# Ex 6.3, 14 - Chapter 6 Class 11 Linear Inequalities

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.3, 14 Solve the following system of inequalities graphically: 3x + 2y 150, x + 4y 80, x 15, y 0, x 0 Now we solve 3x + 2y 150 Lets first draw graph of 3x + 2y = 150 Drawing graph Checking for (0,0) Putting x = 0, y = 0 3x + 2y 150 4(0)+ 3(0) 150 0 150 which is true Hence origin lies in plane 3x + 2y 150 So, we shade left side of line Now we solve x + 4y 80 Lets first draw graph of x + 4y = 80 Drawing graph Checking for (15,0) Putting x = 15, y = 0 y 2x 0 2(15) 0 30 which is false Hence (15,0) does not lie in plane y 2x So, we shade left side of line Also, x 15 So, for all values of y, x = 15 Given x 15, so we shade left side of line Also, given x, y 0 So, the shaded region will lie in 1st 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.