Let $x_1$ and $x_2$ be such that $x_1\not=x_2$ and $3x_i^2-hx_i=b$, $i=1, 2$. Then $x_1+x_2$ equals
$\mathrm{(A)\ } -\frac{h}{3} \qquad \mathrm{(B) \ }\frac{h}{3} \qquad \mathrm{(C) \  } \frac{b}{3} \qquad \mathrm{(D) \  } 2b \qquad \mathrm{(E) \  }-\frac{b}{3}$
Notice that $x_1$ and $x_2$ are the distinct solutions to the quadratic $3x^2-hx-b=0$. By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case $-\frac{-h}{3}=\frac{h}{3}, \boxed{\text{B}}$.