Solve 3 variables from 2 equations?
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 
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
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
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]
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]
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
Invader''s Realm
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
Invader''s Realm
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
2x+4y=28;
x+y=9;
solve for x
x=9-y;
2(9-y)+4y=28;
etc
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.
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.
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]
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]
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).
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).
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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