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प्रश्न
Find graphical solution for following system of linear inequations :
3x + 2y ≤ 180; x+ 2y ≤ 120, x ≥ 0, y ≥ 0
Hence find co-ordinates of corner points of the common region.
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उत्तर
We plot the lines l1 = 3x + 2y = 180 and l2 = x + 2y = 120
For l1
| x | y | Points |
| 60 | 0 | A(60, 0) |
| 0 | 90 | B(0, 90) |
For l2
| x | y | Points |
| 120 | 0 | C( 120, 0 ) |
| 0 | 60 | D( 0, 60 ) |
The shaded region is the quadrilateral OAPD and it is the common solution set, where P is the point of intersection of l1 and l2 Solving
3x + 2y = 180
x + 2y = 120
- - -
∴ 2x = 60
∴ x = 30
∴ 30 + 2y = 120
∴ y = 45
P ( 30, 45 )
Corner points 0(0, 0), A(60, 0). P(30, 45), D(O, 60)
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