# Class 10 NCERT Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.3

Last Updated : 03 Apr, 2024

### (i) x + y = 5 and 2x â€“ 3y = 4Â

Solution:

Here, the two given eqn. are as follows:

x + y = 5 ……….(I)

2x â€“ 3y = 4 ………..(II)

ELIMINATION METHOD:

Multiply equation (I) by 2, and then subtract (II) from it, we get

5y = 6

y = 6/5

Now putting y=6/5 in eqn. (I), we get

x + 6/5 = 5

x = (5â€“(6/5))

x = 19/5

SUBSTITUTION METHOD:

From (I), we get

y=5â€“x…….(III)

Now substituting the value of y in eqn. (II), we get

2x â€“ 3(5â€“x) = 4Â

2x â€“ 15+3x = 4Â

5x = 4+15

x = 19/5

As, putting x = 19/5, in eqn. (III), we get

y = 5 â€“ 19/5

y = 6/5

Hence, by elimination method and substitution method we get,

x = 19/5 and y = 6/5.

### (ii) 3x + 4y = 10 and 2x â€“ 2y = 2

Solution:

Here, the two given eqn. are as follows:

3x + 4y = 10 ……….(I)

2x â€“ 2y = 2 ………..(II)

ELIMINATION METHOD:

Multiply equation (II) by 2, and then add it to (I), we get

7x = 14

x = 14/7Â

x = 2

Now putting x = 2 in eqn. (I), we get

3(2) + 4y = 10

4y = 10 â€“ 6

y = 4/4

y = 1

SUBSTITUTION METHOD:

From (II), we get

x = (2+2y)/2Â

x = y+1 …….(III)

Now substituting the value of x in eqn. (I), we get

3(y+1) + 4y = 10

3y + 3 + 4y = 10

7y = 10 â€“ 3

y = 7/7

y = 1

As, putting y = 1, in eqn. (III), we get

x = 1+1

x = 2

Hence, by elimination method and substitution method we get,

x = 2 and y = 1.

### (iii) 3x â€“ 5y â€“ 4 = 0 and 9x = 2y + 7Â

Solution:

Here, the two given eqn. are as follows:

3x â€“ 5y â€“ 4 = 0

9x = 2y + 7

By rearranging we get,

3x â€“ 5y = 4 ……….(I)

9x â€“ 2y = 7 ………..(II)

ELIMINATION METHOD:

Multiply equation (I) by 3, and then subtract (II) from it, we get

â€“13y = 5

y = -5/13

Now putting y = â€“ 5/13 in eqn. (I), we get

3x â€“ 5(â€“ 5/13) = 4

3x Â = 4 â€“ (25/13)

3x = 27/13

x = 9/13

SUBSTITUTION METHOD:

From (I), we get

3x â€“ 5y = 4

x = (4+5y)/3 …….(III)

Now substituting the value of x in eqn. (II), we get

9((4+5y)/3) â€“ 2y = 7

3(4+5y) â€“ 2y = 7

12+15y â€“ 2y = 7

13y = â€“ 5

y = â€“ 5/13

As, putting y = â€“ 5/13, in eqn. (III), we get

x=(4+5(â€“ 5/13))/3

x = 9/13

Hence, by elimination method and substitution method we get,

x=9/13 and y=â€“ 5/13.

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

Solution:

Here, the two given eqn. are as follows:

x/2 + 2y/3 = â€“1 ………..(A)

x â€“ y/3 = 3 ……………….(B)

by rearranging (multiply (A) by 6 and multiply (B) by 3) we get,

3x + 4y = â€“ 6 ……….(I)

3x â€“ y = 9 Â ………..(II)

ELIMINATION METHOD:

Subtract (II) from (I), we get

5y = â€“ 15

y = â€“ 3

Now putting y = -3 in eqn. (II), we get

3x â€“ (â€“ 3) = 9Â

3x = 9 â€“ 3

x = 6/3

x = 2

SUBSTITUTION METHOD:

From (II), we get

3x â€“ y = 9

y = 3x â€“ 9 …….(III)

Now substituting the value of y in eqn. (I), we get

3x + 4(3x â€“ 9) = â€“ 6Â

3x + 12x â€“ 36 = â€“ 6Â

15x = â€“ 6 + 36

x = 30/15

x = 2

As, putting x = 2, in eqn. (III), we get

y = 3(2) â€“ 9

y = â€“ 3

Hence, by elimination method and substitution method we get,

x=2 and y=â€“ 3.

### (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes Â½ if we only add 1 to the denominator. What is the fraction?

Solution:

Let the fraction be p/q, where p is numerator and q is denominator.

Here, According to the given condition,

(p+1)/(q â€“ 1) = 1 ………………..(A)

and,

p/(q+1) = 1/2 …………………..(B)

Solving (A), we get

(p+1) = q â€“ 1

p â€“ q = â€“ 2 ……………………….(I)

Now, solving (B), we get

2p = (q+1)

2p â€“ q = 1 ……………………….(II)

When equation (I) is subtracted from equation (II) we get,

p = 3

Now putting p = 3 in eqn. (I), we get

3 â€“ q = â€“ 2

q = 3+2

q = 5

So, p = 3 and q = 5.

Hence, the fraction p/q is 3/5.

### (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 us assume, present age of Nuri is x

And present age of Sonu is y.

Here, According to the given condition, the equation formed will be as follows :

x â€“ 5 = 3(y â€“ 5)

x â€“ 3y = â€“ 10 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(I)

Now,

x + 10 = 2(y +10)

x â€“ 2y = 10 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(II)

Subtract eqn. (I) from (II), we get

y = 20

Now putting y = 20 in eqn. (II), we get

x â€“ 2(20) = 10

x = 10+40

x = 50

Hence,Â

Age of Nuri is 50 years

Age of Sonu is 20 years.

### (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.

Solution:

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

Then, Number = 10y + x

And, reverse number = 10x + y

eg: 23Â

x = 3 and y = 2

So, 23 can be represented as = 10(2) + 3 = 23

Here, According to the given condition

x + y = 9 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(I)

and,

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

90y + 9x = 20x + 2y

88y = 11x

x = 8yÂ

x â€“ 8y = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (II)

Subtract eqn. (II) from (I) we get,

9y = 9

y = 1

Now putting y = 1 in eqn. (II), we get

x – 8(1) = 0

x = 8

Hence, the number is 10y + xÂ

=10 Ã— 1 + 8Â

Number = 18

### (iv) Meena went to a bank to withdraw â‚¹ 2000. She asked the cashier to give her â‚¹ 50 and â‚¹ 100 notes only. Meena got 25 notes in all. Find how many notes of â‚¹ 50 and â‚¹ 100 she received.

Solution:

Let the number of â‚¹ 50 notes be x and the number of â‚¹100 notes be y

Here, According to the given condition

x + y = 25 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (I)

50x + 100y = 2000 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(II)

Divide (II) by 50 and then subtract (I) from it.

y = 15

Now putting y = 15 in eqn. (I), we get

x + 15 = 25

x = 10

Hence, Manna has 10 notes of â‚¹ 50 and 15 notes of â‚¹ 100.

### (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid â‚¹ 27 for a book kept for seven days, while Susy paid â‚¹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.Â

Solution:

Let the fixed charge for the first three days be â‚¹ x and,

The charge for each day extra be â‚¹ y.

Here, According to the given condition,

x + 4y = 27 â€¦â€¦.â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (I)

x + 2y = 21 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (II)

Subtract (II) from (I), we get

2y = 6

y = 3

Now putting y = 3 in eqn. (II), we get

x + 4(3) = 27

x = 27 â€“ 12

x = 15

Hence, the fixed charge is â‚¹15

And the Charge per day is â‚¹ 3

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