the pulley is massles and frictionless

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reshown Sep 26, 2014

Therefore, $T_y<M_ag$ and $T_x>M_bg$.
With massless strings and pulley, the tension inside a single string is constant. $Tx=T_y$.
Do a free body diagram of the top pulley. $T_z=T_x+T_y$. Since the two blocks accelerate downward, $T_x+T_y<M_ag+M_bg$.