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.

Note:
$$\frac{x^2-7x+12}{x^2-8x-12}=\frac{x^2-8x+12+x}{x^2-8x-12}=\frac{x^2-8x+12}{x^2-8x-12}+\frac{x}{x^2-8x-12}=1+\frac{x}{x^2-8x-12}=1+\frac{x}{(x-2)(x-6)}$$
Then Partial Fractions:
$$A(x-2)+B(x-6)=x$$
$$x=2\to-4B=2, B=-\frac 12$$
$$x=6\to 4A=6,A=\frac 32$$
Hence $$\frac{x}{(x-2)(x-6)}=-\frac {1}{2(x-2)}+\frac {3}{2(x-6)}$$
Signs are the wrong way around here, but this is simply because $x-2=-(2-x)$. So simply flip them over and flip signs.
$$\frac{x}{(x-2)(x-6)}=\frac 12\bigg(\frac{1}{(2-x)}-\frac {3}{(6-x)}\bigg)$$
as required. 
