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⇔ 3x + 1 ≥ 12x + 4
⇔ 3x - 12x ≥ 4 - 1
⇔ -9x ≥ 3
⇔ x ≤ -1/3
16) √(3x+1) ≥ 4
kuadratkan kedua ruas, get
3x+1 ≥ 16
3x ≥ 16 - 1
3x ≥ 15
x ≥ 15/3
x ≥ 5
syarat :
3x+1 ≥ 0
3x ≥ 0 - 1
3x ≥ -1
x ≥ -1/3
irisan kedua solusi di atas x ≥ 5