NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

Chapter 3 of Class 10 Maths — Pair of Linear Equations in Two Variables — is one of the highest-scoring chapters in the CBSE board exam, carrying close to 11 marks. At Convex Classes, Jaipur, we’ve seen year after year that students who master the three solving methods (Substitution, Elimination, and Cross-Multiplication) along with the graphical approach can comfortably pick up full marks on every question this chapter throws at them.

Below, our faculty has worked through every single question from Exercises 3.1 to 3.7 of the NCERT textbook, step by step, in a way that’s meant to actually build your understanding — not just hand you an answer to copy. Read the working, not just the final number.

Quick Concept Refresher

A linear equation in two variables looks like ax + by + c = 0, where a, b, c are real numbers and a, b are not both zero. When you have two such equations together, you get a pair of linear equations, and their combined solution is the pair of (x, y) values that satisfies both at once.

Graphically, two lines can behave in three ways:

Condition	Type of lines	Number of solutions	Nature
a₁/a₂ ≠ b₁/b₂	Intersecting	Exactly one (unique)	Consistent
a₁/a₂ = b₁/b₂ = c₁/c₂	Coincident	Infinitely many	Consistent (dependent)
a₁/a₂ = b₁/b₂ ≠ c₁/c₂	Parallel	None	Inconsistent

Keep this table in front of you while solving — almost every “consistent or inconsistent” question in this chapter comes straight back to it.

Exercise 3.1

Q1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically.

Convex Classes Solution: Let Aftab’s present age = x years, and his daughter’s present age = y years.

Seven years ago: Aftab’s age = x − 7, daughter’s age = y − 7 Given: x − 7 = 7(y − 7) ⟹ x − 7y = −42 …(i)

Three years hence: Aftab’s age = x + 3, daughter’s age = y + 3 Given: x + 3 = 3(y + 3) ⟹ x − 3y = 6 …(ii)

Subtract (i) from (ii): 4y = 48 ⟹ y = 12, and substituting back gives x = 42.

For the graph, plot points from each equation:

• x − 7y = −42 → points like (0, 6), (−7, 5), (7, 7)
• x − 3y = 6 → points like (0, −2), (6, 0), (−3, −3)

Both lines cross at (42, 12), confirming Aftab is 42 and his daughter is 12.

Q2. The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300. Represent this situation algebraically and geometrically.

Convex Classes Solution: Let cost of one bat = ₹x, cost of one ball = ₹y.

3x + 6y = 3900 x + 3y = 1300

From the first equation, x = (3900 − 6y)/3, giving points such as (100, 600), (300, 500). From the second, x = 1300 − 3y, giving points such as (400, 300), (100, 400).

Plotting both lines shows they represent the two purchase conditions and can be used to read off the cost of a bat and a ball graphically.

Q3. The cost of 2 kg of apples and 1 kg of grapes was ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and geometrically.

Convex Classes Solution: Let cost of 1 kg apples = ₹x, cost of 1 kg grapes = ₹y.

2x + y = 160 → y = 160 − 2x 4x + 2y = 300 → y = (300 − 4x)/2

Tabulate a few points for each and plot — note that these two lines actually turn out to be parallel (check the ratios: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 160/300 ≠ 1/2), a nice example students should notice before jumping into the graph.

Exercise 3.2

Q1(i). 10 students of Class X took part in a Maths quiz. If the number of girls is 4 more than boys, find the number of boys and girls.

Convex Classes Solution: Let girls = x, boys = y. x + y = 10 x − y = 4

Solving graphically, the lines meet at (7, 3) → 7 girls and 3 boys.

Q1(ii). 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen.

Convex Classes Solution: Let pencil = ₹x, pen = ₹y. 5x + 7y = 50 7x + 5y = 46

The two lines intersect at (3, 5) → pencil costs ₹3, pen costs ₹5.

Q2. Using a₁/a₂, b₁/b₂, c₁/c₂, check whether these pairs intersect, are parallel, or coincide:

(i) 5x − 4y + 8 = 0, 7x + 6y − 9 = 0 a₁/a₂ = 5/7, b₁/b₂ = −4/6 = −2/3 → unequal → intersecting, unique solution.

(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0 All three ratios = 1/2 → coincident, infinite solutions.

(iii) 6x − 3y + 10 = 0, 2x − y + 9 = 0 a₁/a₂ = b₁/b₂ = 3, c₁/c₂ = 10/9 → parallel, no solution.

Q3. Check consistency using the ratio test:

(i) 3x + 2y = 5, 2x − 3y = 7 → a₁/a₂ ≠ b₁/b₂ → consistent (unique solution).

(ii) 2x − 3y = 8, 4x − 6y = 9 → ratios equal for a,b but not c → inconsistent.

(iii) (3/2)x + (5/3)y = 7, 9x − 10y = 14 → a₁/a₂ ≠ b₁/b₂ → consistent.

(iv) 5x − 3y = 11, −10x + 6y = −22 → all ratios equal → consistent, infinite solutions.

(v) (4/3)x + 2y = 8, 2x + 3y = 12 → all ratios = 2/3 → consistent, infinite solutions.

Q4. Which pairs are consistent/inconsistent? If consistent, solve graphically:

(i) x + y = 5, 2x + 2y = 10 → coincident → infinite solutions, lines overlap completely.

(ii) x − y = 8, 3x − 3y = 16 → parallel → inconsistent, no solution.

(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0 → intersecting at (2, 2).

(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0 → parallel → inconsistent.

Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions.

Convex Classes Solution: Let width = x, length = y. y − x = 4 y + x = 36

Graphically, the lines meet at (16, 20) → width = 16 m, length = 20 m.

Q6. Given 2x + 3y − 8 = 0, write another equation so that the pair represents:

(i) Intersecting lines: e.g. 2x − 7y + 9 = 0 (a₁/a₂ ≠ b₁/b₂) (ii) Parallel lines: e.g. 6x + 9y + 9 = 0 (a₁/a₂ = b₁/b₂ ≠ c₁/c₂) (iii) Coincident lines: e.g. 4x + 6y − 16 = 0 (all ratios equal to 1/2)

Q7. Draw graphs of x − y + 1 = 0 and 3x + 2y − 12 = 0. Find the triangle formed with the x-axis and shade it.

Convex Classes Solution: The two lines intersect each other at (2, 3), meet the x-axis at (−1, 0) and (4, 0) respectively. These three points are the vertices of the required triangle.

Exercise 3.3 — Substitution Method

Q1(i). x + y = 14, x − y = 4

From the first equation, x = 14 − y. Substituting: 14 − 2y = 4 ⟹ y = 5, so x = 9, y = 5.

Q1(ii). s − t = 3, s/3 + t/2 = 6

s = 3 + t. Substituting gives t = 6, so s = 9, t = 6.

Q1(iii). 3x − y = 3, 9x − 3y = 9

Substituting x = (3 + y)/3 into the second equation gives 9 = 9, a true statement for all y — meaning these equations are actually the same line, so there are infinitely many solutions.

Q1(iv). 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3

Solving gives y = 3, x = 2.

Q1(v). √2x + √3y = 0, √3x − √8y = 0

This pair only satisfies x = 0, y = 0.

Q1(vi). (3x/2) − (5y/3) = −2, (x/3) + (y/2) = 13/6

Solving simultaneously gives x = 2, y = 3.

Q2. Solve 2x + 3y = 11 and 2x − 4y = −24, then find m for y = mx + 3.

Convex Classes Solution: Solving the pair gives x = −2, y = 5. Substituting into y = mx + 3: 5 = −2m + 3 ⟹ m = −1.

Q3(i). The difference between two numbers is 26, and one number is three times the other. Find them.

Let numbers be x and y with y = 3x and y − x = 26. Solving: x = 13, y = 39.

Q3(ii). The larger of two supplementary angles exceeds the smaller by 18°. Find them.

x + y = 180°, x − y = 18°. Solving: x = 99°, y = 81°.

Q3(iii). 7 bats and 6 balls cost ₹3800; 3 bats and 5 balls cost ₹1750. Find the cost of each. Solving the pair: bat = ₹500, ball = ₹50.

Q3(iv). Taxi fare: fixed charge + per-km charge. 10 km costs ₹105, 15 km costs ₹155. Find fixed charge, per-km rate, and the fare for 25 km.

x + 10y = 105, x + 15y = 155 → fixed charge = ₹5, rate = ₹10/km. Fare for 25 km = 5 + 25(10) = ₹255.

Q3(v). A fraction becomes 9/11 when 2 is added to numerator and denominator; it becomes 5/6 when 3 is added to both. Find the fraction.

Solving the two equations gives x = 7, y = 9, so the fraction is 7/9.

Q3(vi). Five years hence, Jacob’s age will be three times his son’s. Five years ago, it was seven times his son’s. Find their present ages.

Solving: Jacob = 40 years, son = 10 years.

Exercise 3.4 — Elimination Method

Q1(i). x + y = 5, 2x − 3y = 4

Multiplying the first by 2 and eliminating x: 5y = 6, y = 6/5, x = 19/5 (verified identically by substitution).

Q1(ii). 3x + 4y = 10, 2x − 2y = 2

Eliminating y: x = 2, y = 1.

Q1(iii). 3x − 5y − 4 = 0, 9x = 2y + 7

Eliminating x: x = 9/13, y = −5/13.

Q1(iv). x/2 + 2y/3 = −1, x − y/3 = 3

Clearing fractions and eliminating: x = 2, y = −3.

Q2(i). Adding 1 to the numerator and subtracting 1 from the denominator makes a fraction equal to 1. Adding 1 to just the denominator makes it 1/2. Find the fraction.

Solving gives a = 3, b = 5, so the fraction is 3/5.

Q2(ii). Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.

Solving: Nuri = 50 years, Sonu = 20 years.

Q2(iii). The digits of a two-digit number add up to 9. Nine times the number is twice the number obtained by reversing the digits. Find the number.

Solving A + B = 9 and −A + 8B = 0: B = 1, A = 8, giving the number 18.

Q2(iv). Meena withdraws ₹2000 in ₹50 and ₹100 notes only, getting 25 notes total. Find how many of each.

Solving: 10 notes of ₹50 and 15 notes of ₹100.

Q2(v). A library charges a fixed amount for the first three days plus extra per day. Saritha paid ₹27 for 7 days, Susy paid ₹21 for 5 days. Find the fixed charge and daily charge.

Solving: fixed charge = ₹15, extra charge = ₹3/day.

Exercise 3.5 — Cross-Multiplication Method

Q1. Determine whether these pairs have a unique, no, or infinite solutions; solve unique cases by cross-multiplication:

(i) x − 3y − 3 = 0, 3x − 9y − 2 = 0

→ ratios show parallel lines, no solution.

(ii) 2x + y = 5, 3x + 2y = 8

→ unique solution: x = 2, y = 1.

(iii) 3x − 5y = 20, 6x − 10y = 40

→ coincident, infinite solutions.

(iv) x − 3y − 7 = 0, 3x − 3y − 15 = 0

→ unique solution: x = 4, y = 1.

Q2(i). Find a and b for infinite solutions: 2x + 3y = 7, (a − b)x + (a + b)y = 3a + b − 2

Solving the resulting equations gives a = 5, b = 1.

Q2(ii). Find k for which the system has no solution:

3x + y = 1, (2k − 1)x + (k − 1)y = 2k + 1

Solving gives k = 2.

Q3. Solve 8x + 5y = 9, 3x + 2y = 4 by substitution and cross-multiplication.

Both methods confirm x = −2, y = 5.

Q4(i). Hostel charges have a fixed part plus a per-day food charge. Student A pays ₹1000 for 20 days, Student B pays ₹1180 for 26 days. Find the fixed charge and daily cost.

Solving: fixed charge = ₹400, cost per day = ₹30.

Q4(ii). A fraction becomes 1/3 when 1 is subtracted from the numerator, and 1/4 when 8 is added to the denominator. Find the fraction.

Solving: x = 5, y = 12, fraction = 5/12.

Q4(iii). Yash scored 40 marks with +3 for correct and −1 for wrong. With +4 and −2 he’d have scored 50. How many questions were there?

Solving: 15 correct + 5 wrong = 20 questions total.

Q4(iv). Two places 100 km apart; cars from each end meet in 5 hours travelling the same direction, or 1 hour travelling towards each other. Find their speeds.

Solving: 60 km/h and 40 km/h.

Q4(v). Reducing a rectangle’s length by 5 and increasing breadth by 3 reduces area by 9 sq units; increasing length by 3 and breadth by 2 increases area by 67 sq units. Find the dimensions.

Solving by cross-multiplication: length = 17 units, breadth = 9 units.

Exercise 3.6 — Equations Reducible to a Pair of Linear Equations

Q1(i). 1/2x + 1/3y = 2, 1/3x + 1/2y = 13/6

Substituting 1/x = m, 1/y = n and solving: x = 1/2, y = 1/3.

Q1(ii). 2/√x + 3/√y = 2, 4/√x − 9/√y = −1

Solving with m = 1/√x, n = 1/√y: x = 4, y = 9.

Q1(iii). 4/x + 3y = 14, 3/x − 4y = 23

Solving: x = 1/5, y = −2.

Q1(iv). 5/(x−1) + 1/(y−2) = 2, 6/(x−1) − 3/(y−2) = 1

Solving: x = 4, y = 5.

Q1(v). (7x − 2y)/xy = 5, (8x + 7y)/xy = 15

Solving: x = 1, y = 1.

Q1(vi). 6x + 3y = 6xy, 2x + 4y = 5xy

Solving: x = 1, y = 2.

Q1(vii). 10/(x+y) + 2/(x−y) = 4, 15/(x+y) − 5/(x−y) = −2

Solving: x = 3, y = 2.

Q1(viii). 1/(3x+y) + 1/(3x−y) = 3/4, 1/[2(3x+y)] − 1/[2(3x−y)] = −1/8

Solving: x = 1, y = 1.

Q2(i). Ritu rows 20 km downstream in 2 hours and 4 km upstream in 2 hours. Find her speed in still water and the current’s speed.

Solving: still water speed = 6 km/h, current = 4 km/h.

Q2(ii). 2 women + 5 men finish embroidery work in 4 days; 3 women + 6 men finish it in 3 days. Find how long one woman and one man would each take alone.

Solving: 1 woman alone = 18 days, 1 man alone = 36 days.

Q2(iii). Roohi travels 300 km partly by train, partly by bus. 60 km by train + rest by bus takes 4 hours; 100 km by train + rest by bus takes 10 minutes longer. Find the speeds.

Solving: train = 60 km/h, bus = 80 km/h.

Exercise 3.7 (Optional / Higher-Order Thinking)

Q1. Ani and Biju’s ages differ by 3 years. Ani’s father Dharam is twice Ani’s age, and Biju is twice his sister Cathy’s age. Cathy and Dharam’s ages differ by 30 years. Find Ani and Biju’s ages.

Convex Classes Solution: There are two valid cases here depending on who is older:

Case 1 (Ani older): Solving A − B = 3 and 4A − B = 60 gives Ani = 19, Biju = 16. Case 2 (Biju older): Solving B − A = 3 and 4A − B = 60 gives Ani = 21, Biju = 24.

Q2. “Give me 100, I’ll be twice as rich as you.” “Give me 10, I’ll be six times as rich.” Find their capitals. (Bhaskara II)

Solving x + 100 = 2(y − 100) and 6(x − 10) = y + 10: x = ₹40, y = ₹170.

Q3. A train at uniform speed would take 2 hours less if 10 km/h faster, and 3 hours more if 10 km/h slower. Find the distance covered.

Solving the system: speed = 50 km/h, time = 12 hours, distance = 600 km.

Q4. Students standing in rows: 3 extra per row means 1 row less; 3 fewer per row means 2 rows more. Find the total number of students.

Solving x = 4 rows, y = 9 students/row → total = 36 students.

Q5. In ∆ABC, ∠C = 3∠B = 2(∠A + ∠B). Find the three angles.

Solving: ∠A = 20°, ∠B = 40°, ∠C = 120°.

Q6. Draw graphs of 5x − y = 5 and 3x − y = 3. Find the vertices of the triangle formed with the y-axis

. The lines intersect the y-axis and each other to form a triangle with vertices at (1, 0), (0, −5) and (0, −3).

Q7. Solve the following pairs:

(i) px + qy = p − q, qx − py = p + q → x = 1, y = −1

(ii) ax + by = c, bx + ay = 1 + c → x = [c(a−b) − b] / (a² − b²), y = [c(a−b) + a] / (a² − b²)

(iii) x/a − y/b = 0, ax + by = a² + b² → x = a, y = b

(iv) (a−b)x + (a+b)y = a² − 2ab − b², (a+b)(x+y) = a² + b² → x = a + b, y = −2ab/(a+b) (v) 152x − 378y = −74, −378x + 152y = −604 → x = 2, y = 1

Q8. ABCD is a cyclic quadrilateral. Find its angles.

Using the property that opposite angles of a cyclic quadrilateral sum to 180°, solving the resulting equations gives x = −15, y = 25, from which each angle of the quadrilateral can be found by substitution.

Convex Classes Jaipur — Tips to Score Full Marks in This Chapter

1. Always write the equations first, clearly labelling what x and y stand for — examiners give marks for correct formulation even if the final arithmetic slips.
2. Practice the ratio test (a₁/a₂, b₁/b₂, c₁/c₂) until it’s automatic. A large chunk of this chapter’s marks come from correctly classifying a pair as consistent, inconsistent, or dependent.
3. Don’t skip the graphical method exercises — Exercise 3.1 and 3.2 questions are frequently asked directly in board papers.
4. For word problems, read twice before writing equations. Misreading “3 more” as “3 times” is the single most common mistake we see in our classroom tests at Convex Classes.
5. Practice elimination and substitution on the same question — knowing both means you can double-check your answer in the exam itself.

If you’d like guided practice on this chapter with doubt-solving, our faculty at Convex Classes, Jaipur conducts regular chapter tests and one-on-one problem-solving sessions for Class 10 CBSE Maths students. Reach out to us to book a demo class.