How safe is to subtract or add equations in a system?

Let's take as example the following system:

$$\begin{cases} 2x + 1 = 0 \\ -x + 1 = 0 \end{cases}$$

I could take x from the first and check on the second which will give me no solution. Taking from one and replacing in other guarantees safety

But if I add them together I will get x + 2 = 0 => x = -2. Checking -2 will fail in both of them. More solutions than needed are shown, and for advanced problems is not easy if possible at all to check the solutions

a = 2, b = 3 => a + b = 5, but a + b = 5 does not mean a = 2, b = 3. Even if there are the same number of equations as the number of variables, since they are on the same domain, I guess some dirty tricks can be applied so the system does not result only true solutions

What rules shall be applied to the system in order to guarantee valid solutions?

We have 2 quadratric equations f(x) and g(x). By using f(x) = g(x) we get all x coordinates of the intersection points. Let's say we want to get all point with y = 2:

$$\begin{cases} f(x) = 2 \\ g(x) = 2 \end{cases}$$

The system results f(x) = g(x) or f(x) - g(x) = 0, and we said earlier that this means we get all intersection points, contrary to our rule of y = 2. In a real example we may have $$f(x) = x^2 + mx + 1$$ (m parameter) and some conditions that our f shall accomplish (so we can determine m). Since the system will not result only true solutions, then the method is quite useless and other ways shall be taken

• Not necessarily. You must first consider whether the conditions are "the same": for example $x+2 = 0, 2x+4=0$ are two conditions but are the same since they wield the same result. ($x=-2$) (we call them linearly dependent). May 17 '20 at 11:48
• For quadratic functions it's a different ball game: If you try to find intersection points that's fine, but if you add a constraint $y=2$ then you force to find only the intersection points which also have the same $y$ value. So you may, again, have more constraints than what is feasible for the function. I have nothing very intelligent to say about quadratic equations in general without going into advanced calculus. May 17 '20 at 11:50
That’s what happened here. The solution set to the original system is a subset of the solutions to the combined equation $$x+2=0$$, just as it’s supposed to be: the empty set is a subset of any set, after all. The moral is that if you solve a derived equation in the process of solving a system of equations, you need to check those solutions for validity against the original system.