In linear optimization, I often see linear inequalities that define the solution space being added up. Obviously if $1 < 4$ and $9 < 10$ then also $1 + 9 < 4 + 10$, so adding up both inequalities does not seem to do much harm (if you're adding up inequalities of the same sign). However if you have an optimization problem:
$\max x_2 \\ s.t.: x_1 \leq 1 \\ \ \ \ \ \ \ \ \ \ x_2 \leq 1\\ \ \ \ \ \ \ \ \ \ x_1, x_2 \geq 0 \ \text{and integer}$
then an optimum solution to this problem is obviously $(x_1,x_2) = (0,1)$.
But if you add both inequality constraints and obtain the following linear program:
$\max x_2 \\ s.t.: x_1 + x_2 \leq 2 \\ \ \ \ \ \ \ \ \ \ x_1, x_2 \geq 0 \ \text{and integer}$
The optimal solution clearly changes to $(x_1, x_2) = (0, 2)$. All we did was adding two linear inequalities (which is often done in linear programming, no?) but the solution space, the optimum objective value and the optimum solution clearly changed...
When are we allowed to add inequalities in linear programming without changing
The two inequalities $$ H_1: x_1 \le 1 \iff (1, 0) \cdot (x_1, x_2) \le 1 \\ H_2: x_2 \le 1 \iff (0, 1) \cdot (x_1, x_2) \le 1 $$ define two affine half spaces (here bounded by lines with given unit normal vectors and distance from the origin) and the solutions are in the intersection $H_1 \cap H_2$.
Together with the other constraints the feasible solutions are in $[0,1]^2\cap \mathbb{Z}^2$.
The inequality $$ H_3: x_1 + x_2 \le 2 \iff (1, 1) / \sqrt{2} \cdot (x_1, x_2) \le \sqrt{2} $$ defines just one affine half space.
The operation of adding the inequalities does not preserve the solution set.
Here is the scene: