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Chapter 3: Pair of Linear Equations in Two Variables

Complete NCERT Solutions | CBSE Class 10 Mathematics (2025-26)

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NCERT Class 10 Maths | Chapter 3: Pair of Linear Equations in Two Variables – Complete Solutions
Exercise 3.1 (Page 28)
1 (i)3 marks
Form the pair of linear equations in the following problems, and find their solutions graphically.
10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Answer:
Let number of boys = x, number of girls = y.
Equations: x + y = 10, y = x + 4.
Solving graphically (or by substitution): x = 3, y = 7.
Boys = 3, Girls = 7.
1 (ii)3 marks
5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
Answer:
Let cost of pencil = ₹x, cost of pen = ₹y.
Equations: 5x + 7y = 50, 7x + 5y = 46.
Solving: x = 3, y = 5.
Pencil = ₹3, Pen = ₹5.
22 marks each
On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x-4y+8=0, 7x+6y-9=0
(ii) 9x+3y+12=0, 18x+6y+24=0
(iii) 6x-3y+10=0, 2x-y+9=0
Answer:
(i) a₁/a₂ = 5/7, b₁/b₂ = -4/6 = -2/3 → a₁/a₂ ≠ b₁/b₂ → Intersecting lines
(ii) a₁/a₂ = 9/18 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2 → a₁/a₂ = b₁/b₂ = c₁/c₂ → Coincident lines
(iii) a₁/a₂ = 6/2 = 3, b₁/b₂ = -3/-1 = 3, c₁/c₂ = 10/9 → a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
32 marks each
On comparing the ratios, find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x+2y=5; 2x-3y=7    (ii) 2x-3y=8; 4x-6y=9
(iii) (3/2)x+(5/3)y=7; 9x-10y=14    (iv) 5x-3y=11; -10x+6y=-22
(v) (4/3)x+2y=8; 2x+3y=12
Answer:
(i) a₁/a₂ = 3/2, b₁/b₂ = 2/-3 → a₁/a₂ ≠ b₁/b₂ → Consistent
(ii) a₁/a₂ = 2/4=1/2, b₁/b₂=-3/-6=1/2, c₁/c₂=-8/-9=8/9 → a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Inconsistent
(iii) After simplification, a₁/a₂ ≠ b₁/b₂ → Consistent
(iv) a₁/a₂=5/-10=-1/2, b₁/b₂=-3/6=-1/2, c₁/c₂=11/-22=-1/2 → a₁/a₂ = b₁/b₂ = c₁/c₂ → Consistent (dependent)
(v) a₁/a₂=(4/3)/2=2/3, b₁/b₂=2/3, c₁/c₂=8/12=2/3 → a₁/a₂ = b₁/b₂ = c₁/c₂ → Consistent (dependent)
53 marks
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Answer:
Let length = l, width = w.
Equations: l = w + 4, (l + w) = 36 → l + w = 36.
Solving: w = 16, l = 20.
Length = 20 m, Width = 16 m.
62 marks each
Given the linear equation 2x+3y-8=0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines    (ii) parallel lines    (iii) coincident lines
Answer:
(i) Intersecting: a₁/a₂ ≠ b₁/b₂ → e.g., 3x + 2y - 7 = 0
(ii) Parallel: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → e.g., 2x + 3y - 12 = 0
(iii) Coincident: a₁/a₂ = b₁/b₂ = c₁/c₂ → e.g., 4x + 6y - 16 = 0
Exercise 3.2 (Page 33) – Substitution Method
1 (i)2 marks
x + y = 14, x – y = 4
Answer: x = 9, y = 5
1 (ii)2 marks
s – t = 3, s/3 + t/2 = 6
Answer: s = 9, t = 6
1 (iii)2 marks
3x – y = 3, 9x – 3y = 9
Answer: Infinitely many solutions (dependent)
1 (iv)2 marks
0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
Answer: x = 2, y = 3
1 (v)2 marks
√2 x + √3 y = 0, √3 x – √8 y = 0
Answer: x = 0, y = 0
1 (vi)2 marks
3x/2 – 5y/3 = –2, x/3 + y/2 = 13/6
Answer: x = 2, y = 3
3 (i)3 marks
The difference between two numbers is 26 and one number is three times the other. Find them.
Answer: Numbers: 13 and 39 (or –13 and –39).
3 (ii)3 marks
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer: Angles: 99° and 81°.
3 (iii)3 marks
The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
Answer: Bat = ₹500, Ball = ₹50.
3 (iv)3 marks
Taxi charges: For 10 km ₹105, for 15 km ₹155. Find fixed charge and charge per km. Also charge for 25 km.
Answer: Fixed = ₹5, per km = ₹10, for 25 km = ₹255.
3 (v)3 marks
A fraction becomes 9/11 if 2 is added to both numerator and denominator. It becomes 5/6 if 3 is added to both. Find the fraction.
Answer: 7/9
3 (vi)3 marks
Five years hence, Jacob's age will be three times that of his son. Five years ago, Jacob's age was seven times his son's age. Find their present ages.
Answer: Jacob = 40 years, Son = 10 years.
Exercise 3.3 (Page 36) – Elimination & Substitution
1 (i)2 marks
Solve: x + y = 5, 2x – 3y = 4
Answer: x = 19/5, y = 6/5 (or x = 3.8, y = 1.2)
1 (ii)2 marks
Solve: 3x + 4y = 10, 2x – 2y = 2
Answer: x = 2, y = 1
1 (iii)2 marks
Solve: 3x – 5y – 4 = 0, 9x = 2y + 7
Answer: x = 9/13, y = -5/13
1 (iv)2 marks
Solve: x/2 + 2y/3 = –1, x – y/3 = 3
Answer: x = 2, y = –3
2 (i)3 marks
If we add 1 to numerator and subtract 1 from denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to denominator. Find the fraction.
Answer: 3/5
2 (ii)3 marks
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their ages.
Answer: Nuri = 50 years, Sonu = 20 years.
2 (iii)3 marks
Sum of digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing digits. Find the number.
Answer: 18
2 (iv)3 marks
Meena withdrew ₹2000 in ₹50 and ₹100 notes. She got 25 notes in all. How many notes of each denomination?
Answer: ₹50 notes = 10, ₹100 notes = 15.
2 (v)3 marks
Lending library: fixed charge for first 3 days, then additional per day. Saritha paid ₹27 for 7 days, Susy paid ₹21 for 5 days. Find fixed charge and per day charge.
Answer: Fixed = ₹15, per extra day = ₹3.
Key Concepts Summary
NoteConsistency
When does a pair of linear equations have a unique solution, no solution, or infinitely many solutions?
Answer:
For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
• Unique solution (intersecting lines): a₁/a₂ ≠ b₁/b₂
• No solution (parallel lines): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
• Infinitely many solutions (coincident lines): a₁/a₂ = b₁/b₂ = c₁/c₂
Algebraic MethodsSubstitution & Elimination
What are the two main algebraic methods to solve a pair of linear equations?
Answer:
Substitution Method: Express one variable in terms of the other from one equation and substitute into the second equation.
Elimination Method: Make coefficients of one variable equal in both equations and then add or subtract to eliminate that variable.

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