# Service for Solving Linear Programming Problems

and other interesting typical problems
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## Example №5. Solving of a System of Linear Equations by the Gauss Jordan Elimination (Many Solutions)

This solution was made using the calculator presented on the site.
Please note that the coefficients will disappear which located in the "red" positions. 2 x1 + 3 x2 - x3 + x4 = 1 8 x1 + 12 x2 - 9 x3 + 8 x4 = 3 4 x1 + 6 x2 + 3 x3 - 2 x4 = 3 2 x1 + 3 x2 + 9 x3 - 7 x4 = 3
( 8 x1 + 2 x1 * ( -4) )
+ ( 12 x2 + 3 x2 * ( -4) )
+ ( -9 x3 + ( - x3) * ( -4) )
+ ( 8 x4 + x4 * ( -4) )
= 3 + 1 * ( -4)
The "red" coefficient is zero. 2 x1 + 3 x2 - x3 + x4 = 1 - 5 x3 + 4 x4 = - 1 4 x1 + 6 x2 + 3 x3 - 2 x4 = 3 2 x1 + 3 x2 + 9 x3 - 7 x4 = 3
( 4 x1 + 2 x1 * ( -2) )
+ ( 6 x2 + 3 x2 * ( -2) )
+ ( 3 x3 + ( - x3) * ( -2) )
+ ( -2 x4 + x4 * ( -2) )
= 3 + 1 * ( -2)
The "red" coefficient is zero. 2 x1 + 3 x2 - x3 + x4 = 1 - 5 x3 + 4 x4 = - 1 5 x3 - 4 x4 = 1 2 x1 + 3 x2 + 9 x3 - 7 x4 = 3
( 2 x1 + 2 x1 * ( -1) )
+ ( 3 x2 + 3 x2 * ( -1) )
+ ( 9 x3 + ( - x3) * ( -1) )
+ ( -7 x4 + x4 * ( -1) )
= 3 + 1 * ( -1)
The "red" coefficient is zero. 2 x1 + 3 x2 - x3 + x4 = 1 - 5 x3 + 4 x4 = - 1 5 x3 - 4 x4 = 1 10 x3 - 8 x4 = 2
( 5 x3 + ( -5 x3) )
+ ( -4 x4 + 4 x4 )
= 1 + ( -1)
The "red" coefficient is zero. 2 x1 + 3 x2 - x3 + x4 = 1 - 5 x3 + 4 x4 = - 1 0 = 0 10 x3 - 8 x4 = 2
( 10 x3 + ( -5 x3) * 2 )
+ ( -8 x4 + 4 x4 * 2 )
= 2 + ( -1) * 2
The "red" coefficient is zero. 2 x1 + 3 x2 - x3 + x4 = 1 - 5 x3 + 4 x4 = - 1 0 = 0 0 = 0 2 x1 + 3 x2 - x3 + x4 = 1 - 5 x3 + 4 x4 = - 1
The equation 2 is divided by -5. 2 x1 + 3 x2 - x3 + x4 = 1 x3 - 4/5 x4 = 1/5
2 x1
+ 3 x2
+ ( - x3 + x3 )
+ ( x4 + ( -4/5 x4) )
= 1 + 1/5
The "red" coefficient is zero. 2 x1 + 3 x2 + 1/5 x4 = 6/5 x3 - 4/5 x4 = 1/5
The equation 1 is divided by 2. x1 + 3/2 x2 + 1/10 x4 = 3/5 x3 - 4/5 x4 = 1/5
Result:
x1 = 3/5 - 3/2 x2 - 1/10 x4
x3 = 1/5 + 4/5 x4