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# Solve 3 variables from 2 equations?

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Is it posible to solve for 3 variables from 2 equations?

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I don''t believe that''s possible MetaKnight. Although, I''m not a mathematician, I have taken many math courses so I''m pretty positive. I guess it could be compared to trying to solve 2 variables from 1 equation, which isn''t possible.. (i hope not

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Nope, it isn''t possible. This is an example of what you would get:

Equations:
x + y + z = 1
2x + 3y + 4z = 2

x = 1 + z
y = -2z
z = z

The Game is Nothing,
The Playing of it Everything

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No, you need n+1 equations where n is the number of unknowns. There are cases where you might seem to have less than n+1 equations.

But it might be possible to generate more equations, but this is usually a trick question type deal which show up on tests.

[edited by - fodd3r on April 27, 2003 12:26:18 AM]

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You need only n (not n + 1) equations to solve for n variables.

x + 1 = 2
1 unknown, 1 equation

x + y = 2
2x + 2y = 8
2 unknowns, 2 equations

Qui fut tout, et qui ne fut rien

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I believe you only need n equations where n is unknown veriables.

2x+4y=28;
x+y=9;

solve for x
x=9-y;
2(9-y)+4y=28;

etc

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the solutions to 2 equations with three variables lies in 3-2=1 dimensional space and therefore forms a line, any point on this line satisfies your equations but the solution is the whole of this line. generally if you are in n dimensional space and have c constraints the the solution space has dimension n-c.

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Saying an equation system with n unknowns need n equations is not the whole truth. The equation system needs to have n linearly independent equations. Take this system for example.
x  + 2y = 5    (a)2x + 4y = 10   (b)

Two equations and two unknowns. Yet the system is not solvable (a single unique solution that is), becuase it only have one linearly independent equation, beacuase (b) = 2*(a).

As a sidenote, one can find the minimum norm solution (making the variables as small as possible) using the following formula.

Ax = y
y = A+x

where A+ is the pseudoinverse, calculated as

A+ = AH(AAH)-1

edit: AH is, the Hermitian transpose of A, that is, the transpose of the complex conjugate of A. For a real valued matrix, this is the same as AT.

[edited by - Brother Bob on April 28, 2003 7:35:03 AM]

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That line thing is correct although sometime with very weird equations you can narrow it down to several points. One such equation is:

x^y = y^x

with solutions of:

2,4
4,2
and pairs of equal integers.

These however do not occur frequently and still leave you with infinitely many solutions (well, aleph0 instead of aleph1 which is a big step down).

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I think n equations for n variables is right for when the variables are real. But for integers I think there will be examples, when this isn''t the case. Not that it''s applicable here.

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