Solve for $y$ in the equation $x = \frac{3y^2 + 14y + 16}{6y^2 + 24y + 24}$ As part of a larger calculation, I came across $x = \frac{3y^2 + 14y + 16}{6y^2 + 24y + 24}$, which I now have to solve for $y$. My initial idea, besides a failed attempt to use the general quadratic formula, was, incorrectly:
\begin{align}
\dots \Leftrightarrow 6xy^2 + 24xy + 24x &= 3y^2 + 14y + 16 \\
\Leftrightarrow \frac{d}{dy} 6xy^2 + 24xy + 24x &= 3y^2 + 14y + 16 \frac{d}{dy} \\
\Leftrightarrow 12xy + 24x &= 6y + 14 \\
\Leftrightarrow (12x-6)y &= 14 - 24x \\
\Leftrightarrow y &= \frac{14 - 24x}{12x-6} \\
\Leftrightarrow y &= \frac{7 - 12x}{6x-3}
\end{align}
This is very "close" to the correct solution as given by WolframAlpha: $y = \frac{8 - 12x}{6x - 3}$. However, the derivation step is incorrect, because $x$ is not independent of $y$ and thus can't be regarded as a constant.
Obviously, the result from WA doesn't help much without a derivation, so that's what I'm looking for here (and maybe some useful tips for dealing with such expressions in general).
Thanks in advance!
 A: $$ (3y+8)(y+2) = 3 y^2 + 14 y + 16  $$
$$ 6 (y+2)^2 =   $$
$$  x = \frac{3y+8}{6y+12} $$
This is a Mobius transformation
$$ y = \frac{12x - 8}{-6x+3}  $$
Audience Request:
If we have constants $a,b,c,d$ with $ad-bc \neq 0,$ and
$$ y = \frac{ax+b}{cx+d}   $$
when $cx+d \neq 0,$ then
$$  x = \frac{dy - b}{-cy + a} $$
A: I do not get, why you try to take the derivative. Why should that help in solving this equation for y?
Anyways:
$$x=\frac{3y^2+14y+16}{6y^2+24+24}=\frac{3y^2+14y+16}{6(y^2+4y+4)}=\frac{3y^2+6y+8y+16}{6(y+2)^2}=\frac{3y(y+2)+8(y+2)}{6(y+2)^2}=\frac{(y+2)(3y+8)}{6(y+2)^2}=\frac{3y+8}{6(y+2)}$$
Now you can proceed:
$$x=\frac{3y+8}{6(y+2)}$$
$$6(y+2)x=3y+8$$
Bring everything with $y$ on one side:
$$6yx-3y=8-12x$$
Factor out $y$:
$$y(6x-3)=8-12x$$
Divide by $6x-3$:
$$y=\frac{8-12x}{6x-3}$$
Note, that it is "lucky" that we can reduce the fraction to linear numerator and denominator and we do not have to deal with quadratic expression, which would lead into a more complicated calculation involving the quadratic formula.
