I know that a system of equations with 3 variables usually has to contain at least 3 equations, but are there any special cases where 2 equations with 3 variables have just one solution? I've researched a bit, and found 2 equations with 3 variables with finite solution sets, but are there any that have exactly one solution? The equations could be linear or quadratic or cubic or sinusoidal or anything... but does such a case exist?

Also, is it possible to find such a case if we know any 2 variables? Like, for example, $a + b + c = w$ and $b + c = d$, given b and d. (of course, this example has more than 1 solution, but are there any like this that have just 1 solution)

Thanks for any help!


To clarify, I'm looking for the general form of such an equation. To be more specific, are there any systems of 2 equations, with 3 variables, that only have 1 solution, but which can be modified to have any 1 solution I desire.


1 Answer 1


Yes there is. $x^2+y^2=0$ and $z=0$.

Edit: $(x-a)^2+(y-b)^2=0$ and $z-c=0$

  • $\begingroup$ Guess I worded my question wrong... see my edit :) $\endgroup$ Dec 2, 2016 at 4:40
  • $\begingroup$ Edited with a general solution. $\endgroup$ Dec 2, 2016 at 4:41
  • $\begingroup$ This gives an unique solution in $\mathbb{R}^3$. Is there any way a system could return an unique solution in $\mathbb{C}^3$? $\endgroup$
    – GLay
    Dec 2, 2016 at 4:44
  • $\begingroup$ Great answer! What about my second question? Are there any 2 equation systems, with a unique solution, with 3 variables and 2 knowns? (in the equation you posted, one needs to known a, b, and c, are there any for which it is only necessary to know a and b, or b and c, etc.?) $\endgroup$ Dec 2, 2016 at 4:53
  • 1
    $\begingroup$ @GLay In $\mathbb{C}^3$ it can't be polynomials, since $\mathbb{C}$ is algebraically closed. But you could take any function $f : \mathbb{C} \to \mathbb{R}$ with the property that $f(z)=0 \iff z=0$ then the equation $f^2(x-a)+f^2(y-b)+f^2(z-c)=0$ would have the unique solution $x=a,y=b,z=c$. For example, $f(w) = |w|$. $\endgroup$
    – dxiv
    Dec 2, 2016 at 4:57

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