NCERT Solutions For Class 10 Maths Chapter – 3 Exercise – 3.4

NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.4

 

Q1. Solve the following pair of linear equation by the elimination method.

  • X + y = 5 and 2x – 3y = 4
  • 3x + 4y = 1- and 2x – 2y = 2
  • 3x  – 5y -4 = 0 and 9x = 2y + 7
  • X/2 + 2y/3 = -1 and x – y/3 = 3

Solution:-

 

  • x + y = 5 and 2x – 3y = 4

By elimination method

X + y  = 5

2x – 3y = 4

Multiplying equation (I) by (ii) we get

2x + 2y = 10

2x

– 3y = 4

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

 

  • 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

Adding(I) and (iii)  we get

7x + 0 =14

Dividing both sides  by 7 we get

X = 14/7

X = 2

Putting equation (I) we get

3x + 4y = 10

3(2) + 4y = 10

Y= 1

 

  • 3x –  5y – 4 = 0 and 9x = 2y + 7

By elimination method

3x – 5y – 4 = 0

3x – 5y = 4

9x = 2y + 7

9x – 2y = 7 (ii)

Multiplying equation (I) by 3 we get

9x – 15y = 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

3x- 5(-5/13) = 4

Multiplying by 13 we get

39x + 25 = 52

X = 9/13

 

  • X/2 + 2y/3 = -1 and x – y/3 = 3

By elimination method

X/2 + 2y/3 = -1

X – y/3 = 3

Multiplying equation (I) by 2 we get

X + 4y/3 = -2

X – y/3 = 3 (ii)

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

5y/3 = -5

Y = -3

 

Putting value of y in this equation

X – y/3 = 3

X – (-3)/3 = 3

X = 2

 

Q2. From the pair of linear equations in the following problems, and find their solutions ( if they exist) by the elimination method:

  • If we add 1 to the numerator and subtract 1 from the the denominator , a fraction reduces to 1. It becomes 1/2 if  we only add 1 to the denominator What is the fraction ?
  • 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 ?
  • The sum of the digits of a two – digit number is 9. Also nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
  • Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
  • A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days while susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge and the charge and the charge for each extra day.

Solution:-

 

  • 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

 

  • X – 3y = 10 (I)

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

X – 2y = 10 (ii)

Subtracting equation (I) from equation (ii) we get

Y = 20

Putting this value in equation (I) we get’

X – 60 = – 10

X = 50

 

  • ATQ

X + y = 9 (I)

9(10 y + x )

= 2(10x + y)

88y – 11x = 0

-x + 8y = 0 (ii)

Adding equation (I) and (ii) we get

9y = 9

Y = 1

Putting the value in equation (I) we get

X = 8

 

  • 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

 

  • ATQ

X + 4y = 27 (I)

X + 2y = 21 (ii)

Subtracting equation (ii) from equation (I) we get

2y = 6

Y = 3

Putting in equation (I) we get

X + 12 = 27

X = 15

 

 

 

NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.4   Q1. Solve the…