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.



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To get a 0 where the lower left red 4 is, multiply R1
by -4 and add it to 1 times R3.  That's written as

-4R1+1R3->R3

Write the multipliers to the left of the two rows you're 
working with; that is, put a -4 by R1 and a 1 by R3




We are going to change only R3. 




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

To get a 0 where the red -2 is, multiply R2
by -2 and add it to 3 times R3.  That's written as

-2R2+3R3->R3

Write the multipliers to the left of the two
rows you're working with; that is, put a -2 by R2 and a 3 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 5 for z in the middle equation:







Substitute 5 for z and 3 for y in the top equation:







So the solution is 

Edwin