minimize	z	=	-(3/2)x1	-x2
subject to			(5/2)x1	+5x2	<=	150
			5x1	+2x2	<=	120
			x1>=0,	x2>=0

introducing non negative slack variables x₃>=0 and x₄>=0, rewrite the problem in standard form:

minimize	z	=	-(3/2)x1	-x2
subject to			(5/2)x1	+5x2	+x3		=150
			5x1	+2x2		+x4	=120
			x1>=0,	x2>=0,	x3>=0,	x4>=0

Select the slack variables x₃ and x₄ as the basic variables, x₁=x₂=0, x₃=150, and x₄=120 are the basic feasible solutions (bfs). So, organize the simplex tableau of cycle 0 with the basic variables x₃ and x₄.