Disgusting Algebra, is this solvable by hand?

[DrummingAtom]

When do I resort to using a CAS to solve an equation? Is this one possible by hand?

I got this from a 2 dimensional GPS problem. Here's the math using the distance formula.

L₁² = (x₁- x)² + (y₁- y)²

L₂² = (x₂- x)² + (y₂- y)²

I tried the substitution method but it gets nasty and I would have 4th power equations. I don't know if there is a clean answer because Wolfram gave me a very long answer. I also used Wolfram to calculate the numerical answer. Thanks for any help.

 

[DrummingAtom]

Ooops.. Solving for x and y.

 

[pwsnafu]

The first thing that came to mind was calculate L₁-L₂ and factorize.

 

[DrummingAtom]

Yeah I did try that too but it still needs a back substitute that uses a very long chain of squared terms. I guess I'm just hoping there is a more elegant method because I don't know much about Linear Algebra.

 

[pwsnafu]

Yeah I did try that too but it still needs a back substitute that uses a very long chain of squared terms.

What? Please post your working so far.

 

[DrummingAtom]

L₁² = (x₁- x)² + (y₁- y)²

L₂² = (x₂- x)² + (y₂- y)²

Then L₁² - L₂² and factor:

L₁² - L₂² = (x₁² - 2xx₁ + x²) - (y₁² - 2yy₁ + y²) - (x₂² - 2xx₂ + x²) + (y₂² - 2yy₂ + y²)

Some algebra:

x = [L₁² - L₂² - (x₁² + y₁² + x₂² - y₂² - 2yy₁ + 2yy₂]/(-2x₁ + 2x₂)

Now I have to back sub to one of the original equations.. That's going to be a mess.

 

[arildno]

Set:

x=Ay+B.

Then it won't be so messy at all.

Determine what A and B is later on, (once you've solved for y and x) from your last line given in your post.

 

[pwsnafu]

Then L₁² - L₂² and factor:

Well you expanded, not factorized. I was thinking using to difference of two squares to obtain
L₁²-L₂² = (x₁+x₂-2x)(x₁-x₂) + (y₁+y₂-2y)(y₁-y₂)
which is just a linear equation. As arildno suggested, redefine the constants.

Now I have to back sub to one of the original equations.. That's going to be a mess.

Why do you need to do that?