AHSME 1974 Problem 1

If $x\not=0$ or $4$ and $y\not=0$ or $6$, then $\frac{2}{x}+\frac{3}{y}=\frac{1}{2}$ is equivalent to $\mathrm{(A)\ } 4x+3y=xy \mathrm{(B) \ }y=\frac{4x}{6-y} \mathrm{(C) \ } \frac{x}{2}+\frac{y}{3}=2$ $\mathrm{(D) \ } \frac{4y}{y-6}=x \mathrm{(E) \ }\text{none of these}$