In the proof of the fact that* after a projective transformation, the equation for any nonsingular cubic curve can be put in the form$\displaystyle y^2z=x(x-z)(x-\lambda z)$, where$\displaystyle \lambda \neq 0,1$, *my notes in a middle-step arrives at the cubic

$\displaystyle (y-\frac{\alpha x+\beta z}{2})^2z=b_2(x,z)$,

and then the projective transformation

$\displaystyle [x,y,z]\mapsto [x,y-\frac{\alpha x+\beta z}{2},z]$

is applied to transform it to

$\displaystyle y^2z=b_3(x,z).$

I can't really see how this works - shouldn't we use the inverse of this projective transformation? To me it seems that replacing $\displaystyle y$ with$\displaystyle y-\frac{\alpha x+\beta z}{2}$ should give us the cubic $\displaystyle (y-\alpha x-\beta z)^2z=b_4(x,z)$, which is not really what we wanted.