How can I show the following relationship algebraically? $$\frac 1 {4-\frac2 x}=\frac1 4+\frac1 {8x-4}\ \ \ \ \forall x\ne\frac1 2$$ I tried to multiply by the conjuage $$\left(\frac 1 {4-\frac2 x}\right)\left(\frac{4+\frac2 x}{4+\frac2 x}\right)=\frac{2+\frac 1 x}{8-\frac2 {x^2}}$$ and ended up nowhere. I realized that partial fraction decomposition must be a valid solution, but I cannot figure out how I would set it up with only the first fraction as information.
To describe where the problem came from, I divided the polynomial $\frac{x}{4x-2}$ in the wrong order, (that is, I calculated $\frac{4x-2}x$) and I figured I could just take the reciprocal of what I came up with to get the right answer.
$\displaystyle\frac1{4-\frac2x}=\frac{x}{4x-2}=\frac{\frac14(4x-2)+\frac12}{4x-2}=\frac14+\frac{\frac12}{4x-2}=\frac14+\frac1{8x-4}$