Examples of systems of linear equations: solution method. Simple and complex methods for solving systems of equations

When solving problems using equations, we were usually looking for one unknown. But there are also problems where there are several unknowns. Such problems are usually solved by constructing systems of equations.

Two cyclists are traveling towards each other from one city to another, the distance between them is 30 km. Suppose that if cyclist 1 leaves 2 hours earlier than his friend, then they will meet 2.5 hours after cyclist 2 leaves; if cyclist 2 leaves 2 hours earlier than cyclist 1, then the meeting will take place 3 hours after the first one leaves. How fast is each cyclist traveling?

Solution.

1. Let us define the speed of cyclist 1 as x km/h, and the speed of cyclist 2 as y km/h.

2. If the first cyclist leaves 2 hours earlier than the second, then, according to the condition, he will ride 4.5 hours to the meeting, while the second will take 2.5 hours. In 4.5 hours the first will cover a distance of 4.5 km, and in 2.5 hours the second will travel a distance of 2.5 km.

3. The meeting of two cyclists means that they have covered a total distance of 30 km, i.e. 4.5x + 2.5 y = 30. This is our first equation.

4. If the second one leaves for 2 hours earlier than first, then, according to the condition, he will travel 5 hours to the meeting, while the first one will take 3 hours. Using reasoning similar to the above reasoning, we arrive at the equation:

5. So, we got a system of equations

(4.5x + 2.5 y = 30,
(3x + 5y = 30.

6. Having solved the resulting system of equations, we will find the roots: x = 5, y = 3.

Thus, the first cyclist travels at a speed of 5 km/h, and the second – 3 km/h.

Answer: 5 km/h, 3 km/h.

After a year, the investor received $6 interest on his savings. By adding $44, the investor left the money for another year. At the end of the year, interest was accrued again, and now the deposit together with interest amounted to $257.5. What was the initial deposit amount and how much interest does the bank charge?

Solution.

1. Let x ($) be the initial deposit, and y (%) be the interest that accrues annually.

2. Then by the end of the year (y/100) ∙ x $ will be added to the initial contribution.
From the condition we obtain the equation (ух/100) = 6.

3. By condition, it is known that at the end of the year the investor contributed another $44, so the contribution at the beginning of the second year was x + 6 + 44, i.e. (x + 50) $. Thus, the amount received at the end of the second year, taking into account accruals, was equal to (x + 50 + (y/100)(x + 50)) $. According to the condition, this amount is equal to $275.5. This allowed us to create a second equation:

x + 50 + (y/100)(x + 50) = 257.5

4. So, we got a system of equations:

((x/100) = 6,
(x + 50 + (y/100)(x + 50) = 257.5

After transforming the system of equations we get:

(xy = 600,
(100x + 50y + xy = 20750.

Having solved the system of equations, we found two roots: 200 and 1.5. Only the first value satisfies our condition.

Substitute the value of x into the equation and find the value of y:
if x = 200, then y = 3.

Thus, the initial deposit was $200, and the bank makes an accrual of 3% per year.

Answer: $200; 3%.

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Systems of equations have been widely used in the economic industry with mathematical modeling 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 common analytical method solutions to such systems, all methods are based on numerical solutions. IN school course mathematics, such methods as permutation, algebraic addition, substitution, as well as graphical and matrix method, solution by Gaussian method.

The main task when teaching solution methods 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 using a particular method

Solving examples of systems of linear equations of the 7th grade program secondary school 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. 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 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 inappropriate.

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, they perform term-by-term addition and multiplication of equations by different 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 arithmetic action 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.

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 the standard one 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 given example a=1, b=16, c=39, therefore D=100. If the discriminant greater than zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant 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.

Visual method for solving systems

Suitable for 3 equation systems. The method is to build on coordinate axis graphs of each equation included in the system. The coordinates of the points of intersection of the curves and will be general decision systems.

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.

IN following example need to find graphic solution systems 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 for short note systems of linear equations. A matrix is a table special type 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. Matrix with units along one of the diagonals and others zero elements called unit.

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 members equations, one equation - one row of the matrix.

A row of a matrix is said to be nonzero 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 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 - 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 variable systems with a large number of linear equations.

Gauss's method is very similar to solutions using substitutions and algebraic addition, but 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 Gaussian 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 students to understand high school, but is one of the most interesting ways to develop the ingenuity of children enrolled in advanced learning programs in math 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 continues to perform the necessary algebraic operations 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.

Being able to solve systems of linear equations is very good, but solving systems of equations in itself is just a method for more complex tasks. Using systems of equations you can solve various tasks that we encounter in life.

Algebra is the science of solving equations and systems of equations. This is exactly the definition that scientists used by the end of the 20th century. The famous scientist Rene Descartes is famous for one of his works, which is called “Descartes Method”. Descartes believed that any problem can be reduced to a mathematical one, any math problem can be reduced to algebraic system equations. And any system can be reduced to solving a single equation.

Unfortunately, Descartes did not have time to fully complete his method and did not write all of its points, but the idea is very good.

And now we, like Descartes, will solve problems using systems of equations, of course, not any, but only those that can be reduced to solving systems of linear equations.

General scheme for solving the problem using systems of equations

Let us describe the general scheme for solving problems using systems of equations:

1. For unknown quantities, we introduce certain notations and compose a system of linear equations.
2. Solve the resulting system of linear equations.
3. I use the entered notations and write down the answer.

Let's try to apply this diagram on a specific task.

It is known that two pencils and three notebooks cost 35 rubles, and two notebooks and three pencils cost 40 rubles. You need to find out how much five pencils and six notebooks cost.

Solution:

We need to find how much one pencil and one notebook cost separately. If we have such data, then it will not be difficult to decide how much five pencils and six notebooks cost.

Let us denote by x the price of one pencil in rubles. And y is the price of one notebook in rubles. Now carefully read the condition and create an equation.

“two pencils and three notebooks cost 35 rubles” means

2*x+3*y = 35;

“two notebooks and three pencils cost 40 rubles” therefore

3*x+2*y = 40;

We get a system of equations:

(2*x+3*y = 35;
(3*x+2*y = 40;

The first point is over. Now it is necessary to solve the resulting system of equations using any of the known methods.

Having solved, we get x=10 and y=5.

Returning to the original notation, we have that the price of one pencil is 10 rubles, and the price of one notebook is 5 rubles.

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Against the tide

Downstream

No. 1193. Mathematics 5th grade. N.Ya.Vilenkin

? km/h

№ 14.1

The distance between two points along the river is 80 km. The boat travels this distance along the river in 4 hours, and against the current in 5 hours. Find the speed of the boat downstream and upstream.

Downstream

4(x+y)

5(x-y)

Answer:

№ 14.4

A boat travels 10 km downstream in 4 hours less than in 6 hours against the current. Find own speed boats, if a raft on the same river in 15 hours floats the same distance in 15 hours as a boat travels on a lake in 2 hours.

Anti-flow

Downstream

4(x+y)

on 10

6(x-y)

4(x+y) +10 =6(x-y)

4x+4y+10=6x-6y

4x-6x+4y+6y=-10

Answer:

№ 14.10

No. ha

in 1 day

Qty

days

Total ha

1 tractor driver

2 tractor driver

№ 14.10

Two tractor drivers plowed 678 hectares together. The first tractor driver worked for 8 days, and the second for 11 days. How many hectares did each tractor driver plow per day, if the first tractor driver plowed 22 hectares less in every 3 days than the second in 4 days?

No. ha

in 1 day

Qty

days

Total ha

1 tractor driver

on 22 hectares

less

2 tractor driver

Answer:

№ 14.5

A motor ship travels 120 km in 5 hours against the flow of the river and 180 km in 6 hours downstream. Find the speed of the river flow and the ship's own speed.

Downstream

6(x+y)

5(x-y)

Answer:

№ 14.11

Qty.

in 1 hour

Qty

hours

Total

brigade

№ 14.11

Two teams worked harvesting potatoes. On the first day, one team worked for 2 hours, and the second for 3 hours, and they collected 23 centners of potatoes. On the second day, the first team collected 2 quintals more in 3 hours of work than the second in 2 hours. How many centners of potatoes did each team harvest in 1 hour of work?

Qty.

in 1 hour

Qty

hours

Total

brigade

by 2 ct

more

brigade

Answer:

№ 14.7

x-1 number

y-2 number

3(x-y)=(x+y)+6

2(x-y)=(x+y)+9

Answer:

№ 14.12

Quantity t

for 1 flight

Qty

flights

Total

tons

car

№ 14.12

On the first day, 27 tons of grain were exported, with one vehicle making 4 trips and the other making 3 trips. The next day, the second car transported 11 tons more in 4 trips than the first car in 3 trips. How many tons of grain were transported on each vehicle in one trip?

Quantity t

for 1 flight

Qty

flights

Total

tons

car

at 11t

more

car

Answer:

№ 14.14

Quantity kg

in 1 box

Qty

boxes

Total

cherries

for 3 drawers

less

cherry

№ 14.14

84 kg of cherries and sour cherries were purchased at the market, and 3 less boxes of cherries were purchased than cherries. How many boxes of cherries and sour cherries were purchased separately, if 1 box contains 8 kg of cherries and 10 kg of sour cherries?

Quantity kg

in 1 box

Qty

boxes

Total

cherries

cherry