How to solve $y = \frac{1}{x-1} +\frac {1}{x-5}$ for $x$ I'm stuck on this one (too long ago for me I guess).
I expanded the fractions coming to $y = \frac{2x-6}{x^2-6x+5}$ and even tried to apply a polynom division (translation?) but this came to nothing.
What's the proper approach on this one?

You have $y (x^2 - 6 x + 5) = 2x - 6$, or $y x^2 - (6y + 2)x + (5y + 6) = 0$, so applying the quadratic formula yields
$$
x = \frac{(6y + 2) \pm \sqrt{(6y +2)^2 - 4 y (5y + 6)}}{2y}.
$$
This simplifies to
$$
x = \frac{(3y + 1) \pm \sqrt{4y^2 + 1}}{y}.
$$
