How do you read this context-free language?
The language defined as $\{a^ib^j\mid 2i=3j+1\}$ is just the set of all strings of $i$ copies of $a$ followed by $j$ copies of $b$, where $2i=3j+1$. Trying various values of $i$ and $j$ and realizing that both $i$ and $j$ must be integers, we see that the admissible solutions are 
$$
(i,j)=(2, 1), (5, 3), (8, 5), (11, 7), \dotsc
$$
For example, with $i=5, j=3$ we see that $2i=2(5) =10=3(3)+1=3j+1$. The corresponding strings are
$$
aab,\quad a^5b^3,\quad a^8b^5,\quad a^{11}b^7 \dotsc
$$
and so on.

With a bit of thought, we can make a PDA for this language by pushing two $a$ markers on the stack for each $a$ we read and then popping off three $a$s for each $b$ we see. If we wind up with a single $a$ on the stack after having read all the $b$s, we pop that off and go to the accept state.



It's worth noting that the PDA you picture does not accept the language described. That one accepts the language $\{a^nb^{2n-1}\mid n&gt;0\}$.