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Solve the Following Systems of Linear Inequations Graphically: 2x + 3y ≤ 35, Y ≥ 3, X ≥ 2, X ≥ 0, Y ≥ 0 - Mathematics

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प्रश्न

Solve the following systems of linear inequations graphically:

2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0 

उत्तर

 Converting the inequations to equations, we obtain:
2x + 3y = 35, x = 0, y = 0

2x + 3y = 35:  This line meets the x-axis at (17.5, 0) and the y-axis at (0, 35/3). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation 2x + 3y ≤ 35 So, the portion containing  the origin represents the solution set of the inequation 2x + 3y ≤ 35

x = 2:  This line is parallel to the x-axis at a distance 2 units from it.
We see that the origin (0, 0) does not satisfy the inequation x\[\geq\]  2 So, the portion that does not contain  the origin represents the solution set of the inequation x\[\geq\]2

y = 3:  This line is parallel to the y-axis at a distance 3 units from it.
We see that the origin (0, 0) does not satisfies the inequation y ≥ 3 So, the portion opposite to the origin represents the solution set of the inequation y ≥ 3

Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
Hence, the shaded region in the figure represents the solution set of the given set of inequations.

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पाठ 15: Linear Inequations - Exercise 15.6 [पृष्ठ ३०]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 15 Linear Inequations
Exercise 15.6 | Q 1.5 | पृष्ठ ३०

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