You know that every ordered pair of numbers corresponds to a specific point on coordinate plane. Since each solution to an equation with two variables x and y is an ordered pair of numbers, all its solutions can be represented by points on the coordinate plane. At these points, the abscissa is the value of the x variable, and the ordinate is the corresponding value of the y variable. Therefore, we get a graph of an equation with two variables.
Remember!
The graph of an equation with two variables is the image on the coordinate plane of all points whose coordinates satisfy the given equation.
Look at Figures 64 and 65. You see a graph of the equation 0.5 x - y = 2, where x is an even single-digit number (Figure 64), and a graph of the equation x 2 + y 2 = 4 (Figure 65). The first graph contains only four points because the variables x and y can only take four values. The second graph is a line on the coordinate plane. It contains many points, since the variable x can take any value from -2 to 2 and there are many such numbers. There are also many corresponding values. They vary from 2 to 2.
Figure 66 shows the graph of the equation x + y = 4. Unlike the graph of the equation x 2 + y 2 = 4 (see Fig. 65), each abscise point of this graph corresponds to a single ordinate. This means that Figure 66 shows the graph of the function. Convince yourself that the graph of the equation in Figure 64 is also the graph of a function.
note
Not every equation has a graph of a function, but every graph of a function is a graph of some equation.
The equation x + y = 4 is a linear equation in two variables. Having solved it for y, we get: y = -x + 4. The resulting equality can be understood as a formula that defines the linear function y = -x + 4. The graph of such a function is a straight line. So, the graph of the linear equation x + y = 4, which is shown in Figure 66, is a straight line.
Can we say that the graph of any linear equation in two variables is a straight line? No. For example, the linear equation 0 ∙ x + 0 ∙ y = 0 is satisfied by any pair of numbers, and therefore the graph of this equation contains all points of the coordinate plane.
Let's find out what is the graph of a linear equation with two variables ax + bу + c = 0 depending on the values of the coefficients a, b and c. Such cases are possible.
Let a ≠ 0, b ≠ 0, c ≠ 0. Then the equation ax + by + c = 0 can be represented as:
We have obtained an equality defining the linear function y(x). Her schedule, and therefore the schedule given equation is a straight line that does not pass through the origin of coordinates (Fig. 67).
2. Let a ≠ 0, b ≠ 0, c = 0. Then the equation ax + by + c = 0 takes the form ax + by + 0 = 0, or y = x.
We have obtained equality, which specifies direct proportionality to y(x). Its graph, and therefore the graph of this equation, is a straight line passing through the origin of coordinates (Fig. 68).
3. Let a ≠ 0, b = 0, c ≠ 0. Then the equation ax + by + c = 0 takes the form ax + 0 ∙ y + c = 0, or x = -.
Received equality does not specify the y() function. This equality is satisfied by such pairs of numbers (x; y), in which x = , and y is any number. On the coordinate plane, these points lie on a straight line parallel to the OY axis. So, the graph of this equation is a straight line parallel to the ordinate axis (Fig. 69).
4. Let a ≠ 0, b = 0, c = 0. Then the equation ax + by + c = 0 takes the form ax + 0 ∙ y + 0 = 0, or x = 0.
This equality is satisfied by such pairs of numbers (x; y), in which x = 0, and y is any number. On the coordinate plane, these points lie on the OY axis. So, the graph of this equation is a straight line coinciding with the ordinate axis.
5. Let a ≠ 0, b ≠ 0, c ≠0. Then the equation ax + bу + c = 0 takes the form 0 ∙ x + by + c = 0, or y = -. This equality defines a function y(x), which takes on the same values for any values of x, that is, it is constant. Its graph, and therefore the graph of this equation, is a straight line parallel to the abscissa axis (Fig. 70).
6. Let a = 0, b ≠ 0, c = 0. Then the equation ax + by + c = 0 takes the form 0 ∙ x + by + 0 = 0, or b = 0. We obtained a constant function y(x), in in which each point of the graph lies on the OX axis. So, the graph of this equation is a straight line coinciding with the abscissa axis.
7. Let a = 0, b = 0, c ≠ 0. Then the equation ax + by + c = 0 takes the form 0 ∙ x + 0 ∙ y + c = 0, or 0 ∙ x + 0 ∙ b = c. And such a linear equation has no solutions, so its graph does not contain a single point on the coordinate plane.
8. Let a = 0, b = 0, c = 0. Then the equation ax + by + c = 0 takes the form 0 ∙ x + 0 ∙ y + 0 = 0, or 0 ∙ x + 0 ∙ y = 0. A such a linear equation has many solutions, so its graph is the entire coordinate plane.
We can summarize the results obtained.
Graph of a linear equation with two variables ax + bу + с = 0:
Is straight if a ≠ 0 or b ≠ 0;
Is the entire plane if a = 0, b = 0 and c = 0;
Does not contain a single point of the coordinate plane if a = 0, b = 0 and c ≠ 0.
Task. Graph the equation 2x - y - 3 = 0
Solutions. The equation 2x - y - 3 = 0 is linear. Therefore, its graph is the line y = 2x - 3. To construct it, it is enough to specify two points belonging to this line. Let's make a table of y values for two arbitrary values of x, for example, for x = 0 and x = 2 (Table 27).
Table 27
On the coordinate plane, we designate points with coordinates (0; -3) and (2; 1) and draw a straight line through them (Fig. 70). This straight line is the desired graph of the equation 2x - y - 3 = 0.
Is it possible to identify the graph of a linear equation with two variables and the graph of a first-degree equation with two variables? No, because there are linear equations that are not equations of the first degree. For example, these are the equation 0 ∙ x + 0 ∙ y + c = 0, 0 ∙ x + 0 ∙ y + 0 = 0.
Note:
The graph of a linear equation in two variables can be a straight line, the entire plane, or not contain a single point on the coordinate plane;
The graph of a first degree equation in two variables is always straight.
Find out more
1. Let a ≠ 0. Then common decision The equations can also be presented in this form: X = - y -. We obtained a linear function x(y). Its graph is a straight line. To construct such a graph, you need to combine the coordinate axes differently: first coordinate axis(independent variable) consider the op-amp axis, and the second (dependent variable)
OX axis. Then it is convenient to position the OU axis horizontally, and the OX axis
Vertically (Fig. 72). The graph of the equation in this case will also be placed differently on the coordinate plane depending on the markings of the coefficients b and c. Explore it yourself.
2. Nikolai Nikolaevich Bogolyubov (1909-1992) - an outstanding domestic mathematician and mechanic, theoretical physicist, founder of scientific schools in nonlinear mechanics and theoretical physics, academician of the Academy of Sciences of the Ukrainian SSR (1948) and the Academy of Sciences of the USSR (since 1953). Born in Nizhny Novgorod Russian Empire. In 1921 the family moved to Kyiv. After graduating from a seven-year school, Bogolyubov independently studied physics and mathematics and, from the age of 14, already took part in a seminar at the Department of Mathematical Physics of Kyiv University under the guidance of Academician D. A. Grave. In 1924, at the age of 15, Bogolyubov wrote his first scientific work, and the following year he was accepted into the graduate school of the ANURSR by academicians. M. Krylov, which he graduated from in 1929, receiving a Doctor of Mathematical Sciences degree at the age of 20.
In 1929 p. MM. Bogolyubov became research fellow Ukrainian Academy of Sciences, in 1934 he began teaching at Kiev University (since 1936 - professor). Since the late 40s of the XX century. At the same time he worked in Russia. He was director of the Joint Institute for Nuclear Research, and later - director of the Mathematical Institute named after. A. Steklova in Moscow, taught at the Moscow state university named after Mikhail Lomonosov. In 1966, he became the first director of the Institute of Theoretical Physics of the Ukrainian Academy of Sciences in Kyiv, which he created, and at the same time (1963-1988) he was an academician and secretary of the Department of Mathematics of the USSR Academy of Sciences.
MM. Bogolyubov - twice Hero of Socialist Labor (1969,1979), awarded the Lenin Prize (1958), USSR State Prize (1947.1953,1984), Gold Medal. M. V. Lomonosov Academy of Sciences of the USSR (1985).
On September 21, 2009, on the facade of the Red building of the Taras Shevchenko Kyiv National University, a Memorial plaque to the brilliant academician Nikolai Bogolyubov in honor of the centenary of his birth.
In 1992, the National Academy of Sciences of Ukraine founded the NAS of Ukraine Prize named after N.M. Bogolyubov, which is awarded by the Department of Mathematics of the NAS of Ukraine for outstanding scientific works in mathematics and theoretical physics. The small planet “22616 Bogolyubov” was named in honor of the scientist.
REMEMBER THE IMPORTANT
1. What is the graph of a linear equation in two variables?
2. In any case, the graph of an equation with two variables is a straight line; plane?
3. In what case does the graph of a linear equation in two variables pass through the origin?
SOLVE PROBLEMS
1078 . Which of Figures 73-74 shows the graph of a linear equation in two variables? Explain your answer.
1079 . At what values of the coefficients a, b and c is the straight line ax + bу + c = 0.
1) passes through the origin;
2) parallel to the x-axis;
3) parallel to the ordinate axis;
4) coincides with the abscissa axis;
5) coincides with the ordinate axis?
1080 . Without performing construction, determine whether the point belongs to the graph of a linear equation with two variables 6x - 2y + 1 = 0:
1)A(-1;2.5); 2)B(0;3.5); 3) C(-2; 5.5); 4)D(1,5;5).
1081 . Without performing construction, determine whether the point belongs to the graph of a linear equation with two variables 3x + 3y - 5 = 0:
1) A (-1; ); 2) B (0; 1).
1082
1) 2x + y - 4 = 0 if x = 0; 3) 3x + 3y - 1 = 0 if x = 2;
2) 4x - 2y + 5 = 0 if x = 0; 4) -5x - y + 6 = 0 if x = 2.
1083 . For a given linear equation in two variables, find the value of y corresponding to set value X:
1)3x - y + 2 = 0 if x = 0; 2) 6x - 5y - 7 = 0 if x = 2.
1084
1) 2x + y - 4 = 0; 4) -x + 2y + 8 = 0; 7) 5x - 10 = 0;
2) 6x - 2y + 12 = 0; 5)-x - 2y + 4 = 0; 8)-2у + 4 = 0;
3) 5x - 10y = 0; 6)x - y = 0; 9) x - y = 0.
1085 . Graph a linear equation with two variables:
1) 4x + y - 3 = 0; 4) 10x - 5y - 1 = 0;
2) 9x - 3y + 12 = 0; 5) 2x + 6 = 0;
3) -4x - 8y = 0; 6) y - 3 = 0.
1086 . Find the coordinates of the point of intersection of the graph of a linear equation with two variables 2x - 3y - 18 = 0 with the axis:
1) axles; 2) axes.
1087 . Find the coordinates of the point of intersection of the graph of a linear equation with two variables 5x + 4y - 20 = 0 with the axis:
1) axles; 2) axes.
1088 . On the straight line, which is the graph of the equation 0.5 x + 2y - 4 = 0, a point is indicated. Find the ordinate of this point if its abscissa is:
5) 4(x - y) = 4 - 4y;
6) 7x - 2y = 2(1 + 3.5 x).
1094 . The graph of a linear equation in two variables passes through point A(3; -2). Find the unknown coefficient of the equation:
1) ax + 3y - 3 = 0;
2) 2x - by + 8 = 0;
3)-x + 3y - c = 0.
1095 . Determine the type of quadrilateral whose vertices are the points of intersection of the graphs of the equations:
x - y + 4 = 0, x - y - 4 = 0, -x - y + 4 = 0, -x - y - 4 = 0
1096 . Plot the equation:
1) a - 4b + 1 = 0; 3) 3a + 0 ∙ b - 12 = 0;
2) 0 ∙ a + 2b + 6 = 0; 4) 0 ∙ a + 0 ∙ b + 5 = 0.
PUT IT IN PRACTICE
1097 . Create a linear equation with two variables based on the following data: 1) 3 kg of sweets and 2 kg of cookies cost 120 UAH; 2) 2 pens are 20 UAH more expensive than 5 pencils. Graph the equation you have created.
1098 . Build a graph of the equation for the problem about: 1) the number of girls and boys in your class; 2) purchasing lined and squared notebooks.
REVIEW PROBLEMS
1099. A tourist walked 12 km in an hour. How many hours will it take for a tourist to cover a distance of 20 km at the same speed?
1100. What should be the speed of the train according to the new schedule so that it can cover the distance between two stations in 2.5 hours, if according to the old schedule, moving at a speed of 100 km/h it covered it in 3 hours?
§ 1 Selection of equation roots in real situations
Let's consider this real situation:
The master and apprentice together made 400 custom parts. Moreover, the master worked for 3 days, and the student for 2 days. How many parts did each person make?
Let's create an algebraic model of this situation. Let the master produce parts in 1 day. And the student is at the details. Then the master will make 3 parts in 3 days, and the student will make 2 parts in 2 days. Together they will produce 3 + 2 parts. Since, according to the condition, a total of 400 parts were manufactured, we obtain the equation:
The resulting equation is called a linear equation in two variables. Here we need to find a pair of numbers x and y for which the equation will take the form of a true numerical equality. Note that if x = 90, y = 65, then we get the equality:
3 ∙ 90 + 65 ∙ 2 = 400
Since the correct numerical equality has been obtained, the pair of numbers 90 and 65 will be a solution to this equation. But the solution found is not the only one. If x = 96 and y = 56, then we get the equality:
96 ∙ 3 + 56 ∙ 2 = 400
This is also a true numerical equality, which means that the pair of numbers 96 and 56 is also a solution to this equation. But a pair of numbers x = 73 and y = 23 will not be a solution to this equation. In fact, 3 ∙ 73 + 2 ∙ 23 = 400 will give us the incorrect numerical equality 265 = 400. It should be noted that if we consider the equation in relation to this real situation, then there will be pairs of numbers that, being a solution to this equation, will not be a solution to the problem. For example, a couple of numbers:
x = 200 and y = -100
is a solution to the equation, but the student cannot make -100 parts, and therefore such a pair of numbers cannot be the answer to the question of the problem. Thus, in each specific real situation it is necessary to take a reasonable approach to the selection of the roots of the equation.
Let's summarize the first results:
An equation of the form ax + bу + c = 0, where a, b, c are any numbers, is called a linear equation with two variables.
The solution to a linear equation in two variables is a pair of numbers corresponding to x and y, for which the equation turns into a true numerical equality.
§ 2 Graph of a linear equation
The very recording of the pair (x;y) leads us to think about the possibility of depicting it as a point with coordinates xy y on a plane. This means that we can obtain a geometric model of a specific situation. For example, consider the equation:
2x + y - 4 = 0
Let's select several pairs of numbers that will be solutions to this equation and construct points with the found coordinates. Let these be points:
A(0; 4), B(2; 0), C(1; 2), D(-2; 8), E(- 1; 6).
Note that all points lie on the same line. This line is called the graph of a linear equation in two variables. It is a graphical (or geometric) model of a given equation.
If a pair of numbers (x;y) is a solution to the equation
ax + vy + c = 0, then the point M(x;y) belongs to the graph of the equation. We can say the other way around: if the point M(x;y) belongs to the graph of the equation ax + y + c = 0, then the pair of numbers (x;y) is a solution to this equation.
From the geometry course we know:
To plot a straight line, you need 2 points, so to plot a graph of a linear equation with two variables, it is enough to know only 2 pairs of solutions. But guessing the roots is not always a convenient or rational procedure. You can act according to another rule. Since the abscissa of a point (variable x) is an independent variable, you can give it any convenient value. Substituting this number into the equation, we find the value of the variable y.
For example, let the equation be given:
Let x = 0, then we get 0 - y + 1 = 0 or y = 1. This means that if x = 0, then y = 1. A pair of numbers (0;1) is the solution to this equation. Let's set another value for the variable x: x = 2. Then we get 2 - y + 1 = 0 or y = 3. The pair of numbers (2;3) is also a solution to this equation. Using the two points found, it is already possible to construct a graph of the equation x - y + 1 = 0.
You can do this: first assign some specific value to the variable y, and only then calculate the value of x.
§ 3 System of equations
Find two natural numbers, the sum of which is 11 and the difference is 1.
To solve this problem, we first create a mathematical model (namely, an algebraic one). Let the first number be x and the second number y. Then the sum of the numbers x + y = 11 and the difference of the numbers x - y = 1. Since both equations deal with the same numbers, these conditions must be met simultaneously. Usually in such cases a special record is used. The equations are written one below the other and combined with a curly brace.
Such a record is called a system of equations.
Now let’s construct sets of solutions to each equation, i.e. graphs of each of the equations. Let's take the first equation:
If x = 4, then y = 7. If x = 9, then y = 2.
Let's draw a straight line through points (4;7) and (9;2).
Let's take the second equation x - y = 1. If x = 5, then y = 4. If x = 7, then y = 6. We also draw a straight line through the points (5;4) and (7;6). We obtained a geometric model of the problem. The pair of numbers we are interested in (x;y) must be a solution to both equations. In the figure we see a single point that lies on both lines; this is the point of intersection of the lines.
Its coordinates are (6;5). Therefore, the solution to the problem will be: the first required number is 6, the second is 5.
List of used literature:
- Mordkovich A.G., Algebra 7th grade in 2 parts, Part 1, Textbook for general education institutions / A.G. Mordkovich. – 10th ed., revised – Moscow, “Mnemosyne”, 2007
- Mordkovich A.G., Algebra 7th grade in 2 parts, Part 2, Problem book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, “Mnemosyne”, 2007
- HER. Tulchinskaya, Algebra 7th grade. Blitz survey: a manual for students of general education institutions, 4th edition, revised and expanded, Moscow, “Mnemosyne”, 2008
- Alexandrova L.A., Algebra 7th grade. Thematic testing work V new form for students of general education institutions, edited by A.G. Mordkovich, Moscow, “Mnemosyne”, 2011
- Alexandrova L.A. Algebra 7th grade. Independent work for students of general education institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, “Mnemosyne”, 2010
Linear equation with two variables - any equation that has the following form: a*x + b*y =с. Here x and y are two variables, a,b,c are some numbers.
The solution to the linear equation a*x + b*y = c is any pair of numbers (x,y) that satisfies this equation, that is, turns the equation with variables x and y into a correct numerical equality. A linear equation has an infinite number of solutions.
If each pair of numbers that are solutions to a linear equation in two variables is depicted on the coordinate plane as points, then all these points form the graph of a linear equation in two variables. The coordinates of the points will be our x and y values. In this case, the x value will be the abscissa, and the y value will be the ordinate.
Graph of a Linear Equation in Two Variables
The graph of a linear equation with two variables is the set of all possible points on the coordinate plane, the coordinates of which will be solutions to this linear equation. It is easy to guess that the graph will be a straight line. That is why such equations are called linear.
Construction algorithm
Algorithm for plotting a linear equation in two variables.
1. Draw coordinate axes, label them and mark the unit scale.
2. In a linear equation, put x = 0, and solve the resulting equation for y. Mark the resulting point on the graph.
3. In a linear equation, take the number 0 as y, and solve the resulting equation for x. Mark the resulting point on the graph
4. If necessary, take an arbitrary value of x and solve the resulting equation for y. Mark the resulting point on the graph.
5. Connect the resulting points and continue the graph beyond them. Sign the resulting straight line.
Example: Graph the equation 3*x - 2*y =6;
Let's put x=0, then - 2*y =6; y= -3;
Let's put y=0, then 3*x = 6; x=2;
We mark the obtained points on the graph, draw a straight line through them and label it. Look at the figure below, the graph should look exactly like this.
We have often come across equations of the form ax + b = 0, where a, b are numbers, x is a variable. For example, bx - 8 = 0, x + 4 = O, - 7x - 11 = 0, etc. The numbers a, b (equation coefficients) can be any, except for the case when a = 0.
The equation ax + b = 0, where a, is called a linear equation with one variable x (or a linear equation with one unknown x). We can solve it, that is, express x through a and b:
We noted earlier that quite often mathematical model the real situation is a linear equation with one variable or an equation that, after transformations, reduces to a linear one. Now let's look at this real situation.
From cities A and B, the distance between which is 500 km, two trains left towards each other, each with its own constant speed. It is known that the first train left 2 hours earlier than the second. 3 hours after the second train left, they met. What are the train speeds?
Let's create a mathematical model of the problem. Let x km/h be the speed of the first train, y km/h be the speed of the second train. The first one was on the road for 5 hours and, therefore, covered a distance of bx km. The second train was on the way for 3 hours, i.e. walked a distance of 3 km.
Their meeting took place at point C. Figure 31 shows a geometric model of the situation. In algebraic language it can be described as follows:
5x + Zu = 500
or
5x + Zu - 500 = 0.
This mathematical model is called a linear equation with two variables x, y.
At all,
ax + by + c = 0,
where a, b, c are numbers, and , is linear the equation with two variables x and y (or with two unknowns x and y).
Let's return to the equation 5x + 3 = 500. We note that if x = 40, y = 100, then 5 40 + 3 100 = 500 is a correct equality. This means that the answer to the question of the problem can be as follows: the speed of the first train is 40 km/h, the speed of the second train is 100 km/h. A pair of numbers x = 40, y = 100 is called a solution to the equation 5x + 3 = 500. It is also said that this pair of values (x; y) satisfies the equation 5x + 3 = 500.
Unfortunately, this solution is not the only one (we all love certainty and unambiguity). In fact, the following option is also possible: x = 64, y = 60; indeed, 5 64 + 3 60 = 500 is a correct equality. And this: x = 70, y = 50 (since 5 70 + 3 50 = 500 is a true equality).
But, say, a pair of numbers x = 80, y = 60 is not a solution to the equation, since with these values a true equality does not work:
In general, a solution to the equation ax + by + c = 0 is any pair of numbers (x; y) that satisfies this equation, that is, turns the equality with the variables ax + by + c = 0 into a true numerical equality. There are infinitely many such solutions.
Comment. Let us return once again to the equation 5x + 3 = 500, obtained in the problem discussed above. Among the infinite number of its solutions there are, for example, the following: x = 100, y = 0 (indeed, 5 100 + 3 0 = 500 is a correct numerical equality); x = 118, y = - 30 (since 5,118 + 3 (-30) = 500 is a correct numerical equality). However, being solutions to the equation, these pairs cannot serve as solutions to this problem, because the speed of the train cannot be equal to zero (then it does not move, but stands still); Moreover, the speed of the train cannot be negative (then it does not travel towards another train, as stated in the problem statement, but in the opposite direction).
Example 1. Draw solutions to a linear equation with two variables x + y - 3 = 0 by points in the xOy coordinate plane.
Solution. Let's select several solutions given equation, i.e. several pairs of numbers that satisfy the equation: (3; 0), (2; 1), (1; 2) (0; 3), (- 2; 5).
A. V. Pogorelov, Geometry for grades 7-11, Textbook for educational institutions
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