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NCERT Solution For Class 10 Maths Chapter 3

 Pair of Linear Equations in Two Variables

3.1 Introduction
3.2 Pair Of Linear Equations In Two Variables
3.3 Graphical Method Of Solution Of A Pair Of Linear Equations
3.4 Algebraic Methods Of Solving A Pair Of Linear Equations
3.4.1 Substitution Method
3.4.2 Elimination Method
3.4.3 Cross-Multiplication Method
3.5 Equations Reducible To A Pair Of Linear Equations In Two Variables
3.6 Summary

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.

Answer

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 -7 0 7
y 5 6 7

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 0 3 6
y -2 -1 0

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.

Answer

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 – 2
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 3900 1300 -1300
y -1300 0 1300

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 3900 1300 -1300
y -1300 0 1300

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


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.

Answer

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 80 40 0
y 0 80 160

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 0 50 100
y 150 50 -50

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

Graphical representation,

Graph 3





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