# Solve the following system of inequalities graphically:

3x + 2 y ≤ 150, x + 4 y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0

**Solution:**

3x + 2 y ≤ 150 ....(1)

x + 4 y ≤ 80 ....(2)

x ≤ 15 ....(3)

The graph of the lines 3x + 2 y = 150 , x + 4 y = 80 and x = 15 , are drawn in the figure below.

Inequality 3x + 2 y ≤ 150 represents the region below the line, 3x + 2 y = 150 (including the line 3x + 2 y = 150)

Inequality x + 4y ≤ 80 represents the region below the line, x + 4 y = 80 (including the line x + 4 y = 80).

Inequality x ≤ 15 represents the region on the left hand side the line, x = 15 (including the line x = 15).

Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities as follows:

NCERT Solutions Class 11 Maths Chapter 6 Exercise 6.3 Question 14

## Solve the following system of inequalities graphically:3x + 2 y ≤ 150, x + 4 y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0

Linear inequations 3x + 2 y ≤ 150, x + 4 y ≤ 80, x ≤ 15, y ≥ 0, x ≥ is given. We have found that x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities

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