Disgusting Algebra, is this solvable by hand?

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Discussion Overview

The discussion revolves around the solvability of a complex algebraic equation derived from a two-dimensional GPS problem using the distance formula. Participants explore methods for solving the equations for variables x and y, considering both manual techniques and the use of computer algebra systems (CAS).

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions when to resort to using a CAS for solving equations, expressing uncertainty about whether the problem can be solved by hand.
  • Another participant suggests calculating the difference L1 - L2 and factoring it as a potential approach.
  • Some participants note that attempts to factor lead to lengthy back substitutions involving squared terms, indicating a desire for a more elegant solution.
  • A participant proposes setting x = Ay + B to simplify the problem, suggesting that determining A and B can be done later.
  • Another participant points out that the expansion of L12 - L22 was not a factorization and suggests using the difference of two squares to obtain a linear equation instead.
  • There is a recurring concern about the complexity of back substitution and whether it is necessary.

Areas of Agreement / Disagreement

Participants express differing opinions on the best approach to solve the equations, with no consensus on a single method or solution. The discussion remains unresolved regarding the most effective technique.

Contextual Notes

Participants mention the complexity of the algebra involved and the potential for lengthy calculations, indicating that assumptions about the equations' structure may affect the approaches discussed.

DrummingAtom
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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.
 
Last edited:
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Ooops.. Solving for x and y.
 
The first thing that came to mind was calculate L1-L2 and factorize.
 
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.
 
DrummingAtom said:
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.
 
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.
 
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.
 
DrummingAtom said:
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?
 
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