# Pair of Linear Equations in two Variables Exercise 3.1

Pair of Linear Equations in two Variables Exercise 3.1

Page No: 44

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.

Let present age of Aftab be       x
And, present age of daughter is represented by y
Then Seven years ago,
Age of Aftab              = -7
Age of daughter       = y-7
According to the question,
(- 7)  = 7 (– 7 )
– 7    = 7 – 49
x- 7 = - 49 + 7
– 7y = - 42                          …(i)
x         = 7y  –  42
Putting y = 5, 6 and 7, we get
x          =  7 × 5  - 42   =  35 -  42       = - 7
x          = 7 × 6 -  42    =  42 – 42       = 0
x          =  7 × 7  – 42 = 49 – 42         = 7

x      -707y      567

Three years from now ,
Age of Aftab              = +3
Age of daughter      = +3
According to the question,
(+ 3) = 3 (+ 3)
+ 3    = 3+ 9
-3=  9-3
-3= 6                              …(ii)
=  3+ 6
Putting, = -2,-1 and 0, we get
=  3 × - 2  + 6 = -6 + 6 =0
=  3 × - 1  + 6 = -3 + 6             = 3
=  3 × 0 +  6 =  0 +  6             = 6

x      036y      -2-10

Algebraic representation
From equation (i)  and (ii)
– 7  = –  42                     …(i)
- 3 = 6                           …(ii)
Graphical representation

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

Let cost of one bat  = Rs x
Cost of one ball      = Rs y
3 bats and 6 balls for Rs 3900 So that
3+ 6y = 3900                     … (i)
Dividing equation by 3, we get
+ 2y = 1300
Subtracting 2y both side we get
x = 1300 – 2y
Putting y = -1300, 0 and 1300  we get
x = 1300 – 2 (-1300)   = 1300 + 2600   =  3900
= 1300 -2(0)                = 1300 - 0          = 1300
x = 1300 – 2(1300)         = 1300 – 2600   = - 1300

x      39001300-1300y      -130001300

Given that  she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x  + 2= 1300                      … (ii)
Subtracting 2y both side we get
= 1300 – 2y
Putting y = - 1300, 0 and 1300  we get
x        = 1300 – 2 (-1300)              = 1300  +  2600        =  3900
= 1300 – 2 (0)                     = 1300 - 0                = 1300
= 1300 – 2(1300)                  = 1300 – 2600           = -1300

x      39001300-1300y      -130001300

Algebraic representation
3+ 6y = 3900                    … (i)
+ 2  = 1300                     … (ii)
Graphical representation,

3. The cost of 2 kg of apples and 1kg 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.

Let cost each kg of apples  = Rs x
Cost of each kg of grapes  = Rs  y
Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160
So that
+  = 160              … (i)
2x          = 160  - y
x            =  (160 – y)/2
Let y = 0 , 80  and 160,  we get
x = (160 – ( 0 )/2   =  80
x = (160- 80 )/2      = 40
x = (160 – 2 × 80)/2 = 0

x      80400y      080160

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x +  2= 300              … (ii)
Dividing by 2 we get
2x + y   = 150
Subtracting 2x both side, we get
= 150 – 2x
Putting x = 0 , 50 , 100  we get
= 150 – 2 × 0                = 150
= 150 – 2 ×  50             =  50
= 150 – 2 × (100)      = -50

x      050100y      15050-50

Algebraic representation,
2y          = 160              … (i)
4x + 2y        = 300              … (ii)

Graphical representation,