(i) Consider x+y≤6 ………………(1)
Draw the graph of x + y = 6 by thick line. It passes through (6, 0) and (0, 6)
Join these points put x = 0 and y = 0 in (1), we get 0 + Q < 6 which is true.
∴ Solution set of (1) is containing the origin.
Consider x + y ≥ 4 ……………… (2)
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 (2), we get 0 + 0 ≥ 4 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.
(ii) Consider 2x + y ≥ 8 ……………… (1)
Draw the graph 2x + y = 8 by thick line. It passes through (4, 0) and (0, 8)
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 ≥ 8 which is false
∴ Solution set of (1) is not containing the origin.
Consider x + 2y > 10 ……………… (2)
Draw the graph x + 2y = 10 by thick line It passes through (10, 0) and (0, 5).
Join these points, put x = 0 and y = 0 in (2), we get 0 + 0 > 10 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.
(iii) Consider 8x + 3y ≤100 ……………… (1)
Draw the graph of 8x + 3y = 100 by thick line
It passes through (25/2, 0) and (0, 100/3)
Join these points, put x = 0 and y = 0 in (1),
we get 0+0 ≤ 100 which is true.
∴ Solution set of (1) contains the origin.
Since x ≥ 0, y ≥ 0, every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of in equations.
(iv) Consider 5x + 4y ≤ 40 ………………… (1)
Draw the graph of 5x + 4.y = 40 by thick line.
It passes through (8, 0) and (0, 10).
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 ≤ 40
which is true.
Solution set of (1) contains the origin.
Consider x ≥ 2 ………………… (2)
Draw the graph of x= 2.
Clearly solution set of (2) is not containing the origin.
Consider y ≥ 3 ………………… (3)
Draw the graph of y = 3
Clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of in equations.
(v) Consider 3x + 2y ≤ 12 ………………… (1)
Draw the graph of 3x + 2y = 12 by thick line.
It passes through (4, 0) and (0, 6).
Join these points put x = 0 and y = 0 in (1),
we get 0+0≤12 which is true.
∴ Solution set of (1) contains the origin
Consider x ≥1 …………….. (2)
Draw the graph of x= 1
Clearly solution set of (2) is not containing the origin.
Consider y ≥ 2 … (3)
Clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of in equations.
(vi) Consider 5x + 4y ≤ 20 …………(1)
Draw the graph of 5x + 4y = 20 by thick line It passes through (4, 0) and (0, 5).
Join these points, put x = 0 and y = 0 in (1),
we get 0 + 0 < 20 which is true.
∴ Solution set of (1) contains the origin.
Consider x > 1 …………….. (2)
Draw the graph of x = 1 clearly solution set of (2) is not containing the origin.
Consider y > 2 …………….. (3)
Draw the graph of y = 2 clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of in equations.
