Question

Determinati numerele naturale care verifica inegalitatile :

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445 ≤ x plus 235 ≤ 475
81 ≤ 3x ≤ 207
1 <(3x plus 2) : 9 ≤ 3
2 ≤ ( 5x-8):5 ≤ 6

Answer 1

1) 445≤x+235≤475 scadem 235
rezulta 210≤x≤240
2) 81≤3x≤207 impartim totul la 3
27≤x≤69
3) 1≤3x+2≤3 scadem 2
-1≤3x≤1 impartim la 3
-1/3≤x≤1/3
4) 2≤(5x-8):5≤6 inmultim cu 5
10≤5x-8≤30 adunam 8
18≤5x≤38 impartim la 5
18/5≤x≤38/5
bafta