Ans: C
$\begin{array}{rcl}
3a+1 & = & 3(b-2) \\
3a+1 & = & 3b-6 \\
3b & = & 3a+7 \\
b & = & \dfrac{3a+7}{3} \\
b & = & a+\dfrac{7}{3}
\end{array}$
$\begin{array}{rcl}
3a+1 & = & 3(b-2) \\
3a+1 & = & 3b-6 \\
3b & = & 3a+7 \\
b & = & \dfrac{3a+7}{3} \\
b & = & a+\dfrac{7}{3}
\end{array}$
