Ex 6.3, 13 - Chapter 6 Class 11 Linear Inequalities (Term 2)
Last updated at Sept. 25, 2018 by Teachoo
Last updated at Sept. 25, 2018 by Teachoo
Transcript
Ex6.3, 13 Solve the following system of inequalities graphically: 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0 Now we solve 4x + 3y ≤ 60 Lets first draw graph of 4x + 3y = 60 Drawing graph Checking for (0,0) Putting x = 0, y = 0 4x + 3y ≤ 60 4(0) + 3(0) ≤ 60 0 ≤ 60 which is true Hence origin lies in plane 3x + 4y ≤ 60 So, we shade left side of line Now we solve y ≥ 2x Lets first draw graph of y = 2x Drawing graph Checking for (0,15) Putting x = 0, y = 15 y ≥ 2x 15 ≥ 2(0) 15 ≥ 0 which is true Hence (0, 15) lies in plane y ≥ 2x So, we shade left side of line Also, x ≥ 3 So, for all values of y, x = 3 Given x ≥ 3, so we shade right side of line Also, given x , y ≥ 0 So, the shaded region will line in 1st quadrant Hence the shaded region represents the given inequality
Ex 6.3
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