**1. Aftab tells his daughter, “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.” (Isn’t this interesting?) Represent this situation algebraically and graphically.**

Solution:

Let the present age of Aftab be x and the present age of his daughter = y

Seven years ago, Age of Aftab = x − 7

Age of his daughter = y − 7

According to the question,

(x – 7) = 7(y – 7)

x – 7 = 7y – 49

x – 7y = -42 ———–(1)

Three years then, Age of Aftab = x + 3

Age of his daughter = y + 3

According to the question

(x + 3) = 3(y + 3)

x = 3 = 3y + 9

x – 3y = 6 ———-(2)

Thus, the algebraic expression is

x – 7y = -42

x – 3y = 6

For, x – 7y = -42

x = -42 + 7y

x | -7 | 0 | 7 |

y | 5 | 6 | 7 |

For x – 3y = 6

x = 6+3y

The solution table is

x | 6 | 3 | 0 |

y | 0 | -1 | -2 |

The graphical representation of the

**2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. **

Solution:

Let the cost of a bat be Rs x and the cost of a ball = Rs y

According to the question, the algebraic representation is

3x + 6y = 3900

x + 2y = 1300

For 3x + 6y = 3900

x = ^{(3900-6y)}/_{3 }

x | 300 | 100 | -100 |

y | 500 | 600 | 700 |

For, x + 2y = 1300

x = 1300 -2y

The solution table is

x | 300 | 100 | -100 |

y | 500 | 600 | 700 |

The graphical representation is,

**3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.**

Solution:

Let the cost of 1 kg of apples be Rs x and the cost of 1 kg of grapes = Rs y

According to the question, the algebraic representation is

2x + y = 160

4x + 3y = 300

For 2x + y = 160

y = 160 – 2x

The solution table is

For 4x + 2y = 300

x | 50 | 60 | 70 |

y | 60 | 40 | 20 |

y = ^{(300-4x)}/_{2}

The solution table is

x | 70 | 80 | 75 |

y | 10 | -10 | 0 |

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This is an interesting way to do it. Normally, I just solve the simultaneous equation

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very much like the idea of showing algebraic and geometric solutions at the same time …

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