Find all extrema for $f(x,y) = 3xy$ subject to the constraint $4x^2 + 2y =48$.

I put it into the form of:

$3xy - \lambda(4x^2 +2y -48) = F(x,y,\lambda)$

$3xy - 4x^2\lambda -2y\lambda + 48\lambda$

$F_x = 3y - 8x\lambda = 0$

$F_y = 3x - 2\lambda = 0$

$F_\lambda = 4x^2 - 2y + 48 = 0$

I'm just learning this so beg my pardon, but am I allowed to solve for $y$ in $F_x$ or must I solve for $y$ in $F_y$ (if so then I don't know how since there is no y in $F_y$).

EDIT:

Thanks to the comments, I understand now:

$y = (8x\lambda) / 3$ and $x = (2\lambda)/3$.

Then using substitution,

$4((2\lambda)/3)^2 - 2((8x\lambda)/3) + 48 = 0$.

and then solve for $\lambda$ and continue to solve for $x$ and $y$

$\blacksquare$