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 $\lvert x\rvert < 1$

Question

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.

Answer 1

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.

Answer 2

We have $$\frac 1 {1 - \frac {x}{3-x} + \frac x{4-x}}=\frac {x^2-7x+12}{x^2-8x+12}=1+\frac {x} {x^2-8x+12}=1+\frac 1 2 \left(\frac 1 {2-x} - \frac 3 {6-x}\right)$$

Answer 3

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}

Answer 4

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.