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

 Pair of Linear Equations in Two Variables


Important Points

  • Two lines a1x + b1y + c1 = 0, a2x + b2y + c = 0 Intersecting, Coincident or Parallel lines if,


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