Technique for solving $\frac{ax+b}{cx+d}=\frac{px+q}{rx+s}$ where the sum of numerators equals the sum of denominators I was looking up some shortcuts to solve quadratic equations. I got a technique that applies when the sum of the numerators and denominators are equal, but I am unable to understand the reasoning behind it. Here I'm showing an example:
$$ \frac{3x + 4}{6x + 7} = \frac{5x + 6}{2x + 3} $$
The solution goes as follows:
"Minute observation of the question helps us to identify that this question falls in a special category of quadratic equations, where the sum of the numerators (N) and the sum of the denominators (D) are found to be equal to 8x + 10.''
For the first root,
$ N_1 + N_2 = D_1 + D_2 = 0$
or, $ 8x + 10 = 0 $
or, $ x = -5/4 $
For the second root
$ N_1 - D_1 = N_2 - D_2 = 0 $
or, $ 3x + 3 = 0 $
or, $ x = -1 $
Can someone explain the reasoning/proof behind this?

Componendo and dividendo (Brilliant) is another method.
Using the third rule with $k=1$, we have:
$$\frac{3x+4+(6x+7)}{3x+4-(6x+7)} = \frac{5x+6+(2x+3)}{5x+6-(2x+3)}$$
$$\Rightarrow \frac{9x+11}{-3x-3} = \frac{7x+9}{3x-3}$$
$$\Rightarrow -9x-11 = 7x+9$$
$$\Rightarrow x = -\frac{5}{4}$$
which is true in general, when we have:
$$\frac{N_1 + D_1}{N_1 - D_1} = \frac{N_2 + D_2}{N_2 - D_2}$$
and $N_1 - D_1 = N_1 + N_2 - D_1 - N_2 = D_1 + D_2 - D_1  - N_2 = D_2 - N_2$, so $N_1 - D_2 = -(N_2 - D_2)$.
Using the fourth rule with $k=1$, we have:
$$\frac{3x+4+(5x+6)}{3x+4+(2x+3)} = \frac{5x+6+(3x+4)}{5x+6+(6x+7)}$$
$$\Rightarrow \frac{8x+10}{5x+5} = \frac{8x+10}{11x+13}$$
$$\Rightarrow 11x+13=5x+5$$
$$\Rightarrow x = -1$$
This can be proven by cross multiplying as well.
