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


2 Answers 2


When you have a system of equations, a linear combination of them produces another equation whose solutions include those of the original system. That’s an important nuance: nothing says that this single equation has exactly the same solution set as the original system, but the original system’s solutions are a subset of those of the new equation. That should make perfect sense if you think about it a bit. For example, if you have a pair of intersecting lines in the plane, every (nontrivial) linear combination of their equations describes some line that passes through the intersection point, but obviously that combined equation has other solutions as well—a whole line’s worth.

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.


For linear equations (such as in the first example) you need to look at how many conditions you apply over how many parameters are in question: in your example you impose 2 different conditions on x (has to equal both 1 and -1/2), therefore there is no solution. Generally speaking you can have one of three outcomes: 1. if the number of parameters exceed the number of conditions you will have infinite solutions. 2. if the number of parameters is exactly the same as the number of conditions you have a single solution. 3. otherwise there is no solution (as in your example). If you studied linear algebra - that has to do with the rank of the matrix involved.

  • $\begingroup$ So this means I shouldn't solve such systems where I have more conditions than parameters? Also, can u also explain the quadratic example I gave? $\endgroup$
    – uIM7AI9S
    Commented May 17, 2020 at 11:32
  • $\begingroup$ 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). $\endgroup$
    – Orenio
    Commented May 17, 2020 at 11:48
  • $\begingroup$ 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. $\endgroup$
    – Orenio
    Commented May 17, 2020 at 11:50

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