Question 150173
In Matrix notation, the problem is,
{{{(
  matrix( 3, 3, 
    1, 2, -1,
    2, -2, 1,
    6, 4, 3
  ))*
(
  matrix( 3, 1, 
    x, 
    y, 
    z
  )
)=
(
  matrix( 3, 1, 
    2, 
    -1, 
    5
  )
)}}}
or simplified
[A][x]=[b]
[Ainv][A][x]=[Ainv][b]
[x]=[Ainv][b]
Find the inverse of matrix A, multiply by the right hand side to find the solution.
In this case,
[A]={{{(
  matrix( 3, 3, 
    1, 2, -1,
    2, -2, 1,
    6, 4, 3
  ))}}}
The inverse of [A] is, 
[Ainv]={{{(1/30)*(
  matrix( 3, 3, 
    10, 10, 0,
    0, -9, 3,
    -20, -8, 6
  ))}}}
Then the matrix multiplication is,
[x]={{{(1/14)*(
  matrix( 3, 3, 
    10, 10, 0,
    0, -9, 3,
    -20, -8, 6
  ))*(
  matrix( 3, 1, 
    2, 
    -1, 
    5
  )
)}}}
{{{x=(1/30)(10*2+(10)(-1)+0(5))}}}
{{{x=(1/30)(20-10)}}}
{{{x=(1/30)(10)}}}
{{{x=1/3}}}
{{{y=(1/30)(0*2+(-9)(-1)+(3)(5))}}}
{{{y=(1/30)(9+15)}}}
{{{y=(1/30)(24)}}}
{{{y=4/5}}}
{{{z=(1/30)(20*2+(8)(-1)+(-6)(5))}}}
{{{z=(1/30)(40-8-30)}}}
{{{z=(1/30)(2)}}}
{{{z=-1/15}}}
Your solution matrix is then
[x]={{{(
  matrix( 3, 1, 
    1/3, 
    4/5, 
    -1/15
  )
)}}}