linear algebra – Is the reduction of this system of nonlinear equations *really* valid?

I’m looking at this post, over on Stack Overflow.

I’ll summarize here.

We’re given the nonlinear system:

$(A_x – P_x)^2 + (A_y – P_y)^2 = (v(t_a – t_0))^2$ ${1}$

$(A_x – P_x)^2 + (A_y – P_y)^2 = (v(t_a – t_0))^2$ ${2}$

$(A_x – P_x)^2 + (A_y – P_y)^2 = (v(t_a – t_0))^2$ ${3}$

Where $P_x, P_y,$ and $t_0$ are unknown. We’re trying to determine $P_x, P_y$.

The answerer says to, “open the brackets, and subtract ${2}$${1}$, ${3}$${2}$, ${1}$${3}$ to discard squares of unknown terms”. Now, we have the linear system:

$B_x^2-A_x^2 -2*(B_x-A_x)*P_x + B_y^2-A_y^2 -2*(B_y-A_y)*P_y = v^2*(t_b^2-t_a^2 -2*(t_b-t_a)*t_0)$

$C_x^2-B_x^2 -2*(C_x-B_x)*P_x + C_y^2-B_y^2 -2*(C_y-B_y)*P_y = v^2*(t_c^2-t_b^2 -2*(t_c-t_b)*t_0)$

$A_x^2-C_x^2 -2*(A_x-C_x)*P_x + A_y^2-C_y^2 -2*(A_y-C_y)*P_y = v^2*(t_a^2-t_c^2 -2*(t_a-t_c)*t_0)$

…which can be solved easily using Gaussian elimination.


Is this a valid solution? I ask as I’ve seen others attempt to solve this very system using much more complex methods, such as Newton’s method. If so, why would anyone attempt anything more complicated than this (simple Gaussian elimination, e.g.)? Surely something must be missing here?