NCERT Class 10 Maths – Exercise 3.3 Solutions

NCERT Class 10 Maths

Chapter 3 – Pair of Linear Equations in Two Variables | Exercise 3.3

(Rationalized Syllabus 2025-26 – Elimination Method)

Q1

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

(i) $x + y = 5$ and $2x – 3y = 4$
By Elimination Method:
$$x + y = 5 \quad \text{…(1)}$$
$$2x – 3y = 4 \quad \text{…(2)}$$
Multiply equation (1) by 3 to make coefficients of $y$ equal:
$$3x + 3y = 15 \quad \text{…(3)}$$
Add equation (2) and (3):
$$(2x – 3y) + (3x + 3y) = 4 + 15$$
$$5x = 19 \Rightarrow x = \frac{19}{5}$$
Substitute $x$ in (1):
$$\frac{19}{5} + y = 5 \Rightarrow y = 5 – \frac{19}{5} = \frac{25 – 19}{5} = \frac{6}{5}$$
✔️ x = 19/5, y = 6/5
(ii) $3x + 4y = 10$ and $2x – 2y = 2$
By Elimination Method:
$$3x + 4y = 10 \quad \text{…(1)}$$
$$2x – 2y = 2 \quad \text{…(2)}$$
Divide equation (2) by 2:
$$x – y = 1 \quad \text{…(3)}$$
Multiply equation (3) by 4:
$$4x – 4y = 4 \quad \text{…(4)}$$
Add (1) and (4):
$$3x + 4x = 10 + 4 \Rightarrow 7x = 14 \Rightarrow x = 2$$
Substitute $x=2$ in (3):
$$2 – y = 1 \Rightarrow y = 1$$
✔️ x = 2, y = 1
(iii) $3x – 5y – 4 = 0$ and $9x = 2y + 7$
Rewrite in standard form:
$$3x – 5y = 4 \quad \text{…(1)}$$
$$9x – 2y = 7 \quad \text{…(2)}$$
Multiply equation (1) by 3:
$$9x – 15y = 12 \quad \text{…(3)}$$
Subtract equation (3) from (2):
$$(9x – 2y) – (9x – 15y) = 7 – 12$$
$$13y = -5 \Rightarrow y = -\frac{5}{13}$$
Substitute $y$ in (1):
$$3x – 5(-\frac{5}{13}) = 4$$
$$3x + \frac{25}{13} = 4 \Rightarrow 3x = 4 – \frac{25}{13} = \frac{52-25}{13} = \frac{27}{13}$$
$$x = \frac{9}{13}$$
✔️ x = 9/13, y = -5/13
(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x – \frac{y}{3} = 3$
Simplify equations by clearing fractions.
Multiply first eq by 6:
$$3x + 4y = -6 \quad \text{…(1)}$$
Multiply second eq by 3:
$$3x – y = 9 \quad \text{…(2)}$$
Subtract (2) from (1):
$$(3x + 4y) – (3x – y) = -6 – 9$$
$$5y = -15 \Rightarrow y = -3$$
Substitute $y=-3$ in (2):
$$3x – (-3) = 9 \Rightarrow 3x + 3 = 9 \Rightarrow 3x = 6 \Rightarrow x = 2$$
✔️ x = 2, y = -3
Q2

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 1/2 if we only add 1 to the denominator. What is the fraction?
Let the fraction be $\frac{x}{y}$.
$$\frac{x+1}{y-1} = 1 \Rightarrow x + 1 = y – 1 \Rightarrow x – y = -2 \quad \text{…(1)}$$
$$\frac{x}{y+1} = \frac{1}{2} \Rightarrow 2x = y + 1 \Rightarrow 2x – y = 1 \quad \text{…(2)}$$
Subtract (1) from (2):
$$(2x – y) – (x – y) = 1 – (-2)$$
$$x = 3$$
Substitute $x=3$ in (1):
$$3 – y = -2 \Rightarrow y = 5$$
✔️ The fraction 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?
Let Nuri’s age = $N$ and Sonu’s age = $S$.
5 years ago:
$$N – 5 = 3(S – 5) \Rightarrow N – 3S = -10 \quad \text{…(1)}$$
10 years later:
$$N + 10 = 2(S + 10) \Rightarrow N – 2S = 10 \quad \text{…(2)}$$
Subtract (1) from (2):
$$(N – 2S) – (N – 3S) = 10 – (-10)$$
$$S = 20$$
Substitute $S=20$ in (2):
$$N – 2(20) = 10 \Rightarrow N = 50$$
✔️ Nuri is 50 years old, Sonu is 20 years old
(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.
Let unit digit be $y$ and tens digit be $x$. Number = $10x + y$.
$$x + y = 9 \quad \text{…(1)}$$
Reversed Number = $10y + x$.
$$9(10x + y) = 2(10y + x)$$
$$90x + 9y = 20y + 2x \Rightarrow 88x – 11y = 0 \Rightarrow 8x – y = 0 \quad \text{…(2)}$$
Add (1) and (2):
$$9x = 9 \Rightarrow x = 1$$
Substitute $x=1$ in (2):
$$8(1) – y = 0 \Rightarrow y = 8$$
✔️ The number is 18
(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.
Let notes of ₹50 = $x$ and notes of ₹100 = $y$.
$$x + y = 25 \quad \text{…(1)}$$
$$50x + 100y = 2000$$
Divide second equation by 50:
$$x + 2y = 40 \quad \text{…(2)}$$
Subtract (1) from (2):
$$y = 15$$
Substitute $y=15$ in (1):
$$x + 15 = 25 \Rightarrow x = 10$$
✔️ ₹50 notes = 10, ₹100 notes = 15
(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.
Let fixed charge (for 3 days) = $x$ and charge per extra day = $y$.
Saritha (7 days = 3 fixed + 4 extra):
$$x + 4y = 27 \quad \text{…(1)}$$
Susy (5 days = 3 fixed + 2 extra):
$$x + 2y = 21 \quad \text{…(2)}$$
Subtract (2) from (1):
$$2y = 6 \Rightarrow y = 3$$
Substitute $y=3$ in (2):
$$x + 2(3) = 21 \Rightarrow x = 15$$
✔️ Fixed charge = ₹15, Charge per day = ₹3
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