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Example 14 - Chapter 6 Class 11 Linear Inequalities (Term 2)

Last updated at July 24, 2019 by

Example 14 - Solve 8x + 3y <= 100 graphically - Chapter 6 - Graph - 2 or more Equation

Example 14 - Chapter 6 Class 11 Linear Inequalities - Part 2
Example 14 - Chapter 6 Class 11 Linear Inequalities - Part 3

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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.