Ex 6.3, 15 - Chapter 6 Class 11 Linear Inequalities
Last updated at Feb. 17, 2020 by Teachoo
Last updated at Feb. 17, 2020 by Teachoo
Transcript
Ex 6.3, 15 Solve the following system of inequalities graphically: x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0 First we solve x + 2y ≤ 10 Lets first draw graph of x + 2y = 10 …(1) Putting x = 0 in (1) 0 + 2y = 10 2y = 10 y = 10/2 y = 5Putting y = 0 in (1) x + 2(0) = 10 x + 0 = 10 x = 10 Points to be plotted are (0,5) , (10,0) Now we solve x + y ≥ 1 Lets first draw graph of x + y = 1 Putting y = 0 in (1) x + 0 = 1 x = 1 Putting x = 0 in (1) 0 + y = 1 y = 1 Points to be plotted are (0,1) , (1,0) Drawing graph Checking for (0,0) Putting x = 0, y = 0 x + y ≥ 1 0 + 0 ≥ 1 0 ≥ 1 which is false Hence origin does not lie in plane x + y ≥ 1 So, we shade right upper side of line Now we solve x – y ≤ 0 Lets first draw graph of x – y = 0 Putting x = 0 in (3) 0 – y = 0 −y = 0 y = 0 Putting y = 2 in (2) x – 2 = 0 x = 0 + 2 x = 2 Points to be plotted are (0,0), (2,2) Drawing graph Checking for (10,0) Putting x = 10, y = 0 x – y ≤ 0 10 – 0 ≤ 0 10 ≤ 0 which is false. Hence (10,0) does not lie in plane x > y So, we shade left side of line Also, given x ≥ 0, y ≥ 0 So, shaded region will lie in first quadrant Hence the shaded region represents the given inequality
Ex 6.3
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