Question 921968
<pre>x+3y+3z=3
x+4y+3z=8
x+4y+4z=2

{{{system(x + 3y + 3z= 3,
x + 4y + 3z= 8,
x + 4y + 4z= 2)}}}

<pre><font size = 4 color = "indigo"><b>
Write that as a matrix by dropping the letters
and putting a vertical line instead of equal signs:

{{{(matrix(3, 5,
     1, 3, 3,"|",3,
     red(1),4, 3,"|",8,
    red(1),red(4), 4,"|",2  ))}}}

The idea is to get three zeros in the three positions
in the lower left corner of the matrix, where the elements
I've colored red are:

To get a 0 where the red 1 on the left of the middle row is,
multiply R1 by -1 and add it to 1 times R2, and put it in place 
of the present R2.  That's written as

-R1+1R2->R2

To make it easy, write the multipliers to the left of the two
rows you're working with; that is, put a -1 by R1 and a 1 by R2

{{{matrix(3,1,-1,1,"")}}}{{{(matrix(3, 5,
     1, 3, 3,"|",3,
     red(1),4, 3,"|",8,
    red(1),red(4), 4,"|",2  ))}}}


We are going to change only R2.  Although R1 gets multiplied
by -1 we are going to just do that mentally and add it to R2, but
not really change R1.

{{{(matrix(3, 5,
     1, 3, 3,"|",3,
     red(0),1, 0,"|",5,
    red(1),red(4), 4,"|",2  ))}}}

-----

To get a 0 where the lower left red 1 is, multiply R1
by -1 and add it to 1 times R3.  That's written as

-1R1+1R3->R3

Write the multipliers to the left of the two rows you're 
working with; that is, put a -1 by R1 and a 1 by R3

{{{matrix(3,1,-1,"",1)}}}{{{(matrix(3, 5,
     1, 3, 3,"|",3,
     red(0),1, 0,"|",5,
    red(1),red(4), 4,"|",2  ))}}}


We are going to change only R3. 

{{{(matrix(3, 5,
     1, 3, 3,"|",3,
     red(0),1, 0,"|",5,
    red(0),red(1), 1,"|",-1  ))}}}


---------------

To get a 0 where the red 1 on the bottom row is,
multiply R2 by -1 and add it to 1 times R3.  That's 
written as

-1R2+1R3->R3

Write the multipliers to the left of the two
rows you're working with; that is, put a -1 by R2 and a 1 by R3

{{{matrix(3,1,"","-1",1)}}}{{{(matrix(3, 5,
     1, 3, 3,"|",3,
     red(0),1, 0,"|",5,
    red(0),red(1), 1,"|",-1  ))}}}

We are going to change only R3. 

{{{(matrix(3, 5,
     1, 3, 3,"|",3,
     red(0),1, 0,"|",5,
    red(0),red(0), 1,"|",-6  ))}}}

Now that we have 0's in the three positions in the
lower left corner of the matrix, we change the matrix
back to equations:

{{{system(1x+3y+3z=3,0x+1y+0z=5,0x+0y+1z=-6)}}}

or

{{{system(x+3y+3z=3,y=5,z=-6)}}}

The third equation is already solved for z, and
The second equation is already solved for y, so

Substitute -6 for z and 5 for y in the top equation:

{{{x+3y+3z=3}}}
{{{x+3(5)+3(-6)=3}}}
{{{x+15-18=3}}}
{{{x-3=3}}}
{{{x=6}}}

So the solution is {{{"(x,y,z)"="(6,5,-6)"}}}

Edwin</pre>