# NCERT solutions for class 10 Maths chapter 3 Pair of Linear Equations in Two Variables (Exercise 3.4)

## Ex 3.4 Question 1.

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

(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

Solutions:

(i) x + y = 5 and 2x – 3y = 4

Elimination method

x + y = 5 …………. (i)

2x – 3y = 4 …………..(ii)

multiplying equation (i) by 2,

2x + 2y = 10 ……………(iii)

When the equation (ii) is subtracted from (iii) ,

5y = 6

y = 6/5

Substituting the value of y in eq. (i) ,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

Substitution method

From the equation (i),

x = 5 – y…….. (iv)

Putting the value of equation (iv) in equation (ii),

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

substituted the value of y equation (iv),

x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

Elimination method:

3x + 4y = 10      …(i)

2x – 2y = 2         …(ii)

Multiplying equation (i) by 3,

4x – 4y = 4         …(iii)

7x = 14

x = 2

Substituting the value of x in equation (1),

6+ 4y = 10

4y = 4

y = 1

Hence, x = 2, y = 1

Substitution method:

From equation (ii),

x = 1 + y ……..(iv)

Putting this value in equation (1),

3(1 + y) + 4y = 10.

7y = 7

y = 1

Substituting the value of y in equation (iv),

x = 1 + 1 = 2

x = 2 and y = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

Elimination method:

3x – 5y – 4 = 0 ………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………(ii)

multiplying equation (i) by (iii)

9x – 15y – 12 = 0 …… (iii)

Subtracting equation (iii from equation (ii),

13y = -5

y = -5/13

Substituting the value of y equation (i)

3x +25/13 −4=0

3x = 27/13

x =9/13

x = 9/13 and y = -5/13

Substitution method:

From the equation (i),

x = (5y+4)/3……… (iv)

Putting the value of equation (iv) in equation (ii),

9(5y+4)/3 −2y −7=0

13y = -5

y = -5/13

Substituting the value of y in equation (v) ,

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

x = 9/13

x =9/13, y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

Elimination method:

3x + 4y = -6 ………. (i)

x-y/3 = 3

3x – y = 9 ……………. (ii)

Subtracting equation (ii)  from equation (i),

-5y = -15

y = 3

Substituting the value of y in equation (i)

3x – 12 = -6

3x = 6

x = 2

x = 2 , y = -3

Substitution method:

From the equation (ii) ,

x = (y+9)/3 ……(iii)

Putting the value of x in equation (i) ,

3(y+9)/3 +4y =−6

5y = -15

y = -3

When y = -3 is substituting the valueof y in equation (iii),

x = (-3+9)/3 = 2

x = 2 and y = -3

## Ex 3.4 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 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.

Solution:

(i) Let the fraction be x/y

(x+1)/(y-1) = 1

⇒ x – y = -2 …………..(i)

x/(y+1) = 1/2

=> 2x-y = 1…………(ii)

Subtracting equation (i) from equation (ii),

x = 3

Substituting this value of x in equation (i1),

3 – y = -2

-y = -5

y = 5

∴The fraction is 3/5

(ii) Let us assume, present age of Nuri is x and Sonu is y.

x – 5 = 3(y – 5)

x – 3y = -10………………..(i)

x + 10 = 2(y +10)

x – 2y = 10…………….(ii)

Subtracting equation (i) from equation (ii),

y = 20

Substituting the value of y in equation (i),

x – 3.20 = -10

x – 60 = -10

x = 50

Therefore,

Present age of Nuri = 50 years

Present age of Sonu = 20 years.

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

Number (n) = 10By + x

Number after reversing order of the digits = 10x + y

According to the question,

x + y = 9……….(i)

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

88 y – 11 x = 0

-x + 8y = 0 …….. (ii)

Adding the equations (i) and (ii),

9B = 9

y=1 ………………….(3)

Substituting this value of y, in the equation (i)

A= 8

∴ The number (N) is 10y + x = 10 x 1 +8 = 18

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

x + y = 25 ……….. (i)

50x + 100y = 2000 …………(ii)

Multiplying equation (i) by 50,

50x + 50 y = 1250           …. (iii)

Subtracting equation (iii) from equation (ii),

50y = 750

y = 15

Substituting the value of y in equation (i),

x = 10

∴Meena received 10 notes of Rs 50 and 15 notes of Rs 100.

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

According to the information given,

x+ 4y = 27 …………. (i)

x + 2y = 21………….. (ii)

Subtracting equation (ii) from equation (i),

2y = 6

y = 3 ………(iii)

Substituting the value of y in equation (i),

x + 12 = 27

x = 15

∴The fixed charge is Rs.15 and the Charge per day is Rs.3