Question: Use the big M method to solve

min z = 3X_{1}

s.t. 2X_{1 }+ X_{2}>= 6

3X_{1 }+ 2X_{2 }= 4

X_{1 }, X_{2}>= 0

initial tableau

BV...... X_{1}......X_{2}.......e_{1}......a_{1}.......a_{2}......RHS...

a_{1}........ 2........1........-1......1........0........6.....

a_{2}........3..........2........0.......0........1.... ....4.....

Z.........-3.........0........0......-M......-M.......0.....

(a_{1})M + (a_{2})M + Z

BV...... X_{1}......X_{2}.......e_{1}......a_{1}.......a_{2}......RHS... Ratio

a_{1}........ 2........1........-1......1........0........6.......3

a_{2}........3..........2........0.......0........1.... ....4..... 4/3

Z.......5M-3......3M.....-M.......0........0.......10M... X1 enters, a_{2}leaves

BV...... X_{1}......X_{2}.......e_{1}......a_{1}.......a_{2}..........RHS...

a_{1}........0.......-1/3.....-1......1.......-2/3.......10/3...

X_{1}........1.......2/3........0......0......1/3...........4/3.....

Z.........0....(-M/3)+2..-M......0...(-5M/3)+1..(10M/3)+4...

no more non-negative coefficients, can't get a_{1 }out of basis, so solution is infeasable. This is what i did so far. If i have made any errors please tell me where i went wrong.