# 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?

# Example (please mention the mistake):

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.