Solve the following system of inequalities graphically. (i) x ≥ 3, y ≥ 2 (ii) x + y ≥ 5,x - y ≤ 3 (iii) 2x + y ≥ 6, 3x + 4y ≤ 12

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Solve the following system of inequalities graphically.

(i) x ≥ 3, y ≥ 2

(ii) x + y ≥ 5,x - y ≤ 3

(iii) 2x + y ≥ 6, 3x + 4y ≤ 12

(iv) x + y ≥ 4, 2x - y > 0

(v) 2x - y > 1, x - 2y < - 1

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(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. 