Question 135271
{{{x+(10)/(x-2)=(x^2+3x)/(x-2)}}}


Put everything on the left:
{{{x+(10)/(x-2)-((x^2+3x)/(x-2))=0}}}


{{{x-2}}} is the common denominator:
{{{(x(x-2))/(x-2)+(10)/(x-2)-((x^2+3x)/(x-2))=0}}}


Expand and collect like terms:
{{{(x^2-2x+10-x^2-3x)/(x-2)=0}}}
{{{(-5x+10)/(x-2)=0}}}


Note that {{{a/b=0}}} if and only if {{{a=0}}} AND {{{b<>0}}}.  So set the numerator equal to 0 and solve:


{{{-5x+10=0}}}
{{{-5x=-10}}}
{{{x=2}}}


However, if {{{x=2}}}, then the denominator {{{x-2}}} equals 0, so {{{x=2}}} must be excluded as a root of this equation, and there are no other roots.  


Therefore the solution set of {{{x+(10)/(x-2)=(x^2+3x)/(x-2)}}} is the empty set.