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प्रश्न
Find the graphical solution of the following system of linear inequations:
2x + 3y ≥ 12, – x + y ≤ 3, x ≤ 4, y ≥ 3
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उत्तर
The given equations in system are 2x + 3y ≥ 12, – x + y ≤ 3, x ≤ 4, y ≥ 3
Consider the equations:
2x + 3y = 12
| x | 6 | 0 |
| y | 0 | 4 |
The two points on the coordinate axes are (6, 0) and (0, 4).
Origin (0, 0) does not satisfy the inequation as
2(0) + 3(0) ≥ 12
i.e. 0 ≥ 12 which is not true.
∴ Solution set is always from the origin
– x + y = 3
| x | – 3 | 0 |
| y | 0 | 3 |
The points on the coordinate axis are
(− 3, 0) and (0, 3)
Origin (0, 0) satisfies the inequation as
– 0 + 0 ≤ 3 i.e. 0 ≤ 3 which is true.
The solution set is towards the origin.
For x ≤ 4, consider x = 4. Equation of line which passes through the point (4, 0) and since 0 ≤ 4 implies solution is towards the origin.
For y ≥ 3, consider y = 3 which is the equation of a line passing through (0, 3) and parallel to the X-axis.
As 0 ≥ 3 is not possible, the solution set is away from the origin,
The shaded portion represents the graphical solution.
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