We add non-negative slack variables to the left side to "take 
up the slack" between the left side and the right side and make 
the inequalities into equations:

3x + 2y + s1 = 18.        

6x + 7y + s2 = 42.

We take the equation for what we want to maximize f(x,y) = 3x + 9y
and get 0 on the right side:

-3x - 9y + f(x,y) = 0

We set the three equations up like this:

  x    y   s1   s2   f(x,y)    k
--------------------------------    
 3x + 2y + s1               = 18
 6x + 7y      + s2          = 42
--------------------------------
-3x - 9y           + f(x,y) =  0

Then we make the tableau matrix by taking off the 
variables and leave only the coefficients, and putting
in partitions like this:

  x    y | s1   s2  |f(x,y)|   k
--------------------------------    
  3    2 |  1    0  |   0  |  18
  6    7 |  0    1  |   0  |  42
--------------------------------
 -3   -9 |  0    0  |   1  |   0

The pivot column is the second one because -9 is the most negative 
number on the bottom row.

[That really isn't necessary to take the most negative number on the bottom row
to indicate the pivot column, because any negative number on the bottom row will
work, but it usually shortens the process to take the lowest negative number]:
                                 ___          ___
To find the pivot row we divide 2)18 = 9 and 7)42 = 6
Since 6 is smaller and it came from row 2, the pivot row is row 2.
So 7 is the pivot element (common to both the pivot column and row) 

So we pivot on 7.  That is, we make the 7 into a 1 and use it to make
to other numbers on the pivot column 0. 

First we get a 1 where the 7 is by dividing that pivot row through by 7

  x    y | s1   s2  |f(x,y)|   k
--------------------------------    
  3    2 |  1    0  |   0  |  18
6/7    1 |  0  1/7  |   0  |   6
--------------------------------
 -3   -9 |  0    0  |   1  |   0

We get a 0 where the 2 is by multiplying the pivot row by -2
and adding it to the top row:

     We take the pivot row  -> 6/7    1 |  0  1/7  |   0  |   6

     multiply it by -2  -->  -12/7   -2 |  0 -2/7  |   0  | -12
     add the top row     -->     3    2 |  1    0  |   0  |  18
     ---------------------------------------------------------------
     get -->                   9/7    0 |  1 -2/7  |   0  |   6

     Replace the top row by that 

  x    y | s1   s2  |f(x,y)|   k
--------------------------------    
9/7    0 |  1 -2/7  |   0  |   6
6/7    1 |  0  1/7  |   0  |   6
--------------------------------
 -3   -9 |  0    0  |   1  |   0

---
We get a 0 where the -9 is by multiplying the pivot row by 9
and adding it to the bottom row:

     We take that pivot row -> 6/7    1 |  0  1/7  |   0  |   6

     multiply it by  9  -->   54/7    9 |  0  9/7  |   0  |  54
     add the bottom row -->     -3   -9 |  0    0  |   1  |   0
     ----------------------------------------------------------
     get -->                  33/7    0 |  0  9/7  |   1  |  54

     Replace the bottom row by that: 

   x    y | s1   s2  |f(x,y)|   k
 --------------------------------    
 9/7    0 |  1 -2/7  |   0  |   6
 6/7    1 |  0  1/7  |   0  |   6
 --------------------------------
33/7    0 |  0  9/7  |   1  |  54

There are no negative numbers left on the bottom row, so we write the
bottom row back as an equations with the variables:

33/7x + 0y + 0s1 + 9/7s2 + 1f(x,y) = 54

Simplify:

33/7x + 9/7s2 + f(x,y) = 54
                f(x,y) = 54 - 33/7x - 9/7s2

So f(x,y) will be a maximum when we don't subtract anything from it,
and that will only be when x and s2 are both 0, since they can't be 
negative.

So the maximum possible value for f(x,y) is 54.

Often a problem will ask for what values of x and y gives this maximum
value.

If you were asked that you would know that it was when x=0 and s2 = 0.
However we want to know what y would have to be.

So we substitute x=0 and s2=0 in

   6x + 7y + s2 = 42
6(0) + 7y + (0) = 42
             7y = 42
              y = 6

So the maximum possible value for f(x,y) is 54 and that's when x=0 and y=6. 

Edwin