Solve
\[\frac{x + 1}{x + 2} + \frac{x + 8}{x + 9} = \frac{x + 2}{x + 3} + \frac{x + 7}{x + 8}.\]
Subtracting 1 from each fraction, we get
\[-\frac{1}{x + 2} - \frac{1}{x + 9} = -\frac{1}{x + 3} - \frac{1}{x + 8}.\]Then
\[\frac{1}{x + 2} + \frac{1}{x + 9} = \frac{1}{x + 3} + \frac{1}{x + 8},\]so
\[\frac{2x + 11}{(x + 2)(x + 9)} = \frac{2x + 11}{(x + 3)(x + 8)}.\]Multiplying both sides by $(x + 2)(x + 9)(x + 3)(x + 8),$ we get
\[(2x + 11)(x + 3)(x + 8) = (2x + 11)(x + 2)(x + 9).\]Then
\[(2x + 11)[(x + 3)(x + 8) - (x + 2)(x + 9)] = (2x + 11)(6) = 0.\]Hence, $x = \boxed{-\frac{11}{2}}.$