|4·x - 1| + |3·x + 2| < 10
Nullstellen der Betragsargumente
4·x - 1 = 0 → x = 1/4
3·x + 2 = 0 → x = - 2/3
Fall 1: x ≤ - 2/3
-(4·x - 1) - (3·x + 2) < 10
- 7·x - 1 < 10
x > - 11/7 → - 11/7 < x ≤ - 2/3
Fall 2: - 2/3 ≤ x ≤ 1/4
-(4·x - 1) + (3·x + 2) < 10
3 - x < 10
x > -7 → - 2/3 ≤ x ≤ 1/4
Fall 3: x ≥ 1/4
(4·x - 1) + (3·x + 2) < 10
7·x + 1 < 10
x < 9/7 → 1/4 ≤ x < 9/7
Zusammenfassung der Lösung
- 11/7 < x < 9/7
