1< 3/x
1) Let's solve in ℝ: 1 < 3x
3x > 1
2) Let's sum the fractions: 3x -1 > 0
ⓘ To add fractions, put them over the same denominator, then add the numerators: 3x -1 = 3 -1·xx
3 -1·xx > 0
ⓘ Multiplying by 1 changes nothing: 1·x = x
3 -xx > 0
We study the sign of the numerator
⇥The expression 3 -x is zero at:
⇥ x = 3
⇥Its coefficient of x, -1, being negative, it is positive before and negative after 3
We study the sign of the denominator, which must not be zero
⇥The expression x is zero at x = 0 (forbidden value), it is negative before and positive after
3) Let's draw up the sign table:
x-∞ 0 3 +∞
3 -x++0-
x-0++
3x -1-❌+0-
Candidates, pending the domain check: x ∈ (0, 3)
For the expression 3x to exist, you must check: x ≠ 0
Domain constraints: x ≠ 0
▶Solution:   x ∈ (0, 3)