How do I use $x-y$ and $x+y$, and to find $x(u,v)$ and $y(u,v)$ for bivariate transformation? Hello I am attempting perform a bi-variate transformation such that 
$(x,y)$ $\rightarrow$ $(u,v)$
where $u = (x+y)/(2xy)$ and $v = \lambda(x+y-(1/u))$
To achieve the transformation, I need to express $x$ and $y$ in terms of $u$ and $v$ (i.e. I need to find $x(u,v)$ and $y(u,v)$)
I already know that $x+y=(uv + \lambda)/(u\lambda)$ and $xy = (uv + \lambda)/(2u^2\lambda)$, but first how do I find $x-y$ using $x+y$ and/or $xy$ and then use both $x-y$ and $x+y$ to find $x(u,v)$ and $y(u,v)$?
 A: You wrote you got that
$$x+y=\frac{uv + \lambda}{u\lambda} \tag{1}\label{eq1A}$$
$$xy = \frac{uv + \lambda}{2u^2\lambda} \tag{2}\label{eq2A}$$
Squaring both sides of \eqref{eq1A} and subtracting $4$ times \eqref{eq2A} gives, on the left hand side
$$\begin{equation}\begin{aligned}
(x+y)^2 - 4xy & = x^2 + 2xy + y^2 - 4xy \\
& = x^2 - 2xy + y^2 \\
& = (x-y)^2
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
and on the right hand side
$$\begin{equation}\begin{aligned}
\left(\frac{uv + \lambda}{u\lambda}\right)^2 - 4\left(\frac{uv + \lambda}{2u^2\lambda}\right) & = \left(\frac{uv + \lambda}{u^2\lambda}\right)\left(\frac{uv + \lambda}{\lambda} - 2\right) \\
& = \left(\frac{uv + \lambda}{u^2\lambda}\right)\left(\frac{uv - \lambda}{\lambda}\right) \\
& = \frac{u^2v^2 - \lambda^2}{u^2\lambda^2}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
You can now equate \eqref{eq3A} with \eqref{eq4A} to take the square root to get the possible values for $x - y$ and, thus, use $x(u,v) = \frac{(x+y) + (x-y)}{2}$ and $y(u,v) = \frac{(x+y) - (x-y)}{2}$. Do you think you can finish the rest yourself?
Update: A somewhat simpler & more direct method is to either note that $(z - x)(z - y) = z^2 - (x + y)z + xy$ or use Vieta's formulas to see that $x$ and $y$ are the roots of
$$z^2 - \left(\frac{uv + \lambda}{u\lambda}\right)z + \frac{uv + \lambda}{2u^2\lambda} = 0 \tag{5}\label{eq5A}$$
so they can be directly determined using the quadratic formula.
