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**Ques 1: Aftab tells his daughter, â€˜â€˜Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.â€™â€™ (Isnâ€™t this interesting ?) Represent this situation algebraically and graphically.****Sol: **At present:

Let Aftabâ€™s age = x years

His daughterâ€™s age = y years

Seven years ago:

Aftabâ€™s age = (x - 7) years

Daughterâ€™s age = (y - 7) years

According to the condition,

[Aftabâ€™s age] = 7 [His daughterâ€™s age]

â‡’ [x - 7] = 7 [y - 7]

â‡’ x - 7 = 7y - 49

â‡’ x - 7y - 7 + 49 = 0

â‡’ x - 7y + 42 = 0 ...(1)

After three years:

Aftabâ€™s age = (x + 3) years

His daughterâ€™s = (y + 3) years

According to the condition,

[Aftabâ€™s age] = 3 [His daughterâ€™s age]

â‡’ [x + 3] = 3 [y + 3]

â‡’ x + 3 = 3y + 9

â‡’ x - 3y + 3 - 9 = 0

â‡’ x - 3y - 6 = 0 ...(2)

Graphical Representation of equation (1) and (2):

From equation (1), we have:

x - 7y + 42 = 0 â‡’

x | 0 | 7 | 35 |

y | 6 | 7 | 11 |

(x, y) | (0. 6) | (7, 7) | (35, 11) |

From equation (2), we have

x - 3y - 6 = 0 â‡’

x | 0 | 6 | 3 |

y | -2 | 0 | 8 |

(x, y) | (0. -2) | (6, 0) | (30, 8) |

Plotting the points (0, 6), (7, 7) and (35, 11), we obtain a straight line l_{1}.

Plotting the points (0, - 2), (6, 0), (30, 8), we obtain a straight line l_{2}. The lines l_{1} and l_{2} intersect at (42, 12).**Ques ****2: The coach of a cricket team buys 3 bats and 6 balls for â‚¹ 3900. Later, she buys another bat and 3 more balls of the same kind for â‚¹ 1300. Represent this situation algebraically and geometrically.Sol: **Let the cost of a bat = â‚¹ x

And the cost of a ball = â‚¹ y

âˆ´ Cost of 3 bats = â‚¹ 3x

Cost of 6 balls = â‚¹ 6y

Again,

Cost of 1 bat = â‚¹ x

Cost of 2 balls = â‚¹ 2y

Algebraic representation:

[Cost of 3 bats] + [Cost of 6 balls] = â‚¹ 3900

â‡’ 3x + 6y = 3900

â‡’ x + 2y = 1300 ...(1)

Also,

[Cost of 1 bat] + [Cost of 3 balls] = â‚¹ 1300

â‡’ x + 3y = 1300 ...(2)

Thus, (1) and (2) are the algebraic representation of the given situation.

We have, for equation (1):

x + 2y = 1300 or y = 1300-x/2

x | 0 | 1300 | 100 |

y | 650 | 0 | 600 |

(x, y) | (0, 650) | (1300, 0) | (100, 600) |

â‡’ and for equation (2):

x + 3y = 1300 â‡’

x | 400 | 100 | 1000 |

y | 300 | 400 | 100 |

(x, y) | (400, 300) | (100, 400) | (1000, 000) |

Plotting the points (0, 650), (100, 600) and (0, 1300), we get a straight line l_{1} and plotting points (100, 400), (400, 300) and (1000, 100), we get another straight line l_{2}. The straight lines l_{1} and l_{2} are the graphical representation of the given situation. We also see from the obtained figures that the straight lines representing the two equations intersect at (1300, 0).**Ques ****3: The cost of 2 kg of apples and 1 kg of grapes on a day was found to be â‚¹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is â‚¹ 300. Represent the situation algebraically and geometrically.Sol: **Let the cost of 1 kg of apples = â‚¹ x

And the cost of 1 kg of grapes = â‚¹ y

Algebraic Representation:

2x + y = 160 ...(1)

and 4x + 2y = 300 â‡’ 2x + y = 150 ...(2)

Graphical Representation:

From equation (1), we have

2x + y = 160 â‡’ y = 160 - 2x

x | 50 | 40 | 30 |

y | 60 | 80 | 100 |

(x, y) | (50, 60) | (40, 80) | (30, 100) |

From the equation (2), we have:

2x + y =150 â‡’ y = 150 - 2x

x | 50 | 30 | 25 |

y | 50 | 90 | 100 |

(x, y) | (50, 50) | (30, 90) | (25, 100) |

Plotting the points (30, 100), (40, 80) and (50, 60), we get a straight line l_{1}.

Also, joining the points (50, 50), (30, 90) and (25, 100), we get a straight line l_{2}.

The straight lines l_{1} and l_{2} are the graphical representation of the equations 1 and 2 respectively. The lines are parallel.

**Graphical Method (Solution of a Pair of Linear Equations)**

Since, the general form of a pair of linear equations is:

a_{1} x + b_{1} y + c_{1} = 0 ...(1)

and a_{2} x + b_{2} y + c_{2} = 0 ...(2)

When we draw their graphs on the same graph paper then we get two straight lines.

âˆ´ Let us suppose that AB and CD represent the graphs of (1) and (2) respectively. Now, three cases are possible:**Case-I: When the lines AB and CD intersect at a point (m, n).**

In this case the given pair of linear equations has a unique solution. The solution is given by:

x = m; y = n

Such pair of linear equations is called consistent system of linear equations.**Case-II: When the lines AB and CD are parallel.**

In this case there is no common point, i.e. There is no solution of the given system of linear equations. Such pair of linear equations is called inconsistent system.**Case-III: When lines AB and CD are coincident.**

In this case every point is a solution i.e., the system has an infinite number of solutions.

Such pair of linear equations is called consistent and dependent.

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