I need some advice for $ \; " \; 3xy+y-6x-2=0 \;"\, $

Solution 1:

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 $$ y={6x+2\over 3x+1}\iff y={2(3x+1)\over 3x+1}\iff y=2. $$

Solution 2:

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 $x$.

Solution 3:

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 $x=-1/3$.