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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!...

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    $\begingroup$ Note $-6x-2=-2(3x+1)$. So, after factoring $(3x+1)(y-2)=0$. $\endgroup$ Apr 26, 2012 at 21:26
  • $\begingroup$ You might consider renaming the question.... $\endgroup$ Apr 26, 2012 at 21:28
  • $\begingroup$ @Donkey_2009: I did my best for the title, please, if you have some other idea for that, make it more clear or better, thank you... :) $\endgroup$ Apr 26, 2012 at 21:33
  • $\begingroup$ @DavidMitra: Yes, David, I see it now... Thank you very much. :) $\endgroup$ Apr 26, 2012 at 21:34
  • $\begingroup$ Are you trying to apply the formula for the solutions of a quadratic to a linear equation? $\endgroup$ Apr 27, 2012 at 13:17

3 Answers 3

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

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  • $\begingroup$ Thank you very much, David, I got my answer. :) $\endgroup$ Apr 26, 2012 at 21:48
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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 $x$.

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  • $\begingroup$ :D So, I did it all wrong except just for the answer... :) I knew that I was just so lucky... :D :) At least, I got a good lesson by trying and sharing it with you... :) Thank you. :) $\endgroup$ Apr 26, 2012 at 21:42
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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 $x=-1/3$.

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  • $\begingroup$ Thank you very much, André. :) Thanks to All... :) $\endgroup$ Apr 26, 2012 at 21:50

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