The solution of the primal is $x_{opt}^{T}=(4,3)$.
From the complementary slackness theorem we know:
$x_j\cdot z_j=0 \ \forall \ \ j=1,2, \ldots , n$
$y_i\cdot s_i=0 \ \forall \ \ j=1,2, \ldots , m$
$s_i$ are the slack variables of the primal problem.
$z_j$ are the slack variabales of the dual problem.
First constraint
$-2x_1-4x_2 +s_1= -12\Rightarrow -8-12+s_1=-12\Rightarrow s_1 >0$
Thus $y_1=0$
Fourth constraint
$x_2 \leq 5$
$3+s_4=5 \Rightarrow s_4=2>0$
Thus $y_4=0$
Both $x_i$ are greater than $0$. Thus $z_1=z_2=0$. We can take the first and second constraints and transform them into equations:
$-2y_1+y_2+y_3 = 6\\
-4y_1+y_2+y_4 = 4\\$
$0+y_2+y_3=6\\
0+y_2+0=4$
Now the solution is obvious.
Note that
$$\begin{array}{rl} x_1 = 0 \lor x_1 \geq 10 &\equiv (x_1 \geq 0 \land x_1 \leq 0) \lor x_1 \geq 10\\\\ &\equiv x_1 \geq 0 \land (x_1 \leq 0 \lor x_1 \geq 10)\end{array}$$
We can handle the disjunction $x_1 \leq 0 \lor x_1 \geq 10$ using the Big M method. We introduce binary variables $z_1, z_2 \in \{0,1\}$ such that $z_1 + z_2 = 1$, i.e., either $(z_1,z_2) = (1,0)$ or $(z_1,z_2) = (0,1)$. We introduce also a large constant $M \gg 10$ so that we can write the disjunction in the form
$$x_1 \leq M z_1 \land x_1 \geq 10 - M z_2$$
If $(z_1,z_2) = (1,0)$, we have $x_1 \leq M$ and $x_1 \geq 10$, which is roughly "equivalent" to $x_1 \geq 10$. If $(z_1,z_2) = (0,1)$, we have $x_1 \leq 0$ and $x_1 \geq 10 - M$, which is roughly "equivalent" to $x_1 \leq 0$.
Thus, we have a mixed-integer linear program (MILP)
$$\begin{array}{ll} \text{maximize} & 1.5x_1 + 2x_2\\ \text{subject to} & x_1, x_2 \leq 300\\ & x_1 \geq 0\\ & x_1 - M z_1\leq 0\\ & x_1 + M z_2 \geq 10\\ & z_1 + z_2 = 1\\ & z_1, z_2 \in \{0,1\}\end{array}$$
For a quick overview of MILP, read Mixed-Integer Programming for Control by Arthur Richards and Jonathan How.
Best Answer
The standard form is sometimes defined with inequalities ($\leq$) and sometimes with equalities. If you want to begin with an algorithm it is useful to have equalities.
If all variables has to be equal or greater than $2016$ then you have to introduce slack variables like you proposed.
$x_1\geq 2016, x_2\geq 2016, x_3\geq 2016, \ldots, x_n\geq 2016$
becomes
$x_1-s_1= 2016, x_2-s_2= 2016, x_3-s_3= 2016, \ldots, x_n-s_n= 2016$
with $x_1, x_2,\dots, x_n, s_1, s_2, s_3, \ldots s_n \geq 0$
Here you get $n$ additional constraint. To get an initial solution artifitial variables have to be intoduced.
$x_1-s_1+a_1= 2016, x_2-s_2+a_2= 2016, x_3-s_3+a_3= 2016, \ldots, x_n-s_n+a_n= 2016$
with $x_1, x_2,\dots, x_n, s_1, s_2, s_3, \ldots s_n, a_1, a_2,\dots, a_n \geq 0$
The initial solution is $x_1=x_2=x_3=\ldots=x_n=s_1= s_2= s_3= \ldots s_n=0$
$a_1=a_2=a_3=\ldots=a_n=2016$
The decision variables are all non-negative.