Exercise 3.4

- 8x + 5y = 9 3x + 2y = 4 Solve the following pair of linear equation by cross -…
- 2x + 3y = 46 3x + 5y = 74 Solve the following pair of linear equation by cross…
- x + 4y + 9 = 0 5x - 1 = 3y Solve the following pair of linear equation by cross…
- 2x + 3y - 7 = 0 6x + 5y - 11 = 0 Solve the following pair of linear equation by…
- 2/x + 3/y = 13 5/x - 4/y = - 2 Solve the following pair of linear equation by…
- x/3 - y/12 = 19/4 x/3 - y/12 = 19/4 Solve the following pair of linear equation…
- ax + by = a - b bx - ay = a + b Solve the following pair of equations by cross…
- a^2 - b^2 x/a^2 + y/b^2 = 2 a not equal 0 , b not equal 0 Solve the following…
- x - y = a + b ax + by = a^2 - b^2 Solve the following pair of equations by…
- 2x/a + y/b = 2 x/a - y/b = 4 a not equal 0 , b not equal 0 Solve the following…
- 2ax + 3by = a + 2b 3ax + 2by = 2a + b Solve the following pair of equations by…
- x/a + y/b = 2 ax + by = a^2 - b^2 Solve the following pair of equations by…
- Solve the following system of equations by cross - multiplication method. a(x +…
- x - 3y - 7 = 0 3x - 3y - 15 = 0 Which of the following pair of linear equations…
- 2x + y = 5 3x + 2y = 8 Which of the following pair of linear equations has…
- 3x - 5y = 20 6x - 10y = 40 Which of the following pair of linear equations has…
- x - 3y - 3 = 0 3x - 9y - 2 = 0 Which of the following pair of linear equations…
- x + y = 2 2x + 2y = 4 Which of the following pair of linear equations has…
- x + y = 2 2x + 2y = 6 Which of the following pair of linear equations has…
- 15/x+y + 7/x-y - 10 = 0 15/x+y + 7/x-y - 10 = 0 [Hint: Let u = 1/x-y and v =…
- ax - ay = 2 (a - 1)x + (a + 1)y = 2(a^2 + 1) [Hint: Let u = 1/x-y and v =…
- If the cost of 2 pencils and 3 erasers is Rs. 9 and the cost of 4 pencils and 6…
- The paths traced by the wheels of two trains are given by equations x + 2y - 4 =…
- The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditure…
- The sum of two - digits number and the number obtained by reversing the digits…
- If we add 1 to the numerator and subtract 1 from the denominator, a fraction…
- The cost of 5 oranges and 3 apples is Rs. 35 and the cost of 2 oranges and 4…
- A part of monthly hostel charges is fixed and the remaining depends on the…
- A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes…

###### Exercise 3.4

Solve the following pair of linear equation by cross - multiplication method:

8x + 5y = 9

3x + 2y = 4

Answer:

Given, pair of equations is

8x + 5y – 9 = 0 and 3x + 2y – 4 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = – 2

On taking II and III ratio, we get

⇒ y = 5

Question 2.

Solve the following pair of linear equation by cross - multiplication method:

2x + 3y = 46

3x + 5y = 74

Answer:

Given, pair of equations is

2x + 3y – 46 = 0 and 3x + 5y – 74 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 8

On taking II and III ratio, we get

⇒ y = 10

Question 3.

Solve the following pair of linear equation by cross - multiplication method:

x + 4y + 9 = 0

5x – 1 = 3y

Answer:

Given, pair of equations is

x + 4y + 9 = 0 and 5x – 3y – 1 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = – 1

On taking II and III ratio, we get

⇒ y = – 2

Question 4.

Solve the following pair of linear equation by cross - multiplication method:

2x + 3y – 7 = 0

6x + 5y – 11 = 0

Answer:

Given, pair of equations is

2x + 3y – 7 = 0 and 6x + 5y – 11 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

Question 5.

Solve the following pair of linear equation by cross - multiplication method:

Answer:

Given, pair of equations is

Let uand v

So, Eq. (1) and (2) reduces to

2u + 3v – 13 = 0

5u – 4v + 2 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ u = 2

On taking II and III ratio, we get

⇒ v = 3

So, u

and v

Question 6.

Solve the following pair of linear equation by cross - multiplication method:

Answer:

Given, pair of equations is

And

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 18

On taking II and III ratio, we get

⇒ y = 15

Question 7.

Solve the following pair of equations by cross - multiplication method.

ax + by = a – b

bx – ay = a + b

Answer:

Given, pair of equations is

ax + by = a – b ⇒ ax + by –(a – b) = 0

bx – ay = a + b ⇒ bx –ay – (a + b) = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 1

On taking II and III ratio, we get

⇒ y = – 1

Question 8.

Solve the following pair of equations by cross - multiplication method.

Answer:

Given, pair of equations is

And

By cross - multiplication method, we have

On taking I and III ratio, we get

x = a2

On taking II and III ratio, we get

⇒ y = b2

Question 9.

Solve the following pair of equations by cross - multiplication method.

x – y = a + b

ax + by = a2 – b2

Answer:

Given, pair of equations is

x – y = a + b ⇒ x – y –(a + b) = 0

ax + by = a2 –b2⇒ ax + by – (a2 –b2 ) = 0

By cross – multiplication method, we have

On taking I and III ratio, we get

⇒ x = a

On taking II and III ratio, we get

⇒ y = – b

Question 10.

Solve the following pair of equations by cross - multiplication method.

Answer:

Given, pair of equations is

And

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒x = 2a

On taking II and III ratio, we get

⇒ y = – 2b

Question 11.

Solve the following pair of equations by cross - multiplication method.

2ax + 3by = a + 2b

3ax + 2by = 2a + b

Answer:

2ax + 3by = a + 2b

3ax + 2by = 2a + b

Given, pair of equations is

2ax + 3by = a + 2b ⇒ 2ax + 3by –(a + 2b) = 0

3ax + 2by = 2a + b ⇒ 3ax + 2by –(2a + b) = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

Question 12.

Solve the following pair of equations by cross - multiplication method.

ax + by = a2 – b2

Answer:

ax + by =

Given, pair of equations is

ax + by = a2 –b2⇒ ax + by – (a2 –b2 ) = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = a

On taking II and III ratio, we get

⇒ y = b

Question 13.

Solve the following system of equations by cross - multiplication method.

a(x + y) + b(x – y) = a2 – ab + b2

a(x + y) – b(x – y) = a2 + ab + b2

Answer:

The given system of equations can be re - written as

ax + ay + bx – by – a2 + ab – b2 = 0

⇒(a + b)x + (a – b)y – (a2 – ab + b2 ) = 0 …(1)

and ax + ay – bx + by – a2 – ab – b2 = 0

⇒ (a – b)x + (a + b)y – (a2 + ab + b2 ) = 0 …(2)

Now, by cross – multiplication method, we have

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

Question 14.

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

x – 3y – 7 = 0

3x – 3y – 15 = 0

Answer:

Given pair of linear equations

x – 3y – 7 = 0

3x – 3y – 15 = 0

⇒ x – y – 5 = 0 …(ii)

As we can see that a1 = 1, b1 = – 3 and c1 = – 7

and a2 = 1, b2 = – 1 and c2 = – 5

∴ Given pair of equations has unique solution

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 4

On taking II and III ratio, we get

⇒ y = – 1

Question 15.

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

2x + y = 5

3x + 2y = 8

Answer:

Given pair of linear equations

2x + y = 5

3x + 2y = 8

As we can see that a1 = 2, b1 = 1 and c1 = – 5

and a2 = 3, b2 = 2 and c2 = – 8

∴ Given pair of equations has unique solution

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 2

On taking II and III ratio, we get

⇒ y = 1

Question 16.

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

3x – 5y = 20

6x – 10y = 40

Answer:

Given pair of linear equations

3x – 5y = 20

6x – 10y = 40

⇒ 3x – 5y = 20 …(ii)

As we can see that a1 = 3, b1 = – 5 and c1 = – 20

and a2 = 3, b2 = – 5 and c2 = – 20

∴ Given pair of equations has infinitely many solutions.

Question 17.

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

x – 3y – 3 = 0

3x – 9y – 2 = 0

Answer:

Given pair of linear equations

x – 3y – 3 = 0

3x – 9y – 2 = 0

As we can see that a1 = 1, b1 = – 3 and c1 = – 3

and a2 = 3, b2 = – 9 and c2 = – 2

∴ Given pair of equations has no solution

Question 18.

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

x + y = 2

2x + 2y = 4

Answer:

Given pair of linear equations

x + y = 2

2x + 2y = 4

⇒ x + y – 2 = 0

As we can see that a1 = 1, b1 = 1 and c1 = – 2

and a2 = 1, b2 = 1 and c2 = – 2

∴ Given pair of equations has infinitely many solution

Question 19.

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

x + y = 2

2x + 2y = 6

Answer:

Given pair of linear equations

x + y = 2

2x + 2y = 6

⇒ x + y – 3 = 0

As we can see that a1 = 1, b1 = 1 and c1 = – 2

and a2 = 1, b2 = 1 and c2 = – 3

∴ Given pair of equations has no solution

Question 20.

Solve the following system of linear equations by cross - multiplication method.

[Hint: Let u = and v = ]

Answer:

Given, pair of equations is

…(1)

…(2)

Let u and v

Now, the Eq. (1) and (2) reduces to

5u + 2v + 1 = 0

15u + 7v – 10 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

⇒ v = 13

So, u …(a)

and v …(b)

On adding Eq. (a) and (b), we get

On putting the value of in Eq. (a), we get

Question 21.

Solve the following system of linear equations by cross - multiplication method.

ax – ay = 2

(a – 1)x + (a + 1)y = 2(a2 + 1)

[Hint: Let u = and v = ]

Answer:

Given, pair of equations is

ax – ay = 2

(a – 1)x + (a + 1)y = 2(a2 + 1)

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

Question 22.

If the cost of 2 pencils and 3 erasers is Rs. 9 and the cost of 4 pencils and 6 erasers is Rs. 18. Find the cost of each pencil and each eraser.

Answer:

Let the cost of one pencil = Rs x

and cost of one eraser = Rs y

According to the question

2x + 3y = 9 …(1)

4x + 6y = 18

⇒2(2x + 3y) = 18

⇒2x + 3y = 9 …(2)

As we can see From Eq. (1) and (2)

∴Given pair of linear equations has infinitely many solutions.

Question 23.

The paths traced by the wheels of two trains are given by equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the paths cross each other?

Answer:

Given paths traced by the wheel of two trains are

x + 2y – 4 = 0 …(i)

2x + 4y – 12 = 0

⇒ x + 2y – 6 = 0 …(ii)

As we can see that a1 = 1, b1 = 2 and c1 = – 4

and a2 = 1, b2 = 2 and c2 = – 6

∴ Given pair of equations has no solution

Hence, two paths will not cross each other.

Question 24.

The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly incomes.

Answer:

Given ratio of incomes = 9:7

And the ratio of their expenditures = 4:3

Saving of each person = Rs. 2000

Let incomes of two persons = 9x and 7x

And their expenditures = 4y and 3y

According to the question,

9x – 4y = 2000

⇒9x – 4y – 2000 = 0 …(i)

7x – 3y = 2000

⇒7x – 3y – 2000 = 0 …(ii)

By cross - multiplication method, we have

On taking I and III ratios, we get

⇒ x = 2000

On taking II and III ratios, we get

⇒ y = 4000

Hence, the monthly incomes of two persons are 9(2000) = Rs18000 and 7(2000) = Rs14000

Question 25.

The sum of two - digits number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

Answer:

Let unit’s digit = y

and the ten’s digit = x

So, the original number = 10x + y

The sum of the number = 10x + y

The sum of the digit = x + y

reversing number = x + 10y

According to the question,

10x + y + x + 10y = 66

⇒11x + 11y = 66

⇒ x + y = 6 …(i)

x – y = 2 …(ii)

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 4

On taking II and III ratio, we get

⇒ y = 2

So, the original number = 10x + y

= 10(4) + 2

= 42

Reversing the number = x + 10y

= 24

Hence, the two digit number is 42 and 24. These are two such numbers.

Question 26.

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we add 1 to the denominator. What is the fraction?

Answer:

Let the numerator = x

and the denominator = y

So, the fraction

According to the question,

Condition I:

⇒ x + 1 = y – 1

⇒ x – y = – 2

⇒ x – y + 2 = 0 …(i)

Condition II:

⇒ 2x = y + 1

⇒ 2x – y = 1

⇒ 2x – y – 1 = 0 …(ii)

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 3

On taking II and III ratio, we get

⇒ y = 5

So, the numerator is 3 and the denominator is 5

Hence, the fraction is

Question 27.

The cost of 5 oranges and 3 apples is Rs. 35 and the cost of 2 oranges and 4 apples is Rs. 28. Find the cost of an orange and an apple.

Answer:

Let the cost of an orange = Rs x

And the cost of an apple = Rs y

According to the question,

5x + 3y = 35

And 2x + 4y = 28

⇒ x + 2y = 14

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 4

On taking II and III ratio, we get

⇒ y = 5

Hence, the cost of an orange is Rs. 4 and cost of an apple is Rs. 5

Question 28.

A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay Rs. 1000 as hostel charges, whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and cost of food per day.

Answer:

Let fixed hostel charge (monthly) = Rs x

and cost of food for one day = Rs y

In case of student A,

x + 20y = 1000

x + 20y – 1000 = 0 …(i)

In case of student B,

x + 26y = 1180

x + 26y – 1180 = 0 …(ii)

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 400

On taking II and III ratio, we get

⇒ y = 30

Hence, monthly fixed charges is Rs. 400 and cost of food per day is Rs. 30

Question 29.

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.

Answer:

Let the numerator = x

and the denominator = y

So, the fraction

According to the question,

Condition I:

⇒ 3(x – 1) = y

⇒ 3x – 3 = y

⇒ 3x – y – 3 = 0 …(i)

Condition II:

⇒ 4x = y + 8

⇒ 4x – y = 8

⇒ 4x – y – 8 = 0 …(ii)

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 5

On taking II and III ratio, we get

⇒ y = 12

So, the numerator is 5 and the denominator is 12

Hence, the fraction is