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3(x + 5) - 8 ≥ 2 (x + 12) - 14
3x + 15 - 8 ≥ 2x + 24 - 14
3x + 7 ≥ 2x + 10
3x - 2x ≥ 10 - 7
x ≥ 3
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3
Solución.
x∈ [3 , infinito)
3x+7_>2x+10
x_>3
C.S.={3;+infinito}