First, we convert the problem into standard form by introducing slack variables:

maximize z = 3X1 + 7X2 + X3 subject to: X1 + 3X2 + 9X3 + X4 = 27 X1 + 6X2 + 3X3 + X5 = 18 X1, X2, X3, X4, X5 >= 0

We can then set up the initial simplex tableau:

	X1	X2	X3	X4	X5	RHS
0	-3	-7	-1	0	0	0
X4	1	3	9	1	0	27
X5	1	6	3	0	1	18

The pivot column is X2, since it has the most negative coefficient in the objective row. To determine the pivot row, we calculate the ratios of the RHS values to the coefficients of X2 in each constraint:

27/3 = 9 18/6 = 3

The minimum ratio is 3, so X5 is the pivot row. We use row operations to make X5 the identity row and eliminate X2 from the other rows:

	X1	X2	X3	X4	X5	RHS
0	0	-1/2	2	0	7/2	63
X4	0	9/2	-6	1	-1/2	9
X2	1	2/3	1	0	1/6	3

The new tableau shows that the optimal solution is X1 = 3, X2 = 0, X3 = 0, with a maximum value of z = 9.