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

Last updated at April 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

Transcript

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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