<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] The first expression can be written as
$$\frac{(3-x)(4-x)}{(3-x)(4-x)-x(4-x)+x(3-x)} = \frac{12-7x+x^2}{12-8x+x^2}.$$
The second can be written as
\begin{align}
1+\frac{1}{2}\cdot\frac{6-x-3(2-x)}{(2-x)(6-x)} & = \frac{24-16x+2x^2+6-x-6+3x}{2(12-8x+x^2)} = \frac{24-14x+2x^2}{2(12-8x+x^2)} \\
& = \frac{12-7x+x^2}{12-8x+x^2}.\end{align}
Since these expressions make sense for $\lvert x \rvert < 1$, then the left hand sides are equal, and your claim follows.
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