Systems of equations received wide application in the economic sector in mathematical modeling of various processes. For example, when solving problems of production management and planning, logistics routes ( transport problem) or equipment placement.
Systems of equations are used not only in mathematics, but also in physics, chemistry and biology, when solving problems of finding population size.
System linear equations name 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 whose value must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving an equation by plotting it will look like a straight line, all points of which are solutions to the polynomial.
Types of systems of linear equations
The simplest examples are considered to be 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 system of equations - this means finding values (x, y) at which the system turns into a true equality or establishing that suitable values of x and y do not exist.
A pair of values (x, y), written as the coordinates of a point, is called a solution to a system of linear equations.
If systems have one common solution or no solution exists, they are called equivalent.
Homogeneous systems of linear equations are systems right side which 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 heterogeneous.
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 or more variables.
When faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not the case. The number of equations in the system does not depend on the variables; there can be as many of them as desired.
Simple and complex methods for solving systems of equations
There is no general analytical method for solving such systems; all methods are based on numerical solutions. IN school course Mathematics describes in detail such methods as permutation, algebraic addition, substitution, as well as graphical and matrix methods, solution by the Gaussian method.
The main task when teaching solution methods is to teach how to correctly analyze the system and find optimal algorithm solutions 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 using a particular method
Solving examples of systems of linear equations in the 7th grade general education curriculum is quite simple and explained in great detail. In any mathematics textbook, this section is given enough attention. Solving examples of systems of linear equations using the Gauss and Cramer method is studied in more detail in the first years of higher education.
Solving systems using the substitution method
The actions of the substitution method are aimed at expressing the value of one variable in terms of the second. The expression is substituted into the remaining equation, then it is reduced to a form with one variable. The action is repeated depending on the number of unknowns in the system
Let us give a solution to an example of a system of linear equations of class 7 using 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. Solution this example does not cause difficulties and allows you to obtain the Y value. Last step This is a check of the received values.
It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and expressing 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, solving by substitution is also impractical.
Solution of an example of a system of linear inhomogeneous equations:
Solution using algebraic addition
When searching for solutions to systems using the addition method, equations are added term by term and multiplied by various numbers. The ultimate goal mathematical operations is an equation with one variable.
For Applications this method practice and observation are required. Solving a system of linear equations using the addition method when there are 3 or more variables is not easy. Algebraic addition is convenient to use when equations contain fractions and decimals.
Solution algorithm:
- Multiply both sides of the equation by a certain number. As a result of the arithmetic operation, one of the coefficients of the variable should 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.
Method of solution by introducing a new variable
A new variable can be introduced if the system requires finding 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 for the introduced unknown, and the resulting value is used to determine the original variable.
The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to a standard quadratic trinomial. You can solve a polynomial by finding the discriminant.
It is necessary to find the discriminant value by well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the factors of the polynomial. In the given example, a=1, b=16, c=39, therefore 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 one solution: x = -b / 2*a.
The solution for the resulting systems is found by the addition method.
Visual method for solving systems
Suitable for 3 equation systems. The method consists in constructing graphs of each equation included in the system on the coordinate axis. The coordinates of the intersection points of the curves will be the general solution of the system.
The graphical method has a number of nuances. Let's look at several examples of solving systems of linear equations in a visual way.
As can be seen from the example, for each line two points were constructed, 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.
The following example requires finding a graphical solution to a 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 a system has a solution or not; it is always necessary to construct a graph.
The matrix and its varieties
Matrices are used to concisely write 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 are equal. A matrix-vector is a matrix of one column with infinite possible number lines. A matrix with ones along one of the diagonals and other zero elements is called identity.
An inverse matrix is a matrix when multiplied by which the original one turns into a unit matrix; such a matrix exists only for the original square one.
Rules for converting a system of equations into a matrix
In relation to systems of equations, the coefficients and free terms of the equations are written as matrix numbers; one equation is one row of the matrix.
A matrix row is said to be nonzero if at least one element of the row is not 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 matrix columns must strictly correspond to the variables. This means that the coefficients of the variable x can be written only in one column, for example the first, the coefficient of the unknown y - only in the second.
When multiplying a matrix, all elements of the matrix 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| is the determinant of the matrix. |K| must not be equal to zero, then the system has a solution.
The determinant is easily calculated for a two-by-two matrix; you just need to multiply the diagonal elements 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 numbers of columns and rows of elements are not repeated in the work.
Solving examples of systems of linear equations using the matrix method
The matrix method of finding a solution allows you to reduce cumbersome entries when solving systems with a large number variables and equations.
In the example, a nm are the coefficients of the equations, the matrix is a vector x n are variables, and b n are free terms.
Solving systems using the Gaussian method
In higher mathematics, the Gaussian method is studied together with the Cramer method, and the process of finding solutions to systems is called the Gauss-Cramer solution method. These methods are used to find variables of systems with a large number of linear equations.
The Gauss method is very similar to solutions by substitution and algebraic addition, but is more systematic. In the school course, the solution by the Gaussian method is used for systems of 3 and 4 equations. The purpose of the method is to reduce 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, while 3 and 4 are, respectively, with 3 and 4 variables.
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 solution by the Gauss method 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. Solving 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 secondary school students to understand, but it is one of the most interesting ways to develop the ingenuity of children enrolled in advanced study programs in mathematics and physics classes.
For ease of recording, calculations are usually done as follows:
The coefficients of the equations 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 left side equations from the right. Roman numerals indicate the numbers of equations in the system.
First, write down the matrix to be worked with, then all the actions carried out with one of the rows. The resulting matrix is written after the "arrow" sign and the necessary algebraic operations are continued until the result is achieved.
The result should be a matrix in which one of the diagonals is equal to 1, and all other coefficients are equal to zero, that is, the matrix is reduced to a unit form. We must not forget to perform calculations with numbers on both sides of the equation.
This recording method is less cumbersome and allows you not to be distracted by listing numerous unknowns.
The free use of any solution method will require care and some experience. Not all methods are of an applied nature. Some methods of finding solutions are more preferable in a particular area of human activity, while others exist for educational purposes.
In this lesson we will continue to study the method of solving systems of equations, namely the method of algebraic addition. First, let's look at the application of this method using the example of linear equations and its essence. Let's also remember how to equalize coefficients in equations. And we will solve a number of problems using this method.
Topic: Systems of equations
Lesson: Algebraic addition method
1. Method of algebraic addition using linear systems as an example
Let's consider algebraic addition method using the example of linear systems.
Example 1. Solve the system
If we add these two equations, then y cancels out, leaving an equation for x.
If we subtract the second from the first equation, the x's cancel each other out, and we get an equation for y. This is the meaning of the algebraic addition method.
We solved the system and remembered the method of algebraic addition. Let's repeat its essence: we can add and subtract equations, but we must ensure that we get an equation with only one unknown.
2. Method of algebraic addition with preliminary equalization of coefficients
Example 2. Solve the system
The term is present in both equations, so the algebraic addition method is convenient. Let's subtract the second from the first equation.
Answer: (2; -1).
Thus, after analyzing the system of equations, you can see that it is convenient for the method of algebraic addition, and apply it.
Let's consider another linear system.
3. Solution of nonlinear systems
Example 3. Solve the system
We want to get rid of y, but the coefficients of y are different in the two equations. Let's equalize them; to do this, multiply the first equation by 3, the second by 4.
Example 4. Solve the system 
Let's equalize the coefficients for x
You can do it differently - equalize the coefficients for y.
We solved the system by applying the algebraic addition method twice.
The algebraic addition method is also applicable to solving nonlinear systems.
Example 5. Solve the system
Let's add these equations together and we'll get rid of y.
The same system can be solved by applying the algebraic addition method twice. Let's add and subtract from one equation another.
Example 6. Solve the system 
Answer: 
Example 7. Solve the system 
Using the method of algebraic addition we will get rid of the xy term. Let's multiply the first equation by .
The first equation remains unchanged, instead of the second we write the algebraic sum.
Answer: 
Example 8. Solve the system
Multiply the second equation by 2 to isolate a perfect square.
Our task was reduced to solving four simple systems.
4. Conclusion
We examined the method of algebraic addition using the example of solving linear and nonlinear systems. In the next lesson we will look at the method of introducing new variables.
1. Mordkovich A.G. et al. Algebra 9th grade: Textbook. For general education Institutions.- 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.
2. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students educational institutions/ A. G. Mordkovich, T. N. Mishustina and others - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill.
3. Makarychev Yu. N. Algebra. 9th grade: educational. 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. Alimov Sh. A., Kolyagin Yu. M., Sidorov Yu. V. Algebra. 9th grade. 16th ed. - M., 2011. - 287 p.
5. Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. — 12th ed., erased. - M.: 2010. - 224 p.: ill.
6. Algebra. 9th grade. In 2 parts. Part 2. Problem book for students of general education 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“I WILL SOLVE THE USE.”
1. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill. No. 125 - 127.
Need to download lesson plan on topic » Algebraic addition method?
Let us analyze two types of solutions to systems of equations:
1. Solving the system using the substitution method.
2. Solving the system by term-by-term addition (subtraction) of the system equations.
In order to solve the system of equations by substitution method you need to follow a simple algorithm:
1. Express. From any equation we express one variable.
2. Substitute. We substitute the resulting value into another equation instead of the expressed variable.
3. Solve the resulting equation with one variable. We find a solution to the system.
To decide system by term-by-term addition (subtraction) method need to:
1. Select a variable for which we will make identical coefficients.
2. We add or subtract equations, resulting in an equation with one variable.
3. Solve the resulting linear equation. We find a solution to the system.
The solution to the system is the intersection points of the function graphs.
Let us consider in detail the solution of systems using examples.
Example #1:
Let's solve by substitution method
Solving a system of equations using 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, which means that it is easiest to express the variable x from the second equation.
x=3+10y
2.After we have expressed it, we substitute 3+10y into the first equation instead of the variable x.
2(3+10y)+5y=1
3. Solve the resulting equation with one variable.
2(3+10y)+5y=1 (open the brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2
The solution to the equation system 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 point where we expressed it, 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 using the term-by-term addition (subtraction) method.
Solving a system of equations using the addition method3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)
1. We choose a variable, let’s say we choose 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. Subtract the second from the first equation 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 into any of the equations, let’s say into 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 intersection point will be x=4.6; y=6.4
Answer: (4.6; 6.4)
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Using this mathematical program, you can solve a system of two linear equations with two variables using the substitution method and the addition method.
The program not only gives the answer to the problem, but also gives detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.
This program may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.
In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.
Rules for entering equations
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.
When entering equations you can use parentheses. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of elements.
For example: 6x+1 = 5(x+y)+2
You can use not only integers in equations, but also fractional numbers in the form of decimals and ordinary fractions.
Rules for entering decimal fractions.
Integer and fractional parts in decimals can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
Whole part separated from the fraction by an ampersand: &
Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)
Solve system of equations
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A little theory.
Solving systems of linear equations. Substitution method
The sequence of actions when solving a system of linear equations using the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression into another equation of the system instead of this variable;
$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$
Let's express y in terms of x from the first equation: y = 7-3x. Substituting the expression 7-3x into the second equation instead of y, we obtain the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$
It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$
Substituting the number 1 instead of x into the equality y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$
Pair (1;4) - solution of the system
Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.
Solving systems of linear equations by addition
Let's consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by substitution, we move from this system to another, equivalent system, in which one of the equations contains only one variable.
The sequence of actions when solving a system of linear equations using the addition method:
1) multiply the equations of the system term by term, selecting factors so that the coefficients of one of the variables become opposite numbers;
2) add the left and right sides of the system equations term by term;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.
Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$
In the equations of this system, the coefficients of y are opposite numbers. By adding the left and right sides of the equations term by term, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$
From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38\) we get an equation with the variable y: \(11-3y=38\). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)
Thus, we found the solution to the system of equations by addition: \(x=11; y=-9\) or \((11;-9)\)
Taking advantage of the fact that in the equations of the system the coefficients of y are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both sides of each of the equations of the original system), in which one of the equations contains only one variable.
Books (textbooks) Abstracts of the Unified State Examination and the Unified State Examination tests online Games, puzzles Plotting graphs of functions Spelling dictionary of the Russian language Dictionary of youth slang Catalog of Russian schools Catalog of secondary educational institutions of Russia Catalog of Russian universities List of tasksThe use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. Only by solving systems of equations of varying complexity on your own will you learn to quickly determine methods for solving any system. Sometimes it can be quite difficult to solve the system quadratic equations. However, the most commonly used method to solve these equations is the substitution/addition method.
Suppose we are given the following system of equations:
\[\left\(\begin(matrix) x^2-xy = 3, \\ y^2-xy = -2 \end(matrix)\right.\]
Let's add the equations of the system:
\[\left\(\begin(matrix) x^2 - xy = 3, \\ x^2 - 2xy + y = 1. \end(matrix)\right.\]
Let's solve the resulting system:
\[\left\(\begin(matrix) x(x -y) = 3, \\ (x - y)^2= 1; \end(matrix)\right.\]
\[(x - y) = -1 \] or \[(x - y) = 1\] - we obtain from 2 equations
Let's substitute 1 or -1 into 1:
\ or \
Since we now know the value of one unknown, we can find the second:
\[-3 - y= -1\] or \
\ or \
Answer: \[(-3; -2); (3; 4)\]
If you need to solve a system of 2 degrees and 1 linear, then you can express 1 of the variables from the linear and substitute this equation into the quadratic one.
Where can I solve a system of quadratic equations online with a calculator?
You can solve the system of equations online on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.
