Question 111206
I'm not sure whether you meant: {{{x+(6/3)=x+(6/x)}}} or {{{(x+6)/3=(x+6)/x}}}.
:
{{{x+(6/3)=x+(6/x)}}}
{{{x+2-x=x-x+(6/x)}}}
{{{2=6/x}}}
{{{12=1/x}}}
{{{x=1/12}}}
:
Or if you meant the second way:
:
{{{(x+6)/3=(x+6)/x}}}.  For this you need to remember the following rule:
if {{{p/q=r/s}}}, then {{{ps=rq}}}.  In this case:
{{{x(x+6)=3(x+6)}}}
{{{x^2+6x=3x+18}}}
{{{x^2+3x-18=0}}}
{{{(x+6)(x-3)=0}}}
Hence:
{{{x=-6}}} or {{{x=3}}}
Both of which are in the domain of the function, since the only real number excluded from the domain is 0 -- the right side of the equation would be undefined for x = 0.
:
To avoid confusion next time, use parentheses, like this: (x+6)/3=(x+6)/x