Question 1184782
{{{x=(y-6)/(2y+k)}}}-------just tried quickly switching x and y

{{{(2y+k)x=y-6}}}

{{{2xy+kx=y-6}}}

{{{2xy-y=-kx-6}}}

{{{y(2x-1)=-kx-6}}}

{{{y=-(kx+6)/(2x-1)}}}



Want  {{{-(kx+6)/(2x-1)=(x-6)/(2x+k)}}}

{{{-(kx+6)(2x+k)=(2x-1)(x-6)}}}

{{{(-kx-6)(2x+k)=(x-6)(2x-1)}}}

Looking at left and right members correspond, {{{highlight(k=-1)}}}.