# Ex 6.3, 10 - Chapter 6 Class 11 Linear Inequalities

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex6.3, 10 Solve the following system of inequalities graphically: 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0 Now we solve 3x + 4y ≤ 60 Lets first draw graph of 3x + 4y = 60 Drawing graph Checking for (0,0) Putting x = 0, y = 0 3x + 4y ≤ 60 3(0) + 4(0) ≤ 60 0 ≤ 60 which is true So, we shade left side of line Hence origin lies in plane 3x + 4y ≤ 60 Now we solve x + 3y ≤ 30 Lets first draw graph of x + 3y = 30 Drawing graph Checking for (0,0) Putting x = 0, y = 0 x + 3y ≤ 30 0 + 3(0) ≤ 30 0 ≤ 30 which is true So, we shade left side of line Hence origin lies in plane x + 3y ≤ 30 Also given that x ≥ 0 , y ≥ 0 Hence the shaded region will lie in the 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

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.