A linear equation with three variables represents a plane in three dimensional geometry.A system of linear equations can be solved using the common methods of substitution and elimination.It is not possible to get a graphical solution for a system of three variables on a two dimensional plane.The system of equations in three variables are generally solved using Gaussian row eliminations.

Number of Solutions of a Linear System

As with a system of linear equations in two variables, the solution(s) of a system of linear equations in more than two variables must fall into one of three categories.
  1. There is exactly one solution.                     
  2. There are infinitely many solutions.             
  3. There is no solution.

Solving a system of linear equations using Elimination method


Solve the system of Linear Equations
$x-2y+3z=9$                       Equation (1)
$-x+3y         =-4$                 Equation (2)
$2x-5y+5z=17$                    Equation (3)
We can eliminate the variable x by manipulating the equations
Adding equations (1) and (2) we get a new equation
$x-2y+3z=9$  
$-x+3y        =-4$ 
----------------------------
        $y+3z =5$                          Equation (4)
We can multiply the first equation by 2 and add it to equation 3 to get rid of x
$2x-5y+5z=17$                           Equation (3)
$-2x+6y      = -8$                    2 x Equation (2)
-------------------------------
           $y+5z = 9$                       Equation (5)
Now equations 4 and 5 are in two variables y and z. We can eliminate the variable y by subtracting equation 4 from 5.
$y+5z = 9$                                  Equation (5)
$y+3z =5$                                   Equation (4)
---------------------
      $2z=4$                                  Subtraction
        $z=2$                                  Equation solved for the variable z
Now we can substitute z=2 back in equation 4 or 5 to solve for y.  Substituting in equation (4)
$y+3(2)=5$           Solving this equation we get $y=-1$
Substituting y =-1 in equation (2) we get,
$-x+3(-1)=-4$         Solving this equation we get $x=1$
Hence the solution for the system is x=1,y=-1 and z=2.

Operations that produce equivalent systems

Two systems of linear equations are said to be equivalent if they have the same solution. Using Gaussian row operations a system can be rewritten into a simpler equivalent form. The second system so written can be solved using simple back substitution.
Each of the following operations on a system of linear equations produces an equivalent system of linear equations.
  1. Interchanging the positions of two equations.
  2. Multiply one of the equations by a non-zero constant.
  3. Add a multiple of one of the equations to another equation and replace the later equation.

Solving a linear system using Gaussian elimination

Let us solve the system solved earlier using row operations.
$x-2y+3z=9$                                Equation (1)
$-x+3y          =-4$                         Equation (2)
$2x-5y+5z=17$                             Equation (3)

$x-2y+3z=9$  
         $   y +3z=5$                             Equation (1) is added to equation (2)
        $  -y-z=-1$                               -2 times equation(1) is added to equation (3).

$x-2y+3z=9$
         $   y +3z=5$    
                $        2z=4$                         Equation (2) in the previous step is added to equation 3.

Now this is an equivalent system of the original system. The equation (3) can be solved easily using division as z =2.
The value of z=2 can be back substituted in equation(2) and the resulting equation can be solved for y.
$y+3(2)=5$  →  $y=-1$.
Finally substituting back the values y =-1 and z=2 in equation (1) we get x =1.
Hence the solution to the given system is x=1,y=-1 and z=2.


Equivalent systems with many solutions and no solution.

  1. If the resulting equivalent system after doing row operations contains one equation as a true statement without variables, then the system has infinitely many solutions.
  2. If the resulting equivalent system contains a false statement in an equation, then the system has no solution.