Looks Unsolvable

$\begin{aligned} \frac 1{x+\color{#3D99F6}9} + \frac 1{x+\color{#3D99F6}7} & = \frac 1{x+\color{#3D99F6}10} + \frac 1{x+\color{#3D99F6}6} & \small \color{#3D99F6} \text{Let }u = x + 8, \text{as }\frac {6+7+9+10}4 = 8 \\ \frac 1{u+1} + \frac 1{u-1} & = \frac 1{u+2} + \frac 1{u-2} & \small \color{#3D99F6} \text{Rearrange} \\ \frac 1{u+1} - \frac 1{u-2} & = \frac 1{u+2} - \frac 1{u-1} \\ \frac {-3}{u^2-u-2} & = \frac {-3}{u^2+u-2} & \small \color{#3D99F6} \text{Divide both sides by }-3 \\ \frac 1{u^2-u-2} & = \frac 1{u^2+u-2} & \small \color{#3D99F6} \text{Taking reciprocal on both sides} \\ u^2-u-2 & = u^2+u-2 \\ \implies u & = 0 & \small \color{#3D99F6} \text{Since }u=x+8 \\ \implies x & = \boxed{-8} \end{aligned}$

The given expression is $\dfrac{1}{x + 9} + \dfrac{1}{x + 7} = \dfrac{1}{x + 10} + \dfrac{1}{x + 6}$

First we will take LCM : $\begin{aligned} \\ \dfrac{x + 9 + x + 7}{(x + 9)(x + 7)} = \dfrac{x + 10 + x +6}{(x + 10)(x + 7)}\\ \\ \dfrac{2x + 16}{(x + 9)(x + 7)} = \dfrac{2x + 16}{(x + 10)(x + 7)} \\ \end{aligned}$ Now, we should cancel $(2x + 16)$ . If we cancel we will not get any real solution.

$\dfrac{2x + 16}{(x + 9)(x + 7)} - \dfrac{2x + 16}{(x + 10)(x + 7)} = 0$

$(2x + 16) \times (\dfrac{1}{(x + 9)(x + 7)} - \dfrac{1}{(x + 10)(x + 7)}) = 0$

\implies 2x + 16 = 0 \space \space (or) \space \space \require {\cancel} \cancel{\dfrac{1}{(x + 9)(x + 7)} - \dfrac{1}{(x + 10)(x + 7)} = 0}

We should ignore the second part because if we solve it we will get $60 = 63$

So, the only part left is $2x + 16 = 0$

$\implies 2x = -16$

$\implies \color{#20A900} x = -8$

Rearrange the expression as $\dfrac{1}{x+9}-\dfrac{1}{x+10} = \dfrac{1}{x+6} - \dfrac{1}{x+7}\implies \\ \,(x+9)\,(x+10) = \,(x+7)\,(x+6)\implies x = \dfrac{42-90}{6} = \boxed{-8}$

Letting z = x+8 gives $\frac{1}{z+1}$ + $\frac{1}{z-1}$ = $\frac{1}{z+2}$ + $\frac{1}{z-2}$ . Making common denominators on both sides gives $\frac{2z}{z^2-1}$ = $\frac{2z}{z^2-4}$ If z is not 0 then we can cross multiply both sides and then divide by 2z to get z^2-1 =z^2-4 which is obviously impossible. So z=0 which means x=-8 and you can see this works .

wolfram one liner