We could also describe the difference in your two approaches in this way. You began with the system $ \ p(x) \ = \ x^2 + 4x + 4 \ = \ 0 \ , \ q(x) \ = \ x^2 + 5x + 6 \ = \ 0 \ \ . $ When you subtract one equation from the other, you have $ \ p(x) - q(x) \ = \ 0 \ \ , $ which is equivalent to the equation we would set up for finding intersections of the curves represented by these functions, $ \ p(x) \ = \ q(x) \ \ . $ You found the single solution $ \ x \ = \ -2 \ \ $ from $ \ -x - 2 \ = \ 0 \ \ , $ which is correct. It so happens for this system that this also locates a common factor of $ \ p(x) \ $ and $ \ q(x) \ $ because both functions are equal to zero at $ \ x \ = \ -2 \ \ , $ but this method would be correct to use in any case (as we'll show shortly for a different system).
When you add the two equations, as you did in your second calculation, you are now solving $ \ p(x) + q(x) \ = \ 0 \ \rightarrow \ p(x) \ = \ -q(x) \ \ , $ which is no longer the original problem. Here, it really is only because both $ \ p(x) \ $ and $ \ q(x) \ $ are equal to zero at $ \ x \ = \ -2 \ $ that this appears as a solution to $ \ 2x^2 + 9x + 10 \ = \ (2x + 5 )·(x + 2) \ = \ 0 \ \ , $ since $ \ (x + 2)·(x + 2) \ + \ (x + 2)·(x + 3) $
$ = \ (x + 2)·[ \ (x + 2) + (x + 3) \ ] \ \ . $ Other systems may not produce any solutions at all by adding the equations.
If we take, for example, the system $ \ p(x) \ = \ x^2 - 5x + 6 \ = \ ( x - 2)·(x - 3) \ = \ 0 \ , $
$ q(x) \ = \ x^2 + 7x + 10 \ = \ (x + 2)·(x + 5) \ = 0 \ \ , $ the two polynomials have no factors in common, but surely must intersect since they both "open upward". We find $ \ p(x) \ = \ q(x) $ $ \Rightarrow \ p(x) - q(x) \ = \ -12x - 4 \ = \ 0 \ \Rightarrow \ x \ = \ -\frac13 \ \ , $ a solution which is not suggested immediately by any of the polynomial factors. [Indeed, we could have chosen two "upward-opening" parabolas represented by polynomials which cannot be factored using real numbers and would still be able to find the intersection(s).]
On the other hand, $ \ p(x) + q(x) \ = \ 2x^2 + 2x + 16 \ \ $ has no real zeroes (or $ \ p(x) \ = \ -q(x) \ $ has no real-number solutions); we see that the function curves do not intersect. So adding the equations together in this system provides no information about the solutions of the original system of equations.