Question 1192827
<pre>
To do this efficiently you need something like a not too old TI-84 
calculator, and to learn how to store a number for a letter, and 
how to scroll up to bring back expressions you have previously used, 
so you won't have to type them in but once.

{{{system(
3x + 2y + 4z = 12,
x + 5y + 2z = -9,
7x + 2y + z = 5)}}}

Rearrange the equations, if possible, so that 

1. the equation that comes 1st is one which has its 
x-coefficient greater in absolute value than either 
its y-coefficient or its z-coefficient.

2. the equation that comes 2nd is one which has its 
y-coefficient greater in absolute value than either 
its x-coefficient or its z-coefficient.  

3. the equation that comes 3rd is one which has its 
z-coefficient greater in absolute value than either 
its x-coefficient or its y-coefficient.

In other words, the largest coefficient in each equation
is on the diagonal.

{{{system(7x + 2y + z = 5,x + 5y + 2z = -9,
3x + 2y + 4z = 12)}}}

Solve the first for x, the second for y, and the third for z

{{{system(
x =(5-2y-z)/7,
y =(-9-x-2z)/5, 
z =(12-3x-2y)/4)}}}

Iteration|    x  |    y  |    z  | 
    1    | 0.0000| 0.0000| 0.0000|   
    2    | 0.7143|-1.9429| 3.4357|
    3    | 0.7786|-3.3300| 4.0811|
    4    | 1.0827|-3.6490| 4.0125|
    5    | 1.1836|-3.6417| 3.9331|
    6    | 1.1848|-3.6011| 3.9120|

Start out with all 0's for the first iteration.
Substitute (x,y,z)=(0,0,0) in the equation for x, get x=0.7143.
Substitute (x,y,z)=(0.7143,0,0) in the equation for y, get y=-1.9429.
Substitute (x,y,z)=(0.7143,-1.9429,0) in the equation for z, get z=-3.4357.
That's the end of the 2nd iteration. Now we begin the 3rd iteration.
Substitute (x,y,z)=(0.7143,-1.9429,-3.4357) in the equation for x, 
get 0.7786.

etc., etc.
    
The actual solution using the methods of 2nd year algebra, gives:

{{{system(x=122/103=1.1845,y=-371/103=-3.6019,z=403/103=3.9126)}}}

You can see that the iterations are getting closer and closer to
the actual solutions.

Edwin</pre>