**Topic**

Representation of the solution domain of the system of inequalities:

a) \(\left\{ {\begin{array}{*{20}{c}}{x – 2y \le 3}\\{x + y \ge – 3}\end{array}} \right .\) b) \(\left\{ {\begin{array}{*{20}{c}}{x + y \le 5}\\{x – 2y \le 2}\\{x \ge – 1}\end{array}} \right.\) c) \(\left\{ {\begin{array}{*{20}{c}}{ – 3x + 2y < 6}\\{x – 2y \ge – 2}\\{2x + y < 4}\end{array}} \right.\)

**Solution method – See details**

- Step 1: Draw a line \(d:x – 2y = 4\).
- Step 2: Take a point \(M\left( {{x_o};{y_o}} \right)\) that is not on d (we usually take the origin O if \(c \ne 0\). (a{x_o} + b{y_o}\) and compare with c
- Step 3: Conclusion
- If \(a{x_o} + b{y_o} < c\) then the half-plane (excluding line d) containing the point M is the solution domain of the inequality \(ax + by < c\)
- If \(a{x_o} + b{y_o} > c\) then the half-plane (excluding d) containing no point M is the solution domain of the inequality \(ax + by > c\)

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**Detailed explanation**

a) We have two lines: \({d_1}:x – 2y = 3;{d_2}:x + y = – 3\)

+) Take O(0; 0) not on the line d_{first} there are 0 – 2.0 = 0 < 3. Therefore, the solution domain of the inequality x – 2y ≤ 3 is a semi-plane containing the point O(0; 0) whose edge is a straight line d_{first}.

+) Take O(0; 0) not on the line d_{2} there are 0 + 0 = 0 > – 3. Therefore, the domain of the solution of the inequality x + y ≥ – 3 is a half-plane containing the point O(0; 0) whose edge is the line d._{2}.

The solution domain of the system of inequalities is the uncrossed region as shown in the following figure:

b) We have b lines: \({d_1}:x + y = 5;{d_2}:x – 2y = 2;{d_3}:x = – 1\)

+) Take O(0; 0) not on the line d_{first} there are 0 + 0 = 0 < 5. Therefore, the domain of the solution of the inequality x + y 5 is a half-plane containing the point O(0; 0) whose edge is a straight line d_{first}.

+) Take O(0; 0) not on the line d_{2} there are 0 – 2.0 = 0 < 2. Therefore, the solution domain of the inequality x – 2y ≤ 2 is a half-plane containing the point O(0; 0) whose edge is a straight line d_{2}.

+) Take O(0; 0) not on the line d_{3} there is 0 – 1. Therefore, the solution domain of the inequality x ≥ – 1 is a half-plane containing the point O(0; 0) and whose edge is a straight line d_{3}.

The solution domain of the system of inequalities is represented as the white region in the following figure:

c) We have three lines: \({d_1}: – 3x + 2y = 6;{d_2}:x – 2y = – 2;{d_3}:2x + y = 4\)

+) Take O(0; 0) not on the line d_{first} yes – 3.0 + 2.0 = 0 < 6. Therefore, the domain of the solution of the inequality – 3x + 2y < 6 is a half-plane containing the point O(0; 0) excluding the line edge d._{first }.

+) Take O(0; 0) not on the line d_{2} there are 0 – 2.0 = 0 > – 2 . Therefore, the solution domain of the inequality x – 2y – 2 is the half-plane containing the point O(0; 0) whose edge is the line d._{2}.

+) Take O(0; 0) not on the line d_{3} has 2.0 + 0 < 4. Therefore, the domain of the solution of the inequality 2x + y < 4 is a half-plane containing the point O(0; 0) and excluding the shore line d._{3}.

The solution domain of the system of inequalities is represented as the uncolored region as shown in the following figure: