Question 3 - Graph - 2 or more Equation - Chapter 5 Class 11 Linear Inequalities

Last updated at Dec. 16, 2024 by

Ex 6.3, 3 - Solve 2x + y >= 6, 3x + 4y <= 12 graphically - Ex 6.3

Ex 6.3, 3 - Chapter 6 Class 11 Linear Inequalities - Part 2
Ex 6.3, 3 - Chapter 6 Class 11 Linear Inequalities - Part 3
Ex 6.3, 3 - Chapter 6 Class 11 Linear Inequalities - Part 4

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Ex6.3, 3 Solve the following system of inequalities graphically: 2x + y 6, 3x + 4y 12 First we solve 2x + y 6 Lets first draw graph of 2x + y = 6 Drawing graph Checking for (0,0) Putting x = 0, y = 0 2x + y 6 2(0) + (0) 6 0 6 which is false So, we shade right side of line Hence origin does not lie in plane 2x + y 6 Now we solve 3x + 4y 12 Lets first draw graph of 3x + 4y = 12 Drawing graph Checking for (0,0) Putting x = 0, y = 0 3x + 4y 12 3(0) + 2(0) 12 0 12 which is true Hence origin lies in plane 3x + 2y > 6 . So, we shade left side of line Hence the shaded region represents the given inequality.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo