We will analyze two types of solving systems of equations:
1. Solution of the system by the substitution method.
2. Solution of the system by term-by-term addition (subtraction) of the equations of the system.
In order to solve the system of equations substitution method you need to follow a simple algorithm:
1. We express. From any equation, we express one variable.
2. Substitute. We substitute in another equation instead of the expressed variable, the resulting value.
3. We solve the resulting equation with one variable. We find a solution to the system.
To solve system by term-by-term addition (subtraction) need:
1. Select a variable for which we will make the same coefficients.
2. We add or subtract the equations, as a result we get an equation with one variable.
3. We solve the resulting linear equation. We find a solution to the system.
The solution of the system is the intersection points of the graphs of the function.
Let us consider in detail the solution of systems using examples.
Example #1:
Let's solve by the substitution method
Solving the system of equations by the substitution method2x+5y=1 (1 equation)
x-10y=3 (2nd equation)
1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, hence it turns out that it is easiest to express the variable x from the second equation.
x=3+10y
2. After expressing, we substitute 3 + 10y in the first equation instead of the variable x.
2(3+10y)+5y=1
3. We solve the resulting equation with one variable.
2(3+10y)+5y=1 (open brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2
The solution of the system of equations is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first paragraph where we expressed we substitute y there.
x=3+10y
x=3+10*(-0.2)=1
It is customary to write points in the first place, we write the variable x, and in the second place the variable y.
Answer: (1; -0.2)
Example #2:
Let's solve by term-by-term addition (subtraction).
Solving a system of equations by the addition method3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)
1. Select a variable, let's say we select x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get a total coefficient of 6.
3x-2y=1 |*2
6x-4y=2
2x-3y=-10 |*3
6x-9y=-30
2. From the first equation, subtract the second to get rid of the variable x. Solve the linear equation.
__6x-4y=2
5y=32 | :5
y=6.4
3. Find x. We substitute the found y in any of the equations, let's say in the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6
The point of intersection will be x=4.6; y=6.4
Answer: (4.6; 6.4)
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Algebraic addition method
You can solve a system of equations with two unknowns in various ways - a graphical method or a variable change method.
In this lesson, we will get acquainted with another way of solving systems that you will surely like - this is the algebraic addition method.
And where did the idea come from - to put something in the systems? When solving systems, the main problem is the presence of two variables, because we cannot solve equations with two variables. So, it is necessary to exclude one of them in some legal way. And such legitimate ways are mathematical rules and properties.
One of these properties sounds like this: the sum of opposite numbers is zero. This means that if there are opposite coefficients for one of the variables, then their sum will be equal to zero and we will be able to exclude this variable from the equation. It is clear that we do not have the right to add only the terms with the variable we need. It is necessary to add the equations as a whole, i.e. separately add like terms on the left side, then on the right. As a result, we will get a new equation containing only one variable. Let's take a look at specific examples.
We see that in the first equation there is a variable y, and in the second the opposite number is y. So this equation can be solved by the addition method.
One of the equations is left as it is. Any one you like best.
But the second equation will be obtained by adding these two equations term by term. Those. Add 3x to 2x, add y to -y, add 8 to 7.
We get a system of equations
The second equation of this system is a simple equation with one variable. From it we find x \u003d 3. Substituting the found value in the first equation, we find y \u003d -1.
Answer: (3; - 1).
Design sample:
Solve the system of equations by algebraic addition
There are no variables with opposite coefficients in this system. But we know that both sides of the equation can be multiplied by the same number. Let's multiply the first equation of the system by 2.
Then the first equation will take the form:
Now we see that with the variable x there are opposite coefficients. So, we will do the same as in the first example: we will leave one of the equations unchanged. For example, 2y + 2x \u003d 10. And we get the second by adding.
Now we have a system of equations:
We easily find from the second equation y = 1, and then from the first equation x = 4.
Design sample:
Let's summarize:
We have learned how to solve systems of two linear equations with two unknowns using the algebraic addition method. Thus, we now know three main methods for solving such systems: the graphical method, the change of variable method, and the addition method. Almost any system can be solved using these methods. In more complex cases, a combination of these techniques is used.
List of used literature:
- Mordkovich A.G., Algebra grade 7 in 2 parts, Part 1, Textbook for educational institutions / A.G. Mordkovich. - 10th ed., revised - Moscow, "Mnemosyne", 2007.
- Mordkovich A.G., Algebra grade 7 in 2 parts, Part 2, Task book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, Mnemosyne, 2007.
- HER. Tulchinskaya, Algebra Grade 7. Blitz survey: a guide for students of educational institutions, 4th edition, revised and supplemented, Moscow, Mnemozina, 2008.
- Alexandrova L.A., Algebra Grade 7. Thematic test papers in a new form for students of educational institutions, edited by A.G. Mordkovich, Moscow, "Mnemosyne", 2011.
- Aleksandrova L.A. Algebra 7th grade. Independent work for students of educational institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, "Mnemosyne", 2010.
Systems of equations are widely used in the economic industry in the mathematical modeling of various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) or equipment placement.
Equation systems are used not only in the field of mathematics, but also in physics, chemistry and biology, when solving problems of finding the population size.
A system of linear equations is a term for two or more equations with several variables for which it is necessary to find a common solution. Such a sequence of numbers for which all equations become true equalities or prove that the sequence does not exist.
Linear Equation
Equations of the form ax+by=c are called linear. The designations x, y are the unknowns, the value of which must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving the equation by plotting its graph will look like a straight line, all points of which are the solution of the polynomial.
Types of systems of linear equations
The simplest are examples of systems of linear equations with two variables X and Y.
F1(x, y) = 0 and F2(x, y) = 0, where F1,2 are functions and (x, y) are function variables.
Solve a system of equations - it means to find such values (x, y) for which the system becomes a true equality, or to establish that there are no suitable values of x and y.
A pair of values (x, y), written as point coordinates, is called a solution to a system of linear equations.
If the systems have one common solution or there is no solution, they are called equivalent.
Homogeneous systems of linear equations are systems whose right side is equal to zero. If the right part after the "equal" sign has a value or is expressed by a function, such a system is not homogeneous.
The number of variables can be much more than two, then we should talk about an example of a system of linear equations with three variables or more.
Faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not so. The number of equations in the system does not depend on the variables, there can be an arbitrarily large number of them.
Simple and complex methods for solving systems of equations
There is no general analytical way to solve such systems, all methods are based on numerical solutions. The school mathematics course describes in detail such methods as permutation, algebraic addition, substitution, as well as the graphical and matrix method, the solution by the Gauss method.
The main task in teaching methods of solving is to teach how to correctly analyze the system and find the optimal solution algorithm for each example. The main thing is not to memorize a system of rules and actions for each method, but to understand the principles of applying a particular method.
The solution of examples of systems of linear equations of the 7th grade of the general education school program is quite simple and is explained in great detail. In any textbook on mathematics, this section is given enough attention. The solution of examples of systems of linear equations by the method of Gauss and Cramer is studied in more detail in the first courses of higher educational institutions.
Solution of systems by the substitution method
The actions of the substitution method are aimed at expressing the value of one variable through the second. The expression is substituted into the remaining equation, then it is reduced to a single variable form. The action is repeated depending on the number of unknowns in the system
Let's give an example of a system of linear equations of the 7th class by the substitution method:
As can be seen from the example, the variable x was expressed through F(X) = 7 + Y. The resulting expression, substituted into the 2nd equation of the system in place of X, helped to obtain one variable Y in the 2nd equation. The solution of this example does not cause difficulties and allows you to get the Y value. The last step is to check the obtained values.
It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and the expression of the variable in terms of the second unknown will be too cumbersome for further calculations. When there are more than 3 unknowns in the system, the substitution solution is also impractical.
Solution of an example of a system of linear inhomogeneous equations:
Solution using algebraic addition
When searching for a solution to systems by the addition method, term-by-term addition and multiplication of equations by various numbers are performed. The ultimate goal of mathematical operations is an equation with one variable.
Applications of this method require practice and observation. It is not easy to solve a system of linear equations using the addition method with the number of variables 3 or more. Algebraic addition is useful when the equations contain fractions and decimal numbers.
Solution action algorithm:
- Multiply both sides of the equation by some number. As a result of the arithmetic operation, one of the coefficients of the variable must become equal to 1.
- Add the resulting expression term by term and find one of the unknowns.
- Substitute the resulting value into the 2nd equation of the system to find the remaining variable.
Solution method by introducing a new variable
A new variable can be introduced if the system needs to find a solution for no more than two equations, the number of unknowns should also be no more than two.
The method is used to simplify one of the equations by introducing a new variable. The new equation is solved with respect to the entered unknown, and the resulting value is used to determine the original variable.
It can be seen from the example that by introducing a new variable t, it was possible to reduce the 1st equation of the system to a standard square trinomial. You can solve a polynomial by finding the discriminant.
It is necessary to find the value of the discriminant using the well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the multipliers of the polynomial. In the given example, a=1, b=16, c=39, hence D=100. If the discriminant is greater than zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant is less than zero, then there is only one solution: x= -b / 2*a.
The solution for the resulting systems is found by the addition method.
A visual method for solving systems
Suitable for systems with 3 equations. The method consists in plotting graphs of each equation included in the system on the coordinate axis. The coordinates of the points of intersection of the curves will be the general solution of the system.
The graphic method has a number of nuances. Consider several examples of solving systems of linear equations in a visual way.
As can be seen from the example, two points were constructed for each line, the values of the variable x were chosen arbitrarily: 0 and 3. Based on the values of x, the values for y were found: 3 and 0. Points with coordinates (0, 3) and (3, 0) were marked on the graph and connected by a line.
The steps must be repeated for the second equation. The point of intersection of the lines is the solution of the system.
In the following example, it is required to find a graphical solution to the system of linear equations: 0.5x-y+2=0 and 0.5x-y-1=0.
As can be seen from the example, the system has no solution, because the graphs are parallel and do not intersect along their entire length.
The systems from Examples 2 and 3 are similar, but when constructed, it becomes obvious that their solutions are different. It should be remembered that it is not always possible to say whether the system has a solution or not, it is always necessary to build a graph.
Matrix and its varieties
Matrices are used to briefly write down a system of linear equations. A matrix is a special type of table filled with numbers. n*m has n - rows and m - columns.
A matrix is square when the number of columns and rows is equal. A matrix-vector is a single-column matrix with an infinitely possible number of rows. A matrix with units along one of the diagonals and other zero elements is called identity.
An inverse matrix is such a matrix, when multiplied by which the original one turns into a unit one, such a matrix exists only for the original square one.
Rules for transforming a system of equations into a matrix
With regard to systems of equations, the coefficients and free members of the equations are written as numbers of the matrix, one equation is one row of the matrix.
A matrix row is called non-zero if at least one element of the row is not equal to zero. Therefore, if in any of the equations the number of variables differs, then it is necessary to enter zero in place of the missing unknown.
The columns of the matrix must strictly correspond to the variables. This means that the coefficients of the variable x can only be written in one column, for example the first, the coefficient of the unknown y - only in the second.
When multiplying a matrix, all matrix elements are sequentially multiplied by a number.
Options for finding the inverse matrix
The formula for finding the inverse matrix is quite simple: K -1 = 1 / |K|, where K -1 is the inverse matrix and |K| - matrix determinant. |K| must not be equal to zero, then the system has a solution.
The determinant is easily calculated for a two-by-two matrix, it is only necessary to multiply the elements diagonally by each other. For the "three by three" option, there is a formula |K|=a 1 b 2 c 3 + a 1 b 3 c 2 + a 3 b 1 c 2 + a 2 b 3 c 1 + a 2 b 1 c 3 + a 3 b 2 c 1 . You can use the formula, or you can remember that you need to take one element from each row and each column so that the column and row numbers of the elements do not repeat in the product.
Solution of examples of systems of linear equations by the matrix method
The matrix method of finding a solution makes it possible to reduce cumbersome notations when solving systems with large quantity variables and equations.
In the example, a nm are the coefficients of the equations, the matrix is a vector x n are the variables, and b n are the free terms.
Solution of systems by the Gauss method
In higher mathematics, the Gauss method is studied together with the Cramer method, and the process of finding a solution to systems is called the Gauss-Cramer method of solving. These methods are used to find the variables of systems with a large number of linear equations.
The Gaussian method is very similar to substitution and algebraic addition solutions, but is more systematic. In the school course, the Gaussian solution is used for systems of 3 and 4 equations. The purpose of the method is to bring the system to the form of an inverted trapezoid. By algebraic transformations and substitutions, the value of one variable is found in one of the equations of the system. The second equation is an expression with 2 unknowns, and 3 and 4 - with 3 and 4 variables, respectively.
After bringing the system to the described form, the further solution is reduced to the sequential substitution of known variables into the equations of the system.
In school textbooks for grade 7, an example of a Gaussian solution is described as follows:
As can be seen from the example, at step (3) two equations were obtained 3x 3 -2x 4 =11 and 3x 3 +2x 4 =7. The solution of any of the equations will allow you to find out one of the variables x n.
Theorem 5, which is mentioned in the text, states that if one of the equations of the system is replaced by an equivalent one, then the resulting system will also be equivalent to the original one.
The Gauss method is difficult for middle school students to understand, but is one of the most interesting ways to develop the ingenuity of children studying in the advanced study program in math and physics classes.
For ease of recording calculations, it is customary to do the following:
Equation coefficients and free terms are written in the form of a matrix, where each row of the matrix corresponds to one of the equations of the system. separates the left side of the equation from the right side. Roman numerals denote the numbers of equations in the system.
First, they write down the matrix with which to work, then all the actions carried out with one of the rows. The resulting matrix is written after the "arrow" sign and continue to perform the necessary algebraic operations until the result is achieved.
As a result, a matrix should be obtained in which one of the diagonals is 1, and all other coefficients are equal to zero, that is, the matrix is reduced to a single form. We must not forget to make calculations with the numbers of both sides of the equation.
This notation is less cumbersome and allows you not to be distracted by listing numerous unknowns.
The free application of any method of solution will require care and a certain amount of experience. Not all methods are applied. Some ways of finding solutions are more preferable in a particular area of human activity, while others exist for the purpose of learning.
A system of linear equations with two unknowns is two or more linear equations for which it is necessary to find all their common solutions. We will consider systems of two linear equations with two unknowns. A general view of a system of two linear equations with two unknowns is shown in the figure below:
( a1*x + b1*y = c1,
( a2*x + b2*y = c2
Here x and y are unknown variables, a1, a2, b1, b2, c1, c2 are some real numbers. A solution to a system of two linear equations with two unknowns is a pair of numbers (x, y) such that if these numbers are substituted into the equations of the system, then each of the equations of the system turns into a true equality. There are several ways to solve a system of linear equations. Consider one of the ways to solve a system of linear equations, namely the addition method.
Algorithm for solving by addition method
An algorithm for solving a system of linear equations with two unknown addition methods.
1. If required, by means of equivalent transformations, equalize the coefficients for one of the unknown variables in both equations.
2. Adding or subtracting the resulting equations to get a linear equation with one unknown
3. Solve the resulting equation with one unknown and find one of the variables.
4. Substitute the resulting expression into any of the two equations of the system and solve this equation, thus obtaining the second variable.
5. Check the solution.
An example of a solution by the addition method
For greater clarity, we solve the following system of linear equations with two unknowns by the addition method:
(3*x + 2*y = 10;
(5*x + 3*y = 12;
Since none of the variables has the same coefficients, we equalize the coefficients of the variable y. To do this, multiply the first equation by three, and the second equation by two.
(3*x+2*y=10 |*3
(5*x + 3*y = 12 |*2
Get the following system of equations:
(9*x+6*y = 30;
(10*x+6*y=24;
Now subtract the first from the second equation. We present like terms and solve the resulting linear equation.
10*x+6*y - (9*x+6*y) = 24-30; x=-6;
We substitute the resulting value into the first equation from our original system and solve the resulting equation.
(3*(-6) + 2*y =10;
(2*y=28; y=14;
The result is a pair of numbers x=6 and y=14. We are checking. We make a substitution.
(3*x + 2*y = 10;
(5*x + 3*y = 12;
{3*(-6) + 2*(14) = 10;
{5*(-6) + 3*(14) = 12;
{10 = 10;
{12=12;
As you can see, we got two true equalities, therefore, we found the right solution.
Very often, students find it difficult to choose a method for solving systems of equations.
In this article, we will consider one of the ways to solve systems - the substitution method.
If a common solution of two equations is found, then these equations are said to form a system. In a system of equations, each unknown represents the same number in all equations. To show that these equations form a system, they are usually written one below the other and combined with a curly bracket, for example
We note that for x = 15, and y = 5, both equations of the system are correct. This pair of numbers is the solution to the system of equations. Each pair of unknown values that simultaneously satisfies both equations of the system is called a solution to the system.
A system can have one solution (as in our example), infinitely many solutions, and no solutions.
How to solve systems using the substitution method? If the coefficients for some unknown in both equations are equal in absolute value (if they are not equal, then we equalize), then by adding both equations (or subtracting one from the other), you can get an equation with one unknown. Then we solve this equation. We define one unknown. We substitute the obtained value of the unknown into one of the equations of the system (in the first or in the second). We find another unknown. Let's look at examples of the application of this method.
Example 1 Solve System of Equations
Here the coefficients at y are equal in absolute value, but opposite in sign. Let's try term by term to add the equations of the system.
The resulting value x \u003d 4, we substitute into some equation of the system (for example, into the first one) and find the value of y:
2 * 4 + y \u003d 11, y \u003d 11 - 8, y \u003d 3.
Our system has a solution x = 4, y = 3. Or the answer can be written in parentheses, as the coordinates of a point, in the first place x, in the second y.
Answer: (4; 3)
Example 2. Solve a system of equations
We equalize the coefficients for the variable x, for this we multiply the first equation by 3, and the second by (-2), we get
Be Careful When Adding Equations
Then y \u003d - 2. We substitute the number (-2) instead of y in the first equation, we get
4x + 3 (-2) \u003d - 4. We solve this equation 4x \u003d - 4 + 6, 4x \u003d 2, x \u003d ½.
Answer: (1/2; - 2)
Example 3 Solve System of Equations
Multiply the first equation by (-2)
Solving the system
we get 0 = - 13.
There is no solution system, since 0 is not equal to (-13).
Answer: There are no solutions.
Example 4 Solve System of Equations
Note that all the coefficients of the second equation are divisible by 3,
let's divide the second equation by three and we get a system that consists of two identical equations.
This system has infinitely many solutions, since the first and second equations are the same (we got only one equation with two variables). How to present the solution of this system? Let's express the variable y from the equation x + y = 5. We get y = 5 - x.
Then answer will be written like this: (x; 5-x), x is any number.
We considered the solution of systems of equations by the addition method. If you have any questions or something is not clear, sign up for a lesson and we will fix all the problems with you.
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