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September 2nd, 2019, 03:58 AM   #1
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How to solve an algebraic equation with fractions with an unknown in the denominator

There must be something simple I'm missing here, but this just isn't working out for me. How do I solve:

(3 / 5x) - (2 / 6x) = (2 / x) + (7 / 3x)

I will probably kick myself when I see the solution... Thanks in advance!
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September 2nd, 2019, 06:18 AM   #2
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Quote:
Originally Posted by brycewaters10 View Post
There must be something simple I'm missing here, but this just isn't working out for me. How do I solve:

(3 / 5x) - (2 / 6x) = (2 / x) + (7 / 3x)

I will probably kick myself when I see the solution... Thanks in advance!
note $x \ne 0$

multiply every term by x ...

$\dfrac{3}{5} - \dfrac{2}{6} = 2 + \dfrac{7}{3}$

$\dfrac{4}{15} = \dfrac{13}{3}$ ... a contradiction.

So, what can you say about the solution?
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September 2nd, 2019, 06:24 AM   #3
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I think there's an error in what you have written, but in general you would multiply by denominators to get all the $x$s in the numerators.

e.g.
\begin{align}&& \frac{3}{5x}-\frac{2}{5} &= \frac{2}{x} + \frac{7}{5} \\
&\text{(multiply by $5x$)}& 3 - 2x &= 10 + 7x \\
&& -7 &= 9x \\
&& x &= -\frac79\end{align}

In your case, multiplying by $x$ will make all of them go away which is why I think you have made a mistake.
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September 2nd, 2019, 12:05 PM   #4
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Quote:
Originally Posted by skeeter View Post
note $x \ne 0$

multiply every term by x ...

$\dfrac{3}{5} - \dfrac{2}{6} = 2 + \dfrac{7}{3}$

$\dfrac{4}{15} = \dfrac{13}{3}$ ... a contradiction.

So, what can you say about the solution?
Thanks, so it means there's no solution...
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September 2nd, 2019, 12:10 PM   #5
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Originally Posted by v8archie View Post
I think there's an error in what you have written, but in general you would multiply by denominators to get all the $x$s in the numerators.

e.g.
\begin{align}&& \frac{3}{5x}-\frac{2}{5} &= \frac{2}{x} + \frac{7}{5} \\
&\text{(multiply by $5x$)}& 3 - 2x &= 10 + 7x \\
&& -7 &= 9x \\
&& x &= -\frac79\end{align}

In your case, multiplying by $x$ will make all of them go away which is why I think you have made a mistake.
No, there was no error in what I wrote (it was a problem my son received in his 9th grade algebra class), but from the first reply to my post, I understand there's simply no solution to this equation... Thank you all for taking the time to answer!
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