How to write a solution to a system of equations. Systems of linear equations
Instruction
Addition method.
You need to write two strictly under each other:
549+45y+4y=-7, 45y+4y=549-7, 49y=542, y=542:49, y≈11.
In an arbitrarily chosen (from the system) equation, insert the number 11 instead of the already found "game" and calculate the second unknown:
X=61+5*11, x=61+55, x=116.
The answer of this system of equations: x=116, y=11.
Graphic way.
It consists in the practical finding of the coordinates of the point at which the lines are mathematically written in the system of equations. You should draw graphs of both lines separately in the same coordinate system. General view: - y \u003d kx + b. To construct a straight line, it is enough to find the coordinates of two points, and x is chosen arbitrarily.
Let the system be given: 2x - y \u003d 4
Y \u003d -3x + 1.
A straight line is built according to the first one, for convenience it needs to be written down: y \u003d 2x-4. Come up with (easier) values for x, substituting it into the equation, solving it, find y. Two points are obtained, along which a straight line is built. (see pic.)
x 0 1
y -4 -2
A straight line is constructed according to the second equation: y \u003d -3x + 1.
Also build a line. (see pic.)
1-5
Find the coordinates of the intersection point of two constructed lines on the graph (if the lines do not intersect, then the system of equations does not have - so).
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If the same system of equations is solved by three different ways, the answer will be the same (if the solution is correct).
Sources:
- Algebra Grade 8
- solve an equation with two unknowns online
- Examples of solving systems of linear equations with two
System equations is a collection of mathematical records, each of which contains a certain number of variables. There are several ways to solve them.
You will need
- -Ruler and pencil;
- -calculator.
Instruction
Consider the sequence of solving the system, which consists of linear equations having the form: a1x + b1y = c1 and a2x + b2y = c2. Where x and y are unknown variables and b,c are free members. When applying this method, each system is the coordinates of the points corresponding to each equation. First, in each case, express one variable in terms of the other. Then set the x variable to any number of values. Two is enough. Plug into the equation and find y. Build a coordinate system, mark the obtained points on it and draw a straight line through them. Similar calculations must be carried out for other parts of the system.
The system has a unique solution if the constructed lines intersect and one common point. It is inconsistent if they are parallel to each other. And it has infinitely many solutions when the lines merge with each other.
This method is considered to be very clear. The main disadvantage is that the calculated unknowns have approximate values. A more accurate result is given by the so-called algebraic methods.
Any solution to a system of equations is worth checking. To do this, substitute the obtained values instead of the variables. You can also find its solution in several ways. If the solution of the system is correct, then everyone should turn out the same.
Often there are equations in which one of the terms is unknown. To solve an equation, you need to remember and do a certain set of actions with these numbers.
You will need
- - paper;
- - Pen or pencil.
Instruction
Imagine that you have 8 rabbits in front of you, and you only have 5 carrots. Think you need to buy more carrots so that each rabbit gets a carrot.
Let's represent this problem in the form of an equation: 5 + x = 8. Let's substitute the number 3 for x. Indeed, 5 + 3 = 8.
When you substituted a number for x, you were doing the same operation as subtracting 5 from 8. Thus, to find unknown term, subtract the known term from the sum.
Let's say you have 20 rabbits and only 5 carrots. Let's compose . An equation is an equality that holds only for certain values of the letters included in it. The letters whose values you want to find are called. Write an equation with one unknown, call it x. When solving our problem about rabbits, the following equation is obtained: 5 + x = 20.
Let's find the difference between 20 and 5. When subtracting, the number from which it is subtracted is reduced. The number that is subtracted is called , and the final result is called the difference. So, x = 20 - 5; x = 15. You need to buy 15 carrots for rabbits.
Make a check: 5 + 15 = 20. The equation is correct. Of course, when we are talking about such simple ones, it is not necessary to perform a check. However, when it comes to equations with three-digit, four-digit, and so on, it is imperative to check to be absolutely sure of the result of your work.
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Helpful advice
To find the unknown minuend, you need to add the subtrahend to the difference.
To find the unknown subtrahend, it is necessary to subtract the difference from the minuend.
Tip 4: How to solve a system of three equations with three unknowns
A system of three equations with three unknowns may not have solutions, despite a sufficient number of equations. You can try to solve it using the substitution method or using the Cramer method. Cramer's method, in addition to solving the system, allows one to evaluate whether the system is solvable before finding the values of the unknowns.
Instruction
The substitution method consists in sequentially one unknown through two others and substituting the result obtained into the equations of the system. Let a system of three equations be given in general view:
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Express x from the first equation: x = (d1 - b1y - c1z)/a1 - and substitute into the second and third equations, then express y from the second equation and substitute into the third. You will get a linear expression for z through the coefficients of the equations of the system. Now go "back": plug z into the second equation and find y, then plug z and y into the first equation and find x. The process is generally shown in the figure until z is found. Further, the record in general form will be too cumbersome, in practice, substituting , you can quite easily find all three unknowns.
Cramer's method consists in compiling the matrix of the system and calculating the determinant of this matrix, as well as three more auxiliary matrices. The matrix of the system is composed of the coefficients at the unknown terms of the equations. The column containing the numbers on the right side of the equations, the column of the right side. It is not used in the system, but is used when solving the system.
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note
All equations in the system must supply additional information independent of other equations. Otherwise, the system will be underdetermined and it will not be possible to find an unambiguous solution.
Helpful advice
After solving the system of equations, substitute the found values into the original system and check that they satisfy all the equations.
By itself the equation with three unknown has many solutions, so most often it is supplemented by two more equations or conditions. Depending on what the initial data are, the course of the decision will largely depend.
You will need
- - a system of three equations with three unknowns.
Instruction
If two of the three systems have only two of the three unknowns, try expressing some variables in terms of the others and plugging them into the equation with three unknown. Your goal with this is to turn it into a normal the equation with the unknown. If this is , the further solution is quite simple - substitute the found value into other equations and find all the other unknowns.
Some systems of equations can be subtracted from one equation by another. See if it is possible to multiply one of by or a variable so that two unknowns are reduced at once. If there is such an opportunity, use it, most likely, the subsequent decision will not be difficult. Do not forget that when multiplying by a number, you must multiply both the left side and the right side. Similarly, when subtracting equations, remember that the right hand side must also be subtracted.
If previous ways did not help, use the general method for solving any equations with three unknown. To do this, rewrite the equations in the form a11x1 + a12x2 + a13x3 \u003d b1, a21x1 + a22x2 + a23x3 \u003d b2, a31x1 + a32x2 + a33x3 \u003d b3. Now make a matrix of coefficients at x (A), a matrix of unknowns (X) and a matrix of free ones (B). Pay attention, multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix, a matrix of free members, that is, A * X \u003d B.
Find the matrix A to the power (-1) after finding , note that it should not be equal to zero. After that, multiply the resulting matrix by matrix B, as a result you will get the desired matrix X, indicating all the values.
You can also find a solution to a system of three equations using the Cramer method. To do this, find the third-order determinant ∆ corresponding to the matrix of the system. Then successively find three more determinants ∆1, ∆2 and ∆3, substituting the values of the free terms instead of the values of the corresponding columns. Now find x: x1=∆1/∆, x2=∆2/∆, x3=∆3/∆.
Sources:
- solutions of equations with three unknowns
Starting to solve a system of equations, figure out what these equations are. The methods of solving linear equations are well studied. Nonlinear equations are most often not solved. There are only one special cases, each of which is practically individual. Therefore, the study of methods of solution should begin with linear equations. Such equations can be solved even purely algorithmically.
the denominators of the found unknowns are exactly the same. Yes, and the numerators are visible some patterns of their construction. If the dimension of the system of equations were greater than two, then the elimination method would lead to very cumbersome calculations. To avoid them, purely algorithmic solutions have been developed. The simplest of them is Cramer's algorithm (Cramer's formulas). For you should know general system equations from n equations.
The system of n linear algebraic equations with n unknowns has the form (see Fig. 1a). In it, aij are the coefficients of the system,
хj – unknowns, bi – free members (i=1, 2, ... , n; j=1, 2, ... , n). Such a system can be compactly written in the matrix form AX=B. Here A is the coefficient matrix of the system, X is the column matrix of unknowns, B is the column matrix of free terms (see Fig. 1b). According to Cramer's method, each unknown xi =∆i/∆ (i=1,2…,n). The determinant ∆ of the matrix of coefficients is called the main determinant, and ∆i is called auxiliary. For each unknown, an auxiliary determinant is found by replacing the i-th column of the main determinant with a column of free terms. Cramer's method for the case of systems of the second and third order is presented in detail in Fig. 2.
A system is a union of two or more equalities, each of which has two or more unknowns. There are two main ways to solve systems of linear equations that are used in the framework school curriculum. One of them is called the method, the other is the addition method.
Standard form of a system of two equations
At standard form the first equation is a1*x+b1*y=c1, the second equation is a2*x+b2*y=c2, and so on. For example, in the case of two parts of the system in both given a1, a2, b1, b2, c1, c2 are some numerical coefficients presented in specific equations. In turn, x and y are unknowns whose values need to be determined. The desired values turn both equations simultaneously into true equalities.Solution of the system by the addition method
In order to solve the system, that is, to find those values of x and y that will turn them into true equalities, you need to take a few simple steps. The first of these is to transform any of the equations in such a way that the numerical coefficients for the variable x or y in both equations coincide in absolute value, but differ in sign.For example, let a system consisting of two equations be given. The first of them has the form 2x+4y=8, the second has the form 6x+2y=6. One of the options for completing the task is to multiply the second equation by a factor of -2, which will lead it to the form -12x-4y=-12. The correct choice of the coefficient is one of the key tasks in the process of solving the system by the addition method, since it determines the entire further course of the procedure for finding unknowns.
Now it is necessary to add the two equations of the system. Obviously, the mutual destruction of variables with equal in value but opposite in sign coefficients will lead it to the form -10x=-4. After that, it is necessary to solve this simple equation, from which it unambiguously follows that x=0.4.
Last step in the process of solving is the substitution of the found value of one of the variables in any of the initial equalities available in the system. For example, substituting x=0.4 into the first equation, you can get the expression 2*0.4+4y=8, from which y=1.8. Thus, x=0.4 and y=1.8 are the roots of the system shown in the example.
In order to make sure that the roots were found correctly, it is useful to check by substituting the found values into the second equation of the system. For example, in this case an equality of the form 0.4*6+1.8*2=6 is obtained, which is true.
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With 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 provides a detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.
This program can be useful for high school students general education schools in preparation for control work and exams, when testing knowledge before the exam, 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 soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.
Thus, you can carry out your own training and/or training of their younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.
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 brackets. 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 the elements.
For example: 6x+1 = 5(x+y)+2
In equations, you can use not only integers, but also fractional numbers in the form of decimal and ordinary fractions.
Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions 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: /
The integer part is 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 a system of equations
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A bit of theory.
Solving systems of linear equations. Substitution method
The sequence of actions when solving a system of linear equations by the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression in another equation of the system instead of this variable;
$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$
Let's express from the first equation y through x: y = 7-3x. Substituting the expression 7-3x instead of y into the second equation, we get the system:
$$ \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 $$
Substituting the number 1 instead of x into the equation y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$
Pair (1;4) - solution of the system
Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.
Solving systems of linear equations by adding
Consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by the substitution method, we pass from a given system to another system equivalent to it, in which one of the equations contains only one variable.
The sequence of actions when solving a system of linear equations by the addition method:
1) multiply the equations of the system term by term, choosing factors so that the coefficients for one of the variables become opposite numbers;
2) add term by term the left and right parts of the equations of the system;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.
Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$
In the equations of this system, the coefficients of y are opposite numbers. Adding term by term the left and right parts of the equations, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \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 adding: \(x=11; y=-9 \) or \((11; -9) \)
Taking advantage of the fact that the coefficients of y in the equations of the system are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both parts of each of the equations of the original symmeme), in which one of the equations contains only one variable.
Books (textbooks) Abstracts of the Unified State Examination and OGE tests online Games, puzzles Construction of graphs of functions Spelling Dictionary of the Russian Language Dictionary of youth slang Directory of Russian schools Catalog of secondary schools in Russia Catalog of Russian universities List of tasksWe 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 to:
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 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 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)
Do you want to prepare for exams for free? Tutor online for free. No kidding.
Let us first recall the definition of a solution to a system of equations in two variables.
Definition 1
A pair of numbers is called a solution to a system of equations with two variables if, when they are substituted into the equation, the correct equality is obtained.
In what follows, we will consider systems of two equations with two variables.
Exist four basic ways to solve systems of equations: substitution method, addition method, graphical method, new variable management method. Let's take a look at these methods concrete examples. To describe the principle of using the first three methods, we will consider a system of two linear equations with two unknowns:
Substitution method
The substitution method is as follows: any of these equations is taken and $y$ is expressed in terms of $x$, then $y$ is substituted into the equation of the system, from where the variable $x.$ is found. After that, we can easily calculate the variable $y.$
Example 1
Let us express from the second equation $y$ in terms of $x$:
Substitute in the first equation, find $x$:
\ \ \
Find $y$:
Answer: $(-2,\ 3)$
Addition method.
Consider this method with an example:
Example 2
\[\left\( \begin(array)(c) (2x+3y=5) \\ (3x-y=-9) \end(array) \right.\]
Multiply the second equation by 3, we get:
\[\left\( \begin(array)(c) (2x+3y=5) \\ (9x-3y=-27) \end(array) \right.\]
Now let's add both equations together:
\ \ \
Find $y$ from the second equation:
\[-6-y=-9\] \
Answer: $(-2,\ 3)$
Remark 1
Note that in this method it is necessary to multiply one or both equations by such numbers that when adding one of the variables "disappears".
Graphical way
The graphical method is as follows: both equations of the system are displayed on the coordinate plane and the point of their intersection is found.
Example 3
\[\left\( \begin(array)(c) (2x+3y=5) \\ (3x-y=-9) \end(array) \right.\]
Let us express $y$ from both equations in terms of $x$:
\[\left\( \begin(array)(c) (y=\frac(5-2x)(3)) \\ (y=3x+9) \end(array) \right.\]
Let's draw both graphs on the same plane:
Picture 1.
Answer: $(-2,\ 3)$
How to introduce new variables
We will consider this method in the following example:
Example 4
\[\left\( \begin(array)(c) (2^(x+1)-3^y=-1) \\ (3^y-2^x=2) \end(array) \right .\]
Solution.
This system is equivalent to the system
\[\left\( \begin(array)(c) ((2\cdot 2)^x-3^y=-1) \\ (3^y-2^x=2) \end(array) \ right.\]
Let $2^x=u\ (u>0)$ and $3^y=v\ (v>0)$, we get:
\[\left\( \begin(array)(c) (2u-v=-1) \\ (v-u=2) \end(array) \right.\]
We solve the resulting system by the addition method. Let's add the equations:
\ \
Then from the second equation, we get that
Returning to the substitution, we get new system exponential equations:
\[\left\( \begin(array)(c) (2^x=1) \\ (3^y=3) \end(array) \right.\]
We get:
\[\left\( \begin(array)(c) (x=0) \\ (y=1) \end(array) \right.\]
Lesson contentLinear Equations with Two Variables
The student has 200 rubles to have lunch at school. A cake costs 25 rubles, and a cup of coffee costs 10 rubles. How many cakes and cups of coffee can you buy for 200 rubles?
Denote the number of cakes through x, and the number of cups of coffee through y. Then the cost of cakes will be denoted by the expression 25 x, and the cost of cups of coffee in 10 y .
25x- price x cakes
10y- price y cups of coffee
The total amount should be 200 rubles. Then we get an equation with two variables x And y
25x+ 10y= 200
How many roots does this equation have?
It all depends on the appetite of the student. If he buys 6 cakes and 5 cups of coffee, then the roots of the equation will be the numbers 6 and 5.
The pair of values 6 and 5 are said to be the roots of Equation 25 x+ 10y= 200 . Written as (6; 5) , with the first number being the value of the variable x, and the second - the value of the variable y .
6 and 5 are not the only roots that reverse Equation 25 x+ 10y= 200 to identity. If desired, for the same 200 rubles, a student can buy 4 cakes and 10 cups of coffee:
In this case, the roots of equation 25 x+ 10y= 200 is the pair of values (4; 10) .
Moreover, a student may not buy coffee at all, but buy cakes for all 200 rubles. Then the roots of equation 25 x+ 10y= 200 will be the values 8 and 0
Or vice versa, do not buy cakes, but buy coffee for all 200 rubles. Then the roots of equation 25 x+ 10y= 200 will be the values 0 and 20
Let's try to list all possible roots of equation 25 x+ 10y= 200 . Let us agree that the values x And y belong to the set of integers. And let these values be greater than or equal to zero:
x∈Z, y∈ Z;
x ≥ 0, y ≥ 0
So it will be convenient for the student himself. Cakes are more convenient to buy whole than, for example, several whole cakes and half a cake. Coffee is also more convenient to take in whole cups than, for example, several whole cups and half a cup.
Note that for odd x it is impossible to achieve equality under any y. Then the values x there will be the following numbers 0, 2, 4, 6, 8. And knowing x can be easily determined y
Thus, we got the following pairs of values (0; 20), (2; 15), (4; 10), (6; 5), (8; 0). These pairs are solutions or roots of Equation 25 x+ 10y= 200. They turn this equation into an identity.
Type equation ax + by = c called linear equation with two variables. A solution or roots of this equation is a pair of values ( x; y), which turns it into an identity.
Note also that if a linear equation with two variables is written as ax + b y = c , then they say that it is written in canonical(normal) form.
Some linear equations in two variables can be reduced to canonical form.
For example, the equation 2(16x+ 3y- 4) = 2(12 + 8x − y) can be brought to mind ax + by = c. Let's open the brackets in both parts of this equation, we get 32x + 6y − 8 = 24 + 16x − 2y . The terms containing unknowns are grouped on the left side of the equation, and the terms free of unknowns are grouped on the right. Then we get 32x - 16x+ 6y+ 2y = 24 + 8 . We bring similar terms in both parts, we get equation 16 x+ 8y= 32. This equation is reduced to the form ax + by = c and is canonical.
Equation 25 considered earlier x+ 10y= 200 is also a two-variable linear equation in canonical form. In this equation, the parameters a , b And c are equal to the values 25, 10 and 200, respectively.
Actually the equation ax + by = c has an infinite number of solutions. Solving the Equation 25x+ 10y= 200, we looked for its roots only on the set of integers. As a result, we obtained several pairs of values that turned this equation into an identity. But on the set of rational numbers equation 25 x+ 10y= 200 will have an infinite number of solutions.
To get new pairs of values, you need to take an arbitrary value for x, then express y. For example, let's take a variable x value 7. Then we get an equation with one variable 25×7 + 10y= 200 in which to express y
Let x= 15 . Then the equation 25x+ 10y= 200 becomes 25 × 15 + 10y= 200. From here we find that y = −17,5
Let x= −3 . Then the equation 25x+ 10y= 200 becomes 25 × (−3) + 10y= 200. From here we find that y = −27,5
System of two linear equations with two variables
For the equation ax + by = c you can take any number of times arbitrary values for x and find values for y. Taken separately, such an equation will have an infinite number of solutions.
But it also happens that the variables x And y connected not by one, but by two equations. In this case, they form the so-called system of linear equations with two variables. Such a system of equations can have one pair of values (or in other words: “one solution”).
It may also happen that the system has no solutions at all. A system of linear equations can have an infinite number of solutions in rare and exceptional cases.
Two linear equations form a system when the values x And y are included in each of these equations.
Let's go back to the very first equation 25 x+ 10y= 200 . One of the pairs of values for this equation was the pair (6; 5) . This is the case when 200 rubles could buy 6 cakes and 5 cups of coffee.
Let us compose the problem so that the pair (6; 5) becomes the only solution for Equation 25 x+ 10y= 200 . To do this, we compose another equation that would connect the same x cakes and y cups of coffee.
Let's put the text of the task as follows:
“A schoolboy bought several cakes and several cups of coffee for 200 rubles. A cake costs 25 rubles, and a cup of coffee costs 10 rubles. How many cakes and cups of coffee did the student buy if it is known that the number of cakes is one more than the number of cups of coffee?
We already have the first equation. This is Equation 25 x+ 10y= 200 . Now let's write an equation for the condition "the number of cakes is one unit more than the number of cups of coffee" .
The number of cakes is x, and the number of cups of coffee is y. You can write this phrase using the equation x − y= 1. This equation would mean that the difference between cakes and coffee is 1.
x=y+ 1 . This equation means that the number of cakes is one more than the number of cups of coffee. Therefore, to obtain equality, one is added to the number of cups of coffee. This can be easily understood if we use the weight model that we considered when studying the simplest problems:
Got two equations: 25 x+ 10y= 200 and x=y+ 1. Since the values x And y, namely 6 and 5 are included in each of these equations, then together they form a system. Let's write down this system. If the equations form a system, then they are framed by the sign of the system. The system sign is a curly brace:
Let's solve this system. This will allow us to see how we arrive at the values 6 and 5. There are many methods for solving such systems. Consider the most popular of them.
Substitution method
The name of this method speaks for itself. Its essence is to substitute one equation into another, having previously expressed one of the variables.
In our system, nothing needs to be expressed. In the second equation x = y+ 1 variable x already expressed. This variable is equal to the expression y+ 1 . Then you can substitute this expression in the first equation instead of the variable x
After substituting the expression y+ 1 into the first equation instead x, we get the equation 25(y+ 1) + 10y= 200 . This is a linear equation with one variable. This equation is quite easy to solve:
We found the value of the variable y. Now we substitute this value into one of the equations and find the value x. For this, it is convenient to use the second equation x = y+ 1 . Let's put the value into it y
So the pair (6; 5) is a solution to the system of equations, as we intended. We check and make sure that the pair (6; 5) satisfies the system:
Example 2
Substitute the first equation x= 2 + y into the second equation 3 x - 2y= 9 . In the first equation, the variable x is equal to the expression 2 + y. We substitute this expression into the second equation instead of x
Now let's find the value x. To do this, substitute the value y into the first equation x= 2 + y
So the solution of the system is the pair value (5; 3)
Example 3. Solve the following system of equations using the substitution method:
Here, unlike the previous examples, one of the variables is not explicitly expressed.
To substitute one equation into another, you first need .
It is desirable to express the variable that has a coefficient of one. The coefficient unit has a variable x, which is contained in the first equation x+ 2y= 11 . Let's express this variable.
After a variable expression x, our system will look like this:
Now we substitute the first equation into the second and find the value y
Substitute y x
So the solution of the system is a pair of values (3; 4)
Of course, you can also express a variable y. The roots will not change. But if you express y, the result is not a very simple equation, the solution of which will take more time. It will look like this:
We see that in this example to express x much more convenient than expressing y .
Example 4. Solve the following system of equations using the substitution method:
Express in the first equation x. Then the system will take the form:
y
Substitute y into the first equation and find x. You can use the original equation 7 x+ 9y= 8 , or use the equation in which the variable is expressed x. We will use this equation, since it is convenient:
So the solution of the system is the pair of values (5; −3)
Addition method
The addition method is to add term by term the equations included in the system. This addition results in a new one-variable equation. And it's pretty easy to solve this equation.
Let's solve the following system of equations:
Add the left side of the first equation to the left side of the second equation. And the right side of the first equation with the right side of the second equation. We get the following equality:
Here are similar terms:
As a result, we obtained the simplest equation 3 x= 27 whose root is 9. Knowing the value x you can find the value y. Substitute the value x into the second equation x − y= 3 . We get 9 − y= 3 . From here y= 6 .
So the solution of the system is a pair of values (9; 6)
Example 2
Add the left side of the first equation to the left side of the second equation. And the right side of the first equation with the right side of the second equation. In the resulting equality, we present like terms:
As a result, we got the simplest equation 5 x= 20, the root of which is 4. Knowing the value x you can find the value y. Substitute the value x into the first equation 2 x+y= 11 . Let's get 8 + y= 11 . From here y= 3 .
So the solution of the system is the pair of values (4;3)
The addition process is not described in detail. It has to be done in the mind. When adding, both equations must be reduced to canonical form. That is, to the mind ac+by=c .
From the considered examples, it can be seen that the main goal of adding equations is to get rid of one of the variables. But it is not always possible to immediately solve the system of equations by the addition method. Most often, the system is preliminarily brought to a form in which it is possible to add the equations included in this system.
For example, the system 
can be solved directly by the addition method. When adding both equations, the terms y And −y vanish because their sum is zero. As a result, the simplest equation is formed 11 x= 22 , whose root is 2. Then it will be possible to determine y equal to 5.
And the system of equations 
the addition method cannot be solved immediately, since this will not lead to the disappearance of one of the variables. Addition will result in Equation 8 x+ y= 28 , which has an infinite number of solutions.
If both parts of the equation are multiplied or divided by the same number that is not equal to zero, then an equation equivalent to the given one will be obtained. This rule is also valid for a system of linear equations with two variables. One of the equations (or both equations) can be multiplied by some number. The result is an equivalent system, the roots of which will coincide with the previous one.
Let's return to the very first system, which described how many cakes and cups of coffee the student bought. The solution of this system was a pair of values (6; 5) .
We multiply both equations included in this system by some numbers. Let's say we multiply the first equation by 2 and the second by 3
The result is a system 
The solution to this system is still the pair of values (6; 5)
This means that the equations included in the system can be reduced to a form suitable for applying the addition method.
Back to the system , which we could not solve by the addition method.
Multiply the first equation by 6 and the second by −2
Then we get the following system:
We add the equations included in this system. Addition of components 12 x and -12 x will result in 0, addition 18 y and 4 y will give 22 y, and adding 108 and −20 gives 88. Then you get the equation 22 y= 88 , hence y = 4 .
If at first it is difficult to add equations in your mind, then you can write down how the left side of the first equation is added to the left side of the second equation, and the right side of the first equation to the right side of the second equation:
Knowing that the value of the variable y is 4, you can find the value x. Substitute y into one of the equations, for example into the first equation 2 x+ 3y= 18 . Then we get an equation with one variable 2 x+ 12 = 18 . We transfer 12 to the right side, changing the sign, we get 2 x= 6 , hence x = 3 .
Example 4. Solve the following system of equations using the addition method:
Multiply the second equation by −1. Then the system will take the following form:
Let's add both equations. Addition of components x And −x will result in 0, addition 5 y and 3 y will give 8 y, and adding 7 and 1 gives 8. The result is equation 8 y= 8 , whose root is 1. Knowing that the value y is 1, you can find the value x .
Substitute y into the first equation, we get x+ 5 = 7 , hence x= 2
Example 5. Solve the following system of equations using the addition method:
It is desirable that the terms containing the same variables are located one under the other. Therefore, in the second equation, the terms 5 y and −2 x change places. As a result, the system will take the form:
Multiply the second equation by 3. Then the system will take the form:
Now let's add both equations. As a result of addition, we get equation 8 y= 16 , whose root is 2.
Substitute y into the first equation, we get 6 x− 14 = 40 . We transfer the term −14 to the right side, changing the sign, we get 6 x= 54 . From here x= 9.
Example 6. Solve the following system of equations using the addition method:
Let's get rid of fractions. Multiply the first equation by 36 and the second by 12
In the resulting system 
the first equation can be multiplied by −5 and the second by 8
Let's add the equations in the resulting system. Then we get the simplest equation −13 y= −156 . From here y= 12 . Substitute y into the first equation and find x
Example 7. Solve the following system of equations using the addition method:
We bring both equations to normal form. Here it is convenient to apply the rule of proportion in both equations. If in the first equation the right side is represented as , and the right side of the second equation as , then the system will take the form:
We have a proportion. We multiply its extreme and middle terms. Then the system will take the form:
We multiply the first equation by −3, and open the brackets in the second:
Now let's add both equations. As a result of adding these equations, we get an equality, in both parts of which there will be zero:
It turns out that the system has an infinite number of solutions.
But we cannot simply take arbitrary values from the sky for x And y. We can specify one of the values, and the other will be determined depending on the value we specify. For example, let x= 2 . Substitute this value into the system:
As a result of solving one of the equations, the value for y, which will satisfy both equations:
The resulting pair of values (2; −2) will satisfy the system:
Let's find another pair of values. Let x= 4. Substitute this value into the system:
It can be determined by eye that y equals zero. Then we get a pair of values (4; 0), which satisfies our system:
Example 8. Solve the following system of equations using the addition method:
Multiply the first equation by 6 and the second by 12
Let's rewrite what's left:
Multiply the first equation by −1. Then the system will take the form:
Now let's add both equations. As a result of addition, equation 6 is formed b= 48 , whose root is 8. Substitute b into the first equation and find a
System of linear equations with three variables
A linear equation with three variables includes three variables with coefficients, as well as an intercept. In canonical form, it can be written as follows:
ax + by + cz = d
This equation has an infinite number of solutions. Giving two variables various meanings, you can find the third value. The solution in this case is the triple of values ( x; y; z) which turns the equation into an identity.
If variables x, y, z are interconnected by three equations, then a system of three linear equations with three variables is formed. To solve such a system, you can apply the same methods that apply to linear equations with two variables: the substitution method and the addition method.
Example 1. Solve the following system of equations using the substitution method:
We express in the third equation x. Then the system will take the form:
Now let's do the substitution. Variable x is equal to the expression 3 − 2y − 2z . Substitute this expression into the first and second equations:
Let's open the brackets in both equations and give like terms:
We have arrived at a system of linear equations with two variables. In this case, it is convenient to apply the addition method. As a result, the variable y will disappear and we can find the value of the variable z
Now let's find the value y. For this, it is convenient to use the equation − y+ z= 4. Substitute the value z
Now let's find the value x. For this, it is convenient to use the equation x= 3 − 2y − 2z . Substitute the values into it y And z
Thus, the triple of values (3; −2; 2) is the solution to our system. By checking, we make sure that these values satisfy the system:
Example 2. Solve the system by addition method
Let's add the first equation with the second multiplied by −2.
If the second equation is multiplied by −2, then it will take the form −6x+ 6y- 4z = −4 . Now add it to the first equation:
We see that as a result of elementary transformations, the value of the variable was determined x. It is equal to one.
Back to main system. Let's add the second equation with the third multiplied by −1. If the third equation is multiplied by −1, then it will take the form −4x + 5y − 2z = −1 . Now add it to the second equation:
Got the equation x - 2y= −1 . Substitute the value into it x which we found earlier. Then we can determine the value y
We now know the values x And y. This allows you to determine the value z. We use one of the equations included in the system:
Thus, the triple of values (1; 1; 1) is the solution to our system. By checking, we make sure that these values satisfy the system:
Tasks for compiling systems of linear equations
The task of compiling systems of equations is solved by introducing several variables. Next, equations are compiled based on the conditions of the problem. From the compiled equations, they form a system and solve it. Having solved the system, it is necessary to check whether its solution satisfies the conditions of the problem.
Task 1. A Volga car left the city for the collective farm. She returned back along another road, which was 5 km shorter than the first. In total, the car drove 35 km both ways. How many kilometers is each road long?
Solution
Let x- length of the first road, y- the length of the second. If the car drove 35 km both ways, then the first equation can be written as x+ y= 35. This equation describes the sum of the lengths of both roads.
It is said that the car was returning back along the road, which was shorter than the first one by 5 km. Then the second equation can be written as x− y= 5. This equation shows that the difference between the lengths of the roads is 5 km.
Or the second equation can be written as x= y+ 5 . We will use this equation.
Since the variables x And y in both equations denote the same number, then we can form a system from them:
Let's solve this system using one of the previously studied methods. In this case, it is convenient to use the substitution method, since in the second equation the variable x already expressed.
Substitute the second equation into the first and find y
Substitute the found value y into the second equation x= y+ 5 and find x
The length of the first road was denoted by the variable x. Now we have found its meaning. Variable x is 20. So the length of the first road is 20 km.
And the length of the second road was indicated by y. The value of this variable is 15. So the length of the second road is 15 km.
Let's do a check. First, let's make sure that the system is solved correctly:
Now let's check whether the solution (20; 15) satisfies the conditions of the problem.
It was said that in total the car drove 35 km both ways. We add the lengths of both roads and make sure that the solution (20; 15) satisfies this condition: 20 km + 15 km = 35 km
Next condition: the car returned back along another road, which was 5 km shorter than the first . We see that the solution (20; 15) also satisfies this condition, since 15 km is shorter than 20 km by 5 km: 20 km − 15 km = 5 km
When compiling a system, it is important that the variables denote the same numbers in all equations included in this system.
So our system contains two equations. These equations in turn contain the variables x And y, which denote the same numbers in both equations, namely the lengths of roads equal to 20 km and 15 km.
Task 2. Oak and pine sleepers were loaded onto the platform, a total of 300 sleepers. It is known that all oak sleepers weighed 1 ton less than all pine sleepers. Determine how many oak and pine sleepers there were separately, if each oak sleeper weighed 46 kg, and each pine sleeper 28 kg.
Solution
Let x oak and y pine sleepers were loaded onto the platform. If there were 300 sleepers in total, then the first equation can be written as x+y = 300 .
All oak sleepers weighed 46 x kg, and pine weighed 28 y kg. Since oak sleepers weighed 1 ton less than pine sleepers, the second equation can be written as 28y- 46x= 1000 . This equation shows that the mass difference between oak and pine sleepers is 1000 kg.
Tons have been converted to kilograms because the mass of oak and pine sleepers is measured in kilograms.
As a result, we obtain two equations that form the system
Let's solve this system. Express in the first equation x. Then the system will take the form:
Substitute the first equation into the second and find y
Substitute y into the equation x= 300 − y and find out what x
This means that 100 oak and 200 pine sleepers were loaded onto the platform.
Let's check whether the solution (100; 200) satisfies the conditions of the problem. First, let's make sure that the system is solved correctly:
It was said that there were 300 sleepers in total. We add up the number of oak and pine sleepers and make sure that the solution (100; 200) satisfies this condition: 100 + 200 = 300.
Next condition: all oak sleepers weighed 1 ton less than all pine . We see that the solution (100; 200) also satisfies this condition, since 46 × 100 kg of oak sleepers are lighter than 28 × 200 kg of pine sleepers: 5600 kg − 4600 kg = 1000 kg.
Task 3. We took three pieces of an alloy of copper and nickel in ratios of 2: 1, 3: 1 and 5: 1 by weight. Of these, a piece weighing 12 kg was fused with a ratio of copper and nickel content of 4: 1. Find the mass of each original piece if the mass of the first of them is twice the mass of the second.
