# Exactly one of the following systems is solvable

1. $Ax\leq b$ and $x\geq 0$
2. $b^{T}y<0$, $A^{T}y\geq 0$ and $y\geq 0$

I can show easily that if (1) is true then (2) is not and converse too. That means both statements can not be true at same time. But I want to show that Exactly one of these is always true.

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 I presume $A$ and $b$ are given and the question concerns the existence of $x$ and $y$? And when you have one vector less than another, does that mean each component of the first is less than the corresponding component of the other? – Gerry Myerson Feb 21 at 23:43 Remains to prove that "neither is true" is impossible. – gt6989b Feb 21 at 23:49 en.wikipedia.org/wiki/Linear_programming#Duality – gt6989b Feb 22 at 0:00