## Solving Lesson 12 Page 29 Math 10 – Kite >

Topic

Solution domain of the system of inequalities $$\left\{ {\begin{array}{*{20}{c}}{2x – 5y > 1}\\{2x + y > – 5}\\{x + y < – 1}\end{array}} \right.$$ is the plane part containing the point with coordinates:

A. $$\left( {0;0} \right)$$ B. $$\left( {1;0} \right)$$ C. $$\left( {0;2} \right)\ ) D. \(\left( {0; – 2} \right)$$

Detailed explanation

Consider the system of inequalities $$\left\{ {\begin{array}{*{20}{c}}{2x – 5y > 1\left( 1 \right)}\\{2x + y > – 5\ left( 2 \right)}\\{x + y < – 1\left( 3 \right)}\end{array}} \right.$$

+) Replace x = 0 and y = 0, we get:

(1) ⇔ 2.0 – 5.0 > 1 ⇔ 0 > 1 (absurd);

=> The point with coordinates (0; 0) is not in the solution domain of the given system of inequalities.

+) Substituting x = 1 and y = 0 in the inequalities (1), (2) and (3) respectively in the system, we get:

(3) ⇔ 1 + 0 < – 1 ⇔ 1 < – 1 (absurd).

Hence the pair (1; 0) is not in the solution domain of the given system of inequalities.

+) Substituting x = 0 and y = 2 in the inequalities (1), (2) and (3) respectively in the system, we get:

(1) ⇔ 2.0 – 5.2 > 1 ⇔ – 10 > 1 (absurd);

(2) ⇔ 2.0 + 2 > – 5 ⇔ 2 > – 5 (always true);

(3) ⇔ 0 + 2 < – 1 ⇔ 2 < – 1 (absurd).

Hence the pair (0; 2) is not in the solution domain of the given system of inequalities.

+) Substituting x = 0 and y = – 2 in the inequalities (1), (2) and (3) respectively in the system, we get:

(1) ⇔ 2.0 – 5.(– 2) > 1 ⇔ 10 > 1 (always true);

(2) ⇔ 2.0 + (– 2) > – 5 ⇔ – 2 > – 5 (always true);

(3) ⇔ 0 + (– 2) < – 1 ⇔ – 2 < – 1 (always true).

Therefore, the pair of numbers (0; – 2) belongs to the solution domain of the given system of inequalities

Choose EASY