# Ex 6.3, 15 - Chapter 6 Class 11 Linear Inequalities

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 6.3, 15 Solve the following system of inequalities graphically: x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0 First we solve x + 2y ≤ 10 Lets first draw graph of x + 2y = 10 Drawing graph Checking for (0,0) Putting x = 0,y = 0 x + 2y ≤ 10 0+2(0)≤10 0 ≤ 10 which is true So, we shade left side of line Hence origin lies in plane x + 2y ≥ 10 Now we solve x + y ≥ 1 Lets first draw graph of x + y = 1 Drawing graph Checking for (0,0) Putting x = 0, y = 0 x + y ≥ 1 0 + 0 ≥ 1 0 ≥ 1 which is false Hence origin does not lie in plane x + y ≥ 1 So, we shade right upper side of line Now we solve x – y ≤ 0 Lets first draw graph of x – y = 0 Drawing graph Checking for (10,0) Putting x = 10, y = 0 x – y ≤ 0 10 – 0 ≤ 0 10 ≤ 0 which is false. Hence (10,0) does not lie in plane x > y So, we shade left side of line Also, given x ≥ 0, y ≥ 0 So, shaded region will lie in first quadrant Hence the shaded region represents the given inequality

Chapter 6 Class 11 Linear Inequalities

Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.