Mark M. answered 12/03/19
Retired math prof. Very extensive Precalculus tutoring experience.
x1 + 3x3 - x4 = 4
x2 - 2x3 + 3x4 = 6
x3 and x4 are parameters (any real numbers)
If we let x3 = s and x4 = t, then:
x1 = -3s + t + 4
x2 = 2s - 3t + 6
x3 = s
x4 = t
Solutions: (x1 , x2, x3, x4) = s(-3, 2, 1, 0) + t(1, -3, 0, 1) + (4, 6, 0, 0), where s and t are any real numbers