Double Constraint!!

Let the acceleration of the block of mass $m$ be $a_2$ , of the wedge be $a_1$ , and of the other block be $a_3$ . Let the normal reaction of the wedge on the block of mass $m$ be $N$ and the tension in the string be $T$ . Then, force balance equations for the block of mass $m$ yield $a_2=\dfrac{g+a_1}{\sqrt 2}, N=\dfrac{m(g-a_1)}{\sqrt 2}$ . Those for the wedge yield $\dfrac{N}{\sqrt 2}-3T=3ma_1\implies T=\dfrac{m(g-7a_1)}{6}$ . For the other block, we have $4T=\dfrac{16m}{3}a_3\implies a_3=\dfrac{3T}{4m}=\dfrac{g-7a_1}{8}$ . The equation of constraint on the system is $a_1=\dfrac{4a_3}{3}$ . Using this we get $a_3=\dfrac{g-\dfrac{28a_3}{3}}{8}\implies 52a_3=30\implies a_3=\dfrac{15}{26}$ . So, $a=15, b=26$ , and $a+b=\boxed {41}$ .