Example 12 - Chapter 6 Class 11 Linear Inequalities (Term 2)
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 12 Solve the following system of linear inequalities graphically. x + y ≥ 5 , x – y ≤ 3 First we solve x + y ≥ 5 Lets first draw graph of x + y = 5 Drawing graph Checking for (0,0) Putting x = 0, y = 0 x + y ≥ 5 0 + 0 ≥ 5 0 ≥ 5 which is false Hence origin does not lie in plane x + y ≥ 5. So, we shade the right side of line Now we solve x – y ≤ 3 Lets first draw graph of x – y = 3 Drawing graph Checking for (0,0) Putting x = 0, y = 0 x – y ≤ 3 0 – 0 ≤ 3 0 ≤ 3 which is true Hence origin lies in plane x + y ≥ 5. So, we shade left side of line Hence the shaded region is the solution of the given inequality
Examples
Example 1
Example 2 Important
Example 3
Example 4 Important
Example 5
Example 6
Example 7
Example 8 Important
Example 9
Example 10 Important
Example 11 Important
Example 12 You are here
Example 13 Important
Example 14
Example 15
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Chapter 6 Class 11 Linear Inequalities (Term 2)
Serial order wise
- Ex 6.1
- Ex 6.2
- Ex 6.3
-
Examples
- Miscellaneous
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.