Answer
$\dfrac{3(x+1)}{x-7}$
Work Step by Step
Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, the given expression, $
\dfrac{3x+3}{x-1}\div \dfrac{x^2-6x-7}{x^2-1}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{3x+3}{x-1}\cdot \dfrac{x^2-1}{x^2-6x-7}
\\\\=
\dfrac{3(x+1)}{x-1}\cdot \dfrac{(x+1)(x-1)}{(x-7)(x+1)}
\\\\=
\dfrac{3(\cancel{x+1})}{\cancel{x-1}}\cdot \dfrac{(x+1)(\cancel{x-1})}{(x-7)(\cancel{x+1})}
\\\\=
\dfrac{3(x+1)}{x-7}
.\end{array}
