Question 9 - Graph - 2 or more Equation - Chapter 5 Class 11 Linear Inequalities

Last updated at August 5, 2026 by

Ex 6.3, 9 - Solve 5x + 4y <= 20, x >= 1, y >= 2 - Class 11 - Graph - 2 or more Equation

Ex 6.3, 9 - Chapter 6 Class 11 Linear Inequalities - Part 2
Ex 6.3, 9 - Chapter 6 Class 11 Linear Inequalities - Part 3

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Ex6.3, 9 Solve the following system of inequalities graphically: 5x + 4y 20, x 1, y 2 First we solve 5x + 4y 20 Lets first draw graph of 5x + 4y = 20 Drawing graph Checking for (0,0) 5x + 4y 20 Putting x = 0, y = 0 5(0) + 4(0) 20 0 20 which is true Hence origin lies in plane 5x + 4y 20 So, we shade left side of line Also, y 2 So, for all values of x, y = 2 Also, x 1 So, for all values of y, x = 1 Hence the shaded region represents the given in equality

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