or rather



Write that as a matrix by dropping the letters
and putting a vertical line instead of equal signs:



Things are easier if it is possible to get a 1 in 
the upper left corner by swapping rows. So let's 
begin by swapping rows 1 and 3.  That operation is
written as 

R1<->R3



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

-1R1+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




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.



-----

To get a 0 where the lower left red 10 is, multiply R1
by -10 and add it to 1 times R3.  That's written as

-10R1+1R3->R3

Write the multipliers to the left of the two rows you're 
working with; that is, put a -10 by R1 and a 1 by R3




We are going to change only R3. 




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

To get a 0 where the red -9 is, multiply R2
by 1 (leave it as it is) and add it to 1 times R3.
We're just adding row 2 as is to row 3 as is. 
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:



We are going to change only R3. 



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:



or just



Solve the third equation for z:





Solving the middle system:




Substitute 12 for z and 0 for y in the top equation:







So the solution is 

Edwin