# 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!

• I understand and I attempted it, but it became pretty ugly pretty quickly. I had hoped for a nicer solution. Commented Jun 17, 2018 at 0:01

$$(3y+8)(y+2) = 3 y^2 + 14 y + 16$$ $$6 (y+2)^2 =$$ $$x = \frac{3y+8}{6y+12}$$ 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 $$x = \frac{dy - b}{-cy + a}$$

• Out of curiosity. Does it help in solving this equation when you notice it is a Mobius transformation? (I barly know what a Mobius transformation is) Commented Jun 17, 2018 at 0:14
• @Cornman added the general pattern. You ought to check, this is good to know....Yes, it helps. Commented Jun 17, 2018 at 0:19
• @WillJagy The lower expression has to have $y$ on the LHS instead of $x$. Commented Jun 17, 2018 at 0:22

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.

• Thanks! I hoped to turn the quadratic equation into a linear one by derivating, and if $x$ was a constant, it would have worked. So you're saying that it isn't generally possible to avoid using the quadratic formula to solve these expressions? Also, how did you recognize the common factor of $(y+2)$? Just experience with the matter? Commented Jun 17, 2018 at 0:08
• It is always easier to see these common factors, when you already factored it out "the hard way". To see the $6(y+2)^2$ is easy. But first I solved $3y^2+14y+16=0$ to get it into linear factors and hoped for the best. For my answer I showed a "fancy" way how one could have seen it, if you just look long enough for it. So yes, it is a lot of expierience involved, but not necessary. And no, it should not be generally possible to solve these type of questions without the quadratic formula. Commented Jun 17, 2018 at 0:12