Let present age of Aftab be x
And, present age of daughter is represented by y
Then Seven years ago,
Age of Aftab = x -7
Age of daughter = y-7
According to the question,
(x - 7)  = 7 (y – 7 )
x – 7 = 7 y – 49
x- 7y = - 49 + 7
x – 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 = x +3
Age of daughter = y +3
According to the question,
(x + 3) = 3 (y + 3)
x + 3 = 3y + 9
x -3y = 9-3
x -3y = 6 …(ii)
x = 3y + 6
Putting, y = -2,-1 and 0, we get
x = 3 × - 2 + 6 = -6 + 6 =0
x = 3 × - 1 + 6 = -3 + 6 = 3
x = 3 × 0 + 6 = 0 + 6 = 6

 

x	0	3	6
y	-2	-1	0

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

 

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

x	3900	1300	-1300
y	-1300	0	1300

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

 

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
2 x + y = 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 + 2y = 300 … (ii)
Dividing by 2 we get
2x + y = 150
Subtracting 2x both side, we get
y = 150 – 2x
Putting x = 0 , 50 , 100 we get
y = 150 – 2 × 0 = 150
y = 150 – 2 ×  50 = 50
y = 150 – 2 × (100) = -50
 

x	0	50	100
y	150	50	-50

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

Graphical representation,

Let number of boys = x
Number of girls = y
Given that total number of student is 10 so that
x + y = 10
Subtract y both side we get
x = 10 – y
Putting y = 0 , 5, 10 we get
x = 10 – 0 = 10
x = 10 – 5 = 5
x = 10 – 10 = 0
 

x	10	5
y	0	5

Given that If the number of girls is 4 more than the number of boys
So that
y = x + 4
Putting x = -4, 0, 4, and we get
y = - 4 + 4 = 0
y = 0 + 4 = 4
y = 4 + 4 = 8
 

x	-4	0	4
y	0	4	8

Graphical representation

Therefore, number of boys = 3 and number of girls = 7.

Let cost of pencil = Rs x
Cost of pens = Rs y
5 pencils and 7 pens together cost Rs 50,
So we get
5x + 7y = 50
Subtracting 7y both sides we get
5x = 50 – 7y
Dividing by 5 we get
x = 10 - 7 y /5
Putting value of y = 5 , 10 and 15 we get
x = 10 – 7 × 5/5 = 10 – 7 = 3
x = 10 – 7 × 10/5 = 10 – 14 = - 4

x = 10 – 7 × 15/5 = 10 – 21 = - 11

 

x	3	-4	-11
y	5	10	15

Given that 7 pencils and 5 pens together cost Rs 46
7x + 5y = 46
Subtracting 7x both side we get
5y = 46 – 7x
Dividing by 5 we get
y = 46/5 - 7x/5
y = 9.2 – 1.4x
Putting x = 0 , 2 and 4 we get
y = 9.2 – 1.4 × 0 = 9.2 – 0 = 9.2
y = 9.2 – 1.4 (2) = 9.2 – 2.8 = 6.4
y = 9.2 – 1.4 (4) = 9.2 – 5.6 = 3.6
 

x	0	2	4
y	9.2	6.4	3.6

Graphical representation

Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.

(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0

Comparing these equation with
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂= 0

We get
a₁ = 5, b₁ = -4, and c₁ = 8
a₂ =7, b₂ = 6 and c₂ = -9
a₁/a₂ = 5/7,
b₁/b₂ = -4/6 and
c₁/c₂ = 8/-9
Hence, a₁/a₂ ≠ b₁/b₂

Therefore, both are intersecting lines at one point.

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂= 0
We get
a₁ = 9, b₁ = 3, and c₁ = 12
a₂ = 18, b₂ = 6 and c₂ = 24
a₁/a₂ = 9/18 = 1/2
b₁/b₂ = 3/6 = 1/2 and
c₁/c₂ = 12/24 = 1/2
Hence, a₁/a₂ = b₁/b₂ = c₁/c₂

Therefore, both lines are coincident

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂= 0

We get
a₁ = 6, b₁ = -3, and c₁ = 10
a₂ = 2, b₂ = -1 and c₂ = 9
a₁/a₂ = 6/2 = 3/1
b₁/b₂ = -3/-1 = 3/1 and
c₁/c₂ = 12/24 = 1/2
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, both lines are parallel

On comparing the ratios a₁/a₂ , b₁/b₂ and c₁/c₂ find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) 3/2x + 5/3y = 7 ; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4/3x + 2y =8 ; 2x + 3y = 12 

Answer

(i) 3x + 2y = 5 ; 2x – 3y = 7
a₁/a₂ = 3/2
b₁/b₂ = -2/3 and
c₁/c₂ = 5/7
Hence, a₁/a₂ ≠ b₁/b₂
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii) 2x – 3y = 8 ; 4x – 6y = 9
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = -3/-6 = 1/2 and
c₁/c₂ = 8/9
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 3/2x + 5/3y = 7 ; 9x – 10y = 14
a₁/a₂ = 3/2/9 = 1/6
b₁/b₂ = 5/3/-10 = -1/6 and
c₁/c₂ = 7/14 = 1/2
Hence, a₁/a₂ ≠ b₁/b₂

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv) 5x – 3y = 11 ; – 10x + 6y = –22
a₁/a₂ = 5/-10 = -1/2
b₁/b₂ = -3/6 = -1/2 and
c₁/c₂ = 11/-22 = -1/2
Hence, a₁/a₂ = b₁/b_2 = c₁/c₂

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

(v) 4/3x + 2y =8 ; 2x + 3y = 12
a₁/a₂ = 4/3/2 = 2/3
b₁/b₂ = /3 and
c₁/c₂ = 8/12 = 2/3
Hence, a₁/a₂ = b₁/b_2 = c₁/c₂

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10

(ii) x – y = 8, 3x – 3y = 16

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Answer

(i) x + y = 5; 2x + 2y = 10
a₁/a₂ = 1/2
b₁/b₂ = 1/2 and
c₁/c₂ = 5/10 = 1/2
Hence, a₁/a₂ = b₁/b_2 = c₁/c₂
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

 

x + y = 5

x = 5 - y 

 

x	4	3	2
y	1	2	3

And, 2x + 2y = 10
x = 10-2y/2
 

x	4	3	2
y	1	2	3

Graphical representation

 

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

 

(ii) x – y = 8, 3x – 3y = 16
a₁/a₂ = 1/3
b₁/b₂ = -1/-3 = 1/3 and
c₁/c₂ = 8/16 = 1/2
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = -1/2 and
c₁/c₂ = -6/-4 = 3/2
Hence, a₁/a₂ ≠ b₁/b₂

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2x + y - 6 = 0
y = 6 - 2x

 

x	0	1	2
y	6	4	2

And, 4x - 2y -4 = 0
y = 4x - 4/2
 

x	1	2	3
y	0	2	4

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = -2/-4 = 1/2 and
c₁/c₂ = 2/5
Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂ 

Therefore, these linear equations are parallel to each other and thus, have no possible solution. Hence, the pair of linear equations is inconsistent.