After the author of the site was able to teach his bot to solve a linear Diophantine equation with two variables, a desire arose to teach the bot to solve similar equations, but with three unknowns. I had to plunge into books.
Having emerged from there two months later, the author realized that he did not understand anything. Extremely smart mathematicians wrote an algorithm for deriving formulas in such a sophisticated way that I was ashamed as a mortal. I was saddened, but I still found a useful thought in the vastness of books, and from this thought came an understanding of how to solve Diophantine equations with three unknowns.
So for everyone who is not a mathematician, but wants to be one :)
A Diophantine equation with three unknowns looks like this:
where are the integers
If we think about what kind of general solution the unknowns might have, then the most banal one looks like this
Let's substitute our general solution into the equation
What is the use of this, the impatient reader will ask? But what, let’s group everything by unknowns, we get
Look, on the right side there is some kind of constant number, designated by the letter d
This means that it does not depend on t (it’s also a variable, you never know what value it wants to become), which means
It is logical to assume that it does not depend on z either, which means
but it directly depends on the constant values of A 3 and B 3, that is
What did we end up with? And we got three typical classical Diophantine equations with two unknowns, which we can solve easily and naturally.
Let's try to decide?
In the first lines of search engines we found this equation:
The first equation will be like this
its roots
Let's get rid of zeros, taking k=-1 for example. (You can take 2 or 100 or -3 if you want) This will not affect the final decision.
Solving the second equation
and its roots
here let k=0 (since X and Y do not coincide even at zero values)
And the last third equation
The roots here are like this
Let us now substitute all the found values into the general form
That's it!
Please note that everything is resolved very easily and transparently! Surely teachers and capable students will adopt this technique, since the author of the bot found it in books.
Another example, already solved using a bot.
Addition: When you solve similar equations using a bot, you may encounter the fact that the bot will give you an error asking you to swap the variables for another attempt to solve the equation. This is due to the fact that during intermediate calculations, an unsolvable equation is obtained
As an example
When trying to solve the equation
in our case
we will get an error, because for any values, the left side will always (!!) have an even number, and the right side, as we see, will have an odd number.
But this does not mean that the original equation is unsolvable. It is enough to change the terms in a different order, for example like this
and we get the answer
A system of linear equations is a set of several considered together linear equations.
A system can have any number of equations with any number of unknowns.
A solution to a system of equations is a set of values of unknowns that satisfies all the equations of the system, that is, turning them into identities.
A system that has a solution is called consistent; otherwise, it is called inconsistent.
Various methods are used to solve the system.
Let 
(the number of equations is equal to the number of unknowns).
Cramer method
Consider solving a system of three linear equations with three unknowns:
(7)
To find unknowns 
Let's apply Cramer's formula:
(8)
Where 
- determinant of the system, the elements of which are the coefficients of the unknowns:
.
obtained by replacing the first column of the determinant 
column of free members:
.
Likewise:
;
.
Example 1. Solve the system using Cramer's formula:
.
Solution: Let's use formulas (8):
;
;
;
;
Answer: 
.
For any system 
linear equations with 
the unknowns can be stated:
Matrix solution
Let us consider solving system (7) of three linear equations with three unknowns using a matrix method.
Using the rules of matrix multiplication, this system of equations can be written as: 
, Where
.
Let the matrix 
non-degenerate, i.e. 
. Multiplying both sides of the matrix equation on the left by the matrix 
, inverse of the matrix 
, we get: 
.
Considering that 
, we have
(9)
Example 2. Solve the system using the matrix method:
.
Solution: Let's introduce the matrices:
- from the coefficients of the unknowns;
- column of free members.
Then the system can be written as a matrix equation: 
.
Let's use formula (9). Let's find the inverse matrix 
according to formula (6):
;
.
Hence,
Received:
.
Answer: 
.
Method of sequential elimination of unknowns (Gauss method)
The main idea of the method used is to sequentially eliminate unknowns. Let us explain the meaning of this method using a system of three equations with three unknowns:
.
Let's assume that 
(If 
, then we change the order of the equations, choosing as the first equation the one in which the coefficient at 
not equal to zero).
First step: a) divide the equation 
on 
; b) multiply the resulting equation by 
and subtract from 
; c) then multiply the result by 
and subtract from 
. As a result of the first step we will have the system:
,
Second step: we deal with the equation 
And 
exactly the same as with equations 
.
As a result, the original system is transformed into the so-called stepwise form:
From the transformed system, all unknowns are determined sequentially without difficulty.
Comment. In practice, it is more convenient to reduce to a stepwise form not the system of equations itself, but a matrix of coefficients, unknowns, and free terms.
Example 3. Solve the system using the Gaussian method:
.
We will write the transition from one matrix to another using the equivalence sign ~.
~
~
~
~
~
.
Using the resulting matrix, we write out the transformed system:
.
Answer: 
.
Note: If the system has a unique solution, then the step system is reduced to a triangular one, that is, to one in which the last equation will contain one unknown. In the case of an uncertain system, that is, one in which the number of unknowns more number linearly independent equations, there will be no triangular system, since the last equation will contain more than one unknown (the system has an infinite number of solutions). When the system is inconsistent, then, after reducing it to stepwise form, it will contain at least one 
value of the form 
, that is, an equation in which all unknowns have zero coefficients, and right side is different from zero (the system has no solutions). The Gauss method is applicable to an arbitrary system of linear equations (for any 
And 
).
Existence theorem for a solution to a system of linear equations
When solving a system of linear equations using the Gaussian method, the answer to the question whether this system is compatible or inconsistent can be given only at the end of the calculations. However, it is often important to solve the question of compatibility or incompatibility of a system of equations without finding the solutions themselves. The answer to this question is given by the following Kronecker-Capelli theorem.
Let the system be given 
linear equations with 
unknown:
(10)
In order for system (10) to be consistent, it is necessary and sufficient that the rank of the system matrix
.
was equal to the rank of its extended matrix
.
Moreover, if 
, then system (10) has a unique solution; if 
, then the system has an infinite number of solutions.
Consider a homogeneous system (all free terms are equal to zero) of linear equations:
.
This system is always consistent since it has a zero solution.
The following theorem gives conditions under which the system also has solutions other than zero.
Terema. In order for a homogeneous system of line equations to have a zero solution, it is necessary and sufficient that its determinant 
was equal to zero:
.
Thus, if 
, then the solution is the only one. If 
, then there are an infinite number of other non-zero solutions. Let us indicate one of the ways to find solutions for a homogeneous system of three linear equations with three unknowns in the case 
.
It can be proven that if 
, and the first and second equations are non-proportional (linearly independent), then the third equation is a consequence of the first two. The solution of a homogeneous system of three equations with three unknowns is reduced to the solution of two equations with three unknowns. A so-called free unknown appears, to which arbitrary values can be assigned.
Example 4. Find all solutions of the system:
.
Solution. Determinant of this system
.
Therefore, the system has zero solutions. You can notice that the first two equations, for example, are not proportional, therefore, they are linearly independent. The third is a consequence of the first two (it turns out if you add twice the second to the first equation). Rejecting it, we obtain a system of two equations with three unknowns:
.
Assuming, for example, 
, we get
.
Solving a system of two linear equations, we express 
And 
through 
:
. Therefore, the solution to the system can be written as: 
, Where 
- arbitrary number.
Example 5. Find all solutions of the system:
.
Solution. It is easy to see that in this system there is only one independent equation (the other two are proportional to it). A system of three equations with three unknowns has been reduced to one equation with three unknowns. Two free unknowns appear. Finding, for example, from the first equation 
for arbitrary 
And 
, we obtain solutions to this system. The general form of the solution can be written, where 
And 
- arbitrary numbers.
Self-test questions
Formulate Cramer's rule for solving the system 
linear equations with 
unknown.
What is the essence of the matrix method of solving systems?
What is Gauss' method for solving a system of linear equations?
State the Kronecker-Capelli theorem.
Formulate a necessary and sufficient condition for the existence of nonzero solutions to a homogeneous system of linear equations.
Examples for self-solution
Find all solutions of the systems:
1.
; 2.
;
3.
; 4.
;
5.
; 6.
;
7.
; 8.
;
9.
; 10.
;
11.
; 12.
;
13.
;
14.
;
15.
.
Determine at what values 
And 
system of equations
a) has a unique solution;
b) has no solution;
c) has infinitely many solutions.
16.
; 17.
;
Find all solutions of the following homogeneous systems:
18.
; 19.
;
20.
; 21.
;
22.
; 23.
;
Answers to examples
1.
; 2.
; 3. Ǿ; 4. Ǿ;
5.
- arbitrary number.
6.
, Where 
- arbitrary number.
7.
; 8.
; 9. Ǿ; 10. Ǿ;
11.
, Where 
- arbitrary number.
12. , where 
And 
- arbitrary numbers.
13.
; 14.
Where 
And 
- arbitrary numbers.
15. Ǿ; 16. a) 
; b) 
; V) 
.
17. a) 
; b) 
; V) 
;
18.
; 19.
; 20., where 
- arbitrary number.
21. , where 
- arbitrary number.
22. , where 
- arbitrary number.
23. , where 
And 
- arbitrary numbers.
The 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. Equations with three unknowns are a common occurrence in mathematics. There are quite a few ways to solve this type of equations, and in most cases the lion’s share of them is supplemented by 2 more equations/conditions. The choice of solution method directly depends on the specific equation.
If your system has only 2 unknowns out of 3, then most likely a convenient solution to this system will be to express some variables in terms of others and substitute them into an equation with 3 unknowns. All this is done in order to transform it into an ordinary equation with only 1 unknown, the solution of which will give a number that can be substituted in place of the unknown and get the final result for all other unknowns.
There are systems of equations that can be solved by subtracting another from one equation. This is possible in in that case, if it is possible to multiply one of the expressions by a variable/value, which allows subtraction to reduce several unknowns. However, it is worth remembering that when multiplying and subtracting by a number, you need to perform operations on both sides of the expression.
Where to solve an equation with 3 unknowns online?
You can solve an equation with three unknown online solvers 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.
Linear equations (first degree equations) with two unknowns
Definition 1. Linear equation (first degree equation) with two unknowns x and y name an equation of the form
Solution . Let us express from equality (2) the variable y through the variable x:
From formula (3) it follows that solutions to equation (2) are all pairs of numbers of the form
where x is any number.
Note. As can be seen from the solution to Example 1, equation (2) has infinitely many solutions. However, it is important to note that not any pair of numbers (x; y) is a solution to this equation. In order to obtain any solution to equation (2), the number x can be taken as any, and the number y can then be calculated using formula (3).
Systems of two linear equations in two unknowns
Definition 3. A system of two linear equations with two unknowns x and y call a system of equations of the form
Where a 1 , b 1 , c 1 , a 2 , b 2 , c 2 – given numbers.
Definition 4. In the system of equations (4) the numbers a 1 , b 1 , a 2 , b 2 are called , and numbers c 1 , c 2 – free members.
Definition 5. By solving the system of equations (4) call a pair of numbers ( x; y) , which is a solution to both one and the other equation of system (4).
Definition 6. The two systems of equations are called equivalent (equivalent), if all solutions of the first system of equations are solutions of the second system, and all solutions of the second system are solutions of the first system.
The equivalence of systems of equations is indicated using the symbol “”
Systems of linear equations are solved using , which we will illustrate with examples.
Example 2. Solve system of equations
Solution . In order to solve system (5) eliminate the unknown from the second equation of the system X .
To this end, we first transform system (5) to a form in which the coefficients for unknown x in the first and second equations of the system become the same.
If the first equation of system (5) is multiplied by the coefficient at x in the second equation (number 7), and the second equation is multiplied by the coefficient at x in the first equation (number 2), then system (5) will take the form
Now let us perform the following transformations on system (6):
- from the second equation we subtract the first equation and replace the second equation of the system with the resulting difference.
As a result, system (6) is transformed into an equivalent system
From the second equation we find y= 3, and substituting this value into the first equation, we get
Answer . (-2 ; 3) .
Example 3. Find all values of the parameter p for which the system of equations
A) has a unique solution;
b) has infinitely many solutions;
V) has no solutions.
Solution . Expressing x through y from the second equation of system (7) and substituting the resulting expression instead of x into the first equation of system (7), we obtain
Let us study solutions to system (8) depending on the values of the parameter p. To do this, first consider the first equation of system (8):
| y (2 - p) (2 + p) = 2 + p | (9) |
If 
, then equation (9) has a unique solution
Thus, in the case when , system (7) has a unique solution
If p= - 2, then equation (9) takes the form
and its solution is any number 
. Therefore, the solution to system (7) is infinite set everyone pairs of numbers
,
where y is any number.
If p= 2, then equation (9) takes the form
and has no solutions, which implies that system (7) has no solutions.
Systems of three linear equations in three unknowns
Definition 7. A system of three linear equations with three unknowns x, y and z call a system of equations having the form
Where a 1 , b 1 , c 1 , d 1 , a 2 , b 2 , c 2 , d 2 , a 3 , b 3 , c 3 , d 3 – given numbers.
Definition 8. In the system of equations (10) the numbers a 1 , b 1 , c 1 , a 2 , b 2 , c 2 , a 3 , b 3 , c 3 called coefficients for unknowns, and the numbers d 1 , d 2 , d 3 – free members.
Definition 9. By solving the system of equations (10) name three numbers (x; y ; z) , when substituting them into each of the three equations of system (10), the correct equality is obtained.
Example 4. Solve system of equations
Solution . We will solve system (11) using method of sequential elimination of unknowns.
To do this first we exclude the unknown from the second and third equations of the system y by performing the following transformations on system (11):
- We will leave the first equation of the system unchanged;
- to the second equation we add the first equation and replace the second equation of the system with the resulting sum;
- from the third equation we subtract the first equation and replace the third equation of the system with the resulting difference.
As a result, system (11) is transformed into an equivalent system
Now eliminate the unknown from the third equation of the system x by performing the following transformations on system (12):
- We will leave the first and second equations of the system unchanged;
- from the third equation we subtract the second equation and replace the third equation of the system with the resulting difference.
As a result, system (12) is transformed into an equivalent system
From the system (13) we consistently find
z = - 2 ; x = 1 ; y = 2 .
Answer . (1; 2; -2) .
Example 5. Solve system of equations
Solution . Note that from this system one can obtain a convenient consequence, adding all three equations of the system:
We compose the main determinant for the system
and calculate it.
Then we compose additional determinants
and calculate them.
According to Cramer's rule, the solution to the system is found using the formulas
; 
;
,If 
1)
Let's calculate:
Using Cramer's formulas we find:
Answer: (1; 2; 3)
2)
Let's calculate:
Since the main determinant 
, and at least one additional one is not equal to zero (in our case 
), then the system has no solution.
3)
Let's calculate:
Since all determinants are equal to zero, the system has an infinite number of solutions, which can be found as follows: 
Solve the systems yourself:
A) 
b) 
Answer: a) (1; 2; 5) b) 
;
;
Practical lesson No. 3 on the topic:
Dot product of two vectors and its application
1. If given 
And 
, That dot product we find by the formula: 
∙
2.If, then the scalar product of these two vectors is found by the formula
1. Given two vectors 
And 
We find their scalar product as follows:
.
2. Two vectors are given:
={2;3;–4}
={1;
–5; 6}
The scalar product is found like this:
3.
,
3.1 Finding the work of a constant force on a straight section of path
1) Under the influence of a force of 15 N, the body moved in a straight line 2 meters. The angle between the force and the direction of movement =60 0. Calculate the work done by a force to move a body.
Given: 
Solution:
2) Given: 
Solution:
3) A body moved from point M(1; 2; 3) to point N(5; 4; 6) under the influence of a force of 60N. The angle between the direction of the force and the displacement vector =45 0. Calculate the work done by this force.
Solution: find the displacement vector 
Finding the module of the displacement vector:
According to the formula 
find a job:
3.2 Determining the orthogonality of two vectors
Two vectors are orthogonal if 
, that is
because 
1) 
– not orthogonal
2) 
–orthogonal
3) Determine at what the vectors 
And 
mutually orthogonal.
Because 
, That 
, Means
Decide for yourself:
A) 
. Find their scalar product.
b) Calculate how much work the force produces 
, if the point of its application, moving rectilinearly, has moved from point M (5; -6; 1) to point N (1; -2; 3)
c) Determine whether the vectors are orthogonal 
And
Answers: a) 1 b) 16 c) yes
3.3. Finding the angle between vectors
1)
. Find 
.
We find 
substitute into the formula:
.
1). Given are the vertices of the triangle A(3; 2; –3), B(5; 1; –1), C(1; –2; 1). Find the angle at vertex A.
Let's put it into the formula: 
Decide for yourself:
Given are the vertices of the triangle A(3; 5; -2), B(5; 7; -1), C(4; 3; 0). Define internal corner at the top of A.
Answer: 90 o
Practical lesson No. 4 on the topic:
VECTOR PRODUCT OF TWO VECTORS AND ITS APPLICATION.
Formula for finding the cross product of two vectors:
|
|
|
|
|
looks like |
||
1) Find the modulus of the vector product:
Let's compose a determinant and calculate it (using Sarrus's rule or the theorem on the expansion of the determinant into the elements of the first row).
1st method: according to Sarrus's rule
Method 2: expand the determinant into the elements of the first row.
2) Find the modulus of the vector product:
4.1. CALCULATION OF THE AREA OF A PARALLELOGRAM BUILT ON TWO VECTORS.
1) Calculate the area of a parallelogram built on vectors
2). Find the vector product and its modulus
4.2. CALCULATING THE AREA OF A TRIANGLE
Example: given are the vertices of the triangle A(1; 0; -1), B(1; 2; 0), C(3; -1; 1). Calculate the area of the triangle.
First, let's find the coordinates of two vectors emanating from the same vertex.
Let's find their vector product
4.3. DETERMINATION OF COLLINEARITY OF TWO VECTORS
If the vector 
And 
are collinear, then
, i.e. the coordinates of the vectors must be proportional.
a) Given vectors:: 
,
.
They are collinear because 
And
after reducing each fraction we get the ratio 
b) Given vectors: 
.
They are not collinear because 
or 
Decide for yourself:
a) At what values m and n of the vector 
collinear?
Answer: 
;
b) Find the vector product and its modulus 
,
.
Answer: 
,
.
Practical lesson No. 5 on the topic:
STRAIGHT LINE ON A PLANE
Problem No. 1. Find the equation of a line passing through point A(-2; 3) parallel to the line 
1. Find the slope of the line 
.
is the equation of a straight line with an angular coefficient and an initial ordinate ( 
). That's why 
.
2. Since the lines MN and AC are parallel, their angular coefficients are equal, i.e. 
.
3. To find the equation of straight line AC, we use the equation of a straight line passing through a point with a given angular coefficient:
. In this formula instead 
And 
substitute the coordinates of point A(-2; 3), instead 
Let’s substitute – 3. As a result of the substitution we get:
Answer: 
Task No. 2. Find the equation of a line passing through the point K(1; –2) parallel to the line.
1. Let's find the slope of the line.
This is the general equation of a line, which in general view is given by the formula. Comparing the equations, we find that A = 2, B = –3. The slope of the straight line given by the equation is found by the formula 
. Substituting A = 2 and B = –3 into this formula, we obtain the slope of the straight line MN. So, 
.
2. Since the lines MN and KS are parallel, their angular coefficients are equal: 
.
3. To find the equation of the straight line KS, we use the formula for the equation of the straight line passing through the point with a given angular coefficient 
. In this formula instead 
And 
Let's substitute the coordinates of the point K(–2; 3), instead of 
Problem No. 3. Find the equation of a line passing through the point K(–1; –3) perpendicular to the line.
1. is the general equation of a straight line, which in general form is given by the formula.
and we find that A = 3, B = 4.
The slope of the straight line given by the equation is found by the formula: 
. Substituting A = 3 and B = 4 into this formula, we obtain the slope of the straight line MN: 
.
2. Since the lines MN and KD are perpendicular, their angular coefficients are inversely proportional and opposite in sign:
.
3. To find the equation of the line KD, we use the formula for the equation of the line passing through the point with a given slope
. In this formula instead 
And 
Let's substitute the coordinates of the point K(–1;–3), instead of 
let's substitute As a result of the substitution we get:
Decide for yourself:
1. Find the equation of the line passing through the point K(–4; 1) parallel to the line 
.
Answer: 
.
2. Find the equation of the line passing through the point K(5; –2) parallel to the line 
.
3. Find the equation of the line passing through the point K(–2, –6) perpendicular to the line 
.
4. Find the equation of the line passing through the point K(7; –2) perpendicular to the line 
.
Answer: 
.
5. Find the equation of the perpendicular dropped from the point K(–6; 7) to the straight line 
.