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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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quote:
Original post by higherspeed
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.

With integer variables ther may not be a solution but you should never need more than n equations to solve for n unknowns. Of course, the solution could be complex.

The difference between us and a computer is that, the computer is blindingly stupid, but it is capable of being stupid many, many million times a second.

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quote:
Original post by higherspeed
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.

n equations for n variables is right when the equations are linear and linearly independent. It is not generally true for equations of real variables. For instance, x²+y²=0 has only one real solution, while x²+y²=-1 has none, and x²+y²=1 has infinitely many.

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z = y + x
|x| = y + z

x = y + z (derivative of absolute value eqtn)
x = -y - z (derivative of absolute value eqtn)

using the first abs val with first prob:

z = y + (y + z)using secnd:
z = 2y + z
2y = 0 (after sub z, switch 0 = 2y)
y = 0

substitution of y

z = y + x
z = 0 + x
z = x

using second abs val w/ first prob
z = y -y - z
z = -z
2z = 0(add z to both sides
z = 0

substitution of z

0 = y + x
x + y = 0 (switch ''x + y'' and ''0'')
x = -y

if z = x, y = 0, z = 0, and x = -y, then all must be equal to 0

z = x
x = -y
y = 0 <--
x = -(0)
x = 0 <--
z = (0)
z = 0 <--

------------------------------------------------------------
Hell is not a bad place to be
It has just been recently added to the attractions list

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quote:
Original post by higherspeed
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.

Hi an example of equation that can be solved in two variables
it''s a diofantine equation for example
:
ax + by = z
where:
a,b,z are relative integers
note that this equation can be solved in Z(so you will find only couples of integers that will solve)
if and only if mcd(a,b) divides z
where i call mcd the greatest common factor of a,b
and i call Z the set of relative integers

to solve this kind of equation the Euclidean algorithm of subsequent division will help
(at this moment i could not write an exhaustive answer if you are interested mail me at: scarcelan@lib.unimib.it and i''ll furnish further explanations )
bye
Andrea Scarcella 1St Year computer science student at University of Bicocca Milano Italy

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You guys are right, I don''t know what I was thinking. Must have been late. *boggle*