Find the equation of all lines having slope 2 and being tangent to the curve $y+\frac{2}{x-3}=0$.

Solution: Eqaution of the curve is,
$y+\frac{2}{x-3}=0$
Differentiating it w.r.t. $x$,
$\Rightarrow \frac{dy}{dx}-\frac{2}{{\left(x-3\right)}^2}=0$
$\Rightarrow \frac{dy}{dx}=\frac{2}{{\left(x-3\right)}^2}$
Now, it is given that slope of the tangent is $2$.
$\therefore \frac{dy}{dx}=2$
$\Rightarrow \frac{2}{{\left(x-3\right)}^2}=2$
$\Rightarrow {\left(x-3\right)}^2=1$
$\Rightarrow x-3=\pm 1$
$\Rightarrow x=4\text{and}x=2$
When $x=4$,
$y+\frac{2}{4-3}=0$
$\Rightarrow y=-2$
So, equation of tangent with this point,$y+2=2\left(x-4\right)\Rightarrow 2x-y=10$
When $x=2$,
$y+\frac{2}{2-3}=0$
$\Rightarrow y=2$
So, equation of tangent with this point,$y-2=2\left(x-2\right)\Rightarrow 2x-y=2.$