Solution
= Solution
Setting the third coordinate to zero and the first to four leaves $6x_2\geq4$ and $4x_2\leq3$. Thus $x_2=2/3$ works. Direct substitution gives
$$
\boxed{x=(4,2/3,0)^T},\qquad Ax=(20/3,-8,0)^T\leq b,\qquad Cx=(-2/3,-4,-4)^T\leq d.
$$
Hence this point lies in both <linear polyhedra>.