# Ex 6.3, 12 - 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, 12 Solve the following system of inequalities graphically: x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1 First we solve x – 2y ≥ 3 Lets first draw graph of x – 2y = 3 Drawing graph Checking for (0,0) Putting x = 0, y = 0 x – 2y ≤ 3 0 - 2(0) ≤ 3 0 ≤ 3 which is true Hence origin lies in plane x – 2y ≤ 3 So, we shade right side of line 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 false Hence origin does not lie in plane 3x + 2y > 6 . So, we shade rights side of line Also, y ≥ 1 So, for all values of x, y = 2 Also, x ≥ 0 So, the shaded region will lie on the right side of y axis 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 9 years. He provides courses for Maths and Science at Teachoo.