If $$2x + \frac{1}{3x} = 6$$,then $$3x + \frac{1}{2x}$$ is equal to

GIven,    $$2x+\ \frac{\ 1}{3x}=6$$

                $$\ \frac{\ \ \ 6x^2+1}{3x}=6$$

            $$\ \ 6x^2+1=18x$$

      $$\ \ \ \ 6x^2-18x+1=0$$ ................(1)

Let       $$3x+\ \frac{\ 1}{2x}=y$$

                 $$\ \frac{\ 6x^2+1}{2x}=y$$

           $$\ \ 6x^2+1=2xy$$

      $$\ \ 6x^2-2xy+1=0$$ ...................(2)

Subracting equation (2) from (1), we get

          $$-18x+2xy=0$$

                          $$2xy=18x$$

                                $$y=9$$

          $$=$$>  $$3x+\ \frac{\ 1}{2x}=9$$