**Download NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.4**

**1. Solve the following pair of linear equation by the elimination method and substitution method.**

**(i) x+y = 5 and 2x-3y = 4**

**Solution:**

x+y = 5………(1)

2x-3y = 4……..(2)

**Elimination method**: Multiplying both side of equation (1) by 3 and equation (2) by 1 we get the equations as

3x+3y = 15……(3)

2x-3y = 4……….(4)

Adding equation (3) and equation (4) we have,

5x = 19

x =

Substituting this value of x in equation (1)

+y = 5

y = 5-

=

=

Hence the solution is x = and y =

**Substitution method: **

From equation (1) we have

y = 5-x ………..(3)

Substituting the above value of y in equation (2) we get

2x-3(5-x) = 4

2x-15+3x = 4

5x = 19

x =

And Substituting the value of x = in equation (3),

y = 5-

=

Hence the required solution is x = and y =

**(ii) 3x+4y = 10 and 2x-2y = 2**

**Solution:**

3x+4y = 10…….(1)

2x-2y = 2………(2)

**Elimination method: **Multiplying both side of equation (1) by 1 and equation (2) by 2 we get the equations as

3x+4y = 10……(3)

4x-4y = 4……….(4)

Adding equation (3) and equation (4) we have,

7x = 14

x =

x = 2

Substituting this value of x in equation (1)

2×2-2y = 2

4-2y = 2

2y = 2

y = 1

Hence the solution is x = 2 and y = 1

**Substitution method: **

From equation (2) we have

x-y = 1

x = y+1…….(5)

Substituting the above value of x in equation (1) we get

3(y+1)+4y = 10

7y+3 = 10

7y = 7

y = 1

And substituting the value of y in equation (5),

x= 1+1 = 2

Hence the required solution is x =2 and y = 1

**(iii) 3x-5y-4 = 0 and 9x = 2y+7**

**Solution:** The given equations can be written as

3x-5y = 4……(1)

9x-2y = 7…….(2)

**Elimination method:**

Multiply both side of equation (1) by 2 and equation (2) by 5 the equations becomes

6x-10y = 8……….(3)

45x-10y = 35……….(4)

Subtracting the above two equations,

-39x = -27

x =

x =

Substituting x = in the equation (1)

3. – 5y = 4

5y = – 4

5y = –

y = –

Hence the solution is x = and y = –

**Substitution method: **

The equation (2) can be written as

x = ……….(5)

Hence from equation (1)

3× -5y-4 = 0

= 0

-39y-15 = 0

y = –

y = –

Substituting the above value of x in equation (5) we get

x =

x =

x =

x =

Hence the required solution is x = and y = –

**(iv) + = -1 and x- = 3**

Solution: The given equations can be written as

= -1

3x+4y = -6……(1) and

= 3

3x-y = 9…….(2)

**Elimination method:**

Subtracting equation (1) by equation (2)

5y = -15

y = -3

Substituting y = -3 in equation (2)

3x-(-3) = 9

3x = 9-3

3x = 6

x = 2

Hence the solution is x = 2 and y = -3

**Substitution method: **

The equation (2) can be written as

y = 3x-9 ……….(3)

Hence in equation (1) substituting the above value of y,

3x+4(3x-9) = -6

3x+12x-36= -6

15x = 30

x =

x = 2

Substituting the above value of x in equation (3) we get

y = 3×2-9

y = 6-9

y = -3

Hence the required solution is x = 2 and y = -3

**2. Form the pair of linear equations in the following problems and find their solutions (if they exists) by the elimination of method:**

**(i) If we add 1 to the numerator and subtract 1 from the denominator, the fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?**

**Solution:** Let the numerator be x and the denominator be y of the fraction. i.e., the fraction be

Then by given condition,

= 1……….(1)

And = ………..(2)

The equation (1) can be written as,

x+1 = y-1

x-y = -2………(3)

And the equation (2) can be written as,

2x = y+1

2x-y = 1……….(4)

Subtracting equation (4) from equation (3),

x= 3

Putting x = 3 in equation (4)

2×3-y = 1

6-y = 1

y = 5

Hence the fraction is = .

**(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?**

**Solution:** Let Nuri is x years old and Sonu is y years old.

Then by using the given condition,

x-5 = 3(y-5)………….(1) and

x+10 = 2(y+10)………(2)

The above two equations can be written as,

x-3y = 5-15

x-3y = -10 ………..(3)

And

x-2y = 20-10

x-2y = 10………….(4)

Subtracting equation (4) from equation (3),

y= 20

Substituting y = 20 in equation (4)

x = 10+2× 20

= 50

Therefore Nuri is 50 years old and Sonu is 20 years old.

**(iii) The sum of the digits of a two digits number is 9. Also nine times this number is twice the number obtained by reversing the order of the digits. Find the number.**

**Solution:** Let the 1^{st} digit of the number is x and the 2^{nd} digit is y.

That is the number is 10x+y.

Since the sum of the digits is 9

Then x+y = 9………(1)

Now if we reverse the order of the digit, the 1^{st} digit will be y and the 2nd digit will be x and hence the number is 10y+x.

Using the given condition,

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

90x+9y = 20y+2x

88x-11y = 0

8x-y = 0…………(2)

Adding equation (1) and equation (2)

9x = 9

x = 1

Substituting this value in equation (2)

8× 1-y = 0

y = 8

Hence the number is 10× 1+8 = 18.

**(iv) Meena went to a bank to with ₹ She asked the cashier to give her ₹50 and ₹100 notes only. Meena get 25 notes in all. Find how many notes of ₹50 and ₹100 she recived.**

**Solution:** Let the number of notes of ₹50 is x and that of ₹100 is y.

Then from the given condition we have,

x+y = 25……..(1) and

50x+100y = 2000………..(2)

Multiplying both side of equation (1) by 50 and equation (2) by 1

50x+50y = 1250……….(3) and

50x+100y = 2000………..(4)

Subtracting equation (4) from equation (3),

50y = 750

y =

y = 15

Substituting this value in equation (1)

x = 25-15 = 10

Therefore the number of notes of ₹50 is 10 and ₹100 is 15.

**(v) A landing library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for 7 days, while Susy paid ₹21 for the book she kept for five days. Find fixed charge and the charge for each extra day.**

**Solution:** Let x be the fixed charge for three days and y be the charge for each extra day.

Therefore Saritha keep the book extra 7-3 = 4 days and Susy keep the book extra 5-3 = 2 days.

Then using given conditions we get the equations as,

x+4y = 27….(1) and

x+2y = 21…..(2)

Subtracting equation (1) by equation (2) we have,

2y = 6

y = 3

Substituting the value of y in equation (2)

x+2×3 = 21

x+6 = 21

x = 15

Hence the fixed charge for first 3 days is ₹15 and the charge for each extra day is ₹3.

**Download NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.4**