7.1. SYSTEMS OF EQUATIONS, ALGEBRAIC PROCEDURES 107

To solve this system replace the second equation by (−2) times the first equation addedto the second. This yields the system

x+3y+6z = 25y+2z = 8

2y+5z = 19(7.5)

Now take (−2) times the second and add to the third. More precisely, replace the thirdequation with (−2) times the second added to the third. This yields the system

x+3y+6z = 25y+2z = 8

z = 3(7.6)

At this point, you can tell what the solution is. This system has the same solution as theoriginal system and in the above, z = 3. Then using this in the second equation, it followsy+ 6 = 8 and so y = 2. Now using this in the top equation yields x+ 6+ 18 = 25 and sox = 1. This process is called back substitution.

Alternatively, in 7.6 you could have continued as follows. Add (−2) times the bottomequation to the middle and then add (−6) times the bottom to the top. This yields

x+3y = 7, y = 2, z = 3

Now add (−3) times the second to the top. This yields

x = 1, y = 2, z = 3,

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

7.1.2 Gauss EliminationA less cumbersome way to represent a linear system is to write it as an augmented matrix.For example the linear system, 7.4 can be written as 1 3 6 | 25

2 7 14 | 580 2 5 | 19

 .

It has exactly the same information as the original system but here it is understood there

is an x column,

 120

 , a y column,

 372

 and a z column,

 6145

 . The rows corre-

spond to the equations in the system. Thus the top row in the augmented matrix correspondsto the equation,

x+3y+6z = 25.

Now when you replace an equation with a multiple of another equation added to itself, youare just taking a row of this augmented matrix and replacing it with a multiple of another