1. Find dual from primal conversion
MAX z = x1 - x2 + 3x3
subject to
x1 + x2 + x3 <= 10
2x1 - x2 - x3 <= 2
2x1 - 2x2 - 3x3 <= 6
and x1,x2,x3 >= 0

Solution:
Primal is (Solution steps of Primal by Simplex method)

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

subject to

`` `` `x_1` ` + ` `` `x_2` ` + ` `` `x_3` ≤ `10` `` `2` `x_1` ` - ` `` `x_2` ` - ` `` `x_3` ≤ `2` `` `2` `x_1` ` - ` `2` `x_2` ` - ` `3` `x_3` ≤ `6`

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

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=-1,c_3=3` becomes right hand side constants in dual

In primal, The right hand side constants `b_1=10,b_2=2,b_3=6` becomes coefficient of objective function in dual

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

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

Dual is (Solution steps of Dual by Simplex method)

MIN `z_y` `=` `` `10` `y_1` ` + ` `2` `y_2` ` + ` `6` `y_3`

subject to

`` `` `y_1` ` + ` `2` `y_2` ` + ` `2` `y_3` ≥ `1` `` `` `y_1` ` - ` `` `y_2` ` - ` `2` `y_3` ≥ `-1` `` `` `y_1` ` - ` `` `y_2` ` - ` `3` `y_3` ≥ `3`

and `y_1,y_2,y_3 >= 0; `