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Solve Lesson 22 Page 31 Math 10 SBT – Kite>


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



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