How would you split up the fraction $$\frac{x}{a+b}$$ (or$$\frac{1}{a+b}$$) so one fraction has $x$ and $a$ in it, only and another one has $x$ and $b$, only?
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$\begingroup$If you have $x/(a+b)$, and you want to split it up then you can rewrite it like this. I assume here that the split will look something like $x/a + Y/b$:
$$\frac{x}{a} + \frac{Y}{b} = \frac{x}{a+b} $$
We solve for $Y$ to obtain:
$$Y = -\frac{b^2 x}{a (a+b)} $$
This given us:
$$ \frac{x}{a + b} = \frac{x}{a} - \frac{b x}{a (a+b)} $$
That may or may not be what you're after bt it one way to do it.
$\endgroup$ 1 $\begingroup$$\frac c{a + b} = \frac xa + \frac yb=$
Let $x$ or $y$ be whatever you want and solve for the other.
So if I say $x = 31$ we then have
$y = \frac {cba - 31(a+b)}{a(a+b)}$
And $\frac c{a+b} = \frac {31}a + \frac {\frac {cba - 31(a+b)}{a(a+b)}}b$
which is kind of stupid and pointless.
If you have a criteria that $x$ and $y$ must both be integers then
you have the condition that $ {\frac {cba - x(a+b)}{a(a+b)}}$ be an integer.
.... which can't always happen. If $\gcd(a,b)=1$ and $c = 1$ the would require $(a+b)|ab$ which can't usually happen.
And if you don't have the criteria that $x, y$ be integers then... well, what's the point.
Now, we can probably do $\frac c{a+b} = \frac x{ak} + \frac y{bk}$ all integers but... why would we have a need to do that?
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