# Solve the following system of inequalities graphically. (i) x + y ≤ 9, y > x, x ≥ 0 (ii) x + 2y ≤ 8, 2x + y ≤ 8, x ≥ 0, y ≥ 0

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Solve the following system of inequalities graphically.

(i) x + y ≤ 9, y > x, x ≥ 0

(ii) x + 2y ≤ 8, 2x + y ≤ 8, x ≥ 0, y ≥ 0

(iii) 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0

(iv) 2x + y ≥ 4, x + y ≤ 3, 2x - 3y ≤ 6

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(i) Consider x + y ≤ 9 …………….. (1)

Draw the graph of x + y = 9 by thick line.

It passes through (9, 0) and (0, 9).

Join these points, put x = 0 and y = 0 in (1),

we get 0 + 0 < 9 which is true.

Consider y > x …………….. (2)

Draw the graph of y = x by dotted line.

Clearly solution set of (2) does not contain (1,0).

Since x ≥ 0, every point in the shaded region in the first quadrant, including the points on the line x + y = 9, excluding the points on the line y ≥ x and y-axis, represents the solution of the given system of in equations.

(ii) Consider x + 2y ≤ 8 …………….. (1)

Draw the graph of x + 2y = 8 by thick line.

It passes through (8, 0) and (0,4).

Join these points, put x = 0 and y = 0 in (1),

we get 0 + 0 < 8 which is true.

Solution set of (1) is containing the origin.

Consider 2x + y≤8 …………….. (2)

Draw the graph of 2x + y = 8 by thick line.

It passes through (4, 0) and (0, 8)

Join these points, put x = 0 and y = 0 in (2),

we get 0 + 0 < 8 is true.

Solution set of (2) is containing the origin.

Since x > 0, y > 0, every point in the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of in equations.

(iii) Consider 3x + 4y ≤ 60 …………. (1)

Draw the graph of 3x + 4y = 60 by thick line.

It passes through (20, 0) and (0, 15).

Join these points, put x = 0 and y = 0 in (1),

we get 0 + 0 < 60 is true.

∴ Solution set of (1) contains (0, 0).

Consider x + 3y < 30 ………………… (2)

Draw the graph of x + 3y = 30.

It passes through (30,0) and (0,10)

Join these points, put x = 0 and y = 0 in (2),

we get 0 + 0 ≤ 30 is true.

∴ Solution set of (2) contains (0, 0).

Since x ≥ 0, y ≥ 0 every point in the shaded region in the quadrant, including the points on the lines, represents the solution of the given solution of in equations.

(iv) Consider 2x + y > 4  …………. (1)

Draw the graph of 2x + y = 4 by thick line

It passes through (2, 0) and (0, 4)

Join these points, put x = 0 and y = 0 in (1),

we get 0 + 0 ≥ 4 is not true.

∴ Solution set of (1) does not contain (0, 0).

Consider x + y ≤ 3 …………. (2)

Draw the graph of x + y = 3, by thick line.

It passes through (3, 0) and (0, 3).

Join these points, put x = 0 and y = 0 in (2),

we get 0 + 0 < 3, which is true

Solution set of (2) contains (0, 0)

Consider 2x-3y≤ 6 …………. (3)

Draw the graph of 2x – 3y = 6 by thick line

It passes through (3, 0) and (0, -2)

Join these points, put x = 0 and y = 0 in (3),

we get 0 – 0 < 6, which is true.

∴ Solution set of (3) contains (0, 0)

Hence, the shaded region represents the solution set of the given system of in equations.

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