Solve:
$\displaystyle y'=\dfrac{2}{(x+2y-3)}$ by letting $\displaystyle x+2y-3=v$
I always get confused up as to what we're meant to do with the $\displaystyle y'$
$\displaystyle y'=\dfrac{2}{v}$
Is $\displaystyle \dfrac{dv}{dx}=1+2\dfrac{dy}{dx}$?