Technique for solving $\frac{ax+b}{cx+d}=\frac{px+q}{rx+s}$ where the sum of numerators equals the sum of denominators

Your equation has the form $$ N_1/D_1 = N_2/D_2 $$ that implies $$ N_1 D_2 = N_2 D_1 $$ Add $N_2 D_2$ to both sides, $$ (N_1+N_2) D_2 = N_2 (D_1+D_2) \qquad \qquad (*) $$ but $(N_1+N_2) = (D_1+D_2) $, so you cam simplify the terms in the parenthesis, and remain with $$ D_2 = N_2 \quad \Rightarrow \quad D_2 -N_2 = 0 $$ Similarly you can show that an equivalent condition is $D_1-N_1 = 0$.

Note that equation $(*)$ is satisfied also if $(N_1+N_2)=(D_1+D_2)=0$, which gives the other solution.

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.