Solving a Problem with 4 Simultaneous Equations and 6 Unknown Variables

[KingBigness]

I have a problem that involves 4 simulationious equations with 6 unknown variables.
Been trying to solve it using elimination but getting stuck in a loop.
Any advice on how to solve this? I know there will be multiple answers.
Is elimination the best method?

Can't upload the problem now cause I have writing this from a phone, so if you need the question I'll upload it as soon as I can.

Thank you for any help.

 

[micromass]

Yes, elimination seems to be the right way to go.
So we'll try to help once you uploaded the problem and what you tried...

 

[KingBigness]

Thank you sorry for the useless post haha I'll get it up as soon as I get home

 

[Ray Vickson]

Such systems are met with millions of time per day when solving _linear programming_ problems. There, the concept of a *basic* solution arises; this is a solution in which four of the variables are solved for as functions of the other two, then setting those two to zero, assuming that the 4x4 submatrix of those 4 variables is nonsingular. If every choice of 4x4 matrix is allowed (i.e., all are nonsingular), the number of basic solutions is C(6,4) = 6*5/2 = 15. If some of the 4x4 matrices are singular, the corresponding basic solution is non-existent (by definition).

While there could be as many as 15 different basic solutions there are infinitely many non-basic solutions, simply by assigning arbitrary non-zero values to the right-hand-side variables in each basic system.

For more on basic solutions, see, eg.,
http://www2.isye.gatech.edu/~spyros/LP/LP.html .

RGV

 

[KingBigness]

sorry for the ridiculously late reply. Finally got home so here is the question and my attempt at a solution.

 

[HallsofIvy]

Well the first thing I notice is that if you add twice the second equation to the first equation, you eliminate three unknowns at a stroke! $x_3$, $x_4$, and $x_6$ all cancel leaving $4x_1+ 4x_2- 7x_5= -2$. Since you know you will want to solve for four of the unknowns in terms of the other two, I would choose the two to be $x_1$ and $x_2$ so that I already have $x_5= (4/7)x_1+ (4/7)x_2+ 2/7$.

Replace $x_5$ in each of the equations by that and continue.

 

[KingBigness]

Well the first thing I notice is that if you add twice the second equation to the first equation, you eliminate three unknowns at a stroke! $x_3$, $x_4$, and $x_6$ all cancel leaving $4x_1+ 4x_2- 7x_5= -2$. Since you know you will want to solve for four of the unknowns in terms of the other two, I would choose the two to be $x_1$ and $x_2$ so that I already have $x_5= (4/7)x_1+ (4/7)x_2+ 2/7$.

Replace $x_5$ in each of the equations by that and continue.

Of course. Thank you, will continue working and let you guys know how I go.

 

[KingBigness]

So I left this problem and I have come back to it.
I have ended up with
x1=-3-x2
x3=-1+x6
x4=3
x5=-2

I am stuck...is this the answer?
I know 2 will be defined, 2 will be infinite and 2 will be defined by the infinite variables.
I am just not sure how to get to that conclusion.