Réponse :
Bonjour,
1) 2x + 1 > 3
⇔ 2x > 2
⇔ x > 1
S = ] 1 ; +∞ [
2) 3x – 2 ≤ 7
⇔ 3x ≤ 9
⇔ x ≤ 3
S = ] -∞ ; 3 ]
3) 7x + 3 > 2x – 5
⇔ 7x – 2x > –5 – 3
⇔ 5x > –8
⇔ x > –8/5
S = ] –8/5 ; +∞ [
4) 5x – 3 ≤ 8x – 6
⇔ 5x – 8x ≤ –6 + 3
⇔ –3x ≤ –3
⇔ x ≥ 1
S = [ 1 ; +∞ [