Using a trick on P.6 of L. Vandenberghe's notes.
System 1 is equivalent to
\begin{align}
\exists t &\ge 1 \text{ s.t. } \\
A\mathbf{x} &\leq \mathbf{b} \\
C\mathbf{x} &\ge \mathbf{d} + \frac1t\sum_{k=1}^r {\bf e}_k \\
\mathbf{x} &\geq 0
\end{align}
$\forall k \in \{1,\dots,r\}, {\bf e}_k$ is a standard unit vector. Let ${\bf u} = t{\bf x}$. Write the above system into the canonical form.
\begin{align}
\exists t &\ge 1 \text{ s.t. } \\
A\mathbf{u} &\le t\mathbf{b} \\
-C\mathbf{u} &\le -t\mathbf{d} - \sum_{k=1}^r {\bf e}_k \\
\mathbf{u} &\geq 0
\end{align}
Add slack variables ${\bf s}_1, {\bf s}_2$.
\begin{align}
& \exists t \ge 1 \text{ s.t. } \\
& \begin{bmatrix}
A & I_m & 0 \\
-C & 0 & I_r
\end{bmatrix}
\begin{bmatrix}
{\bf u} \\ {\bf s}_1 \\ {\bf s}_2
\end{bmatrix} =
\begin{bmatrix}
t\mathbf{b} \\ -t\mathbf{d} - \sum\limits_{k=1}^r {\bf e}_k
\end{bmatrix} \tag1\label1 \\
& \mathbf{u},{\bf s}_1, {\bf s}_2 \geq 0
\end{align}
To make life easier, let's consider a variant of Farkas' Lemma.
A variant of Farkas' lemma
Let $A$ be $m \times n$ matrix and $b \in \mathbb{R}^m$ $m$-dimensional vector.
Then, exactly one of the following holds:
- there exists some $x \in \mathbb{R}^n$, $x \geq 0$, such that $Ax = b$
- there exists some vector $y \in \mathbb{R}^m$ such that $y^TA \geq 0$ and $y ^Tb = -1$.
From Farkas' lemma, either \eqref{1} or \eqref{2} has a solution.
\begin{align}
& \exists t \ge 1 \exists {\bf y} \in \Bbb R^m \exists {\bf s} \in \Bbb R^r \text{ s.t. } \\
& \begin{bmatrix}
{\bf y} \\ {\bf s}
\end{bmatrix}^T
\begin{bmatrix}
A & I_m & 0 \\
-C & 0 & I_r
\end{bmatrix} \ge {\bf 0} \tag2\label2 \\
& \begin{bmatrix}
{\bf y} \\ {\bf s}
\end{bmatrix}^T
\begin{bmatrix}
t\mathbf{b} \\ -t\mathbf{d} - \sum\limits_{k=1}^r {\bf e}_k
\end{bmatrix} = -1 \\
& \mathbf{u},{\bf s}_1, {\bf s}_2 \geq 0
\end{align}
From the inequality in \eqref{2}, we immediately have
\begin{align}
\mathbf{y}^T A &\geq \mathbf{s}^T C \\
\mathbf{y}, \mathbf{s} &\geq 0
\end{align}
From the equality in \eqref{2}, we have
\begin{align}
t{\bf y}^T {\bf b} - {\bf s}^T \left( t{\bf d} + \sum\limits_{k=1}^r {\bf e}_k \right) &= -1 \quad \forall t \ge 1 \quad \text{(to be justified below)}\\
t({\bf y}^T {\bf b} - {\bf s}^T {\bf d}) - \sum\limits_{k=1}^r s_k &= -1 \tag#\label# \\
{\bf y}^T {\bf b} - {\bf s}^T {\bf d} - \frac1t \sum\limits_{k=1}^r s_k &= -\frac1t.
\end{align}
As $t \to +\infty$, ${\bf y}^T {\bf b} = {\bf s}^T {\bf d}$, which is OP's system 2's second constraint. Substitute this equality to \eqref{#} to get $\sum\limits_{k=1}^r s_k = 1$.
Justification for taking $t \to +\infty$
- If system 1 is infeasible, $\forall t \ge 1$, \eqref{1} doesn't hold.
- For each $t \ge 1$, apply Farka's lemma to conclude that \eqref{2} has a solution.
- Since \eqref{2} is solvable for all $t \ge 1$, equation \eqref{#} derived from \eqref{2} holds for all $t \ge 1$.
We finally get OP's system 2.
\begin{align}
\mathbf{y}^T A &\geq \mathbf{s}^T C \\
\mathbf{y}^T \mathbf{b} &= \mathbf{s}^T\mathbf{d} \\
\sum_{k=1}^r s_k &= 1 \\
\mathbf{y}, \mathbf{s} &\geq 0
\end{align}
Note that the steps above can be reversed: if we assume the feasibility of OP's system 2, we have \eqref{#} without the need to take limits. Thus, \eqref{2} is equivalent to system 2.
Best Answer
Your solution is correct. Your equation is $A^T y_1 + B^T y_2 - B^T y_3 = c$, which is equivalent to to $A^T y_1 + B^T( y_2 - y_3) = c$. You can indeed replace $y_2-y_3$ with $v$. What can you say about the sign of $v$?