Consider the following example.
Example 4.1.1 Find x and y such that
 (4.1) 
The set of ordered pairs,
You can verify that
Example 4.1.3 To illustrate the third of these operations on this particular system, consider the following.

The system has the same solution set as the system

To obtain the second system, take the second equation of the first system and add −2 times the first equation to obtain

Now, this clearly shows that y = 2 and so it follows from the other equation that x + 2 = 7 and so x = 5.
Of course a linear system may involve many equations and many variables. The solution set is still the collection of solutions to the equations. In every case, the above operations of Definition 4.1.2 do not change the set of solutions to the system of linear equations.
Theorem 4.1.4 Suppose you have two equations, involving the variables,

 (4.2) 
where E_{1} and E_{2} are expressions involving the variables and f_{1} and f_{2} are constants. (In the above example there are only two variables, x and y and E_{1} = x + y while E_{2} = 2x − y.) Then the system E_{1} = f_{1},E_{2} = f_{2} has the same solution set as
 (4.3) 
Also the system E_{1} = f_{1},E_{2} = f_{2} has the same solutions as the system, E_{2} = f_{2},E_{1} = f_{1}. The system E_{1} = f_{1},E_{2} = f_{2} has the same solution as the system E_{1} = f_{1},aE_{2} = af_{2} provided a≠0.
Proof: If
The second assertion of the theorem which says that the system E_{1} = f_{1},E_{2} = f_{2} has the same solution as the system, E_{2} = f_{2},E_{1} = f_{1} is seen to be true because it involves nothing more than listing the two equations in a different order. They are the same equations.
The third assertion of the theorem which says E_{1} = f_{1},E_{2} = f_{2} has the same solution as the system E_{1} = f_{1},aE_{2} = af_{2} provided a≠0 is verified as follows: If
Stated simply, the above theorem shows that the elementary operations do not change the solution set of a system of equations.
Here is an example in which there are three equations and three variables. You want to find values for x,y,z such that each of the given equations are satisfied when these values are plugged in to the equations.
To solve this system replace the second equation by
 (4.5) 
Now take
 (4.6) 
At this point, you can tell what the solution is. This system has the same solution as the original system and in the above, z = 3. Then using this in the second equation, it follows y + 6 = 8 and so y = 2. Now using this in the top equation yields x + 6 + 18 = 25 and so x = 1. This process is called back substitution.
Alternatively, in 4.6 you could have continued as follows. Add

Now add

a system which has the same solution set as the original system. This avoided back substitution and led to the same solution set.