p^2(1 − x) − 2pqx = q^2(1 + x)
How would you make x the subject in this one?
$\displaystyle \displaystyle p^2(1-x) - 2pq\,x = q^2(1 + x)$
$\displaystyle \displaystyle p^2 - p^2x - 2pq\,x = q^2 + q^2x$
$\displaystyle \displaystyle p^2 - q^2 = p^2x + 2pq\,x + q^2x$
$\displaystyle \displaystyle p^2 - q^2 = (p^2 + 2pq + q^2)x$
$\displaystyle \displaystyle (p-q)(p+q) = (p+q)^2x$
$\displaystyle \displaystyle \frac{(p-q)(p+q)}{(p+q)^2} = x$
$\displaystyle \displaystyle x = \frac{p-q}{p+q}$.