(i) Consider x – 2y ≤ 3 ………….. (1)
Draw the graph of x - 2y = 3 by thick line.
It passes through
Join these points, put x = 0 and y = 0 in (1),
we get 0 – 0 < 3, which is true.
∴Solution of (1) contains the origin.
Consider 3x + 4y > 12 …………. (2)
Draw the graph of 3x + 4y = 12 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 false.
∴ Solution set of (2) does not contain the origin.
Consider y>1 ………….. (3)
Draw the graph of y = 1
clearly solution set of (3) does not contain (0, 0).
Since x > 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.
(ii) Consider 4x + 3y ≤ 60 ……….. (1)
Draw the graph of 4x+3y = 60 by the thick line.
It passes through (15, 0) and (0, 20)
Join these points, put x = 0 and y = 0 in (1),
we get 0 – 0 ≤ 60 ,which is true.
∴ Solution set of (1) contains(0,0)
Consider y ≥2x ……………. (2)
Draw the graph of y = 2x by thick line
It passes through (0, 0) and (1,2)
Join these points,
clearly solution set of (2) is not containing (1,1)
Consider x ≥ 3 ……………. (3)
Draw the graph of x = 3
clearly solution set of (3) does not contain the origin.
Since 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 + 2y ≤ 150 ……………. (1)
Draw the graph of 3x + 2y = 150
It passes through (50, 0) and (0, 75) put x = 0 and y = 0 in (1),
we get 0 +0 < 150, which is true.
∴ Solution set of (1) contains (0, 0)
Consider x + 4y < 80 ……………. (2)
Draw the graph of x + 4y = 80.
It passes through (80, 0) and (0, 20).
Join these points, put x = 0 and y = 0 in (2),
we get 0 + 0 < 80, which is true.
∴ Solution set of (2) contains (0, 0)
Consider x < 15 ……………. (3)
Clearly solution set of (3) does not contain 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.
(iv) Consider x + 2y ≤10 …………(1)
Draw the graph of x+2y=10
It passes through (10, 0) and (0, 5) put x – 0 and y = 0 in (1),
we get 0 + 0 < 10, which is true.
∴ Solution set of (1), contain (0, 0)
Consider x + y > 1 ………… (2)
Draw the graph of x + y = 1
It passes through (1,0) and (0, 1) 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
Consider x – y < 0 ………… (3)
Draw the graph of x – y = 0.
It passes through (0, 0) and (1, 1) put x = 2 and y = 0 in (3)
we get 2 – 0 < 0 which is false
∴ Solution set of (3) does not contain (2, 0)
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.
