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 is, multiply R3
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 R3 gets multiplied
by 2 we are going to just do that mentally and add it to R2, but
not really change R3.



-----

To get a 0 where the lower left red -1 is, multiply R1
by 1 and add it to 5 times R3.  That's written as

1R1+5R3->R3

Write the multipliers to the left of the two rows you're 
working with; that is, put a 1 by R1 and a 5 by R3



We are going to change only R3. 



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

To get a 0 where the remaining red -1 is, multiply R2
by 1 and add it to -4 times R3.  That's written as

1R2-4R3->R3

Write the multipliers to the left of the two
rows you're working with; that is, put a +1 by R2 and a -4 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:



Solve the third equation for z:





Substitute 3 for z in the middle equation:







Substitute 3 for z and 1 for y in the top equation:









So the solution is 

Edwin