# Example of Solving a System of Linear Equations by Gauss Jordan 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. - 4 x1 + 5 x2 - 3 x3 + 3 x4 = 20 4 x1 + 2 x2 + 3 x3 + 4 x4 = 10 5 x1 + 4 x2 + 4 x3 + 3 x4 = 20
( -4 x1 + 5 x1 )
+ ( 5 x2 + 4 x2 )
+ ( -3 x3 + 4 x3 )
+ ( 3 x4 + 3 x4 )
= 20 + 20
This transformation will allow us to count without fractions for some time. x1 + 9 x2 + x3 + 6 x4 = 40 4 x1 + 2 x2 + 3 x3 + 4 x4 = 10 5 x1 + 4 x2 + 4 x3 + 3 x4 = 20
( 4 x1 + x1 * ( -4) )
+ ( 2 x2 + 9 x2 * ( -4) )
+ ( 3 x3 + x3 * ( -4) )
+ ( 4 x4 + 6 x4 * ( -4) )
= 10 + 40 * ( -4)
The "red" coefficient is zero. x1 + 9 x2 + x3 + 6 x4 = 40 - 34 x2 - x3 - 20 x4 = - 150 5 x1 + 4 x2 + 4 x3 + 3 x4 = 20
( 5 x1 + x1 * ( -5) )
+ ( 4 x2 + 9 x2 * ( -5) )
+ ( 4 x3 + x3 * ( -5) )
+ ( 3 x4 + 6 x4 * ( -5) )
= 20 + 40 * ( -5)
The "red" coefficient is zero. x1 + 9 x2 + x3 + 6 x4 = 40 - 34 x2 - x3 - 20 x4 = - 150 - 41 x2 - x3 - 27 x4 = - 180
( -41 x2 + ( -34 x2) * ( -41/34) )
+ ( - x3 + ( - x3) * ( -41/34) )
+ ( -27 x4 + ( -20 x4) * ( -41/34) )
= -180 + ( -150) * ( -41/34)
The "red" coefficient is zero. x1 + 9 x2 + x3 + 6 x4 = 40 - 34 x2 - x3 - 20 x4 = - 150 7/34 x3 - 49/17 x4 = 15/17
The equation 3 is divided by 7/34. x1 + 9 x2 + x3 + 6 x4 = 40 - 34 x2 - x3 - 20 x4 = - 150 x3 - 14 x4 = 30/7
- 34 x2
+ ( - x3 + x3 )
+ ( -20 x4 + ( -14 x4) )
= -150 + 30/7
The "red" coefficient is zero. x1 + 9 x2 + x3 + 6 x4 = 40 - 34 x2 - 34 x4 = - 1020/7 x3 - 14 x4 = 30/7
x1
+ 9 x2
+ ( x3 + x3 * ( -1) )
+ ( 6 x4 + ( -14 x4) * ( -1) )
= 40 + 30/7 * ( -1)
The "red" coefficient is zero. x1 + 9 x2 + 20 x4 = 250/7 - 34 x2 - 34 x4 = - 1020/7 x3 - 14 x4 = 30/7
The equation 2 is divided by -34. x1 + 9 x2 + 20 x4 = 250/7 x2 + x4 = 30/7 x3 - 14 x4 = 30/7 x1 + 11 x4 = - 20/7 x2 + x4 = 30/7 x3 - 14 x4 = 30/7