# Example of Solving a System of Linear Equations by Gaussian Elimination.

### This solution has been done by the calculator presented on the site.

Please note that the coefficients will disappear which located in the "red" positions. 3 x1 + 2 x2 + x3 + x4 = - 2 x1 - x2 + 4 x3 - x4 = - 1 - 2 x1 - 2 x2 - 3 x3 + x4 = 9 x1 + 5 x2 - x3 + 2 x4 = 4
The equation 2 and equation 1 are reversed. x1 - x2 + 4 x3 - x4 = - 1 3 x1 + 2 x2 + x3 + x4 = - 2 - 2 x1 - 2 x2 - 3 x3 + x4 = 9 x1 + 5 x2 - x3 + 2 x4 = 4
( 3 x1 + x1 * ( -3) )
+ ( 2 x2 + ( - x2) * ( -3) )
+ ( x3 + 4 x3 * ( -3) )
+ ( x4 + ( - x4) * ( -3) )
= -2 + ( -1) * ( -3)
The "red" coefficient is zero. x1 - x2 + 4 x3 - x4 = - 1 5 x2 - 11 x3 + 4 x4 = 1 - 2 x1 - 2 x2 - 3 x3 + x4 = 9 x1 + 5 x2 - x3 + 2 x4 = 4
( -2 x1 + x1 * 2 )
+ ( -2 x2 + ( - x2) * 2 )
+ ( -3 x3 + 4 x3 * 2 )
+ ( x4 + ( - x4) * 2 )
= 9 + ( -1) * 2
The "red" coefficient is zero. x1 - x2 + 4 x3 - x4 = - 1 5 x2 - 11 x3 + 4 x4 = 1 - 4 x2 + 5 x3 - x4 = 7 x1 + 5 x2 - x3 + 2 x4 = 4
( x1 + x1 * ( -1) )
+ ( 5 x2 + ( - x2) * ( -1) )
+ ( - x3 + 4 x3 * ( -1) )
+ ( 2 x4 + ( - x4) * ( -1) )
= 4 + ( -1) * ( -1)
The "red" coefficient is zero. x1 - x2 + 4 x3 - x4 = - 1 5 x2 - 11 x3 + 4 x4 = 1 - 4 x2 + 5 x3 - x4 = 7 6 x2 - 5 x3 + 3 x4 = 5
( 5 x2 + ( -4 x2) )
+ ( -11 x3 + 5 x3 )
+ ( 4 x4 + ( - x4) )
= 1 + 7
This transformation will allow us to count without fractions for some time. x1 - x2 + 4 x3 - x4 = - 1 x2 - 6 x3 + 3 x4 = 8 - 4 x2 + 5 x3 - x4 = 7 6 x2 - 5 x3 + 3 x4 = 5
( -4 x2 + x2 * 4 )
+ ( 5 x3 + ( -6 x3) * 4 )
+ ( - x4 + 3 x4 * 4 )
= 7 + 8 * 4
The "red" coefficient is zero. x1 - x2 + 4 x3 - x4 = - 1 x2 - 6 x3 + 3 x4 = 8 - 19 x3 + 11 x4 = 39 6 x2 - 5 x3 + 3 x4 = 5
( 6 x2 + x2 * ( -6) )
+ ( -5 x3 + ( -6 x3) * ( -6) )
+ ( 3 x4 + 3 x4 * ( -6) )
= 5 + 8 * ( -6)
The "red" coefficient is zero. x1 - x2 + 4 x3 - x4 = - 1 x2 - 6 x3 + 3 x4 = 8 - 19 x3 + 11 x4 = 39 31 x3 - 15 x4 = - 43
( 31 x3 + ( -19 x3) * 31/19 )
+ ( -15 x4 + 11 x4 * 31/19 )
= -43 + 39 * 31/19
The "red" coefficient is zero. x1 - x2 + 4 x3 - x4 = - 1 x2 - 6 x3 + 3 x4 = 8 - 19 x3 + 11 x4 = 39 56/19 x4 = 392/19
We will find the variable x4 from equation 4 of the system.
56/19 x4 = 392/19
x4 = 7
We will find the variable x3 from equation 3 of the system.
- 19 x3 + 11 x4 = 39
- 19 x3 = 39 - 11 x4
- 19 x3 = 39 - 11 * ( 7 )
x3 = 2
We will find the variable x2 from equation 2 of the system.
x2 - 6 x3 + 3 x4 = 8
x2 = 8 + 6 x3 - 3 x4
x2 = 8 + 6 * ( 2 ) - 3 * ( 7 )
x2 = - 1
We will find the variable x1 from equation 1 of the system.
x1 - x2 + 4 x3 - x4 = - 1
x1 = - 1 + x2 - 4 x3 + x4
x1 = - 1 + ( - 1 ) - 4 * ( 2 ) + ( 7 )
x1 = - 3
Result:
x1 = - 3
x2 = - 1
x3 = 2
x4 = 7