Well-designed AP Board Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3 offers step-by-step explanations to help students understand problem-solving strategies.

## Pair of Linear Equations in Two Variables Class 10 Exercise 3.3 Solutions – 10th Class Maths 3.3 Exercise Solutions

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) \(\frac{x}{2}+\frac{2 y}{3}\) = -1 and x – \(\frac{y}{3}\) = 3

Solution:

i) Given linear equations are

x + y = 5 …… (1)

and 2x – 3y = 4 ……. (2)

Multiply by;2 of (1) to make the x coefficient are equal.

2x + 2y = 2 × 5

2x + 2y = 10

Subtract (2) from (3)

Therefore, x = \(\frac{19}{5}\) and y = \(\frac{6}{5}\)

Check :

Put x = \(\frac{19}{5}\) and y = \(\frac{6}{5}\) in (2)

ii) Given linear equations are

3x + 4y = 10 ……… (1)

and 2x – 2y = 2 ……. (2)

Multiply by 2 of (2) to make the y coefficients are equal.

4x – 4y = 4 ……. (3)

by adding (1) and (3)

Put x = 2 in (1)

3(2) + 4y = 10

4y = 10 – 6 = 4

y = \(\frac{4}{4}\)

y = 1

Therefore, x = 2 and y = 1

Check :

Put x = 2 and y = 1 in (2)

2(2) – 2(1) = 2

4 – 2 = 2

2 = 2

LHS = RHS

So, our solution is correct.

iii) Given linear equations are

3x – 5y – 4 = 0 and 9x = 2y + 7

3x – 5y = 4 ……(1) 9x – 2y = 7 …….(2)

Multiply by 3 of (1) to make the co-efficients of x equal.

9x – 15y = 12 ……..(3)

Subtract (2) from (3)

Put y = \(\frac{-5}{13}\) in (1)

3x – 5(-\(\frac{5}{13}\)) = 4

3x + \(\frac{25}{13}\) = 4

3x = \(\frac{4}{1}\) – \(\frac{25}{13}\)

= \(\frac{4 \times 13-25}{13}\)

= \(\frac{52-25}{13}\)

3x = \(\frac{27}{13}\)

∴ x = \(\frac{9}{13}\)

Therefore, x = \(\frac{9}{13}\) and y = \(\frac{-5}{13}\)

Check :

Put x = \(\frac{9}{13}\) and y = \(\frac{-5}{13}\) in (2)

9(\(\frac{9}{13}\)) – 2(\(\frac{-5}{13}\)) = 7

\(\frac{81}{13}+\frac{10}{13}\) = 7

\(\frac{81+10}{13}\) = 7

7 = 7

LHS = RHS

So, our solution is correct.

iv) Given linear equations are

Put y = -3 in (1)

3x + 4(-3) = -6

3x – 12 = -6

3x = -6 + 12 = 6

Therefore, x = 2 and y = -3

Check :

Put x = 2 and y = -3 in (2)

3(2) – (-3) = 9

6 + 3 = 9

9 = 9

LHS = RHS

So, our solution is correct.

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 \(\frac{1}{2}\) if we only add 1 to the deonominator. 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 the numerator x and denominator y.

Then fraction = \(\frac{x}{y}\)

If we add 1 to numerator and subtract 1 from denominator.

Then \(\frac{x+1}{y-1}\) = 1

x + 1 = y – 1

x – y = -2 ………….(1)

If 1 added to denominator, then

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

2x = y + 1

2x – y = 1 ……….(2)

By subtracting (1) from (2)

Put x = 3 in (1)

3 – y = -2

-y = -2 – 3

-y = -5

y = 5

Fraction = \(\frac{x}{y}\) = \(\frac{3}{5}\)

ii) Let the present ages of Nuri and Sonu are x and y.

Five years ago Nuri age = 3 times of Sonu

x – 5 = 3(y – 5)

x – 5 = 3y – 15

x = 3y – 10 ….. (1)

10 years later Nuri age = 2 times of Sonu

x + 10 = 2(y + 10)

x + 10 = + 20

x – 2y = 20 – 10

x – 2y = 10 …… (2)

Put (1) in (2)

3y – 10 – 2y = 10

3y – 2y = 10 + 10

y = 20

Put y = 20 in (1)

x = 3(20) – 10 = 60 – 10

x = 50

Therefore, Nuri’s present age (x) = 50 years

Sonu’s present age (y) = 20 years

iii) Let the digits in the tens and units place are x and y representively then x + y = 9 ……..(1)

Then two-digited number = 10 × x + 1 × y = 10x + y

If we reversed the digits then number = 10 × y + 1 × x = 10y + x

9 times of number = 2 times of reversed number

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

90x + 9y = 20y + 2x

90x – 2x = 20y – 9y

88x = 11y

8x – y = 0 ……….(2)

By adding (1) and (2)

Put x = 1 in (2)

8(1) – y = 0

8 – y = 0

y = 8

Therefore, the tens digit x = 1 and ones digit (y) = 8

So, number = 10x + y

= 10(1) + 8

= 10 + 8

= 18.

iv) Let the number of ₹ 50 and ₹ 100 notes are x and y respectively.

x + y = 25 ………. (1)

50x + 100y = 2000

50(x + 2y) = 2000

By subtracting (1) from (2)

Put y = 15 in (1)

x + 15 = 25

x = 25 – 15 = 10

Number of ₹ 50 notes (x) = 10

Number of ₹ 100 notes (y) = 15.

v) Let fixed chargees for first three days is ₹ x and additional charges per each day ₹ y.

Sarita paid ₹ 27 per 7 days

= first 3 days + 4 days additional

x + 4y = 27 ……(1)

Susi paid ₹ 1 per 5 days

x + 2y = 21 ……..(2)

By subtracting (2) from (1)

y = \(\frac{6}{2}\) = 3

Put y = 3 in (1)

x + 4(3) = 27

x = 27 – 12 = 15

Therefore, fixed charges for first three days (x) = ₹ 15

additional charges per each day (y) = ₹ 3.