Question 9 - Solving Pair of Linear Inequalities - Chapter 5 Class 11 Linear Inequalities

Last updated at April 16, 2024 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

Transcript

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

Go Ad-free
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.