How do you solve #((5, 1, 4), (2, -3, -4), (7, 2, -6))X=((5), (2), (5))#?

Starting with the augmented matrix:

#((5,1,4,|,5),(2,-3,-4,|,2),(7,2,-6,|,5))#

Perform a sequence of row operations until the left hand side #3xx3# matrix is the identity matrix. Then the right hand column will be #X#.

Subtract row 1 and 2 from row 3 to get:

#((5,1,4,|,5),(2,-3,-4,|,2),(0,4,-6,|,-2))#

Divide row 1 by #5# to get:

#((1,1/5,4/5,|,1),(2,-3,-4,|,2),(0,4,-6,|,-2))#

Subtract twice row 1 from row 2 to get:

#((1,1/5,4/5,|,1),(0,-17/5,-28/5,|,0),(0,4,-6,|,-2))#

Multiply row #2# by #-5/17# to get:

#((1,1/5,4/5,|,1),(0,1,28/17,|,0),(0,4,-6,|,-2))#

Subtract #4# times row 2 from row 3 to get:

#((1,1/5,4/5,|,1),(0,1,28/17,|,0),(0,0,-214/17,|,-2))#

Multiply row #3# by #-17/214# to get:

#((1,1/5,4/5,|,1),(0,1,28/17,|,0),(0,0,1,|,17/107))#

Subtract #1/5# row 2 from row 1 to get:

#((1,0,8/17,|,1),(0,1,28/17,|,0),(0,0,1,|,17/107))#

Subtract #8/17# row 3 from row 1 to get:

#((1,0,0,|,99/107),(0,1,28/17,|,0),(0,0,1,|,17/107))#

Subtract #28/17# row 3 from row 2 to get:

#((1,0,0,|,99/107),(0,1,0,|,-28/107),(0,0,1,|,17/107))#

So:

#X = ((99/107),(-28/107),(17/107))#