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$$\frac{x}{m} - \frac{1}{n} = \frac{3x}{mn}$$

I've tried $$\frac{x}{m} - \frac{1}{n} = \frac{3x}{mn} // \frac{x}{m} x \frac{n}{n} = \frac{3x}{mn} // \frac{xn - m}{mn} = \frac{3x}{mn}$$ This is where I get stuck. I need a tutor!

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    $\begingroup$ Try putting the terms on the LHS over a common denominator and see what you can do after this. (If this is what you intended in the original post, then yes, this is good) $\endgroup$ – ÍgjøgnumMeg Mar 15 '16 at 20:59
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    $\begingroup$ Ok I changed it around a bit. $\endgroup$ – William Zlacki Mar 15 '16 at 21:07
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Your original equation is this and you are asked to isolate $x$;

$$\frac{x}{m} - \frac{1}{n} = \frac{3x}{mn}.$$

If you place the terms on the left hand side over a common denominator you will get this;

$$\frac{nx}{mn} - \frac{m}{mn} = \frac{3x}{mn}.$$

Since the terms on the left hand side have a common denominator, you can rewrite this as follows;

$$\frac{nx - m}{mn} = \frac{3x}{mn}.$$

You now have the left hand side and right hand side over a common denominator and can multiply through by $mn$ to get

$$nx - m = 3x.$$

Now, this is a lot simpler; collect the $x$ terms, factorise and isolate.

$$nx - m = 3x \implies nx - 3x = m$$

Factorising gives;

$$x(n - 3) = m.$$

Finally, divide through by $(n - 3)$;

$$x = \frac{m}{n -3}.$$

It is important to note that this is only valid for the case where $n \neq 3$ since this would give us an undefined answer.

Happy mathing!

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    $\begingroup$ Something I like to do when I'm confused is to push things back into simple terms. How would you subtract fractions with the same denominator, for example $\frac{3}{4} - \frac{1}{4}$? $\endgroup$ – ÍgjøgnumMeg Mar 15 '16 at 21:10
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    $\begingroup$ Exactly, so $$\frac{nx}{mn} - \frac{m}{mn} = \frac{nx-m}{mn}$$ and hence $$\frac{nx - m}{mn} = \frac{3x}{mn}.$$ Can you see what to do from here? $\endgroup$ – ÍgjøgnumMeg Mar 15 '16 at 21:19
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    $\begingroup$ Then you may need to look a bit more at rearranging equations. I will edit my answer to provide a full solution so that you know what this kind of problem looks like and perhaps use the solution in future problems. $\endgroup$ – ÍgjøgnumMeg Mar 15 '16 at 21:24
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    $\begingroup$ No problem, if this helped feel free to mark the question as solved. Hopefully you benefit from the answer and can use it for other similar problems. ;) $\endgroup$ – ÍgjøgnumMeg Mar 15 '16 at 21:32
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    $\begingroup$ Sorry to revive a dead horse just beat it, or whatever. Someone upvoted my answer and reading through this gave me another thought: you say $n$ cannot be -3. Wouldn't we also have to say that neither $n$ nor $m$ can be $0$ as well? Thanks $\endgroup$ – Elem-Teach-w-Bach-n-Math-Ed Apr 16 '16 at 18:19
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Once you get that second term fixed, think normal fractions with the same denominator. $3/4-x/4=1/4$ means I just need to solve $3-x=1$. Hope that helps without giving too much away.

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You can easily see that $x= \frac{m}{n-3}$.

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  • $\begingroup$ Am I on the right path with the way I was solving it? $\endgroup$ – William Zlacki Mar 15 '16 at 20:59
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    $\begingroup$ Just do this : $\frac{xn-3x}{mn}=\frac{1}{n}$ then you can see $x= \frac{m}{n-3}$. $\endgroup$ – drxy Mar 15 '16 at 21:04
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    $\begingroup$ You were sort of on the right track. You did the first term right getting $xn/mn$. You should have multiplied your second term by $m/m$ however to get all three to the same denominator. $\endgroup$ – Elem-Teach-w-Bach-n-Math-Ed Mar 15 '16 at 21:05
  • $\begingroup$ $$\frac{ x(n - 3) }{mn} = \frac{1}{n}$$ $\endgroup$ – William Zlacki Mar 15 '16 at 21:13

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