Find dual from primal conversion
MAX z = x1 - 2x2 + 3x3
subject to
-2x1 + x2 + 3x3 = 2
2x1 + 3x2 + 4x3 = 1
and x1,x2,x3 >= 0

Solution:
Primal problem is

MAX `z_x` `=` `` `` `x_1` ` - ` `2` `x_2` ` + ` `3` `x_3`

subject to

` - ` `2` `x_1` ` + ` `` `x_2` ` + ` `3` `x_3` = `2` `` `2` `x_1` ` + ` `3` `x_2` ` + ` `4` `x_3` = `1`

and `x_1,x_2,x_3 >= 0; `

In primal, There are `3` variables and `2` constraints, so in dual there must be `3` constraints and `2` variables

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

In primal, The right hand side constants `b_1=2,b_2=1` becomes coefficient of objective function in dual

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

Let `y1,y2` be the dual variables

Since `1^(st),2^(nd)` constraints in the primal are equalities, the corresponding dual variables `y_1,y_2` will be unrestricted in sign.

Now, Dual problem is

MIN `z_y` `=` `` `2` `y_1` ` + ` `` `y_2`

subject to

` - ` `2` `y_1` ` + ` `2` `y_2` ≥ `1` `` `` `y_1` ` + ` `3` `y_2` ≥ `-2` `` `3` `y_1` ` + ` `4` `y_2` ≥ `3`

and ```y_1,y_2` unrestricted in sign

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