Prove for all p > 1, $x * y = (p + 1)^{-2}$

Question

We have 2 equations with 3 unknowns x, y and Real parameter p:

1. \begin{equation*} px + y = 1 \end{equation*}
2. \begin{equation*}x + py = 1\end{equation*}

How to prove for all p > 1, $x * y = (p + 1)^{-2}$

Answer 1

A naive approach.

From the first, $y = 1-px$.

Substituting in the second, $1 =x+py =x+p(1-px) =x+p-p^2x =x(1-p^2)+p $ or $x =\dfrac{1-p}{1-p^2} $.

If $p \ne 1$, $x =\dfrac1{1+p} $. Then $y =1-px =1-\dfrac{p}{1+p} =\dfrac{1+p-p}{1+p} =\dfrac1{1+p} =x $ so $xy =\dfrac1{(1+p)^2} $.