I need some advice for $ \; " \; 3xy+y-6x-2=0 \;"\, $ PS: I edited the title again and I think, it's better now... :) Actually, almost the whole solving way is wrong but, I understood why... :D :) 
You can check that or add some more useful links about my method, unfortunately, which I applied it wrong... Thank you. :)
Here is a problem which I've encountered and found the answer luckily, I think. But, actually, I need some better or faster ways to solve that:
The problem: $ \; \large{ 3xy+y-6x-2=0 \; , \; y=\,?  } \; $
$$ My \; method:  $$
$$ \large{ y(3x+1)-6x-2=0 } \\ $$ 
$$ \large{ X_{1}+X_{2}=\frac{-b}{a} \; \land \; X_{1}-X_{2}=\frac{ \sqrt{\Delta} } {|a|} \; \land \; \Delta=(b)^{2}-((4) \times (a) \times (c)) } $$
$$ \large{ X_{1}+X_{2}=\frac{-(-6)}{(3x+1)} \; , \; \; X_{1}-X_{2}=\frac{ \sqrt{44+24x} } {|3x+1|} \; , \; \; \Delta=44+24x  } $$
$$ \large{ (X_{1}+X_{2})+(X_{1}-X_{2})=2X_{1} \\ 
2X_{1}=\frac{-(-6)}{(3x+1)}+\frac{ \sqrt{44+24x} } {|3x+1|} \\ 
\frac{ \sqrt{44+24x}+6 } {|3x+1|}=2X_{1} \\ 
8X_{1}=\sqrt{44+24x}+6 \\ 
(8X_{1})^{2}=(\sqrt{44})^{2}+(\sqrt{24})^{2}+(6)^{2} \\
64X_{1}=44+24x+36 \\ 
40X_{1}=80 \\ 
X_{1}=2 } $$
$$ \text{ Then, I recalled the first equation with } \, ''X_{1}'' \, \text{ and rewrote it down: } $$
$$ \large{ 3 \times (2) \times y + y - 6 \times (2) = 0 \\
7y-14=0 \\
y-2=0 \\
y=2 } $$
My other question is, if you would encounter this problem in a test, which method you would try to do it as fast as you could?
Thank you very much!...
 A: I would isolate $y$ on one side of the equation (that is, solve for $y$ in terms of $x$; nothing at the outset tells you $y$ has a particular value).
$$\tag{1}
y(3x+1)-6x-2 \iff y(3x+1) =6x+2.
$$
The next step would be to divide both sides by $3x+1$. But we can't do that if $x=-1/3$. However, if $x=-1/3$, then   equation $(1)$ holds, and $y$ can be anything.
If $x\ne-1/3$, then  equation $(1)$ is equivalent to
$$
y={6x+2\over 3x+1}\iff y={2(3x+1)\over 3x+1}\iff y=2.
$$
A: Frankly, I have no idea what you have done above.
How about the following: Factor your equation once more to get $(3x+1)(y-2)=0$.
At least one term must be zero. So the solutions are $x=-\frac{1}{3}$ with arbitrary $y$, or $y=2$ with arbitrary $x$.
A: The expression factors as $(3x+1)(y-2)$.
Alternately, we have
$$y(3x+1)=6x+2.$$
Divide, which is fine unless $3x+1=0$. We get
$$y=\frac{6x+2}{3x+1}=2.$$
If $3x+1=0$, then any $y$ will do. So the solutions of the given equation are (i) all pairs $(x,y)$ such that $y=2$ and (ii) all pairs $(x,y)$ such that $x=-1/3$.
