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

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

Solve graphically

Question 1.

x < y

Solution:

x < y

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

X | 0 | 1 |

y | 0 | 1 |

Step – 2: Let us take a point say (1, 0) which is not on the line. Putting x = 1, y = 0 in the equation we get 1 < 0 (false).

⇒ (1, 0) does not satisfy the inequality.

⇒ The solution is the half-plane that does not contain (1, 0)

Step – 3: The shaded region is the solution region.

Question 2.

3x + 4y ≥ 12

Solution:

3x + 4y ≥ 12

Step – 1: Let us draw the line 3x + 4y = 12

X | 4 | 0 |

y | 0 | 3 |

Step – 2: Let us consider the point (0, 0) which is not on the line. Putting x = 0, y = 0 in the inequality we have 0 ≥ 12 (false).

∴ (0, 0) does not satisfy the inequality.

⇒ The half-plane that does not contain (0, 0) is the solution region.

Step – 3: The shaded region is the solution region.

Question 3.

x – y > 0

Solution:

x – y > 0

Step – 1: Let us draw the dotted line x = y.

X | 0 | 1 |

y | 0 | 1 |

Step – 2: Let us consider (1, 0) which is not on the line.

Putting x = 1, y = 0 in the inequation we get 1 > 0 (True)

⇒ (1, 0) satisfies the inequation.

⇒ The half-plane containing (1, 0) is the solution region.

Question 4.

x + 2y – 5 ≤ 0

Solution:

x + 2y – 5 ≤ 0

Step – 1: Let us draw the line x + 2y – 5 = 0

⇒ y = \(\frac{5-x}{2}\)

X | 5 | 1 |

y | 0 | 2 |

Step – 2: Let us consider the point (0, 0) which does not lie on the line putting x = 0, y = 0 in the inequation we get – 5 < 0 (True).

⇒ The point satisfies the inequation.

⇒ The half-plane containing (0, 0) is the solution region.

Step – 3: The shaded region is the solution region.

Question 5.

7x – 4y < 14

Solution:

7x – 4y < 14

Step – 1: Let us draw the dotted line 7x – 4y = 14

X | 2 | 6 |

y | 0 | 7 |

Step – 2: Let us consider the point (0, 0) which does not lie on the line. Putting x = 0, y = 0 in the inequality we get 0 < 14 (True).

⇒ (0, 0) satisfies the inequation.

⇒ The half-plane including (0, 0) is the solution region.

Step – 3: The solution region is the shaded region.

Question 6.

x + 8y + 10 > 0

Solution:

x + 8y + 10 > 0

Step – 1: Let us draw the dotted line x + 8y + 10 = 0

X | -10 | -2 |

y | 0 | -1 |

Step – 2: Let us consider the point (0, 0) which does not lie on the line. Putting x = 0, y = 0 in the inequation we get 10 > 0 (True)

⇒ (0, 0) satisfies the inequality.

⇒ The half-plane containing origin is the solution region.

Step – 3: The solution region’ is the shaded region.

Question 7.

5x + 6y < 12

Solution:

5x + 6y < 12

Step – 1: Let us draw the dotted line 5x + 6y = 12

x | -6 | 0 | 6 |

y | 7 | 2 | -3 |

Step – 2: Putting x = 0, y = 0 in the equation we get, 0 < 12 (True)

⇒ The (0, 0) satisfies the equation.

Step – 3: The shaded region is the required solution.

Question 8.

– 3x + y > 0

Solution:

Step – 1: Let us draw the dotted line – 3x + y = 0

x | 0 | 1 | -1 |

y | 0 | 3 | -3 |

Step – 2: Putting x – 1, y = 0 in the equation we get, – 3 > 0 (false)

∴ Point (1, 0) does not satisfy the in the equation.

Step – 3: The shaded half-plane is the solution.

Question 9.

3x + 8y > 24

Solution:

Step- 1: Let us draw the dotted graph of 3x + 8y = 24

x | 8 | 0 |

y | 0 | 3 |

Step- 2: Putting x = 0, y- 0 we get, 0 > 24 (false)

∴ 0, (0, 0) does not satisfy the in equality.

Question 10.

x + y > 1

Solution:

Step – 1: Let us draw the graph x + y = 1

x | 1 | 0 |

y | 0 | 1 |

Step – 2: Putting x = 0, y = 0 in the equation we get, 0 > 1 (false)

∴ 0 (0, 0) does not satisfy the in equation

Step – 3: The shaded region is the solution.

Question 11.

x ≤ 0

Solution:

Step – 1: Let us draw the graph x = 0

Step – 2: Putting x = – 1 we get, – 1 ≤ 0 (True)

Thus, the shaded region is the solution.

Question 12.

y > 5

Solution:

Step – 1: Let us draw the dotted graph of y = 5

x | 0 | 1 | -1 |

y | 5 | 5 | 5 |

Step- 2: Putting x = 0, y = 0 we have 0 > 5 (false)

we 0(0, 0) does not satisfy. the inequality.

Step – 3: The shaded region is the solution.