Ex 6.3, 10 - Solve: 3x + 4y <= 60, x + 3y <= 30, x > 0, y > 0

Ex 6.3,  10 - Chapter 6 Class 11 Linear Inequalities - Part 2
Ex 6.3,  10 - Chapter 6 Class 11 Linear Inequalities - Part 3 Ex 6.3,  10 - Chapter 6 Class 11 Linear Inequalities - Part 4 Ex 6.3,  10 - Chapter 6 Class 11 Linear Inequalities - Part 5

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Question 11 Solve the following system of inequalities graphically: 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6 First we solve 2x + y ≥ 4 Lets first draw graph of 2x + y = 4 Putting x = 0 in (1) 2(0) + y = 4 0 + y = 4 y = 4 Putting y = 0 in (1) 2x + (0) = 4 2x = 4 x = 4/2 x = 2 Points to be plotted are (0, 4), (2, 0) Drawing graph Checking for (0,0) Putting x = 0, y = 0 2x + y ≥ 6 2(0) + (0) ≥ 6 0 ≥ 6 which is false So, we shade right side of line Hence origin does not lie in plane 2x + y ≥ 6 First we solve x + y ≤ 3 Lets first draw graph of x + y = 3 Putting y = 0 in (1) x + 0 = 3 x = 3 Putting x = 0 in (1) 0 + y = 3 y = 3 Points to be plotted are (0, 3), (3, 0) Drawing graph Checking for (0,0) Putting x = 0, y = 0 x + y ≤ 3 0 + 0 ≤ 3 0 ≤ 3 which is true So, we shade left side of line Hence origin does not lie in plane x + y ≤ 3 Now we solve 2x – 3y ≤ 6 Lets first draw graph of 2x – 3y = 6 Putting x = 0 in (1) 3(0) – 3y = 6 0 – 3y = 6 –3y = 6 y = ( 6)/(−3) y = –2 Putting y = 0 in (1) 2x – 3(0) = 6 2x – 0 = 6 2x = 6 x = ( 6)/2 x = 3 Drawing graph Checking for (0,0) Putting x = 0, y = 0 2x - 3y ≤ 6 2(0) – 3(0) ≤ 6 0 ≤ 6 which is true So, we shade left upper side of line 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.