jmc

Cross multiplying, we obtain $x+1=2(3x-1)$. (This is the same thing as multiplying both sides by both $3x-1$ and by $x+1$.) Then, we solve for $x$: $$\begin{array}{rl}x+1& =2(3x-1)\\ \Rightarrow x+1& =6x-2\\ \Rightarrow -5x& =-3\\ \Rightarrow x& =\boxed{\frac{3}{5}}.\end{array}$$