# Service for Solving Linear Programming Problems

and other interesting typical problems
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## Example №2. Solving of a System of Linear Equations by the Gauss 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
We will find the variable x3 from equation 2 of the system.
- 5 x3 + 4 x4 = - 1
x3 = 1/5 + 4/5 x4
We will find the variable x1 from equation 1 of the system.
2 x1 + 3 x2 - x3 + x4 = 1
2 x1 = 1 - 3 x2 + x3 - x4
2 x1 = 1 - 3 x2 + ( 1/5 + 4/5 x4 ) - x4
x1 = 3/5 - 3/2 x2 - 1/10 x4
Result:
x1 = 3/5 - 3/2 x2 - 1/10 x4
x3 = 1/5 + 4/5 x4