Example 14 - Chapter 6 Class 11 Linear Inequalities (Term 2)
Last updated at July 24, 2019 by Teachoo
Transcript
Example 14 Solve the following system of inequalities : 8x + 3y ≤ 100 , x ≥ 0 , y ≥ 0 First we solve 8x + 3y ≥ 100 Lets first draw graph of 8x + 3y = 100 Drawing graph Checking for (0,0) Putting x = 0, y = 0 8x + 3y ≤ 100 8(0) + 3(0) ≤ 100 0 ≤ 100 which is true So, we shade left side of line Hence origin lies in plane 8x + 3y ≤ 100 Also x ≥ 0 & y ≥ 0 So, our shaded region will lie in first quadrant. Hence the shaded region represents the given inequality
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Chapter 6 Class 11 Linear Inequalities (Term 2)
Serial order wise
- Ex 6.1
- Ex 6.2
- Ex 6.3
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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.