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.

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.
