Given:

#[ (2,-1,3,|,24), (0,2,-1,|, 14), (7,-5,0,|, 6) ]#

Swap Row 3 with Row 1:

#R_3harrR_1#

#[ (7,-5,0,|, 6), (0,2,-1,|, 14), (2,-1,3,|,24) ]#

We want position #(1,1) to be 1, therefore we multiply Row 2 by -3, add it to Row 1 and and put the result in Row 1:

#R_1-3R_3 to R_1#

#[ (1,-2,-9,|, -66), (0,2,-1,|, 14), (2,-1,3,|,24) ]#

We want #(3,1) to be 0 therefore we multiply Row 1 by -2, add it to Row 3 and put the result in Row_3 :

#R_3-2R_1 to R_3#

#[ (1,-2,-9,|, -66), (0,2,-1,|, 14), (0,3,21,|,156) ]#

Divide Row 3 by 3:

#1/3R_3 to R_3#

#[ (1,-2,-9,|, -66), (0,2,-1,|, 14), (0,1,7,|,52) ]#

We want position #(2,2)# to be 1 so we swap Row 2 and Row 3:

#R_2harrR_3#

#[ (1,-2,-9,|, -66), (0,1,7,|,52), (0,2,-1,|, 14) ]#

We want postion #(3,2) to be 0, so we multiply Row 2 by -2, add to Row 3 and put the result in Row 3:

#R_3-2R_2 to R_3#

#[ (1,-2,-9,|, -66), (0,1,7,|,52), (0,0,-15,|, -90) ]#

Divide Row 3 by -15:

#-1/15R_3 to R_3#

#[ (1,-2,-9,|, -66), (0,1,7,|,52), (0,0,1,|, 6) ]#

I shall only use the short notation from here to the end.

#R_2-7R_3 to R_2#

#[ (1,-2,-9,|, -66), (0,1,0,|,10), (0,0,1,|, 6) ]#

#R_1+9R_3 to R_1#

#[ (1,-2,0,|, -12), (0,1,0,|,10), (0,0,1,|, 6) ]#

#R_1+2R_2 to R_1#

#[ (1,0,0,|, 8), (0,1,0,|,10), (0,0,1,|, 6) ]#

We have a square identity matrix on the left, therefore, the solution is in the column vector on the right:

#x = 8, y = 10, and z = 6#