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I have the equation
$$\left(\dfrac{x+1}{x-2}\right)^2 + \left(\dfrac{x+1}{x-3}\right) = 12\left(\dfrac{x-2}{x-3}\right)^2$$ I want to solve this equation in the set of all real numbers.

First way. Put $t = \dfrac{x-2}{x-3}$, and then $x = \dfrac{3t-2}{t - 1}$ Substitute $t$ onto the given equation, we have \begin{equation*} 12t^4-4t^3-13t^2+24t-9 = 0. \end{equation*} This equation has two roots $t = -\dfrac{3}{2}$ and $t=\dfrac{1}{2}$. And then, the given equation has roots $x= \dfrac{13}{5}$ and $x=1$.

Please solve for me the given equation with another way. Thank you very much.

I have just found another way. The given equation equavalent to \begin{equation*} \left(\dfrac{x+1}{x-2}\right)^2 + \left(\dfrac{x+1}{x-2} \right)\cdot\left(\dfrac{x-2}{x-3} \right) = 12\left(\dfrac{x-2}{x-3}\right)^2. \end{equation*} Put $a = \dfrac{x+1}{x-2}$ and $b = \dfrac{x-2}{x-3}$, we get \begin{equation*} a^2 + ab - 12b^2 = 0. \end{equation*} Solve this equation, we get $a = -4b$ and $a = 3b$.

With $a = -4b$, we have \begin{equation*} \dfrac{x+1}{x-2} = -4 \dfrac{x-2}{x-3}. \end{equation*} This equation has two roots $x= \dfrac{13}{5}$ and $x=1$.

With $a = 3b$, we have \begin{equation*} \dfrac{x+1}{x-2} =3 \dfrac{x-2}{x-3}. \end{equation*} has no real solution.

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    $\begingroup$ Is there a reason why you didn't just multiply through by denominators to end up with a quartic in $x$? I'm not sure why you made a substitution. $\endgroup$
    – Michael Albanese
    Oct 4, 2012 at 3:07
  • $\begingroup$ Thank you. Solve your way, we get $$\dfrac{10x^4-86x^3+281x^2-400x+195}{(x-2)^2 \cdot (x-3)^2}=0.$$ $\endgroup$
    – minthao_2011
    Oct 4, 2012 at 3:10
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    $\begingroup$ I like the new way, the one which introduces the extra factor in the 2nd term. $\endgroup$
    – Gerry Myerson
    Oct 4, 2012 at 3:54
  • $\begingroup$ You can close a question when you feel you've already found your answer. $\endgroup$
    – Simply Beautiful Art
    Dec 28, 2015 at 20:35

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