Why does this trick for solving this equation work? The question is to

solve the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$

Now if we solve $ \left | 2x-2 \right | = \left | 4x-5 \right |$ first (that is, setting the denominators equal) we find that $x=\frac{3}{2},\frac{7}{6}.$ If we now go back to substitute these values of $x$ in LHS of the original equation, we get that $$\frac{ 4x\left ( \left ( 4x-5 \right )^{2}+3 \right ) +  12x\left ( \left ( 2x-2 \right )^{2}+3 \right ) }{\left ( \left ( 4x-5 \right )^{2}+3 \right )\left ( \left ( 2x-2 \right )^{2}+3 \right )} = 6.$$
Thus, we see that $x=\frac{3}{2},\frac{7}{6}$ both satisfy the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 6.$$
However, I don't know how this is relevant to the fact that $x=\frac{1}{2},\frac{7}{2}$ are solutions to the original equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$
That is, what is the relationship between the respective solutions of $\text{original LHS}=6$ and $\text{original LHS}=1$?
Please tell me the mystery behind this process, and can we use this technique on other equations like this? Thank you.
 A: In problems like these, the first thing to do is to clear denominators. If we do so and expand everything, we will be left with
$$16x^4-100x^3+200x^2-175x+49=0.$$
If you substitute values (with smart guesses using the rational roots theorem), you will be able to find that $x=1/2$ and $x=7/2$ are solutions. Then by the Factor theorem , $(2x-1)$ and $(2x-7)$ are factors of the polynomial on the LHS. Then use long division to factor the LHS. You should obtain 
$$(4x^2-9x+7)(2x-1)(2x-7)=0,$$
and the rest of the solution is easy to complete.
A: Note that$$\frac{4x}{(2x-2)^2+3}+\frac{12x}{(4x-5)^2+3}-1=\frac{-16 x^4+100 x^3-200 x^2+175 x-49}{\left(4 x^2-10 x+7\right) \left(4x^2-8 x+7\right)},$$the solutions of your equations are the roots of the polynomial$$-16 x^4+100 x^3-200 x^2+175 x-49.$$Using the rational root theorem, you can deduce that $\frac12$ and $\frac72$ are roots of this polynomial. On the other hand$$\frac{-16 x^4+100 x^3-200 x^2+175 x-49}{\left(x-\frac12\right)\left(x-\frac72\right)}=-4 \left(4 x^2-9 x+7\right).$$So, there are no more real roots and there are two complex non-real roots.
A: As the LHS is not a homogeneous function, there is no particular relation between the solutions of 
$$f(x)=1$$ and $$f(x)=6.$$
By the way, there is no clear relation between $(\frac32,\frac76)$ and $(\frac12,\frac72)$ besides the coincidental equality of the products.
I doubt that this pseudo-property generalizes.
