1. Find solution using dual-simplex method
MAX Z = -2x1 - x2
subject to
-3x1 - x2 <= -3
-4x1 - 3x2 <= -6
-x1 - 2x2 <= -3
and x1,x2 >= 0

Solution:
Problem is

Max `Z` `=` ` - ` `2` `x_1` ` - ` `` `x_2`

subject to

` - ` `3` `x_1` ` - ` `` `x_2` ≤ `-3` ` - ` `4` `x_1` ` - ` `3` `x_2` ≤ `-6` ` - ` `` `x_1` ` - ` `2` `x_2` ≤ `-3`

and `x_1,x_2 >= 0; `

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '`<=`' we should add slack variable `S_1`

2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`

3. As the constraint-3 is of type '`<=`' we should add slack variable `S_3`

After introducing slack variables

Max `Z` `=` ` - ` `2` `x_1` ` - ` `` `x_2` ` + ` `0` `S_1` ` + ` `0` `S_2` ` + ` `0` `S_3`

subject to

` - ` `3` `x_1` ` - ` `` `x_2` ` + ` `` `S_1` = `-3` ` - ` `4` `x_1` ` - ` `3` `x_2` ` + ` `` `S_2` = `-6` ` - ` `` `x_1` ` - ` `2` `x_2` ` + ` `` `S_3` = `-3`

and `x_1,x_2,S_1,S_2,S_3 >= 0`

Iteration-1		`C_j`	`-2`	`-1`	`0`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`
`S_1`	`0`	`-3`	`-3`	`-1`	`1`	`0`	`0`
`S_2`	`0`	`-6`	`-4`	`(-3)`	`0`	`1`	`0`
`S_3`	`0`	`-3`	`-1`	`-2`	`0`	`0`	`1`
`Z=0`		`Z_j`	`0`	`0`	`0`	`0`	`0`
		`Z_j-C_j`	`2`	`1`	`0`	`0`	`0`
		`"Ratio"=(Z_j-C_j)/(S_2,j)` and `S_2,j<0`	`-0.5`	`-0.3333``uarr`	---	---	---

Minimum negative `X_B` is `-6` and its row index is `2`. So, the leaving basis variable is `S_2`.

Maximum negative ratio is `-0.3333` and its column index is `2`. So, the entering variable is `x_2`.

`:.` The pivot element is `-3`.

Entering `=x_2`, Departing `=S_2`, Key Element `=-3`

`R_2`(new)`= R_2`(old) `-: (-3)`

`R_2`(old) =	`-6`	`-4`	`-3`	`0`	`1`	`0`
`R_2`(new)`= R_2`(old) `-: (-3)`	`2`	`4/3`	`1`	`0`	`-1/3`	`0`

`R_1`(new)`= R_1`(old) + `R_2`(new)

`R_1`(old) =	`-3`	`-3`	`-1`	`1`	`0`	`0`
`R_2`(new) =	`2`	`4/3`	`1`	`0`	`-1/3`	`0`
`R_1`(new)`= R_1`(old) + `R_2`(new)	`-1`	`-5/3`	`0`	`1`	`-1/3`	`0`

`R_3`(new)`= R_3`(old) + `2 R_2`(new)

`R_3`(old) =	`-3`	`-1`	`-2`	`0`	`0`	`1`
`R_2`(new) =	`2`	`4/3`	`1`	`0`	`-1/3`	`0`
`2 xx R_2`(new) =	`4`	`8/3`	`2`	`0`	`-2/3`	`0`
`R_3`(new)`= R_3`(old) + `2 R_2`(new)	`1`	`5/3`	`0`	`0`	`-2/3`	`1`

Iteration-2		`C_j`	`-2`	`-1`	`0`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`
`S_1`	`0`	`-1`	`(-5/3)`	`0`	`1`	`-1/3`	`0`
`x_2`	`-1`	`2`	`4/3`	`1`	`0`	`-1/3`	`0`
`S_3`	`0`	`1`	`5/3`	`0`	`0`	`-2/3`	`1`
`Z=-2`		`Z_j`	`-4/3`	`-1`	`0`	`1/3`	`0`
		`Z_j-C_j`	`2/3`	`0`	`0`	`1/3`	`0`
		`"Ratio"=(Z_j-C_j)/(S_1,j)` and `S_1,j<0`	`-0.4``uarr`	---	---	`-1`	---

Minimum negative `X_B` is `-1` and its row index is `1`. So, the leaving basis variable is `S_1`.

Maximum negative ratio is `-0.4` and its column index is `1`. So, the entering variable is `x_1`.

`:.` The pivot element is `-5/3`.

Entering `=x_1`, Departing `=S_1`, Key Element `=-5/3`

`R_1`(new)`= R_1`(old) `xx(-3/5)`

`R_1`(old) =	`-1`	`-5/3`	`0`	`1`	`-1/3`	`0`
`R_1`(new)`= R_1`(old) `xx(-3/5)`	`3/5`	`1`	`0`	`-3/5`	`1/5`	`0`

`R_2`(new)`= R_2`(old) - `4/3 R_1`(new)

`R_2`(old) =	`2`	`4/3`	`1`	`0`	`-1/3`	`0`
`R_1`(new) =	`3/5`	`1`	`0`	`-3/5`	`1/5`	`0`
`4/3 xx R_1`(new) =	`4/5`	`4/3`	`0`	`-4/5`	`4/15`	`0`
`R_2`(new)`= R_2`(old) - `4/3 R_1`(new)	`6/5`	`0`	`1`	`4/5`	`-3/5`	`0`

`R_3`(new)`= R_3`(old) - `5/3 R_1`(new)

`R_3`(old) =	`1`	`5/3`	`0`	`0`	`-2/3`	`1`
`R_1`(new) =	`3/5`	`1`	`0`	`-3/5`	`1/5`	`0`
`5/3 xx R_1`(new) =	`1`	`5/3`	`0`	`-1`	`1/3`	`0`
`R_3`(new)`= R_3`(old) - `5/3 R_1`(new)	`0`	`0`	`0`	`1`	`-1`	`1`

Iteration-3		`C_j`	`-2`	`-1`	`0`	`0`	`0`
`B`	`C_B`	`X_B`	`x_1`	`x_2`	`S_1`	`S_2`	`S_3`
`x_1`	`-2`	`3/5`	`1`	`0`	`-3/5`	`1/5`	`0`
`x_2`	`-1`	`6/5`	`0`	`1`	`4/5`	`-3/5`	`0`
`S_3`	`0`	`0`	`0`	`0`	`1`	`-1`	`1`
`Z=-12/5`		`Z_j`	`-2`	`-1`	`2/5`	`1/5`	`0`
		`Z_j-C_j`	`0`	`0`	`2/5`	`1/5`	`0`
		Ratio	---	---	---	---	---

Since all `Z_j-C_j >= 0` and all `X_(Bi) >= 0` thus the current solution is the optimal solution.

Hence, optimal solution is arrived with value of variables as :
`x_1=3/5,x_2=6/5`

Max `Z=-12/5`