Hi I was wondering how one can get from
$\displaystyle 2x = \lambda(2x-2) $ to $\displaystyle x = \frac{\lambda}{\lambda - 1}$
and
$\displaystyle 2y = \lambda(2y-4) $ to $\displaystyle y = \frac{2\lambda}{\lambda - 1}$
thanks
Hi FalconPUNCH!,
We have $\displaystyle 2x=\lambda(2x-2)$ . Divide throughout by $\displaystyle 2$ hence $\displaystyle x=\lambda x-\lambda$ . Take the $\displaystyle x$ and the $\displaystyle \lambda$ over therefore $\displaystyle \lambda = x(\lambda-1)$ and thus dividing throughout by $\displaystyle (\lambda -1)$ gives the required results.
The seconds is similar. Hope this helps.
$\displaystyle 2x = \lambda(2x-2) $
$\displaystyle 2x=2x\lambda-2\lambda$
$\displaystyle x-\lambda x=-\lambda$
$\displaystyle x(1-\lambda)=-\lambda$
$\displaystyle x=\frac{-\lambda}{1-\lambda}$
$\displaystyle x=\frac{\lambda}{\lambda-1}$
The other one is done in a similar manner.