# Service for Solving Linear Programming Problems

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## Example №4. Solving of a System of Linear Equations by the Gauss Jordan Elimination (One Solution)

This solution has been made using the calculator presented on the site.
Please note that the coefficients will disappear which located in the "red" positions. 4 x1 + 2 x2 - 3 x3 = - 3 5 x1 + 3 x2 - 5 x3 = - 8 4 x1 + x2 + 5 x3 = 22
( 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. 4 x1 + 2 x2 - 3 x3 = - 3 x1 + x2 - 2 x3 = - 5 4 x1 + x2 + 5 x3 = 22
The equation 2 and equation 1 are reversed. x1 + x2 - 2 x3 = - 5 4 x1 + 2 x2 - 3 x3 = - 3 4 x1 + x2 + 5 x3 = 22
( 4 x1 + x1 * ( -4) )
+ ( 2 x2 + x2 * ( -4) )
+ ( -3 x3 + ( -2 x3) * ( -4) )
= -3 + ( -5) * ( -4)
The "red" coefficient is zero. x1 + x2 - 2 x3 = - 5 - 2 x2 + 5 x3 = 17 4 x1 + x2 + 5 x3 = 22
( 4 x1 + x1 * ( -4) )
+ ( x2 + x2 * ( -4) )
+ ( 5 x3 + ( -2 x3) * ( -4) )
= 22 + ( -5) * ( -4)
The "red" coefficient is zero. x1 + x2 - 2 x3 = - 5 - 2 x2 + 5 x3 = 17 - 3 x2 + 13 x3 = 42
( -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. x1 + x2 - 2 x3 = - 5 x2 - 8 x3 = - 25 - 3 x2 + 13 x3 = 42
( -3 x2 + x2 * 3 )
+ ( 13 x3 + ( -8 x3) * 3 )
= 42 + ( -25) * 3
The "red" coefficient is zero. x1 + x2 - 2 x3 = - 5 x2 - 8 x3 = - 25 - 11 x3 = - 33
The equation 3 is divided by -11. x1 + x2 - 2 x3 = - 5 x2 - 8 x3 = - 25 x3 = 3
x2
+ ( -8 x3 + x3 * 8 )
= -25 + 3 * 8
The "red" coefficient is zero. x1 + x2 - 2 x3 = - 5 x2 = - 1 x3 = 3
x1
+ x2
+ ( -2 x3 + x3 * 2 )
= -5 + 3 * 2
The "red" coefficient is zero. x1 + x2 = 1 x2 = - 1 x3 = 3
x1
+ ( x2 + x2 * ( -1) )
= 1 + ( -1) * ( -1)
The "red" coefficient is zero. x1 = 2 x2 = - 1 x3 = 3
Result:
x1 = 2
x2 = - 1
x3 = 3