Find the equation of all lines having slope $1$that are tangents to the curve $y=\frac{1}{x-1},x\ne 1$.

Solution: Equation of the curve,
$y=\frac{1}{x-1}$
$\therefore$ Slope of tangent,$\left(\frac{dy}{dx}\right)=-\frac{1}{{\left(x-1\right)}^2}$
Slope of tangent to this curve is given $-1$.
$\therefore -\frac{1}{{\left(x-1\right)}^2}=-1$
$\therefore x=0\text{and}x=2$
$\Rightarrow y=-1\text{and}y=1$
Therefore, equation of tangents to the given curve,
$\frac{y+1}{x-0}=-1\text{and}\frac{y-1}{x-2}=-1$
$\Rightarrow x+y+1=0\text{and}x+y-3=0$, which are the required equations.