Service for Solving Linear Programming Problems

and other interesting typical problems		 Русский

Example №1. Solving of a System of Linear Equations by the Gauss Elimination (One Solution)

This solution was made using the calculator presented on the site.
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Please note that the coefficients will disappear which located in the "red" positions.
Знак системы4x1+2x2- 3 x3 = - 3
5x1+3x2- 5 x3 = - 8
4x1+x2+5x3 = 22
The equation 1 multiplied by -1 is added to the equation 2.   more info
( 5 x1 + 4 x1 * ( -1) )
+ ( 3 x2 + 2 x2 * ( -1) )
+ ( -5 x3 + ( -3 x3) * ( -1) )
= -8 + ( -3) * ( -1)
This transformation will allow us to count without fractions for some time.
Знак системы4x1+2x2- 3 x3 = - 3
x1 +x2- 2 x3 = - 5
4x1+x2+5x3 = 22
The equation 2 and equation 1 are reversed.
Знак системыx1 +x2- 2 x3 = - 5
4x1+2x2- 3 x3 = - 3
4x1+x2+5x3 = 22
The equation 1 multiplied by -4 is added to the equation 2.   more info
( 4 x1 + x1 * ( -4) )
+ ( 2 x2 + x2 * ( -4) )
+ ( -3 x3 + ( -2 x3) * ( -4) )
= -3 + ( -5) * ( -4)
The "red" coefficient is zero.
Знак системыx1 +x2- 2 x3 = - 5
- 2 x2+5x3 = 17
4x1+x2+5x3 = 22
The equation 1 multiplied by -4 is added to the equation 3.   more info
( 4 x1 + x1 * ( -4) )
+ ( x2 + x2 * ( -4) )
+ ( 5 x3 + ( -2 x3) * ( -4) )
= 22 + ( -5) * ( -4)
The "red" coefficient is zero.
Знак системыx1 +x2- 2 x3 = - 5
- 2 x2+5x3 = 17
- 3 x2+13x3 = 42
The equation 3 multiplied by -1 is added to the equation 2.   more info
( -2 x2 + ( -3 x2) * ( -1) )
+ ( 5 x3 + 13 x3 * ( -1) )
= 17 + 42 * ( -1)
This transformation will allow us to count without fractions for some time.
Знак системыx1 +x2- 2 x3 = - 5
x2 - 8 x3 = - 25
- 3 x2+13x3 = 42
The equation 2 multiplied by 3 is added to the equation 3.   more info
( -3 x2 + x2 * 3 )
+ ( 13 x3 + ( -8 x3) * 3 )
= 42 + ( -25) * 3
The "red" coefficient is zero.
Знак системыx1 +x2- 2 x3 = - 5
x2 - 8 x3 = - 25
- 11 x3 = - 33
We will find the variable x3 from equation 3 of the system.
- 11 x3 = - 33
x3 = 3
We will find the variable x2 from equation 2 of the system.
x2 - 8 x3 = - 25
x2 = - 25 + 8 x3
x2 = - 25 + 8 * ( 3 )
x2 = - 1
We will find the variable x1 from equation 1 of the system.
x1 + x2 - 2 x3 = - 5
x1 = - 5 - x2 + 2 x3
x1 = - 5 - ( - 1 ) + 2 * ( 3 )
x1 = 2
Result:
x1 = 2
x2 = - 1
x3 = 3

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