(i) x ≥ 3 ……………………… (1)
First draw the line x = 3 by thick line.
Put x = 0 in (1), we get 0 > 3 is not true.
∴ Solution set of (1) is not containing the origin.
y ≥ 2 ………………….. (2)
Now draw the line y = 2 by thick line
Put y = 0 in (2), we get, 0 > 2 is not true.
∴ Solution set of (2) is not containing the origin.
∴ Shaded region is the required solution region.
(ii) Consider x + 3 ≥ 5 ……………….. (1)
Draw the graph of x + y = 5 by thick line.
It passes through (5, 0) and (0, 5) Join these points put x = 0 and y = 0 in (1), we get 0 + 0 ≥ 5 which is false
∴ Solution set of (1) is not containing the origin.
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) is containing the origin.
Hence, the shaded region represents the solution set of the given system of inequalities
(iii) Consider 2x + y ≥ 6 ……………….(1)
Draw the graph of 2x + y = 6 i.e. x/3 + y/6 = 1 by thick line.
It passes through (3, 0) and (0, 6).
Join these points put x = 0 and y = 0 in (1), we get 0 + 0 ≥ 6 which is false.
∴ Solution set of (1) is not containing the origin.
Consider 3x + 4y ≤ 12 ……………….. (2)
Draw the graph of 3x + 4y ≤ 12 i.e x/4 + y/3 = 1 by thick line.
It passes through (4, 0) and (0, 3).
Join these points put x = 0 and y = 0 in (2), we get 0 + 0 < 12 which is true.
∴ Solution set of (2) is containing the origin.
Hence, the shaded region represents the solution set of the given system of inequalities.
Draw the graph of x+y=4 by thick line.
It passes through (4, 0) and (0,4)
Join these points put x = 0 and y = 0 in (1), we get 0 + 0 > 4 which is false.
∴ Solution set of (1) is not containing the origin.
Consider 2x – y > 0 ………………(2)
Draw the graph of 2x – y = 0 by dotted line.
It passes through (0, 0) and (1,2)
Join these points put x = 2 and y = 1 in (2), we get
2(2) – 1 > 0 which is true.
∴ Solution set of (2) is containing the point (2, 1)
Hence, the shaded region represents the solution set of the given system of in equations.
(v) Consider 2x - y > 1 ………………(1)
Draw the graph of 2x – y = 1 by dotted line
It passes through (1/2, 0) and (0, -1).
Join these points, put x = 0 and y = 0 in (1), we get 0 – 0 > 1 which is false.
∴ Solution set of (1) is not containing the origin.
Consider x – 2y < -1 ……………… (2)
Draw the graph of x – 2y = -1 by dotted line.
It passes through (-1, 0) and Join these points, put x = 0 and y = 0 in (2) we get 0 – 0 < -1 which is false.
∴ Solution set of (2) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of in equations.
