Question

3. Resolver ::::::::::::::::::

Answer 1

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$6 + 4(x - 2) \leqslant 3(x - 5) + 14 < 4(x + 1) \\ 6 + 4x - 8 \leqslant 3x - 15 + 14 < 4x + 4 \\ 4x - 2 \leqslant 3x - 1 < 4x + 4 \\ - 2 \leqslant - x - 1 < 4 \\ - 1 \leqslant - x < 5 \\ - 5 < x \leqslant 1$

C.S = 《-5; 1]