$\displaystyle \left\frac{1}{x}+\frac{1}{3}=\right\frac{3}{2x}-1$

So the lowest common multiple is 6? As they both already have an x in them. So... $\displaystyle 6\times\left(\frac{1}{x}+\frac{1}{3}\right)=\left( \frac{3}{2x}-1\right)\times6$

So then I have $\displaystyle \frac{6}{6x}+2=\frac{18}{12x}-6$

$\displaystyle x+2=\frac{3}{2x}-6$

is this correct so far? What next?