Solving multiple linear equations I'm a bit rusty on my linear algebra.
I have the following equations:
$$\operatorname{weight}_C = \frac{\frac{P_y - A_y}{B_y - A_y} - \frac{P_x - A_x}{B_x - A_x}}{ \frac{A_x-C_x}{B_x-A_x} -  \frac{A_y-C_y}{B_y-A_y}};$$
$$\operatorname{weight}_B = \frac{P_x - A_x + \operatorname{weight}_C\cdot(A_x-C_x)}{B_x-A_x};$$
$$\operatorname{weight}_A = 1-\operatorname{weight}_B-\operatorname{weight}_C;$$
And I have values for all variables except $P_x$ and $P_y$.
How do I go about solving for $P_x$ and $P_y$ given the equations above?
Is it possible given this information?
 A: Your broader problem seems easier than the actual question you asked.
You know $P$'s distance from $A,B,C$. Let us call them $a,b,c$.
Now use good old Pythagoras.
I will borrow your terminology, so $(A_x,A_y)$ are the coordinates of $A$, etc.
You get the equations
$(P_x-A_x)^2+(P_y-A_y)^2=a^2$ and
$(P_x-B_x)^2+(P_y-B_y)^2=b^2$.
Subtracting those equations gives you a linear relation between $P_x$ and $P_y$ which allows you
to express $P_x$ as an easy function of $P_y$.
Substituting this in one of the equations gives you a quadratic equation in $P_y$,
that will give you 0, 1 or 2 possibilities for $P_y$
and consequently the same number of possibilities for $P_x$.
You really have drawn a circle around $A$ and a circle around $B$ and computed the points of intersection.
Substituting this in a third equation
$(P_x-C_x)^2+(P_y-C_y)^2=c^2$ will show you which of the choices is correct.
This makes clear that you cannot just prescribe arbitrary values for the distances of $P$ to $A$, $B$ and $C$:
the circles may decide not to intersect at all!
A: You have two equations in two unknowns-the third has no unknowns in it.  Isolate $P_x$ in the second by multiplying by $B_x-A_x$ then adding the appropriate terms.  It should just be a number because you know everything else.  Then just put that into the first equation and solve for $P_y$
