Graphical solutions of systems of linear inequalities in two variables
- 3. 2𝑥 − 𝑦 > 3 𝑥 + 2𝑦 ≤ 4 Procedure: Graph both equations in the same Cartesian Plane!
- 4. 2𝑥 − 𝑦 > 3 2𝑥 − 𝑦 = 3 −𝑦 = −2𝑥 + 3 𝑦 = 2𝑥 − 3 (−1) 2𝑥 − 𝑦 > 32𝑥 − 𝑦 > 3 𝑥 + 2𝑦 ≤ 4 (−1)
- 5. (0, 0) 2𝑥 − 𝑦 > 3 2𝑥 − 𝑦 > 3 2(0) − (0) > 3 0 − 0 > 3 0 > 3 2𝑥 − 𝑦 > 3 𝑥 + 2𝑦 ≤ 4
- 6. 𝑥 + 2𝑦 ≤ 4 2𝑥 − 𝑦 > 32𝑥 − 𝑦 > 3 𝑥 + 2𝑦 ≤ 4 𝑥 + 2𝑦 = 4 2𝑦 = −𝑥 + 4 2 2 𝑦 = − 1 2 𝑥 + 2 𝑥 + 2𝑦 ≤ 4
- 7. (0) + 2(0) ≤ 4 2𝑥 − 𝑦 > 32𝑥 − 𝑦 > 3 𝑥 + 2𝑦 ≤ 4 𝑥 + 2𝑦 ≤ 4 (0, 0) 0 + 0 ≤ 4 0 ≤ 4
- 8. 2𝑥 − 𝑦 > 32𝑥 − 𝑦 > 3 𝑥 + 2𝑦 ≤ 4 𝑥 + 2𝑦 ≤ 4
- 9. 2𝑥 − 𝑦 > 32𝑥 − 𝑦 > 3 𝑥 + 2𝑦 ≤ 4 𝑥 + 2𝑦 ≤ 4 (2, −3) 2𝑥 − 𝑦 > 3 2(2) − (−3) > 3 4 − (−3) > 3 4 + 3 > 3 7 > 3 𝑥 + 2𝑦 ≤ 4 2 + 2(−3) ≤ 4 2 − 6 ≤ 4 −4 ≤ 4
- 10. 𝑦 > −2𝑥 + 3 𝑦 ≥ 𝑥 − 2 𝑦 > −2𝑥 + 3 𝑦 > −2𝑥 + 3 0 > −2(0) + 3 0 > 0 + 3 0 > 3
- 11. 𝑦 > −2𝑥 + 3 𝑦 ≥ 𝑥 − 2 𝑦 > −2𝑥 + 3 𝑦 ≥ 𝑥 − 2 0 ≥ 0 − 2 0 ≥ −2 𝑦 ≥ 𝑥 − 2
- 12. 𝑦 > −2𝑥 + 3 𝑦 ≥ 𝑥 − 2 𝑦 > −2𝑥 + 3 𝑦 > −2𝑥 + 3 (1, 3) 3 > −2(1) + 3 3 > −2 + 3 3 > 1 𝑦 ≥ 𝑥 − 2 3 ≥ 1 − 2 3 ≥ −1 𝑦 ≥ 𝑥 − 2
- 13. Procedure: 1. Graph both equations in the same Cartesian Plane 2. Solution points are those located in the overlapping shaded region 3. Check sample solution points on the overlapping shaded region
- 15. 𝑦 > −2𝑥 + 3 𝑦 ≥ 𝑥 − 2 Find out who are my favorite students!
- 17. Solve this system of linear inequalities by graphing. Identify a sample solution point. Check this point by substitution. 𝑦 > −2𝑥 + 3 𝑦 ≥ 𝑥 − 2
- 18. Try to solve the same system of linear inequalities algebraically. Use substitution or elimination. Can you find the solutions? What are the problems and difficulties you encountered? Do you think you can solve these systems algebraically? Defend your answer. 𝑦 > −2𝑥 + 3 𝑦 ≥ 𝑥 − 2