Answer
$\dfrac{6+x}{x-1}; x \ne 1$
Work Step by Step
The given expression is equivalent to:
$=\dfrac{6}{x-1}-\dfrac{x}{-x+1}
\\=\dfrac{6}{x-1} - \dfrac{x}{-1(x-1)}
\\=\dfrac{6}{x-1} - \dfrac{-x}{x-1}$
The expressions are similar so subtract the numerators and copy the denominator to obtain:
$=\dfrac{6-(-x)}{x-1}
\\=\dfrac{6+x}{x-1}; x \ne 1$
($x$ cannot be $1$ as it makes the expression undefined.)
