1. Algebra/Simplifcation

Hi I was wondering how one can get from

$2x = \lambda(2x-2)$ to $x = \frac{\lambda}{\lambda - 1}$

and

$2y = \lambda(2y-4)$ to $y = \frac{2\lambda}{\lambda - 1}$

thanks

2. Hi FalconPUNCH!,

We have $2x=\lambda(2x-2)$ . Divide throughout by $2$ hence $x=\lambda x-\lambda$ . Take the $x$ and the $\lambda$ over therefore $\lambda = x(\lambda-1)$ and thus dividing throughout by $(\lambda -1)$ gives the required results.

The seconds is similar. Hope this helps.

3. Originally Posted by FalconPUNCH!
Hi I was wondering how one can get from

$2x = \lambda(2x-2)$ to $x = \frac{\lambda}{\lambda - 1}$

and

$2y = \lambda(2y-4)$ to $y = \frac{2\lambda}{\lambda - 1}$

thanks
$2x = \lambda(2x-2)$

$2x=2x\lambda-2\lambda$

$x-\lambda x=-\lambda$

$x(1-\lambda)=-\lambda$

$x=\frac{-\lambda}{1-\lambda}$

$x=\frac{\lambda}{\lambda-1}$

The other one is done in a similar manner.

4. Thank you guys