<s>[INST] Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$ I have been asked to show that $$\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$$ is equivalent to writing $$1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$$
From here I just tried to work out the bottom of the first fraction which I found to be $\frac{(3-x)(4-x)-x}{(3-x)(4-x)}$ now taking the reciprocal gives me $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$.
I did try factorising the bottom to get to $\frac{(3-x)(4-x)}{(x-6)(x-2)}$ and then using partial fractions gets me to $\frac{-15}{2(x-6)}+\frac{1}{2(x-2)}$ which is definitely not where I want to be. 
I feel like I need to work from $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$ and somehow pull out a $1$ here but not entirely sure how. 
Would really appreciate if anyone could help me. 
Thank you.
 [/INST] Hint:
\begin{align}
\frac{(3-x)(4-x)}{(3-x)(4-x)-x}&=\frac{(3-x)(4-x)}{x^2-8x+12}=\frac{(3-x)(4-x)}{(x-2)(x-6)} \\
&=1+\frac x{(x-2)(x-6)} \qquad\text{(by Euclidean division)}\\
&=1+\frac A{x-2}+\frac B{x-6}\qquad\text{(partial fractions)}
\end{align}
</s>