If y is not equal to 4, x is not equal to 0, and
$$\frac{y^2-16}{3x}=\frac{y-4}{6}$$
then in terms of x, y equals:

$$\left(A\right)\ \frac{\left(x+8\right)}{2}$$

$$\left(B\right)\ \frac{\left(x-8\right)}{2}$$

$$\left(C\right)\ -\frac{3x}{2}$$

$$\left(D\right)\ \frac{\left(-3x+8\right)}{2}$$

$$\left(E\right)\ \frac{\left(3x-8\right)}{2}$$

The OA is B.

Can any expert help me with this PS question please? I don't have it clear. Thanks.