solve for x in the following Quadratic equations I am not able to solve this problem for my son. Is ther any error in the question itself.
$\dfrac{1}{(x+1)} + \dfrac{1}{(x+5)}=\dfrac{1}{(x+2)}+\dfrac{1}{(x+4)}$.
Thanks in advance..

$\dfrac{1}{(x+1)} + \dfrac{1}{(x+5)}=\dfrac{1}{(x+2)}+\dfrac{1}{(x+4)}$
$\implies \dfrac{1}{(x+5)}-\dfrac{1}{(x+4)}=\dfrac{1}{(x+2)}-\dfrac{1}{(x+1)} $
$\implies \dfrac{x+4-x-5}{(x+4)(x+5)}=\dfrac{x+1-x-2}{(x+1)(x+2)}$
$\implies (x+4)(x+5)=(x+1)(x+2)$
$\implies x^2+9x+20=x^2+3x+2$
$\implies 9x-3x+20-2=0$
Or,$$x=-3$$
