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$$m \cdot (a - b) + c = a$$

I can't figure out how to isolate $a$ in the equation above. I can't seem to detach $m$ so that $a$ can be left alone. What steps are required to isolate $a$?

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  • $\begingroup$ Is anything given about m? $\endgroup$ – Qwerty May 22 '16 at 20:18
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We can expand and group the $a$ terms together. Let us first assume $m\ne 1$: $$m(a-b)+c=a\\ma-mb+c=a\\ma-a=mb-c\\a(m-1)=mb-c\\a=\frac{mb-c}{m-1}$$ If $m=1$ we see that $0=b-c$ or $b=c$.

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$$m (a - b) + c = a \Rightarrow \\ ma - a = mb - c \Rightarrow \\ a = \frac{mb-c}{m-1}$$

The last passage is valid only if $m \neq 1$, otherwise the denominator turns to be to $0$ (thanks to egreg for this). In that very special case, you should start from beginning, getting: $$a-b +c = a\Rightarrow \\ b = c.$$

This means that if you want $m=1$, then you must also have $b=c$, otherwise $a$ can't be calculated.

To understand this, take $m=1$, $b = 2$ and $c=3$. Then:

$$1 \cdot (a - 2) + 3 = a \Rightarrow \\ a-a-2+3 = 0 \Rightarrow \\ 1 = 0...$$

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    $\begingroup$ For $m\ne 1$ $\quad$ $\endgroup$ – egreg May 22 '16 at 20:28
  • $\begingroup$ @egreg yeah, that's important, really forgot about it $\endgroup$ – the_candyman May 22 '16 at 20:29
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First, expand the first term to get $ma-mb+c=a$. Now the key is to get all the terms involving $a$ together; this requires that you move the $ma$ term to the right side by subtracting to get $-mb+c=a-ma$. You can then pull out the $a$ term to get $-mb+c=a(1-m)$. Can you finish?

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You can distribute $m*(a-b)= ma -mb$. The get all $a$'s on the same side. You now have $a-ma = -mb + c$. Factor out $a$ from the left side to get $a * (1-m) = c - mb$. Now divide and you get $a = \frac{c-mb}{1-m}$

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