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



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 2 on the left of the middle row is,
multiply R1 by -2 and add it to 1 times R2, and put it in place 
of the present R2.  That's written as

-2R1+1R2->R2

To make it easy, write the multipliers to the left of the two
rows you're working with; that is, put a -2 by R1 and a 1 by R2



We are going to change only R2.  Although R1 gets multiplied
by -2 we are going to just do that mentally and add it to R2, but
not really change R1.



Since it just happens that row 2 can be divided through by 3,
without getting fractions, let's do it.  We multiply row 2 by
1/3.  That operation is written

R2->R2



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




We are going to change only R3. 



It will be more convenient to get a 1 where the 2 is,
so we swap rows 2 and 3.  This is written

R2<->R3



To get a 0 where the red 2 is, multiply R2
by -2 and add it to R3.  That's written as

-2R2+R3->R3

Write the multipliers to the left of the two
rows you're working with; that is, put a -2 by R2 and a 1 by R3



We are going to change only R3. 



Since it just happens that row 3 can be divided through by 5,
without getting fractions, let's do it.  We multiply row 3 by
1/5.  That operation is written

R3<->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:



he third equation gives us f



Substitute -1 for z in the middle equation:






Substitute -1 for z and 2 for y in the top equation:







So the solution is 

Edwin