Given that the [ratio](https://artofproblemsolving.com/wiki/index.php/Ratio) of $3x - 4$ to $y + 15$ is constant, and $y = 3$ when $x = 2$, then, when $y = 12$, $x$ equals:
$\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 8$
Let $k$ be the constant ratio. Then $k = \frac{3(2)-4}{(3)+15} = \frac{1}{9} = \frac{3x - 4}{(12) + 15}$. Solving gives $3x - 4 = 3 \Longrightarrow x = \frac 73 \Rightarrow \mathrm{(B)}$.