Question

What are the solutions of the compound inequality 2d + 3 ≤ –11 or 3d – 9 > 15?

Answer 1

Step-by-step explanation:

1.

$2d + 3 \leqslant - 11 \\ 2d \leqslant - 11 - 3 \\ 2d \leqslant - 14 \\ \frac{2d}{2} \leqslant \frac{ - 14}{2} \\ d \leqslant - 7$

2.

$3d - 9 > 15 \\ 3d > 15 + 9 \\ 3d > 24 \\ \frac{3d}{3} > \frac{24}{3} \\ d = 8$

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