If I have an equation like this

$\displaystyle [2f+(p+q) ]*[2f-(p+q) ]=4pq$


Are any of the three methods below any better than the other?

Are any of them incorrect?
Is there a 'correct' way that should always be used?

Many thanks


$\displaystyle ------------------$


Method 1.


$\displaystyle [2f+(p+q) ]*[2f-(p+q) ]=4pq$

$\displaystyle let\ \ \ 4pq=rs$


$\displaystyle 2f+(p+q)=r$


$\displaystyle 2f-(p+q)=s$


$\displaystyle 4f=r+s\ \ \ \ \ \ \ \ \ \ f=\dfrac{r+s}{4}$


$\displaystyle 2(p+q)=r-s\ \ \ \ \ \ \ \ p+q=\dfrac{r-s}{2}$


$\displaystyle ------------------$


Method 2.


$\displaystyle [2f+(p+q) ]*[2f-(p+q) ]=4pq$

$\displaystyle let\ \ \ 4pq=4rs$

$\displaystyle 2f+(p+q)=2r$

$\displaystyle 2f-(p+q)=2s$

$\displaystyle 4f=2r+2s\ \ \ \ \ \ \ \ \ \ \ f=\dfrac{r+s}{2}$

$\displaystyle 2(p+q)=2r-2s\ \ \ \ \ \ \ \ p+q=r-s$

$\displaystyle ------------------$


Method 3.


$\displaystyle [2f+(p+q) ]*[2f-(p+q) ]=4pq$

$\displaystyle let\ \ \ 4pq=4rs$

$\displaystyle 2f+(p+q)=4r$

$\displaystyle 2f-(p+q)=s$

$\displaystyle 4f=4r+s\ \ \ \ \ \ \ \ \ \ \ \ \ f=\dfrac{4r+s}{4}$

$\displaystyle 2(p+q)=4r-s\ \ \ \ \ \ \ \ p+q=\dfrac{4r-s}{2}$