Question 1203645

if you have 


{{{y=1/(2x) +3}}}....swap variables


{{{x=1/ 2y +3}}}....solve for{{{  y}}}


{{{x-3=1/ 2y }}}


{{{2y (x-3)=1 }}}


{{{2y =1/ (x-3)}}}


{{{y =1/2(x-3)}}}


so, inverse is


{{{y}}}'={{{ 1/2(x-3)}}}