1. ## Algebra/Simplifcation

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

2. 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.

3. Originally Posted by FalconPUNCH!
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
$\displaystyle 2x = \lambda(2x-2)$

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

$\displaystyle x-\lambda x=-\lambda$

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

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

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

The other one is done in a similar manner.

4. Thank you guys