---
title: "NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3"
url: https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-3
dateModified: 2026-10-07T15:50:31+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

Chapter 3: Pair of Linear Equations in Two Variables. Every question from Exercise 3.3, with full working and the final answer.

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## EXERCISE 3.3

### Question 1

*3 marks · Short answer*

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

**Part (i)**

1. Given equations are $x + y = 5$ and $2x - 3y = 4$.
2. Multiplying the first equation by $3$, we get $3x + 3y = 15$.
3. Adding this to $2x - 3y = 4$ gives $5x = 19$, so $x = \frac{19}{5}$.
4. Substituting $x = \frac{19}{5}$ in $x + y = 5$ gives $y = 5 - \frac{19}{5} = \frac{6}{5}$.
5. The solution is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

Answer (i): $x = \frac{19}{5}, y = \frac{6}{5}$

**Part (ii)**

1. Given equations are $3x + 4y = 10$ and $2x - 2y = 2$.
2. Dividing the second equation by $2$, we get $x - y = 1$, or $x = y + 1$.
3. Substituting $x = y + 1$ in the first equation gives $3(y + 1) + 4y = 10$.
4. This simplifies to $7y + 3 = 10$, which gives $7y = 7$, or $y = 1$.
5. Substituting $y = 1$ gives $x = 1 + 1 = 2$.
6. The solution is $x = 2$ and $y = 1$.

Answer (ii): $x = 2, y = 1$

**Part (iii)**

1. Given equations are $3x - 5y - 4 = 0$ (or $3x - 5y = 4$) and $9x - 2y = 7$.
2. Multiplying the first equation by $3$, we get $9x - 15y = 12$.
3. Subtracting this from $9x - 2y = 7$ gives $13y = -5$, so $y = -\frac{5}{13}$.
4. Substituting $y = -\frac{5}{13}$ in $3x - 5y = 4$ gives $3x - 5\left(-\frac{5}{13}\right) = 4$.
5. This simplifies to $3x = 4 - \frac{25}{13} = \frac{27}{13}$, so $x = \frac{9}{13}$.
6. The solution is $x = \frac{9}{13}$ and $y = -\frac{5}{13}$.

Answer (iii): $x = \frac{9}{13}, y = -\frac{5}{13}$

**Part (iv)**

1. Given equations are $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$.
2. Multiplying the first equation by $6$, we get $3x + 4y = -6$. Multiplying the second by $3$, we get $3x - y = 9$.
3. Subtracting the second equation from the first gives $5y = -15$, so $y = -3$.
4. Substituting $y = -3$ in $3x - y = 9$ gives $3x - (-3) = 9$, which means $3x = 6$, so $x = 2$.
5. The solution is $x = 2$ and $y = -3$.

Answer (iv): $x = 2, y = -3$

**Answer:** Solutions obtained by elimination and substitution methods.

> Common mistake: Making errors with signs when eliminating or substituting fractions.

### Question 2

*5 marks · Case-based*

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 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.

**Part (i)**

1. Let the numerator be $x$ and the denominator be $y$, so the fraction is $\frac{x}{y}$.
2. According to the question, $\frac{x + 1}{y - 1} = 1 \implies x - y = -2$ and $\frac{x}{y + 1} = \frac{1}{2} \implies 2x - y = 1$.
3. Subtracting the first equation from the second gives $x = 3$.
4. Substituting $x = 3$ into $x - y = -2$ gives $y = 5$.
5. Therefore, the required fraction is $\frac{3}{5}$.

Answer (i): The fraction is $\frac{3}{5}$.

**Part (ii)**

1. Let Nuri's present age be $x$ years and Sonu's present age be $y$ years.
2. Five years ago, $x - 5 = 3(y - 5) \implies x - 3y = -10$.
3. Ten years later, $x + 10 = 2(y + 10) \implies x - 2y = 10$.
4. Subtracting the first equation from the second gives $y = 20$.
5. Substituting $y = 20$ into $x - 2y = 10$ gives $x = 50$.
6. Therefore, Nuri's age is $50$ years and Sonu's age is $20$ years.

Answer (ii): Nuri is $50$ years old and Sonu is $20$ years old.

**Part (iii)**

1. Let the ten's digit be $x$ and the unit's digit be $y$. The number is $10x + y$.
2. The sum of the digits is $9$, so $x + y = 9$.
3. Nine times the number is twice the number obtained by reversing the digits: $9(10x + y) = 2(10y + x)$.
4. Simplifying gives $88x - 11y = 0 \implies 8x - y = 0$.
5. Adding $x + y = 9$ and $8x - y = 0$ gives $9x = 9 \implies x = 1$, and $y = 8$.
6. Therefore, the number is $18$.

Answer (iii): The number is $18$.

**Part (iv)**

1. Let the number of ₹ 50 notes be $x$ and ₹ 100 notes be $y$.
2. Total notes: $x + y = 25$.
3. Total amount: $50x + 100y = 2000 \implies x + 2y = 40$.
4. Subtracting the first equation from the second gives $y = 15$.
5. Substituting $y = 15$ into $x + y = 25$ gives $x = 10$.
6. Therefore, Meena received $10$ notes of ₹ 50 and $15$ notes of ₹ 100.

Answer (iv): $10$ notes of ₹ 50 and $15$ notes of ₹ 100.

**Part (v)**

1. Let the fixed charge for the first three days be ₹ $x$ and the charge for each extra day be ₹ $y$.
2. Saritha paid ₹ 27 for 7 days: $x + 4y = 27$.
3. Susy paid ₹ 21 for 5 days: $x + 2y = 21$.
4. Subtracting the second equation from the first gives $2y = 6 \implies y = 3$.
5. Substituting $y = 3$ into $x + 2y = 21$ gives $x = 15$.
6. Therefore, the fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.

Answer (v): Fixed charge is ₹ 15 and extra day charge is ₹ 3.

**Answer:** Solutions to all five parts found using elimination method.

> Common mistake: Forming incorrect algebraic equations from the given word statements or making sign errors during subtraction in the elimination method.

## Related pages

- [All Chapter 3 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions)
- [Exercise 3.1](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-1)
- [Exercise 3.2](https://www.swavid.com/maths/class/10/chapter/pair-of-linear-equations-in-two-variables/ncert-solutions/exercise-3-2)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student