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Set up an augmented matrix and use Gaussian elimination to solve the system:
3 2 | 15
6 4 | 30
~2 * R1 -> R1
6 4 | 30
6 4 | 30
~ -1 * R1 + R2-> R2
6 4 | 30
0 0 | 0
~ 1/6 * R1 -> R1
1 2/3 | 5
0 0 | 0
We can conclude two things from this:
1) The system is consistent, because there are no "bad" rows (no row reduces down to 0 ... | 1)
2) There is a free variable.
The solution to the system is x + 2/3y = 5, where 'y' is free.
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