# Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.1 pdf || UP Board

# __Pair of Linear Equations in Two
Variables__

__Pair of Linear Equations in Two Variables__

__Important Points__

- Two lines a
_{1}x + b_{1}y + c_{1}= 0, a_{2}x + b_{2}y + c = 0 Intersecting, Coincident or Parallel lines if,

__Exercise – 3.1__

__Exercise – 3.1__

**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 Aftab's present age = x years

Let the present age of the daughter
= y years

Aftab's age 7 years ago = x – 7
years

and daughter’s age 7 years ago = y –
7 years

x – 7 = 7(y – 7)

x -7 = 7y – 49

x – 7y = -42 ………………(1)

Aftab's present age after 3 years =
x + 3 years

Present age of daughter after 3
years = y + 3 years

x + 3 = 3(y + 3)

x + 3 = 3y + 9

x – 3y = 6 ………………….…(2)

So, in algebraic form,

x – 7y = -42

x – 3y = 6

Now to express graphically three
solutions,

From equation (1),

x – 7y = -42

From equation (2),

x – 3y = 6

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

**Solution****:**

Let cost of one bat = ₹x

Let cost of one ball = ₹y

according to the first condition,

3x + 6y = 3900 ……………. (1)

according to the second condition,

x + 2y = 1300 …………….(2)

So, in algebraic form,

3x + 6y = 3900

x + 2y = 1300

Now to express graphically three
solutions,

From equation (1),

3x + 6y = 3900

x |
300 |
100 |
-100 |

y |
500 |
600 |
700 |

From equation (2),

x + 2y = 1300

x |
300 |
100 |
-100 |

y |
500 |
600 |
700 |

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

**Solution****:**

Let cost of 1 kg apple = ₹x

Let cost of 1 kg of grapes = ₹ y

according to the first condition,

2x + y = 160 ……………. (1)

according to the second condition,

4x + 2y = 300 …………….(2)

So, in algebraic form,

2x + y = 160

4x + 2y = 300

Now to express graphically three
solutions,

From equation (1),,

2x + y = 160

x |
50 |
60 |
70 |

y |
60 |
40 |
20 |

From equation (2),

4x + 2y = 300

x |
70 |
80 |
75 |

y |
10 |
-10 |
0 |

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