# Solving an equation of terms containing a variable in their denominator

I am having issues solving the following equation: $$\frac{x}{2x-3} - \frac{1}{2x} = \frac{3}{4x-6}$$

The resolution of this is 1.

It is in the section of parabolas and it should be pretty easy to solve.

My steps are as follows:

$$\frac{x}{2x-3} - \frac{1}{2x} = \frac{3}{4x-6}$$ Multiply by $2x(4x-6)$

$$4x^2 -1(4x-6) = 6x$$ gets me here, remove $6x$

$$4x^2 - 10x + 6 = 0$$

Solving this with the parabola formula I ended up getting $\{4,6\}$.

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You've misapplied the quadratic formula; if you try substituting 4 and 6 into your original equation, you do not have an equality. Post what you did with the quadratic formula and maybe we can debug from there. –  Ｊ. Ｍ. Dec 8 '10 at 3:22
@J.M.: He divided by $2$ instead of $2a$ (that is, instead of $8$, or instead of $4$ if he first divided the whole equation by $2$). –  Arturo Magidin Dec 8 '10 at 3:31
Shaharyar, you might want to post that as an answer and accept it so we have less unanswered questions in this site. :) –  Ｊ. Ｍ. Dec 8 '10 at 3:43
Done, will have to wait 2 days to be allowed to accept it :) –  Shaharyar Dec 8 '10 at 3:45

HINT $\rm\displaystyle\quad \frac{1}{2\ x}\ =\ \frac{2\ x}{4\ x-6} - \frac{3}{4\ x-6}\ =\ \frac{1}2\ \ \Rightarrow\ \ x\ =\ \ldots\quad$ Note: no quadratic formula needed.

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@J. M. :

After checking out the formula I found out that I dividing by $2$ instead of $4a$/$8$

$$\frac{10 \pm \sqrt{100-96}}{2}$$

With the fixed form I get the results $1.5$ and $1$ of which $1$ is matching result.

Thanks for the kind support!

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And do you see why you need to discard the solution $x=\frac{3}{2}$? –  Arturo Magidin Dec 8 '10 at 4:35
Yup, evil division by $0$ of doom. –  Shaharyar Dec 8 '10 at 7:14