Answer
$(-\infty,6)\cup(6,+\infty))$
Work Step by Step
$f(x)+g(x)=3x+7+\frac{2}{x-6}$ simplify
$\frac{(3x+7)(x-6)+2}{(x-6)}=\frac{3x^2-11x-40}{x-6}$ since the denominator can't be equal to 0 $ x\ne6$
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