x − 3y + z = 1

x + y + 2z = 14

1 -3 1 | 1

1 1 2 | 14

1 -3 1 | 1

0 4 1 | 13

1 -3 1 | 1

0 1 1/4 | 13/4

1 0 7/4 | 43/4

0 1 1/4 | 13/4

So x = 43/4 y = 13/4 and z = 0. Which is one solution.

But because there is 1 variable in my last reuction does this mean there are infinite solutions?