These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1.

Solve the following pair of linear equations by the elimination method and the substitution

(i) x + y = 5 and 2x – 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

(iv) x/2 + 2y/3 = -1 and x – y/3 = 3

Solution:

(i) By Elimination Method:

Fquations are x + y = 5

and 2x – 3y = 4

Multiply equation (i) by 2 and subtract equation (ii) from it, we have

(ii) By Elimination method:

Equations are 3x + 4y = 10

and 2x – 2y = 2

Multiplying equation (ii) by 2 and adding to equation (i), we

(iii) By Elimination Method:

(iv) By Elimination Method:

Substracting equation (i) from (iii) we get

5y = – 15

⇒ y = – 3

Subtracting the value of y = – 3 in (ii), we get

3x – (- 3) = 9

3x + 3 = 9

3x = 6 x = 2

∴ x = 2 and y = 3

Substituting method

From equation (ii) we have

3x – y = 9

⇒ 3x = 9 + y

⇒ x = \(\frac { 9 + y }{ 3 }\)

Substituting the value of x = equation, (i) we get

3(\(\frac { 9 + y }{ 3 }\)) + 4y = – 6

⇒ 9 + y + 4y = – 6

⇒ 5y = – 15

⇒ y = – 3

Again, substituting the value of y = -3 in equation (i) we get

3x – (-3) = 9

⇒ 3x + 3 = 9

⇒ 3x = 6

⇒ x = 2

∴ x = 2 and y = – 3

Question 2.

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

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

(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?

(iii) 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.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

(v) A lending 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 seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

(i) Let numerator be x and denominator be y.

\(\frac { x + 1 }{ y – 1 }\) = 1

x – y = – 2 … (i)

According to question’s second condition

\(\frac { x }{ y + 1 }\) = \(\frac { 1 }{ 2 }\)

2x – y = 1 … (i)

Substract equation (i) from equation we eliminate y

x = 3

Substitute this value in equation (i) we get

3 – y = – 2

y = 5

Hence the fraction is \(\frac { 3 }{ 5 }\)

(ii) Let the present age of Nuri be x

and let the present age of Sonu be y.

According to questions first condition

(x – 5) = 3 (y – 5)

x – 3y = – 10 … (i)

According to eq. question’s second condition

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

x – 2y = 10 … (ii)

Substracting equation (i) from equation (ii)

We eliminate for x.

y = 20

Putting this value in equation (i)

x – 60 = – 10

x = 50

Present age of Nuri is 50 years old and Sonu is 20 years old.

(iii) Let the ten’s digit of the number be x

and the unit’s digit of the number by y.

According to question’s first condition

x + y = 9 … (i)

According to question’s second condition

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

90x + 9y – 20y + 2x

88x = 11y

8x – y = 0 (dividing by 11)

8x – y = 0 … (ii)

Adding equation (i) and (ii) we eleminate for y, we get

9x = 9

x = 1

Putting the value in equation (i)

1 + y = 9

y = 8

Hence the units digit is 8 and the ten’s digit is 1. Then the number is 18.

(iv) Let the notes of ₹ 50 be x

and let the notes of ₹ 100 be y.

According to question’s first condition

50x + 100y = 2000

x + 2 y= 40 … (i)

According to question second condition

x + y = 25

Substract equation (i) from equation (ii) we eleminate for x.

– y = – 15

y = 15

Putting this value in equation (i)

x + 30 = 40

x = 40

Hence ₹ 50 notes is 10 and the notes of ₹ 100 is 15.

(v) Let the fixed charge be ₹ x and let the additional charge per day be ₹ y.

According to questions first condition

x + 4y = 27 … (i)

According to questions second condition

x + 2y = 21 … (ii)

Eliminating x for equation (i) and (ii)

We subtract equation (ii) from equation (1)

2y = 6

y = 3

Putting this value in equation (ii)

x + 6 = 21

x = 21 – 6

x = 15

Hence the fixed charge is ₹ 15 and the additional charge per day is ₹ 3.