Solving second degree simultaneous equation

Question

What would be the solution to the following two equations ?

A₁uv + B₁u + C₁v + D₁ = 0

A₂uv + B₂u + C₂v + D₂ = 0

u, v in [0, 1]

The solution needs to be blazing fast because it needs to be solved for each pixel, hopefully a direct rather than iterative solution.

This is basically trying to find the inverse of a coons patch where the boundaries are straight lines.

Answer 1

Solve equation 1 for u, you get u = (-C_1v -D_1)/(A_1v+B_1). Substitute that into equation 2, multiply through by (A_1v+B_1), and you should get a quadratic in v. Use the quadratic equation to solve for v.

Bonus points for figuring out what happens when A_1v+B_1 is zero.

Answer 2

I used wolfram alpha.

The generic form timed out while calculating, but I did get a solution substituting one constant with a number.

The resulting solution was pages long :P.

I thing I will have to go with some other solution that approximates u,v, the direct solution will be too slow for a per pixel approach.