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Example 13 - Solve 5x + 4y <= 40, x >= 2, y >= 3 graphically

Example  13 - Chapter 6 Class 11 Linear Inequalities - Part 2
Example  13 - Chapter 6 Class 11 Linear Inequalities - Part 3

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Transcript

Example 13 Solve the following system of inequalities graphically: 5x + 4y ≤ 40, x ≥ 2, y ≥ 3 First we solve 5x + 4y ≤ 40 Lets first draw graph of 5x + 4y = 40 Putting x = 0 in (1) 5(0) + 4y = 40 0 + 4y = 40 4y = 40 y = 40/( 4) y = 10 Putting y = 0 in (1) 5x + 4(0) = 40 5x + 0 = 40 5x = 40 x = 40/( 5) x = 8 Points to be plotted are (0,10), (8,0) Drawing graph Checking for (0,0) 5x + 4y ≤ 40 Putting x = 0, y = 0 5(0) + 4(0) ≤ 40 0 ≤ 40 which is true Hence origin lies in plane 5x + 4y ≤ 40 So, we shade left side of line Points to be plotted are (0, 3) , (3, 3) , (6, 3) Points to be plotted are (0, 3) , (3, 3) , (6, 3) Points to be plotted are (2, 0) , (2, 3) , (2, 6) 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.