Find dual from primal conversion
MIN z = x1 - 3x2 - 2x3
subject to
3x1 - x2 + 2x3 <= 7
2x1 - 4x2 >= 12
-4x1 + 3x2 + 8x3 = 10
and x1,x2 >= 0 and x3 unrestricted in sign

Solution:
Primal problem is

MIN `z_x` `=` `` `` `x_1` ` - ` `3` `x_2` ` - ` `2` `x_3`

subject to

`` `3` `x_1` ` - ` `` `x_2` ` + ` `2` `x_3` ≤ `7` `` `2` `x_1` ` - ` `4` `x_2` ≥ `12` ` - ` `4` `x_1` ` + ` `3` `x_2` ` + ` `8` `x_3` = `10`

and `x_1,x_2 >= 0; ``x_3` unrestricted in sign

Since objective function is minimizing, all `<=` constraints (1) can be converted to `>=` type by multipling both sides by -1

MIN `z_x` `=` `` `` `x_1` ` - ` `3` `x_2` ` - ` `2` `x_3`

subject to

` - ` `3` `x_1` ` + ` `` `x_2` ` - ` `2` `x_3` ≥ `-7` `` `2` `x_1` ` - ` `4` `x_2` ≥ `12` ` - ` `4` `x_1` ` + ` `3` `x_2` ` + ` `8` `x_3` = `10`

and `x_1,x_2 >= 0; ``x_3` unrestricted in sign

In primal, There are `3` variables and `3` constraints, so in dual there must be `3` constraints and `3` variables

In primal, The coefficient of objective function `c_1=1,c_2=-3,c_3=-2` becomes right hand side constants in dual

In primal, The right hand side constants `b_1=-7,b_2=12,b_3=10` becomes coefficient of objective function in dual

In primal, objective function is minimizing, so in dual objective function must be maximizing

The `x_3` variable in the primal is unrestricted in sign, therefore the `3^(rd)` constraint in the dual shall be equality.

Let `y1,y2,y3` be the dual variables

Since `3^(rd)` constraint in the primal is equality, the corresponding dual variable `y_3` will be unrestricted in sign.

Now, Dual problem is

MAX `z_y` `=` ` - ` `7` `y_1` ` + ` `12` `y_2` ` + ` `10` `y_3`

subject to

` - ` `3` `y_1` ` + ` `2` `y_2` ` - ` `4` `y_3` ≤ `1` `` `` `y_1` ` - ` `4` `y_2` ` + ` `3` `y_3` ≤ `-3` ` - ` `2` `y_1` ` + ` `8` `y_3` = `-2`

and `y_1,y_2 >= 0; ``y_3` unrestricted in sign

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