Does the adage: N (distinct) equations and N unknowns gives 1 solution always hold true?

The problem: I have a signal reciever and 3 signal beacons in a 2d plane. The reciever does not know its initial position, nor does it know any of the positions of the beacons. The reciever can move (by an unknown amount); or take a measurement of its distances from the beacons.

Initially, the reciever takes one measurement. Let its starting position be $$(x_{t0},y_{t0})$$, and the positions of the beacons be $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$, with the range measurements being $$r_{1t0},r_{2t0},r_{3t0}$$. This yields: $$r_{1t0}^2 = (x_{t0}-x_1)^2 + (y_{t0}-y_1)^2$$ $$r_{2t0}^2 = (x_{t0}-x_2)^2 + (y_{t0}-y_2)^2$$ $$r_{3t0}^2 = (x_{t0}-x_3)^2 + (y_{t0}-y_3)^2$$ This is 3 equations and 8 unknowns (2 initial position coordinates + 6 beacon coordinates).

Next, suppose I move to a new (unknown) position $$(x_{t1},y_{t1})$$, and take a new measurement $$r_{1t0},r_{2t0},r_{3t0}$$. This yields: $$r_{1t1}^2 = (x_{t1}-x_1)^2 + (y_{t1}-y_1)^2$$ $$r_{2t1}^2 = (x_{t1}-x_2)^2 + (y_{t1}-y_2)^2$$ $$r_{3t1}^2 = (x_{t1}-x_3)^2 + (y_{t1}-y_3)^2$$ Thus, I've added 3 equations and only 2 unknowns! Then, extrapolating, I could have 6 measurements, for a total of (3x2 beacon coordinates + 6x2 position coordinates) = 18 unknowns and 3x6 equations from the readings above. According to the adage of N equations N unknowns, I should be able to solve for all of my unknowns $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$, and $$x_{tn},y_{tn}$$ uniquely...

But we know this is impossible, because suppose I have a set of solutions $$x,y$$. Then I could have an equally good set of solutions $$x+1,y$$ which obey all the equations above!

So my question is then: is the adage broken? N equations N unknowns does not yield a unique solution? Or does it only apply to linear systems? Or are some of my equations not unique?

Thanks.

• Your "always" is incorrect, even for linear equations. There could be redundant equations, which yield no new information. Apr 11 '21 at 0:02
• I am aware of the redundant thing - i think that's what I meant by 'unique' equations. Thanks Apr 11 '21 at 0:02
• In general, no. Consider $$1=3x+6y~~,~~2=2x+4y$$ Apr 11 '21 at 0:04
• Well $x^2 + y^2 = -7$ and $y= 2$ has no solutions. "unique" doesn't eliminate redundant. $a +b+ c = 3$ and $3a - 2b + 4c = 7$ and $4a-b+5c = 2$ are three "unique" equations with no solution. Apr 11 '21 at 0:24
• And it does only apply to linear equations. Her is one equation, on unknown, and two solutions. $x^2 = 49$. Two solutions $x = -7$ and $x = 7$. Less trivial $ax^2 + bx = c= 0$ has two solutions. And $ax^2 + bxy + cy^2 = C_1$ and $dx^2 + exy + fy^2 = C_2$ will give you 2 acceptable expression with $x$ in terms of $y$ and for $2$ fourth degree polynomials with in terms of $y$ we could have $4$ potential solutions each so hypothetically $8$ solutions. (ALthough maybe we can prove there must be redundancies) Apr 11 '21 at 0:32

It only applies to linear equations.

$$x^2 = -7$$ has no real solutions (although it has complex solutions)

And $$x^2 + 5x-6$$ is one equation with one unknown and $$2$$ solutions.

Sand $$(x- 7)(y-3)(x+8)(y+2) =0$$ and $$(x-5)(y-4)(x-2)(x+3) =0$$ is two equations with $$2$$ unknowns and $$4$$ solutions

Think about the case of three equations in three unknowns. Then each equation represents a surface in 3D space, so we have a situation where geometric intuition can help us. A solution to the system of equations is a point where all three surfaces intersect.

If the three equations are linear, then we have three planes. The set of intersection points could be empty, or infinite, or it might consist of a single point.

Now suppose one of the equations represents a circular cylinder, and the other two are still planes. The intersection of the cylinder with one of the planes could be an ellipse, a pair of lines, a single line, or the empty set. Intersecting each of these with the other plane might give you one point, two points, an infinite number of points, or the empty set.

You can try a few more simple examples, too, but I think it’s already clear that your adage is flawed.

In your beacon problem, all of your equations represent spheres. It might help you to think about the different ways that spheres can intersect each other. In general, ignoring special cases, the intersection of two spheres is a circle, and this circle will intersect a third sphere in two points. And so on.