An actuary question but more of an algebraic manipulation

Question

I'm stuck in this actuary question.

If $a_{\overline{n|}}=x$ and $a_{\overline{2n|}}=y$, express $d$ as a function of $x$ and $y$

Hints:

$a_{\overline{n|}}=\frac{1-v^n}{i} $, $v=\frac{1}{1+i}$, and ofcourse $a_{\overline{2n|}}=\frac{1-v^{2n}}{i} $

Some identities

$v=1-d$

$d=\frac{i}{1+i}$

$d=iv$

Note: $i$ is some rate of interest. The answer key says $d=\frac{2x-y}{x^2+2x-y}$. How did this happen?

Answer 1

Thanks, I already saw how it's done.

"We are given $a_{\overline{n|}}=\frac{1-v^n}{i} =x$ so that $v^n=1-ix$. Also we are given $a_{\overline{2n|}}=\frac{1-v^{2n}}{i} =y$, so that $v^{2n}=1-iy$. But $(v^n)^2$ so that $ 1-iy=(1-ix)^2 $. This equation is the quadratic $x^2i^2-(2x-y)i=0$ so that $i=\frac{2x-y}{x^2}$. Then applying the formula $d=\frac{i}{1+i}$, we have $d=\frac{2x-y}{x^2+2x-y}$. "