is the set of solutions to this inequality.
  4. To solve a system of inequalities, it is necessary to find the area of ​​intersection of all half-planes that are the solution to each inequality in the system.

This area may turn out to be empty, then the system of inequalities has no solutions, it is inconsistent. Otherwise, the system is said to be consistent.
Solutions can be a finite number and an infinite set. The area can be a closed polygon or it can be unlimited.

Let's look at three relevant examples.

Example 1. Graphically solve the system:
x + y- 1 ≤ 0;
–2x- 2y + 5 ≤ 0.

  • consider the equations x+y–1=0 and –2x–2y+5=0 corresponding to the inequalities;
  • let us construct the straight lines given by these equations.

Figure 2

Let us define the half-planes given by the inequalities. Take an arbitrary point, let (0; 0). Consider x+ y– 1 0, we substitute the point (0; 0): 0 + 0 – 1 ≤ 0. hence, in the half-plane where the point (0; 0) lies, x + y – 1 ≤ 0, i.e. the half-plane lying below the straight line is the solution to the first inequality. Substituting this point (0; 0) into the second one, we get: –2 ∙ 0 – 2 ∙ 0 + 5 ≤ 0, i.e. in the half-plane where the point (0; 0) lies, -2 x – 2y+ 5≥ 0, and we were asked where -2 x – 2y+ 5 ≤ 0, therefore, in another half-plane - in the one above the straight line.
Find the intersection of these two half-planes. The lines are parallel, so the planes do not intersect anywhere, which means that the system of these inequalities has no solutions, it is inconsistent.

Example 2. Find graphically solutions to the system of inequalities:

Figure 3
1. Write down the equations corresponding to the inequalities and construct straight lines.
x + 2y– 2 = 0

x 2 0
y 0 1

y – x – 1 = 0
x 0 2
y 1 3

y + 2 = 0;
y = –2.
2. Having chosen the point (0; 0), we determine the signs of inequalities in the half-planes:
0 + 2 ∙ 0 – 2 ≤ 0, i.e. x + 2y– 2 ≤ 0 in the half-plane below the straight line;
0 – 0 – 1 ≤ 0, i.e. y –x– 1 ≤ 0 in the half-plane below the straight line;
0 + 2 =2 ≥ 0, i.e. y+ 2 ≥ 0 in the half-plane above the line.
3. The intersection of these three half-planes will be an area that is a triangle. It is not difficult to find the vertices of the region as the points of intersection of the corresponding lines


In this way, BUT(–3; –2), AT(0; 1), FROM(6; –2).

Let us consider one more example, in which the resulting domain of the solution of the system is not limited.

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