Please use the Gauss-Jordan method to solve system of equations
23.   6x-3y = 1
     -12x-6y= -2
I thought this one is no solution but the answer (back of the book)
is ((23y + 1)/6,y).  Why?

The system you typed has solution (1/6, 0).  Either you mistyped the
problem or the book is wrong.

A similar system that would have the solution you gave would be:

      6x-23y = 1
      -12x+46y= -2



29.  4x + 4y - 4z = 24
     2x -  y +  z = -9
      x - 2y + 3z =  1

[4   4  -4 | 24]
[2  -1   1 | -9]
[1  -2   3 |  1]

Divide row 1 thru by 4

[4   4  -4 | 24]÷(4)
[2  -1   1 | -9]
[1  -2   3 |  1]
 
[1   1  -1 |  6]
[2  -1   1 | -9]
[1  -2   3 |  1]

Get a 0 where the 2 is by multiplying row 1 by -2
and adding it to row 2

[2 -1 1|-9]+(-2)·[1 1 -1|6] = [2 -1 1|-9]+[-2 -2 2|-12] = [0 -3 3|-21] 

[1   1  -1 |   6]
[0  -3   3 | -21]
[1  -2   3 |   1]

Divide row 2 thru by -3

[1   1  -1 |   6]
[0   1  -1 |   7]
[1  -2   3 |   1]

Get a 0 where the 1 is in row 3 column 1 by multiplying 
row 1 by -1 and adding to row 3:

[1 -2 3|1]+(-1)[1 1 -1|6] = [1 -2 3|1]+[-1 -1 1|-6] = [0 -3 4|-5]

[1   1  -1 |   6]
[0   1  -1 |   7]
[0  -3   4 |  -5]

Get a 0 where the -3 is by multiplying row 2 by 3 and adding to
row 3

[0 -3 4|-5]+(3)[0 1 -1|7] = [0 -3 4|-5]+[0 3 -3|21] = [0 0 1|16]

[1   1  -1 |   6]
[0   1  -1 |   7]
[0   0   1 |  16]

This matrix represents this system:

1x + 1y - 1z =  6
0x + 1y - 1z =  7
0x + 0y + 1z = 16

or

 x +  y -  z =  6
      y -  z =  7
           z = 16
 
Now back-substitute:

Substitute 16 for z in the 2nd equation:

       y - z = 7
      y - 16 = 7
           y = 23

Substitute 23 for y and 16 for z in the 1st equation:

   x + y - z = 6
 x + 23 - 16 = 6
       x + 7 = 6
           x = -1

So the solution is (x, y, z) = (-1, 23, 16)

Edwin