$A$ is $p\times n$, $B$ is $q\times n$, what can we say about Systems $1$ and $2$ of equations? 
Problem: Let $A$ be a $p\times n$ matrix, and $B$ be a $q\times n$ matrix. System 1: $Ax < 0, Bx= 0$ for some $x\in\mathbb R^n$. System 2: $A^Tu + B^Ty = 0$ for some non-zero $(u,y)$ with $u\ge 0$.
Is it true that if System $1$ has no solution, System $2$ has a solution? Prove or disprove.
Further, show that if $B$ has full rank, exactly one of the systems has a solution. Is this necessarily true if $B$ is not of full rank? Prove or disprove.

Let's say that system $1$ has no solution. Then, $(Ax)_i\ge 0$ for some $1\le i\le p$, for all $x\in\mathbb R^n$. I don't know if System $2$ has a solution or not, so let's suppose it doesn't (if it does, we should get a contradiction). So, for all $u \ne 0, y\ne 0, u\ge 0$ we have $A^Tu + B^Ty \neq 0$. What's next? I'm stuck here.
It would be nice if you could help me solve it or at least point me in the right direction!

Clarification: For $x\in\Bbb R^n$, $x>0$ is interpreted element wise, i.e. $x_i > 0$ for all $1\le i \le n$. Similarly for other symbols, i.e. $\le, \ge, <, \not <, \not >, =$ etc.
 A: Let $e$ be the all-one vector.
Consider the optimization problem $(P)$: $$\max 0$$
$$Ax \le - e $$$$ Bx=0$$
The corresponding dual $(D)$ is
$$\min - e^Tu$$
$$A^Tu+B^Ty=0$$
$$u \ge 0$$
If $(P)$ is infeasible, then $(D)$ is either infeasible or unbounded. Since $(0,0)$ is a solution to $(D)$, hence $(D)$ must be unbounded. At least one component of $u$ can be chosen to be positive.
That is we have proven that if System $1$ has no solution, then System $2$ has a solution.

If $B$ has full rank and  suppose System $1$ has a solution. Then the optimal value of $(P)$ is $0$. Hence by strong duality, the optimal value of $(D)$ is $0$ as well. That is we must have $u=0$ and we end up wiht $B^Ty=0$. Since $B$ is of full rank, the unique solution of $B^Ty=0$ is $y=0$. That is to say, the feasible point of $(D)$ is the zero vector. System $2$ is not feasible.
This is not true if $B$ is not of full rank. For example, if $B=0$ and let $A=1$, then System $1$ is feasible since $x$ can take any negative number. System $2$ remains feasible since we can let $u=0$ and $y=1$.
