Ex 6.3, 12 - Chapter 6 Class 11 Linear Inequalities (Term 2)
Last updated at Feb. 17, 2020 by Teachoo
Ex 6.3
Last updated at Feb. 17, 2020 by Teachoo
Ex 6.3, 12 Solve the following system of inequalities graphically: x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1 First we solve x – 2y ≥ 3 Lets first draw graph of x – 2y = 3 Putting x = 0 in (1) 0 – 2y = 3 −2y = 3 y = ( 3)/( −2) y = –1.5 Putting y = 0 in (1) x – 2(0) = 3 x – 0 = 3 x = 3 Drawing graph Checking for (0,0) Putting x = 0, y = 0 x – 2y ≤ 3 0 - 2(0) ≤ 3 0 ≤ 3 which is true Hence origin lies in plane x – 2y ≤ 3 So, we shade right side of line Points to be plotted are (0, 6) , (3, 0) Now we solve 3x + 4y ≥ 12 Lets first draw graph of 3x + 4y = 12 Putting x = 0 in (1) 3(0) + 4y = 12 0 + 4y = 12 4y = 12 y = 12/4 y = 3 Putting y = 0 in (1) 3x + 4(0) = 12 3x + 0 = 12 3x = 12 x = 12/3 x = 4 Points to be plotted are (0, 3) , (4, 0) Drawing graph Checking for (0,0) Putting x = 0, y = 0 3x + 4y ≥ 12 3(0) + 2(0) ≥ 12 0 ≥ 12 which is false Hence origin does not lie in plane 3x + 2y > 6 . So, we shade rights side of line Points to be plotted are (0, 1), (–1, 1), (4, 1) Also, y ≥ 1 So, for all values of x, y = 2 Also, x ≥ 0 So, the shaded region will lie on the right side of y axis Hence the shaded region represents the given inequality