Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(c) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(c)

Solve the following systems of linear inequalities graphically.

Question 1.

2x – y ≥ 0, x – 2y ≤ 0, x ≤ 2, y ≤ 2 [Hint: You may consider the point (2, 2) to determine the SR of the first two inequalities.]

Solution:

2x – y ≥ 0

x – 2y ≤ 0

x ≤ 2

y ≤ 2

Step – 1: Let us draw the lines.

2x – y = 0, x – 2y = 0, x = 2, y = 2

2x – y = 0

X | 0 | 1 |

y | 0 | 2 |

x – 2y = 0

X | 0 | 2 |

y | 0 | 1 |

Step – 2: Let us consider point (1, 0) which does not line on any of these lines.

Putting x = 1, y = 0 in the inequations we get

2 ≥ 0 (True)

1 ≤ 0 (False)

1 ≤ 2 (True)

0 ≤ 2 (True)

Point (1, 0) satisfies all inequality except x – 2y < 0.

∴ Thus the shaded region is the solution region.

Question 2.

x – y < 1, y – x < 1

Solution:

x – y < 1

y – x < 1

Step – 1: Let us draw the dotted lines.

x – y = 1 and y – x = 1

x – y = 1 ⇒ y = x – 1

X | 1 | 0 |

y | 0 | -1 |

y – x = 1 ⇒ y = x + 1

X | 0 | -1 |

y | 1 | 0 |

Step – 2: Let us consider the point (0, 0) which does not lie on these lines.

Putting x = 0, y = 0 in the inequations

we get

0 < 1 (True)

0 < 1 (True)

∴ (0, 0) satisfies both the inequations.

∴ Thus the shaded region is the feasible region.

Question 3.

x – 2y + 2 < 0, x > 0

Solution:

x – 2y + 2 < 0, x > 0

Step – 1: Let us draw the dotted line x – 2y + 2 = 0

⇒ y = \(\frac{x+2}{2}\)

X | -2 | 0 |

y | 0 | 1 |

Step – 2: Let us consider the point (1, 0) that does not lie on the lines putting x = 0, y = 0 in the inequation, we get

2 < 0 (false)

1 > 0 (True)

⇒ (1, 0) satisfies x > 0 and does not satisfy x – 2y + 2 < 0.

∴ Thus the shaded region is the solution region.

Question 4.

x – y + 1 ≥ 0, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0

Solution:

x – y + 1 ≥ 0

3x + 4y ≤ 12

x ≥ 0, y ≥ 0

Step – 1: Let us draw the lines.

x – y + 1 = 0

3x + 4y = 12

Now, x – y + 1 = 0 ⇒ y = x + 1

X | 0 | -1 |

y | 1 | 0 |

3x + 4y = 12

X | 4 | 0 |

y | 0 | 3 |

Step – 2: Let us consider the point (0, 0) which does not lie on these lines.

Putting x = 0, y = 0 in the inequations

we get

1 ≥ 0 (True)

0 ≤ 12 (False)

∴ (0, 0) satisfies both the inequations and x > 0, y > 0 is the first quadrant.

∴ Thus the shaded region is the solution region.

Question 5.

x + y > 1, 3x – y < 3, x – 3y + 3 > 0

Solution:

x + y > 1

3x – y < 3

x – 3y + 3 > 0

Step – 1: Let us draw the lines.

x + y = 1

3x – y = 3

x – 3y + 3 = 0

Now x + y = 1

⇒ y = 1 – x

X | 1 | 0 |

y | 0 | 1 |

3x – y = 3

⇒ y = 3x – 3

X | 1 | 0 |

y | 0 | -3 |

x – 3y + 3 = 0

⇒ y = \(\frac{x+3}{3}\)

X | -3 | 0 |

y | 0 | 1 |

Step – 2: Let us consider the point (0, 0) that does not lie on these lines. Putting x = 0, y = 0 in the inequations we get,

0 > 1 (False)

0 < 3 (True)

3 > 0 (True)

Thus (0, 0) satisfies 3x – y < 3 and x – 3y + 3 > 0 but does not satisfy x + y > 1

∴ The shaded region is the solution region.

Question 6.

x > y, x < 1, y > 0

Solution:

x > y, x < 1, y > 0

Step – 1: Let us draw the dotted lines.

Step – 2: Let us consider a point (2, 1) that does not lie on any of the lines.

Putting x = 2, y = 2 in the inequations

we get,

2 > 1 (True)

2 < 1 (False)

1 > 0 (True)

⇒ (2, 1) satisfies x > y and y > 0 but does not satisfy x < 1.

∴ Thus the shaded region is the solution region.

Question 7.

x < y, x > 0, y < 1

Solution:

x < y

x > 0

y < 1

Step – 1: Let us draw the dotted lines.

x = y

x = 0

and y = 1

Step – 2: Let us consider point (1, 0) that does not lie on these lines.

Putting x = 0, y = 0 in the inequations

we get

1 < 0 (False)

1 > 0 (True)

0 < 1 (True)

Clearly (1, 0) satisfies x > 0, y < 1 but does not satisfy x < y.

∴ The shaded region is the solution region.