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Solve the Following Systems of Linear Inequation Graphically: 2x + 3y ≤ 6, 3x + 2y ≤ 6, X ≥ 0, Y ≥ 0 - Mathematics

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प्रश्न

Solve the following systems of linear inequation graphically:

 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0 

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उत्तर

 Converting the inequations to equations, we obtain:
2x + 3y = 6, 3x + 2y = 6, x = 0, y = 0

2x + 3y =6:  This line meets the x-axis at (3,0) and the y-axis at (0, 2). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation 2x + 3y ≤ 6
So, the portion containing  the origin represents the solution set of the inequation 2x + 3y ≤ 6

3x+2y =6:  This line meets the x-axis at (2, 0) and the y-axis at (0, 3). Draw a thick line joining these points.
We see that the origin (0,0) satisfies the inequation 3x + 2y ≤ 6.
So, the portion containing  the origin represents the solution set of the inequation 3x + 2y ≤ 6

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

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 15 Linear Inequations
Exercise 15.6 | Q 1.1 | पृष्ठ ३०

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