= Solution
Only $z,w$ have printed nonnegativity restrictions; $x,y$ are unrestricted. In four-dimensional space, a <vertex of a linear polyhedron> needs four independent <active constraints>. There are exactly four available constraints, so all must be active. Solving them gives
$$
z=w=0,\qquad y=3,\qquad 2x-y=1,
$$
and therefore the only possible <vertex of a linear polyhedron> is \b[$(x,y,z,w)=(2,3,0,0)$]. It is feasible, and its four active normals are independent, so it is indeed a <vertex of a linear polyhedron>.
One can also see uniqueness geometrically from the <slack variables> $s_1=1-2x+y-5w$ and $s_2=3-y-z-w$. The <feasible region> is in affine bijection with the <nonnegative orthant> in $(s_1,s_2,z,w)$:
$$
y=3-s_2-z-w,\qquad x=2-\tfrac12s_1-\tfrac12s_2-\tfrac12z-3w.
$$
That <orthant> has only the origin as an <extreme point>, giving the same unique <vertex of a linear polyhedron>.
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