Solve a fractional equation!

$\begin{aligned} \frac {x+\blue 1}{x+2} + \frac {x+6}{x+\red 7} & = \frac {x+2}{x+3} + \frac {x+5}{x+6} & \small \blue{\text{Let }u = x + \frac {1 + \red 7}2 = x + 4} \\ \frac {u-3}{u-2} + \frac {u+2}{u+3} & = \frac {u-2}{u-1} + \frac {u+1}{u+2} \\ 1 - \frac 1{u-2} + 1 - \frac 1{u+3} & = 1 - \frac 1{u-1} + 1 - \frac 1{u+2} \\ \frac 1{u-2} + \frac 1{u+3} & = \frac 1{u-1} + \frac 1{u+2} \\ \frac 1{u-2} - \frac 1{u+2} & = \frac 1{u-1} - \frac 1{u+3} \\ \frac 4{u^2 - 4} & = \frac 4{u^2 + 2u - 3} \\ u^2 - 4 & = u^2 + 2u - 3 \\ 2 u & = - 1 \\ u & = - \frac 12 \\ \implies x & = u - 4 = \boxed{-4.5} \end{aligned}$

From the given equation, $1-\frac{1}{x+2} + 1 - \frac{1}{x+7} = 1-\frac{1}{x+3} + 1 - \frac{1}{x+6}$ $\frac{1}{x+6} - \frac{1}{x+7} = \frac{1}{x+2} - \frac{1}{x+3}$ $(x+2)(x+3)=(x+6)(x+7)$ $x^2+5x+6=x^2+13x+42$ $8x = -36$ $x=-4.5$