4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

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Solution

4x + 3y ≤ 60 … (1)

y ≥ 2x … (2)

x ≥ 3 … (3)

The graph of the lines, 4x + 3y = 60, y = 2x, and x = 3, are drawn in the figure below.

Inequality (1) represents the region below the line, 4x + 3y = 60 (including the line 4x + 3y = 60). Inequality (2) represents the region above the line, y = 2x (including the line y = 2x). Inequality (3) represents the region on the right hand side of the line, x = 3 (including the line x = 3).

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

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Solve the following systems of inequalities graphically:

$4 x + 3 y \leq 60, y \geq 2 x, x \geq 3, x, y \geq 0$

[[Q]]

Solve the following system of inequalities graphically:

4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

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4 x - 3 y = 0

2 x - y = 3

2 x + 3 y = 0

4 x - 3 y = 0

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x - y \geq 0 - x + 2 y \geq 2 x \geq 3 y \leq 4 x, y \geq 0

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