## Linear Algebra - System of 2 Equations with 3 Variables--possible?

1. The problem statement, all variables and given/known data
Solve: x1-3x2+4x3=-4
3x1-7x2+7x3=-8
-4x1+6x2-x3=7

3. The attempt at a solution
I was able to make it to:
1 -3 4 -4
0 -10 25 -11
0 0 0 0
So the third row goes away, and I am left with:
1 -3 4 -4
0 -10 25 -11
I am pretty sure that cannot be solved, or am I overlooking something? Thank you in advance!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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 It looks like you have a free variable. So you can solve for the other two in terms of it.

Recognitions:
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 Quote by chrisdapos I am pretty sure that cannot be solved, or am I overlooking something? Thank you in advance!!
The solution isn't unique, you can try plugging in a value for $x_1$ and solving the rest.
$$x_1=1,2,3,4,5,6....\pi, e,...$$

## Linear Algebra - System of 2 Equations with 3 Variables--possible?

whenever you have more unknowns than equations, you get infinitely many solutions and one or more variables become free varaibles

Recognitions:
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 Quote by proton whenever you have more unknowns than equations, you get infinitely many solutions and one or more variables become free varaibles
Sometimes there can still be zero solutions:
$$x+y+z=0$$
$$x+y+z=1$$

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