<s>[INST] Solving $y^2 - yx - y + x = 0$ for $y$? I solved this equation for $y$ by inspection and confirmed it with Wolfram Alpha -
$y^2 - yx - y + x = 0$
I got the values $y = 1$ and $y = x$
However I was wondering is there a formal method for solving it? I expressed it as a polynomial -
$y^2 + (-1 - x)y + x = 0$
and used the quadratic formula but it just left me with an awkward expression involving powers of $x$ and a square root...no $y = 1$ and $y = x$ which is what I was looking for...
So how would I go about solving this formally?
 [/INST] Well, check again.
For $y^2+(−1−x)y+x=0$ with $y$ as the quadratic variable.
The quadratic formula with $b$ as $(-1-x)$, $a$ as $1$, and $c$ as $x$ tells that the roots are
$$y_{1,2}=\displaystyle \frac{(1+x) \pm \sqrt{1+x^2+2x - 4x}}{2}=\frac{(1+x)\pm\sqrt{1+x^2-2x}}{2}=\frac{(1+x) \pm{1-x}}{2}$$
Hence the solutions are $$y_1=\frac{1+x+1-x}{2}=1 \ , \ y_2=\frac{1+x-1+x}{2}=x$$
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