Answer: (c) $x=3, y=4$

Solution:
The given system of equations are
$\begin{array}{l}
4 x+6 y=3 x y\dots (i) \\
8 x+9 y=5 x y\dots (ii)
\end{array}$
Dividing equation(i) and equation(ii) by $x y$, we obtain
$\frac{6}{x}+\frac{4}{y}=3\dots \dots(iii)$
$\frac{9}{x}+\frac{8}{y}=5\dots \dots(iv)$
Multiplying equation(iii) by 2 and subtracting equation(iv) from it, we obtain
$\frac{12}{x}-\frac{9}{x}=6-5$
$\Rightarrow \frac{3}{x}=1$
$\Rightarrow x=3$
Substituting $\mathrm{x}=3$ in equation(iii), we get
$\frac{6}{3}+\frac{4}{y}=3$
$\Rightarrow \frac{4}{y}=1$
$\Rightarrow y=4$
Therefore, $x=3$ and $y=4$