RD Chapter 15- Linear Inequations Ex-15.1 |
RD Chapter 15- Linear Inequations Ex-15.2 |
RD Chapter 15- Linear Inequations Ex-15.3 |
RD Chapter 15- Linear Inequations Ex-15.4 |
RD Chapter 15- Linear Inequations Ex-15.5 |

Solve the following systems of linear inequations graphically.

(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0

(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0

(iii) x – y ≤ 1, x + 2y ≤ 8, 2x + y ≥ 2, x ≥ 0, y ≥ 0

(iv) x + y ≥ 1, 7x + 9y ≤ 63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0

(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

**Answer
1** :

(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y ≤ 6

So when,

x | 0 | 1 | 3 |

y | 2 | 1.33 | 0 |

3x + 2y ≤ 6

So when,

x | 0 | 1 | 2 |

y | 3 | 1.5 | 0 |

x ≥ 0, y ≥ 0

(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y ≤ 6

So when,

x | 0 | 1 | 3 |

y | 2 | 1.33 | 0 |

x + 4y ≤ 4

So when,

x | 0 | 2 | 4 |

y | 1 | 0.5 | 0 |

x ≥ 0, y ≥ 0

(iii) x – y ≤ 1, x + 2y ≤ 8,2x + y ≥ 2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x – y ≤ 1

So when,

x | 0 | 2 | 1 |

y | -1 | 1 | 0 |

x + 2y≤ 8

So when,

x | 0 | 4 | 8 |

y | 4 | 2 | 0 |

2x + y ≥ 2

So when,

x | 0 | 2 | 1 |

y | 2 | -2 | 0 |

x ≥ 0, y ≥ 0

(iv) x + y ≥ 1, 7x + 9y ≤63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x + y ≥ 1

So when,

x | 0 | 2 | 1 |

y | 1 | -1 | 0 |

7x + 9y ≤ 63

So when,

x | 0 | 5 | 9 |

y | 7 | 3.11 | 0 |

x ≤ 6, y ≤ 5 and x ≥0, y ≥ 0

(v) 2x + 3y ≤ 35, y ≥ 3, x≥ 2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

2x + 3y ≤ 35

So when,

x | 0 | 5 | 17.5 |

y | 11.667 | 8.33 | 0 |

y ≥ 3, x ≥ 2, x ≥ 0, y≥ 0

Show that the solution set of the following linear inequations is empty set:

(i) x – 2y ≥ 0, 2x – y ≤ –2, x ≥ 0, y ≥ 0

(ii) x + 2y ≤ 3, 3x + 4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

**Answer
2** :

(i) x – 2y ≥ 0, 2x – y ≤–2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of the inequality,

x – 2y ≥ 0

So when,

x | 0 | 2 | 4 |

y | 0 | 1 | 2 |

2x – y ≤ –2

So when,

x | 0 | 1 | -1 |

y | 2 | 4 | 0 |

x ≥ 0, y ≥ 0

The lines do notintersect each other for x ≥ 0, y ≥ 0

Hence, there is nosolution for the given inequations.

(ii) x + 2y ≤ 3, 3x +4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

x + 2y ≤ 3

So when,

x | 0 | 1 | 3 |

y | 1.5 | 1 | 0 |

3x + 4y ≥ 12

So when,

x | 0 | 2 | 4 |

y | 3 | 1.5 | 0 |

y ≥ 1, x ≥ 0, y ≥ 0

Find the linear inequations for which the shaded area in Fig. 15.41 isthe solution set. Draw the diagram of the solution set of the linearinequations.

**Answer
3** :

Here, we shall applythe concept of a common solution area to find the signs of inequality by usingtheir given equations and the given common solution area (shaded part).

We know that,

If a line is in theform ax + by = c and c is positive constant. (In case of negative c, therule becomes opposite), so there are two cases which are,

If a line is above theorigin:

(i) If the shaded areais below the line then ax + by < c

(ii) If the shadedarea is above the line then ax + by > c

If a line is below theorigin then the rule becomes opposite.

So, according to therules

**Answer
4** :

Here, we shall applythe concept of a common solution area to find the signs of inequality by usingtheir given equations and the given common solution area (shaded part).

We know that,

If a line is in theform ax + by = c and c is positive constant.

So, according to therules