it has infinite numbers of solutions, for any value of z u have different values of x,y
it is clear from the beginning since u have three variables and two equations
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?