Question 331575
<pre><b>
I can't tell if the x is under the radical or not, so I'll do it
both ways:

{{{2+sqrt(3)*x = x/3}}}

{{{6+3sqrt(3)x=x}}}

{{{3sqrt(3)x - x=-6}}}

{{{x(3sqrt(3)-1)=-6}}}

{{{x=-6/(3sqrt(3)-1)}}}{{{((3sqrt(3)+1))/((3sqrt(3)+1))}}}

Rationalize the denominator:

{{{x=(-6/(3sqrt(3)-1))}}}{{{""*""}}}{{{((3sqrt(3)+1))/((3sqrt(3)+1))}}}

{{{x=    (  -6(3sqrt(3)+1)  )/(9*3-1)  }}}

{{{x=(-6(3sqrt(3)+1))/(27-1)}}}

{{{x=(-6(3sqrt(3)+1))/26}}}

{{{x=(-3(3sqrt(3)+1))/13}}}



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Assuming the x is under the radical:

{{{2+sqrt(3x) = x/3}}}

{{{6+3sqrt(3x)=x}}}

{{{3sqrt(3x)=x-6}}}

{{{9(3x)=x^2-12x+36}}}

{{{27x=x^2-12x+36}}}

{{{0=x^2-39x+36}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-(-39) +- sqrt( (-39)^2-4*1*36 ))/(2*1) }}}

{{{x = (39 +- sqrt( 1521-144 ))/2 }}}

{{{x = (39 +- sqrt(1377))/2 }}}

{{{x = (39 +- sqrt(81*17))/2 }}}

{{{x = (39 +- 9sqrt(17))/2 }}}

{{{x = (3(13 +- 3sqrt(17)))/2 }}}

Edwin</pre>