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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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