And, present age of his daughter is represented as = y

Seven years ago,

Aftab’s age = x – 7

Age of Aftab’s 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

Three years from now,

Aftab’s age = x + 3

Age of Aftab’s 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

Algebraic representation

From equation (i) and (ii)

x – 7y = – 42 … **(i)**

x – 3y = 6 … **(ii)**

Graphical representation

And, cost of one ball be = Rs y

3 bats and 6 balls are for Rs 3900

Therefore, 3x + 6y = 3900 … **(i)**

Dividing equation by 3, we get

x + 2y = 1300

Subtracting 2y from 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

Given that she buys another bat and 2 more balls of the same kind for Rs 1300

We get

x + 2y = 1300 … **(ii)**

Subtracting 2y from both sides 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

Algebraic representation

3x + 6y = 3900 … **(i)**

x + 2y = 1300 … **(ii)**

Graphical representation,

And, 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

Therefore,

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

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300

Therefore,

4x + 2y = 300 … **(ii)**

Dividing by 2 we get

2x + y = 150

Subtracting 2x from both the sides, 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

Algebraic representation,

2x + y = 160 … **(i)**

4x + 2y = 300 … **(ii)**

Graphical representation,

Let the number of girls be = y

Given that total number of students is 10

Therefore x + y = 10

Subtracting y from both the sides we get

x = 10 – y

Putting y = 0, 5, 10 we get

x = 10 – 0 = 10

x = 10 – 5 = 5

x = 10 – 10 = 0

Given: the number of girls is 4 more than the number of boys

Therefore

y = x + 4

Putting x = – 4, 0, 4, and we get

y = – 4 + 4 = 0

y = 0 + 4 = 4

y = 4 + 4 = 8

Graphical representation

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

Let the cost of one pen = Rs y

5 pencils and 7 pens together cost = Rs 50,

Therefore

5x + 7y = 50

Subtracting 7y from both the 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

Given: 7 pencils and 5 pens together cost Rs 46

7x + 5y = 46

Subtracting 7x from both the sides 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

Graphical representation

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

7x + 6y – 9 = 0

On comparing these equation with

a_{1}x + b_{1}y + c_{1} = 0

a_{2}x + b_{2}y + c_{2}= 0

We get

a_{1} = 5, b_{1} = – 4, and c_{1} = 8

a_{2} =7, b_{2} = 6 and c_{2} = – 9

a_{1}/a_{2} = 5/7,

b_{1}/b_{2} = – 4/6 and

c_{1}/c_{2} = 8/–9

Hence, a_{1}/a_{2} ≠ b_{1}/b_{2}

Therefore, both the lines intersect at one point.

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

Comparing these equations with

a_{1}x + b_{1}y + c_{1} = 0

a_{2}x + b_{2}y + c_{2}= 0

We get

a_{1} = 9, b_{1} = 3, and c_{1} = 12

a_{2} = 18, b_{2} = 6 and c_{2} = 24

a_{1}/a_{2} = 9/18 = 1/2

b_{1}/b_{2} = 3/6 = 1/2 and

c_{1}/c_{2} = 12/24 = 1/2

Hence, a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Therefore, both the lines are coincident

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Comparing these equations with

a_{1}x + b_{1}y + c_{1} = 0

a_{2}x + b_{2}y + c_{2}= 0

We get

a_{1} = 6, b_{1} = – 3, and c_{1} = 10

a_{2} = 2, b_{2} = – 1 and c_{2} = 9

a_{1}/a_{2} = 6/2 = 3/1

b_{1}/b_{2} = – 3/–1 = 3/1 and

c_{1}/c_{2} = 12/24 = 1/2

Hence, a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Therefore, both lines are parallel

Ans. (i) 3x + 2y = 5 ; 2x – 3y = 7

a_{1}/a_{2} = 3/2

b_{1}/b_{2} = – 2/3 and

c_{1}/c_{2} = 5/7

Hence, a_{1}/a_{2} ≠ b_{1}/b_{2}

These linear equations intersect each other at one point and therefore have only one possible solution. Hence, the pair of linear equations is consistent.

(ii) 2x – 3y = 8; 4x – 6y = 9

a_{1}/a_{2} = 2/4 = 1/2

b_{1}/b_{2} = – 3/–6 = 1/2 and

c_{1}/c_{2} = 8/9

Hence, a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

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

a_{1}/a_{2} = 3/2/9 = 1/6

b_{1}/b_{2} = 5/3/–10 = – 1/6 and

c_{1}/c_{2} = 7/14 = 1/2

Hence, a_{1}/a_{2} ≠ b_{1}/b_{2}

Therefore, these linear equations intersect 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

a_{1}/a_{2} = 5/–10 = – 1/2

b_{1}/b_{2} = – 3/6 = – 1/2 and

c_{1}/c_{2} = 11/–22 = – 1/2

Hence, a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Therefore, these linear equations are coincident and have infinite number of possible solutions.

Therefore, the pair of linear equations is consistent.

(v) 4/3x + 2y =8; 2x + 3y = 12

a_{1}/a_{2} = 4/3/2 = 2/3

b_{1}/b_{2} = 2 /3 and

c_{1}/c_{2} = 8/12 = 2/3

Hence, a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Therefore, these linear equations are coincident and have infinite number of possible solutions.

Therefore, the pair of linear equations is consistent.

Ans. (i) x + y = 5; 2x + 2y = 10

a_{1}/a_{2} = 1/2

b_{1}/b_{2} = 1/2 and

c_{1}/c_{2} = 5/10 = 1/2

Hence, a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Therefore, these linear equations are coincident and have infinite number of possible solutions.

Therefore, the pair of linear equations is consistent.

x + y = 5

x = 5 – y

And, 2x + 2y = 10

x = 10 – 2y/2

Graphical representation

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

(ii) x – y = 8, 3x – 3y = 16

a_{1}/a_{2} = 1/3

b_{1}/b_{2} = – 1/–3 = 1/3 and

c_{1}/c_{2} = 8/16 = 1/2

Hence, a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

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

a_{1}/a_{2} = 2/4 = 1/2

b_{1}/b_{2} = – 1/2 and

c_{1}/c_{2} = – 6/–4 = 3/2

Hence, a_{1}/a_{2} ≠ b_{1}/b_{2}

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

And, 4x – 2y – 4 = 0

y = 4x – 4/2

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

a_{1}/a_{2} = 2/4 = 1/2

b_{1}/b_{2} = – 2/–4 = 1/2 and

c_{1}/c_{2} = 2/5

Hence, a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Therefore, these linear equations are parallel to each other and thus, have no possible solution.

Hence, the pair of linear equations is inconsistent.

Ans. Let length of the rectangle be = x m

Let Width of the rectangle be = y m

According to the question,

y – x = 4 ... **(i)**

y + x = 36 ... **(ii)**

y – x = 4

y = x + 4

y + x = 36

Graphical representation

From the figure, it can be seen that these lines intersect each other at only one point i.e., (16, 20).

Therefore, the length and width of the given garden is 20 m and 16 m respectively.

Ans. (i) Intersecting lines:

Condition,

a_{1}/a_{2} ≠ b_{1}/b_{2}

The second line such that it is intersecting the given line is

2x + 4y – 6 = 0 as

a_{1}/a_{2} = 2/2 = 1

b_{1}/b_{2} = 3/4 and

a_{1}/a_{2} ≠ b_{1}/b_{2}

(ii) Parallel lines

Condition,

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Hence, the second line can be

4x + 6y – 8 = 0 as

a_{1}/a_{2} = 2/4 = 1/2

b_{1}/b_{2} = 3/6 = 1/2 and

c_{1}/c_{2} = – 8/–8 = 1

and a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

(iii) Coincident lines

Condition,

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Hence, the second line can be

6x + 9y – 24 = 0 as

a_{1}/a_{2} = 2/6 = 1/3

b_{1}/b_{2} = 3/9 = 1/3 and

c_{1}/c_{2} = – 8/–24 = 1/3

and a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Ans. x – y + 1 = 0

x = y – 1

3x + 2y – 12 = 0

x = 12 – 2y/3

Graphical representation

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at ( – 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( – 1, 0), and (4, 0).

Ans. (i) x + y = 14 ... **(i)**

x – y = 4 ... **(ii)**

From equation (i), we get

x = 14 – y ... **(iii)**

Putting this value in equation **(ii)**, we get

(14 – y) – y = 4

14 – 2y = 4

10 = 2y

y = 5 ... **(iv)**

Putting this in equation **(iii)**, we get

x = 9

∴ x = 9 and y = 5

(ii) s – t = 3 ... **(i)**

s/3 + t/2 = 6 ... **(ii)**

From equation (i), we get s = t + 3

Putting this value in equation **(ii)**, we get

t + 3/3 + t/2 = 6

2t + 6 + 3t = 36

5t = 30

t = 30/5 ... **(iv)**

Putting in equation **(iii)**, we get

s = 9

∴ s = 9, t = 6

(iii) 3x – y = 3 ... (i)

9x – 3y = 9 ... **(ii)**

From equation **(i)**, we get

y = 3x – 3 ... **(iii)**

Putting this value in equation **(ii)**, we get

9x – 3(3x – 3) = 9

9x – 9x + 9 = 9

9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by y = 3x – 3

Therefore, one of its possible solutions is x = 1, y = 0.

(iv) 0.2x + 0.3y = 1.3 ... **(i)**

0.4x + 0.5y = 2.3 ... **(ii)**

0.2x + 0.3y = 1.3

Solving equation (i), we get

0.2x = 1.3 – 0.3y

Dividing by 0.2, we get

x = 1.3/0.2 – 0.3/0.2

x = 6.5 – 1.5 y ... **(iii)**

Putting the value in equation **(ii)**, we get

0.4x + 0.5y = 2.3

(6.5 – 1.5y) × 0.4x + 0.5y = 2.3

2.6 – 0.6y + 0.5y = 2.3

–0.1y = 2.3 – 2.6

y = – 0.3/–0.1

y = 3

Putting this value in equation **(iii)** we get

x = 6.5 – 1.5 y

x = 6.5 – 1.5(3)

x = 6.5 – 4.5

x = 2

∴ x = 2 and y = 3

(vi) 3/2x – 5/3y = – 2 ... **(i)**

x/3 + y/2 = 13/6 ... **(ii)**

From equation **(i)**, we get

9x – 10y = – 12

x = – 12 + 10y/9 ... **(iii)**

Putting this value in equation **(ii)**, we get

Ans. 2x + 3y = 11 ... **(i)**

Subtracting 3y both side we get

2x = 11 – 3y … **(ii)**

Putting this value in equation second we get

2x – 4y = – 24 … **(iii)**

11- 3y – 4y = – 24

7y = – 24 – 11

–7y = – 35

y = – 35/–7

y = 5

Putting this value in equation **(iii)** we get

2x = 11 – 3 × 5

2x = 11 – 15

2x = – 4

Dividing by 2 we get

x = – 2

Putting the value of x and y

y = mx + 3.

5 = – 2m + 3

2m = 3 – 5

m = – 2/2

m = – 1

Ans. Let the larger number be = x

Let the smaller number be = y

The difference between the two numbers is 26

x – y = 26

x = 26 + y

Given that one number is three times the other

So x = 3y

Putting the value of x we get

26 + y = 3y

–2y = – 2 6

y = 13

So value of x = 3y

Putting value of y, we get

x = 3 × 13 = 39

Therefore the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Ans. Let the first angle be= x

And the second angle be = y

As both angles are supplementary so that sum will 180

x + y = 180

x = 180 – y ... **(i)**

Given; difference is 18 degree

Therefore

x – y = 18

Putting the value of x we get

180 – y – y = 18

– 2y = – 162

y = – 162/–2

y = 81

Putting the value back in equation **(i)**, we get

x = 180 – 81 = 99 Hence, the angles are 99° and 81°.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats

and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Ans. Let the cost of each bat be = Rs x

Let the cost of each ball be = Rs y

Given: coach of a cricket team buys 7 bats and 6 balls for Rs 3800.

7x + 6y = 3800

6y = 3800 – 7x

Dividing by 6, we get

y = (3800 – 7x)/6 … **(i)**

Given that she buys 3 bats and 5 balls for Rs 1750 later.

3x + 5y = 1750

Putting the value of y

3x + 5 ((3800 – 7x)/6) = 1750

Multiplying by 6, we get

18x + 19000 – 35x = 10500

–17x =10500 – 19000

–17x = – 8500

x = – 8500/–17

x = 500

Putting this value in equation **(i)** we get

y = ( 3800 – 7 × 500)/6

y = 300/6

y = 50

Hence the cost of each bat = Rs 500 and the cost of each balls = Rs 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km?

How much does a person have to pay for traveling a distance of 25 km?

Ans. Let the fixed charge for taxi = Rs x

And variable cost per km = Rs y

Total cost = fixed charge + variable charge

Given that for a distance of 10 km, the charge paid is Rs 105

x + 10y = 105 … **(i)**

x = 105 – 10y

Given that for a journey of 15 km, the charge paid is Rs 155

x + 15y = 155

Putting the value of x we get

105 – 10y + 15y = 155

5y = 155 – 105

5y = 50

Dividing by 5, we get

y = 50/5 = 10

Putting this value in equation **(i)** we get

x = 105 – 10 × 10

x = 5

Cost for traveling a distance of 25 km

= x + 25y

= 5 + 25 × 10

= 5 + 250

=255

A person has to pay Rs 255 for 25 Km.

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Ans. Let the Numerator be = x

Let the Denominator be = y

Fraction = x/y

A fraction becomes 9/11, if 2 is added to both the numerator and the denominator

(x + 2)/y + 2 = 9/11

On Cross multiplying,

11x + 22 = 9y + 18

Subtracting 22 from both sides,

11x = 9y – 4

Dividing by 11, we get

x = 9y – 4/11 … **(i)**

Given: if 3 is added to both the numerator and the denominator it becomes 5/6.

(x+3)/y +3 = 5/6 … **(ii)**

On Cross multiplying,

6x + 18 = 5y + 15

Subtracting the value of x, we get

6(9y – 4 )/11 + 18 = 5y + 15

Subtracting 18 from both the sides

6(9y – 4 )/11 = 5y – 3

54 – 24 = 55y – 33

–y = – 9

y = 9

Putting this value of y in equation **(i)**, we get

x = 9y – 4

11 … **(i)**

x = (81 – 4)/77

x = 77/11

x = 7

Hence our fraction is 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Ans. Let the present age of Jacob be = x year

And the present Age of his son be = y year

Five years from now,

Jacob’s age will be = x + 5 year

Age of his son will be = y + 5year

Given; the age of Jacob will be three times that of his son

x + 5 = 3(y + 5)

Adding 5 to both side,

x = 3y + 15 – 5

x = 3y + 10 … **(i)**

Five years ago,

Jacob’s age= x – 5 year

His son’s age = y – 5 year

Jacob’s age was seven times that of his son

x – 5 = 7(y – 5)

Putting the value of x from equation **(i)** we get

3y + 10 – 5 = 7y – 35

3y + 5 = 7y – 35

3y – 7y = – 35 – 5

–4y = – 40

y = – 40/–4

y = 10 year

Putting the value of y in equation we get

x = 3 × 10 + 10

x = 40 years

Hence, Present age of Jacob = 40 years and present age of his son = 10 years.

Ans. **(i)** x + y =5 and 2x –3y = 4

By elimination method

x + y =5 ... **(i)**

2x –3y = 4 ... **(ii)**

Multiplying equation **(i)** by **(ii)**, we get

2x + 2y = 10 ... **(iii)**

2x – 3y = 4 ... **(ii)**

Subtracting equation **(ii)** from equation **(iii)**, we get

5y = 6

y = 6/5

Putting the value in equation **(i)**, we get

x = 5 – (6/5) = 19/5

Hence, x = 19/5 and y = 6/5

By substitution method

x + y = 5 ... **(i)**

Subtracting y from both side, we get

x = 5 – y ... **(iv)**

Putting the value of x in equation **(ii)** we get

2(5 – y) – 3y = 4

–5y = – 6

y = – 6/–5 = 6/5

Putting the value of y in equation **(iv)** we get

x = 5 – 6/5

x = 19/5

Hence, x = 19/5 and y = 6/5 again

(ii) 3x + 4y = 10 and 2x – 2y = 2

By elimination method

3x + 4y = 10 .... **(i)**

2x – 2y = 2 ... **(ii)**

Multiplying equation **(ii)** by 2, we get

4x – 4y = 4 ... **(iii)**

3x + 4y = 10 ... **(i)**

Adding equation **(i)** and **(iii)**, we get

7x + 0 = 14

Dividing both side by 7, we get

x = 14/7 = 2

Putting in equation **(i)**, we get

3x + 4y = 10

3(2) + 4y = 10

6 + 4y = 10

4y = 10 – 6

4y = 4

y = 4/4 = 1

Hence, answer is x = 2, y = 1

By substitution method

3x + 4y = 10 ... **(i)**

Subtract 3x both side, we get

4y = 10 – 3x

Divide by 4 we get

y = (10 – 3x )/4

Putting this value in equation **(ii)**, we get

2x – 2y = 2 ... **(i)**

2x – 2(10 – 3x )/4) = 2

Multiply by 4 we get

8x – 2(10 – 3x) = 8

8x – 20 + 6x = 8

14x = 28

x = 28/14 = 2

y = (10 – 3x)/4

y = 4/4 = 1

Hence, answer is x = 2, y = 1 again.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By elimination method

3x – 5y – 4 = 0

3x – 5y = 4 ...**(i)**

9x = 2y + 7

9x – 2y = 7 ... **(ii)**

Multiplying equation **(i)** by 3, we get

9 x – 15 y = 11 ... **(iii)**

9x – 2y = 7 ... **(ii)**

Subtracting equation **(ii)** from equation **(iii)**, we get

–13y = 5

y = – 5/13

Putting value in equation **(i)**, we get

3x – 5y = 4 ... **(i)**

3x – 5(–5/13) = 4

Multiplying by 13 we get

39x + 25 = 52

39x = 27

x =27/39 = 9/13

Hence our answer is x = 9/13 and y = – 5/13

By substitution method

3x – 5y = 4 ... **(i)**

Adding 5y on both sides we get

3x = 4 + 5y

Dividing by 3 we get

x = (4 + 5y )/3 ... **(iv)**

Putting this value in equation **(ii)** we get

9x – 2y = 7 ... **(ii)**

9 ((4 + 5y )/3) – 2y = 7

On solve we get

3(4 + 5y ) – 2y = 7

12 + 15y – 2y = 7

13y = – 5

y = – 5/13

(iv) x/2 + 2y/3 = – 1 and x – y/3 = 3

By elimination method

x/2 + 2y/3 = – 1 ... **(i)**

x – y/3 = 3 ... **(ii)**

Multiplying equation **(i)** by 2, we get

x + 4y/3 = – 2 ... **(iii)**

x – y/3 = 3 ... **(ii)**

Subtracting equation **(ii)** from equation **(iii)**, we get

5y/3 = – 5

Dividing by 5 and multiplying by 3, we get

y = – 15/5

y = – 3

Putting this value in equation **(ii)**, we get

x – y/3 = 3 ... **(ii)**

x – (–3)/3 = 3

x + 1 = 3

x = 2

Therefore x = 2 and y = – 3.

By substitution method

x – y/3 = 3 ... **(ii)**

Adding y/3 on both sides, we get

x = 3 + y/3 ... **(iv)**

Putting the value in equation **(i)** we get

x/2 + 2y/3 = – 1 ... **(i)**

(3+ y/3)/2 + 2y/3 = – 1

3/2 + y/6 + 2y/3 = – 1

Multiplying by 6,

9 + y + 4y = – 6

5y = – 15

y = – 3

Therefore x = 2 and y = – 3.

Ans. **(i)** Let the fraction be x/y

According to the question, x + 1/y – 1 = 1

⇒ x – y = – 2 ... **(i)** x/y + 1 = 1/2

⇒ 2x – y = 1 ... **(ii)**

Subtracting equation **(i)** from equation **(ii)**, we get

x = 3 ... **(iii)**

Putting this value in equation **(i)**, we get

3 – y = – 2

–y = – 5

y = 5

Hence, the fraction is 3/5

(ii) Let the present age of Nuri be = x

Let the present age of Sonu be = y

According to the given information,

(x – 5) = 3(y – 5)

x – 3y = – 10 ... **(i)**

(x + 10y) = 2(y + 10)

x – 2y = 10 ... **(ii)**

Subtracting equation **(i)** from equation **(ii)**, we get

y = 20 ... **(iii)**

Putting this value in equation **(i)**, we get

x – 60 = – 10

x = 50

Hence, age of Nuri = 50 years and age of Sonu = 20 years.

(iii) Let the unit digit and tens digits of the number be x and y respectively.

Then, number = 10y + x

Number after reversing the digits = 10x + y

According to the question,

x + y = 9 ... **(i)**

9(10y + x) = 2(10x + y)

88y – 11x = 0

–x + 8y =0 ... **(ii)**

Adding equation **(i)** and **(ii)**, we get

9y = 9

y = 1 ... **(iii)**

Putting the value in equation **(i)**, we get

x = 8

Hence, the number is 10y + x = 10 × 1 + 8 = 18.

(iv) Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.

According to the question,

x + y = 25 ... **(i)**

50x + 100y = 2000 ... **(ii)**

Multiplying equation **(i)** by 50, we get

50x + 50y = 1250 ... **(iii)**

Subtracting equation **(iii)** from equation **(ii)**, we get

50y = 750

y = 15

Putting this value in equation **(i)**, we have x = 10

Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v) Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.

According to the question,

x + 4y = 27 ... **(i)**

x + 2y = 21 ... **(ii)**

Subtracting equation **(ii)** from equation **(i)**, we get

2y = 6

y = 3 ... **(iii)**

Putting in equation **(i)**, we get

x + 12 =27

x = 15

Hence, fixed charge = Rs 15 and Charge per day = Rs 3.

Ans. **(i)** x – 3y – 3 = 0

3x – 9y – 2 =0

a_{1}/a_{2} = 1/3

b_{1}/b_{2} = – 3/-9 = 1/3 and

c_{1}/c_{2} = – 3/-2 = 3/2

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5

3x + 2y = 8

a_{1}/a_{2} = 2/3

b_{1}/b_{2} = 1/2 and

c_{1}/c_{2} = – 5/–8 = 5/8

a_{1}/a_{2} ≠ b_{1}/b_{2}

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

x/b_{1}c_{2}– b_{2}c_{1} = y/c_{1}a_{2} – c_{2}a_{1} = 1/a_{1}b_{2} – a_{2}b_{1}

x/-8–(–10) = y/–15 + 16 = 1/4 – 3

x/2 = y/1 = 1

x/2 = 1, y/1 = 1

∴ x = 2, y = 1.

(iii) 3x – 5y = 20

6x – 10y = 40

a_{1}/a_{2} = 3/6 = 1/2

b_{1}/b_{2} = – 5/–10 = 1/2 and

c_{1}/c_{2} = – 20/–40 = 1/2

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0

3x – 3y – 15= 0

a_{1}/a_{2} = 1/3

b_{1}/b_{2} = – 3/–3 = 1 and

c_{1}/c_{2} = – 7/–15 = 7/15

a_{1}/a_{2} ≠ b_{1}/b_{2}

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,

x/45 – (21) = y/–21 – (–15) = 1/–3 – (–9)

x/24 = y/–6 = 1/6

x/24 = 1/6 and y/–6 = 1/6

x = 4 and y = – 1

∴ x = 4, y = – 1.

Ans. 2x + 3y –7 = 0

(a – b)x + (a + b)y – (3a + b –2) = 0

a_{1}/a_{2} = 2/a – b = 1/2

b_{1}/b_{2} = – 7/a + b and

c_{1}/c_{2} = – 7/–(3a + b – 2) = 7/(3a + b – 2)

For infinitely many solutions,a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

2/a – b = 7/3a + b – 26a + 2b – 4 = 7a – 7b

a – 9b = – 4 ... **(i)**

2/a – b = 3/a+b

2a + 2b = 3a – 3b

a – 5b = 0 ... **(ii)**

Subtracting equation **(i)** from **(ii)**, we get

4b = 4

b = 1

Putting this value in equation **(ii)**, we get

a – 5 × 1 = 0

a = 5

Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k –1)x + (k –1)y = 2k + 1

Ans. 3x + y –1 = 0

(2k –1)x + (k –1)y – (2k + 1) = 0

a_{1}/a_{2} = 3/2k – 1

b_{1}/b_{2} = 1/k – 1 and

c_{1}/c_{2} = – 1/–2k – 1 = 1/2k + 1

For no solutions,

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

3/2k – 1 = 1/k-1 ≠ 1/2k + 1

3/2k – 1 = 1/k – 1

3k – 3 = 2k – 1

k = 2

Hence, for k = 2, the given equation has no solution.

Ans. 8x + 5y = 9 ... **(i)**

3x + 2y = 4 ... **(ii)**

From equation **(ii)**, we get

x = 4 – 2y/3 ... **(iii)**

Putting this value in equation **(i)**, we get

8(4 – 2y/3) + 5y = 9

32 – 16y + 15y = 27

–y = – 5

y = 5 ... **(iv)**

Putting this value in equation **(ii)**, we get

3x + 10 = 4

x = – 2

Hence, x = – 2, y = 5

By cross multiplication again, we get

8x + 5y – 9 = 0

3x + 2y – 4 = 0

x/–20 – (–18) = y/–27 – (–32) = 1/16 – 15

x/–2 = y/5 = 1/1

x/–2 = 1 and y/5 = 1

x = – 2 and y = 5

Ans. Let x be the fixed charge of the food and y be the charge for food per day.

According to the question,

x + 20y = 1000 ... **(i)**

x + 26y = 1180 ... **(ii)**

Subtracting equation **(i)** from equation **(ii)**, we get

6y = 180

y = 180/6 = 30

Putting this value in equation **(i)**, we get

x + 20 × 30 = 1000

x = 1000 – 600

x = 400

Hence, fixed charge = Rs 400 and charge per day = Rs 30

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4

when 8 is added to its denominator. Find the fraction.

Ans. Let the fraction be x/y

According to the question,

x–1/y = 1/3

⇒ 3x – y = 3... **(i)**

x/y + 8 = 1/4

⇒ 4x – y = 8 ... **(ii)**

Subtracting equation **(i)** from equation **(ii)**, we get

x = 5 ... **(iii)**

Putting this value in equation **(i)**, we get

15 – y = 3

y = 12

Hence, the fraction is 5/12.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Ans. Let the number of right answers and wrong answers be x and y respectively.

According to the question,

3x – y = 40 ... **(i)**

4x – 2y = 50

⇒ 2x – y = 25 ... **(ii)**

Subtracting equation **(ii)** from equation **(i)**, we get

x = 15 ... **(iii)**

Putting this value in equation **(ii)**, we get

30 – y = 25

y = 5

Therefore, number of right answers = 15

And number of wrong answers = 5

Total number of questions = 20

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Ans. Let the speed of 1st car and 2nd car be u km/h and v km/h.

Respective speed of both cars while they are travelling in same direction = (u – v) km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (u + v) km/h

According to the question,

5(u – v) = 100

⇒ u – v = 20 ... **(i)**

1(u + v) = 100 ... **(ii)**

Adding both the equations, we get

2u = 120

u = 60 km/h ... **(iii)**

Putting this value in equation **(ii)**, we obtain

v = 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Ans. Let length and breadth of rectangle be x unit and y unit respectively.

Area = xy

According to the question,

(x – 5) (y + 3) = xy – 9

⇒ 3x – 5y – 6 = 0 ... **(i)**

(x + 3) (y + 2) = xy + 67

⇒ 2x – 3y – 61 = 0 ... **(ii)**

By cross multiplication, we get

x/305 – (–18) = y/–12 – (–183) = 1/9 – (–10)

x/323 = y/171 = 1/19

x = 17, y = 9

Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

Ans. **(i)** 1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Let 1/x = p and 1/y = q, then the equations changes as below:

p/2 + q/3 = 2

⇒ 3p + 2q – 12 = 0 ... **(i)**

p/3 + q/2 = 13/6

⇒ 2p + 3q – 13 = 0 ... **(ii)**

By cross-multiplication method, we get

p/–26 – (–36) = q/–24 – (–39) = 1/9–4

p/10 = q/15 = 1/5

p/10 = 1/5 and q/15 = 1/5

p = 2 and q = 3

1/x = 2 and 1/y = 3

Hence, x = 1/2 and y = 1/3

2p + 3q = 2 ... **(i)**

4p – 9q = – 1 ... **(ii)**

Multiplying equation **(i)** by 3, we get

6p + 9q = 6 ... **(iii)**

Adding equation **(ii)** and **(iii)**, we get

10p = 5

p = 1/2 ... **(iv)**

Putting in equation **(i)**, we get

2 × 1/2 + 3q = 2

3q = 1

q = 1/3

p = 1/vx = 1/2

vx = 2

x = 4

and

q = 1/vy = 1/3

vy = 3

y = 9

Hence, x = 4, y = 9

(iii) 4/x + 3y = 14

3/x – 4y = 23

Putting 1/x = p in the given equations, we get

4p + 3y = 14

⇒ 4p + 3y – 14 = 0

3p – 4y = 23

⇒ 3p – 4y – 23 = 0

By cross-multiplication, we get

p/–69 – 56 = y/–42 – (–92) = 1/–16 – 9

⇒ – p/125 = y/50 = – 1/25

Now,

–p/125 = – 1/25 and y/50 = – 1/25

⇒ p = 5 and y = – 2

Also, p = 1/x = 5

⇒ x = 1/5

So, x = 1/5 and y = – 2 is the solution.

(iv) 5/x – 1 + 1/y – 2 = 2

6/x – 1 – 3/y – 2 = 1

Putting 1/x – 1 = p and 1/y – 2 = q in the given equations, we obtain

5p + q = 2 ... **(i)**

6p – 3q = 1 ... **(ii)**

Now, by multiplying equation **(i)** by 3 we get

15p + 3q = 6 ... **(iii)**

Now, adding equation **(ii)** and **(iii)**

21p = 7

⇒ p = 1/3

Putting this value in equation **(ii)** we get,

6×1/3 – 3q =1

⇒ 2 – 3q = 1

⇒ – 3q = 1 – 2

⇒ – 3q = – 1

⇒ q = 1/3

Now,

p = 1/x – 1 = 1/3

⇒1/x – 1 = 1/3

⇒ 3 = x – 1

⇒ x = 4

Also,

q = 1/y – 2 = 1/3

⇒ 1/y – 2 = 1/3

⇒ 3 = y – 2

⇒ y = 5

Hence, x = 4 and y = 5 is the solution.

(v) 7x – 2y/xy = 5

⇒ 7x/xy – 2y/xy = 5

⇒ 7/y – 2/x = 5 ... **(i)**

8x + 7y/xy = 15

⇒ 8x/xy + 7y/xy = 15

⇒ 8/y + 7/x = 15 ... **(ii)**

Putting 1/x = p and 1/y = q in **(i)** and **(ii)** we get,

7q – 2p = 5 ... **(iii)**

8q + 7p = 15 ... **(iv)**

Multiplying equation **(iii)** by 7 and multiplying equation **(iv)** by 2 we get,

49q – 14p = 35 ... **(v)**

16q + 14p = 30 ... **(vi)**

Now, adding equation **(v)** and **(vi)** we get,

49q – 14p + 16q + 14p = 35 + 30

⇒ 65q = 65

⇒ q = 1

Putting the value of q in equation **(iv)**

8 + 7p = 15

⇒ 7p = 7

⇒ p = 1

Now,

p = 1/x = 1

⇒ 1/x = 1

⇒ x = 1

also, q = 1 = 1/y

⇒ 1/y = 1

⇒ y = 1

Hence, x =1 and y = 1 is the solution.

(vi) 6x + 3y = 6xy

⇒ 6x/xy + 3y/xy = 6

⇒ 6/y + 3/x = 6 ... **(i)**

2x + 4y = 5xy

⇒ 2x/xy + 4y/xy = 5

⇒ 2/y + 4/x = 5 ... **(ii)**

Putting 1/x = p and 1/y = q in **(i)** and **(ii)** we get,

6q + 3p – 6 = 0

2q + 4p – 5 = 0

By cross multiplication method, we get

p/–30 – (–12) = q/–24 – (–15) = 1/6 – 24

p/–18 = q/–9 = 1/–18

p/–18 = 1/–18 and q/–9 = 1/–18

p = 1 and q = 1/2

p = 1/x = 1 and q = 1/y = 1/2

x = 1, y = 2

Hence, x = 1 and y = 2

(vii) 10/x+y + 2/x – y = 4

15/x + y - 5/x – y = – 2

Putting 1/x + y = p and 1/x – y = q in the given equations, we get:

10p + 2q = 4

⇒ 10p + 2q – 4 = 0 ... **(i)**

15p – 5q = – 2

⇒ 15p – 5q + 2 = 0 ... **(ii)**

Using cross multiplication, we get

p/4 – 20 = q/–60 – (–20) = 1/–50 – 30

p/–16 = q/–80 = 1/–80

p/–16 = 1/–80 and q/–80 = 1/–80

p = 1/5 and q = 1

p = 1/x + y = 1/5 and q = 1/x – y = 1

x + y = 5 ... **(iii)**

and x – y = 1 ... **(iv)**

Adding equation **(iii)** and **(iv)**, we get

2x = 6

x = 3 .... (v)

Putting value of x in equation **(iii)**, we get

y = 2

Hence, x = 3 and y = 2

(viii) 1/3x + y + 1/3x – y = 3/4

1/2(3x – y) – 1/2(3x – y) = – 1/8

Putting 1/3x + y = p and 1/3x – y = q in the given equations, we get

p + q = 3/4 ... **(i)**

p/2 – q/2 = – 1/8

p – q = – 1/4 ... **(ii)**

Adding **(i)** and **(ii)**, we get

2p = 3/4 – 1/4

2p = 1/2

p = 1/4

Putting the value in equation **(ii)**, we get

1/4 – q = – 1/4

q = 1/4 + 1/4 = 1/2

p = 1/3x + y = 1/4

3x + y = 4 ... **(iii)**

q = 1/3x – y = 1/2

3x – y = 2 ... **(iv)**

Adding equations **(iii)** and **(iv)**, we get

6x = 6

x = 1 ... **(v)**

Putting the value in equation **(iii)**, we get

3(1) + y = 4

y = 1

Hence, x = 1 and y = 1

Ans. Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.

Speed of Ritu while rowing

Upstream = (x – y) km/h

Downstream = (x + y) km/h

According to question,

2(x + y) = 20

⇒ x + y = 10 ... **(i)**

2(x – y) = 4

⇒ x – y = 2 ... **(ii)**

Adding equation **(i)** and **(ii)**, we get

Putting this equation in **(i)**, we get

y = 4

Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Ans. Let the number of days taken by a woman and a man be x and y respectively.

Therefore, work done by a woman in 1 day = 1/x

According to the question,

4(2/x + 5/y) = 1

2/x + 5/y = 1/4

3(3/x + 6/y) = 1

3/x + 6/y = 1/3

Putting 1/x = p and 1/y = q in these equations, we get

2p + 5q = 1/4

By cross multiplication, we get

p/–20 – (–18) = q/–9 – (–18) = 1/144 – 180

p/–2 = q/–1 = 1/–36

p/–2 = – 1/36 and q/–1 = 1/–36

p = 1/18 and q = 1/36

p = 1/x = 1/18 and q = 1/y = 1/36

x = 18 and y = 36

Hence, number of days taken by a woman = 18 and number of days taken by a man = 36

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Ans. Let the speed of train and bus be u km/h and v km/h respectively.

According to the given information,

60/u + 240/v = 4 ... **(i)**

100/u + 200/v = 25/6 ... **(ii)**

Putting 1/u = p and 1/v = q in the equations, we get

60p + 240q = 4 ... **(iii)**

100p + 200q = 25/6

600p + 1200q = 25 ... **(iv)**

Multiplying equation **(iii)** by 10, we get

600p + 2400q = 40 .... **(v)**

Subtracting equation **(iv)** from **(v)**, we get1200q = 15

q = 15/200 = 1/80 ... **(vi)**

Putting equation **(iii)**, we get

60p + 3 = 4

60p = 1

p = 1/60

p = 1/u = 1/60 and q = 1/v = 1/80

u = 60 and v = 80

Hence, speed of train = 60 km/h and speed of bus = 80 km/h.