Question

Determinati x apartine R pentru care 0 ≤ $\frac{x+1}{2}$ ≤ 3

Answer 1

$0 \leqslant \frac{x + 1}{2} \leqslant 3 \\ 0 \leqslant x + 1 \leqslant 6 \\ - 1 \leqslant x \leqslant 5$
=> x apartine intervalului [ -1 ; 5 ]

Answer 2

inmultim dubla inegalitate cu 2
0≤x+1≤6

scadem 1 de peste tot
-1≤x≤5

x∈[-1;5]