Let $2\sin x+3\cos y=3$and $3\sin y+2\cos x=4$, then

$2\sin x+3\cos y=3$and $3\sin y+2\cos x=4$ Squaring and adding, we get

 $\begin{array}{l}4+9+12\sin (x+y)=25\\ \Rightarrow \sin (x+y)=1\Rightarrow \sin (x+y)=\sin \frac{\pi }{2}\\ \Rightarrow x+y=(4n+1)\frac{\pi }{2},n\in I\end{array}$

If $x+y=\frac{\pi }{2}\Rightarrow y=\frac{\pi }{2}-x$

$\Rightarrow 5\sin x=3\Rightarrow \sin x=\frac{3}{5}$ or $\cos x=\frac{4}{5}$

Also, $\cos y=\frac{3}{5}$ and $\sin y=\frac{4}{5}$

Hence,$y > x.$