Separating addition terms in denominator

Question

If I have a fraction such as:

$\frac{1+d(6-4a)}{1-a+d(7-4a)}$

then how can I separate it so I have it as $\frac{1}{1-a}+(some-term)$

Thanks.

Answer 1

You initially have the fraction

$$\frac{1 + d(6 - 4a)}{1 - a + d(7 - 4a)}.$$

You can rewrite this as:

$$\frac{1 - d + d(7 - 4a)}{1 - a + d(7 - 4a)}.$$

Furthermore, you can rewrite the last expression as:

$$\frac{(a - d) + (1 - a + d(7 - 4a))}{1 - a + d(7 - 4a)}$$

which simplifies to

$$1 + \frac{a - d}{1 - a + d(7 - 4a)}.$$

Without further information about the allowable values for $d$ (or perhaps, a relationship between $a$ and $d$), I don't think this last quantity can be further streamlined to suit your purposes.

What sort of simplification did you have in mind?

Answer 2

I write what i wrote in a comment.

You want $\frac{1+d(6−4a)}{1−a+d(7−4a)}=\frac{1}{1-a}+q$

so $q=\frac{1+d(6−4a)}{1−a+d(7−4a)}-\frac{1}{1-a}=\frac{d(4a^2−6a−1)}{(a−1)(a(4⋅d+1)−7d−1)}$

ergo

$$\frac{1+d(6−4a)}{1−a+d(7−4a)}=\frac{1}{1-a}+\frac{d(4a^2−6a−1)}{(a−1)(a(4d+1)−7d−1)}$$