# Thread: Linear equality system with 5 variables

1. ## Linear equality system with 5 variables

So here is the system:

V+W+X+Y+Z=7
X+2y+3z=8
x+2y+5z=10

I used Gaussian elimination method and got this:

V+W+X+Y+Z=7
X+2Y+3Z=8
-2Z=-2

So from here I can get z=1 but couldn't go up from here as the next equation (x+2y+3z)=8 has two other unknowns and the other is even worse with 5 variables. Anyone help please?

2. Originally Posted by Keep
So here is the system:

V+W+X+Y+Z=7
X+2y+3z=8
x+2y+5z=10

I used Gaussian elimination method and got this:

V+W+X+Y+Z=7
X+2Y+3Z=8
-2Z=-2

So from here I can get z=1 but couldn't go up from here....
You have five variables but only three equations. Where, "from here", were you needing to "go"? (Such systems cannot typically be solved for a unique solution.)

3. Originally Posted by stapel
You have five variables but only three equations. Where, "from here", were you needing to "go"? (Such systems cannot typically be solved for a unique solution.)
I was just wondering if there was a way I can find a solution to the variables even if it is something like 0=2 which means the system is unsolvable etc. So it is not possible to do anything beyond the point of getting the value of z as 1? So I just leave it that way and write 'system unsolvable?'

4. The system may be solveable. It's just not uniquely solveable. It's like a "system" of one equation in two variables: All you can do is solve "in terms of" the extra variables. You cannot find unique numerical solutions.

5. Originally Posted by stapel
The system may be solveable. It's just not uniquely solveable. It's like a "system" of one equation in two variables: All you can do is solve "in terms of" the extra variables. You cannot find unique numerical solutions.
Is it possible to do that then since that is what I want i.e. solving for the extra variables such that I can find the values of v,w,x,y,z or until I get some nonsense result like 0=3 etc thereby 'proving' it is not a uniquely solvable system?