Question 854878
The most efficient method I know is a modification of Gauss:

The procedure is long so I will summarize:


r1 <-> r2

-2r1+r2 -> r2
-3r1+r3 -> r3
-5r2+r3 -> r3

yields:

{{{
(
 matrix( 3, 4, 
   1,2,-2,1,
   0,-1,-2,-5,
    0,0,17,27
 )
)
}}}


Next using back substitution we arrive at


x=9/17
y=31/17
z=27/17