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.
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.
Solve the first for x, the second for y, and the third for z
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:
You can see that the iterations are getting closer and closer to
the actual solutions.
Edwin