Solving $y^2 - yx - y + x = 0$ for $y$?

Question

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?

Answer 1

$$y^2-(x+1)y+x=0$$ First method - using the quadratic formula: $$y_{1,2}=\frac{x+1\pm\sqrt{(x+1)^2-4x}}{2}=\frac{x+1\pm\sqrt{(x-1)^2}}{2}=\frac{(x+1)\pm(x-1)}{2}$$ Thus the solutions are $y=x$ or $y=1$.

Second method - taking common factors: $$y^2-yx-y+x=y(y-x)-1(y-x)=(y-1)(y-x)=0$$ Thus the solutions are $y=x$ or $y=1$.

Answer 2

Factor by grouping: $y^2 - yx - y + x = y(y-x)-(y-x)=(y-x)(y-1)=0$ So either $y-x=0$ or $y-1=0$. The result follows.

Answer 3

You can complete the square to get $$\left( y-\frac{1+x}2\right)^2=\frac{(1-x)^2}4$$ from which the solutions are obvious.

Answer 4

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 $$y_1=\frac{1+x+1-x}{2}=1 \ , \ y_2=\frac{1+x-1+x}{2}=x$$