$$\frac{72x}{3}+\frac{9}{3}\ge\frac{17}{3}$$
$x\geq \frac{1}{9}$
$$72x+9\geq 17$$
$$72x\geq 17-9$$
$$72x\geq 8$$
$$x\geq \frac{8}{72}$$
$$x\geq \frac{1}{9}$$
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