Question

решите неравенство
2х
----
3 ≥ 9

Answer 1

2x-3≥9

2x≥12

x≥6

x є [6; + бесконечность)

Answer 2

$\dfrac{2x}{3} \geqslant 9, ~~~~~~ | \cdot \dfrac{2}{3}\\\\\\\dfrac{2x}{3}\cdot \dfrac{2}{3}\geqslant 9 \cdot \dfrac{2}{3}\\\\\\x \geqslant \dfrac{27}{2}$