If $x=2$ and $y=-3$ satisfy the equation $2 x^{2}+k x y=4$, then the value of $k$ is
(A) $\frac{2}{3}$
(B) 0
(C) $\frac{4}{3}$
(D) $-\frac{2}{3}$
(E) -2
Substituting $x=2$ and $y=-3$, we obtain

$$
\begin{aligned}
2(2)^{2}+k(2)(-3) & =4 \\
8-6 k & =4 \\
4 & =6 k \\
k & =\frac{2}{3}
\end{aligned}
$$

ANSWER: (A)