# 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?

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Try to write the what you've got in the form of quadratic equation with parameter $y$ and then solve this quadratic equatiom for $x$. – SnowAngel6147 Dec 4 '13 at 14:27

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 $$x = \frac{(3y + 1) \pm \sqrt{4y^2 + 1}}{y}.$$

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Right. Thanks a lot, now it seems too simple ;-) (will accept as soon as possible). – Alfe Dec 4 '13 at 14:31

sometimes it is instructive to explore an approach which makes use of some specific features of a problem. in this case suppose we make the substitution:

$$x = 3 + 2t$$ where $t=tan\theta$

after a little algebra you will see that $$-2y = \frac{2t}{1-t^2} = tan 2\theta$$ hence $$x = 3 + 2tan(\frac12\; arctan(-2y))$$ in the general case this will give both values of $x$ because $tan \alpha = tan (\alpha+\pi)$

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