Topic
The uncrossed part (including d) in Figure 11 is the solution domain of the inequality:
A. \(2x – 3y \le – 12\) B. \(2x – 3y \ge – 12\) C. \(3x – 2y \le 12\) D. \(3x – 2y \ge 12\) 

Figure 11 

Solution method – See details
Determine the equation of the line
Consider points in the solution domain and conclude
Detailed explanation
Let the line d have the form: y = ax + b
d goes through (6;0) and (0;4) so we have: \(\left\{ \begin{array}{l}0 = – 6a + b\\4 = 0.a + b\end {array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{2}{3}\\b = 4\end{array} \right \Leftrightarrow y = \frac{ 2}{3}x + 4 \Leftrightarrow 2x – 3y = – 12\)
Taking the point O(0; 0) not in d, we have 2.0 – 3.0 = 0 > – 12, but point O is not in the solution domain
Therefore, the required inequality is \(2x – 3y \le – 12\)
Choose A