NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables (NCERT 2026–27)

These Class 10 Maths Chapter 3 solutions cover Pair of Linear Equations in Two Variables from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 3.1, Exercise 3.2 and Exercise 3.3 is solved step by step — by the graphical method, the substitution method and the elimination method — with each answer cross-checked against the book’s answer key, so you can revise quickly and score full marks.

Class: 10 Subject: Mathematics Chapter: 3 Chapter Name: Pair of Linear Equations in Two Variables Exercises: 3.1, 3.2, 3.3 Session: 2026–27
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Chapter 3 Overview

Chapter 3, Pair of Linear Equations in Two Variables, studies two linear equations of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 considered together. Geometrically each equation is a straight line, so a pair of equations corresponds to two lines that may intersect at one point (a unique solution — consistent), be coincident (infinitely many solutions — dependent and consistent), or be parallel (no solution — inconsistent). The chapter teaches you to recognise these three cases from the ratios of coefficients, and to find solutions by the graphical method, the substitution method and the elimination method, along with framing real-life word problems as a pair of linear equations.

Key Concepts & Definitions

Linear equation in two variables: an equation of the form ax + by + c = 0, where a, b, c are real numbers and a, b are not both zero. Its graph is a straight line.

Pair of linear equations: two such equations taken together, written as a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

Consistent pair: a pair that has at least one solution (lines intersect or coincide).

Inconsistent pair: a pair that has no solution (the lines are parallel).

Dependent pair: a consistent pair whose two equations are equivalent — the lines coincide and there are infinitely many solutions.

Methods of solution: graphical method (draw both lines and read the common point) and algebraic methods — substitution and elimination.

Important Formulas & Conditions (Chapter 3)

General pair: a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

Intersecting lines (unique solution, consistent): a1/a2 ≠ b1/b2.

Coincident lines (infinitely many solutions, dependent & consistent): a1/a2 = b1/b2 = c1/c2.

Parallel lines (no solution, inconsistent): a1/a2 = b1/b2 ≠ c1/c2.

Substitution method: express one variable from one equation and substitute it into the other, reducing it to one variable.

Elimination method: make the coefficients of one variable numerically equal, then add or subtract the equations to eliminate that variable.

NCERT Solutions — Exercise 3.1

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 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. (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 that of one pen.

SOLUTION (i) Let the number of boys = x and girls = y. Total: x + y = 10. Girls are 4 more than boys: y = x + 4, i.e. x − y = −4. For x + y = 10: points (0, 10) and (10, 0). For x − y = −4: points (0, 4) and (−4, 0). Plotting both lines, they intersect at (3, 7). So boys = 3, girls = 7 (check: 3 + 7 = 10 and 7 − 3 = 4). ✓ (ii) Let cost of one pencil = ₹ x and one pen = ₹ y. 5x + 7y = 50 and 7x + 5y = 46. For 5x + 7y = 50: points (10, 0) and (3, 5). For 7x + 5y = 46: points (8, −2) and (3, 5). Both lines pass through (3, 5), so they intersect there. So one pencil = ₹ 3 and one pen = ₹ 5. ✓

2. 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

SOLUTION (i) a1/a2 = 5/7, b1/b2 = −4/6 = −2/3. Since 5/7 ≠ −2/3, the lines intersect at a point (unique solution). (ii) a1/a2 = 9/18 = 1/2, b1/b2 = 3/6 = 1/2, c1/c2 = 12/24 = 1/2. All three are equal, so the lines are coincident. (iii) a1/a2 = 6/2 = 3, b1/b2 = −3/−1 = 3, c1/c2 = 10/9. Here a1/a2 = b1/b2 ≠ c1/c2, so the lines are parallel.

3. On comparing the ratios a1/a2, b1/b2 and c1/c2, 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

SOLUTION (i) a1/a2 = 3/2, b1/b2 = 2/−3 = −2/3. Since 3/2 ≠ −2/3, lines intersect → unique solution → consistent. (ii) a1/a2 = 2/4 = 1/2, b1/b2 = −3/−6 = 1/2, c1/c2 = 8/9. Here a1/a2 = b1/b2 ≠ c1/c2 (parallel), so the pair is inconsistent. (iii) a1/a2 = (3/2)/9 = 1/6, b1/b2 = (5/3)/−10 = −1/6. Since 1/6 ≠ −1/6, lines intersect → unique solution → consistent. (iv) a1/a2 = 5/−10 = −1/2, b1/b2 = −3/6 = −1/2, c1/c2 = 11/−22 = −1/2. All equal → coincident → infinitely many solutions → consistent. (v) a1/a2 = (4/3)/2 = 2/3, b1/b2 = 2/3, c1/c2 = 8/12 = 2/3. All equal → coincident → consistent.

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) x + y = 5 ;  2x + 2y = 10 (ii) x − y = 8 ;  3x − 3y = 16 (iii) 2x + y − 6 = 0 ;  4x − 2y − 4 = 0 (iv) 2x − 2y − 2 = 0 ;  4x − 4y − 5 = 0

SOLUTION (i) a1/a2 = 1/2, b1/b2 = 1/2, c1/c2 = 5/10 = 1/2. All equal → coincident → consistent with infinitely many solutions. Both equations are x + y = 5; every point on it (e.g. (0, 5), (5, 0), (2, 3)) is a solution. (ii) a1/a2 = 1/3, b1/b2 = −1/−3 = 1/3, c1/c2 = 8/16 = 1/2. Since a1/a2 = b1/b2 ≠ c1/c2, lines are parallel → inconsistent (no solution). (iii) a1/a2 = 2/4 = 1/2, b1/b2 = 1/−2 = −1/2. Since 1/2 ≠ −1/2, lines intersect → consistent. For 2x + y = 6: points (0, 6), (3, 0). For 4x − 2y = 4: points (1, 0), (0, −2). The lines meet at (2, 2), so x = 2, y = 2 (check: 2(2) + 2 = 6 ✓ and 4(2) − 2(2) = 4 ✓). (iv) a1/a2 = 2/4 = 1/2, b1/b2 = −2/−4 = 1/2, c1/c2 = −2/−5 = 2/5. Since a1/a2 = b1/b2 ≠ c1/c2, lines are parallel → inconsistent (no solution).

5. 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.

SOLUTION Let length = x m and width = y m. Half the perimeter = x + y = 36. Length is 4 m more than width: x = y + 4, i.e. x − y = 4. Adding the two equations: 2x = 40 → x = 20. Then y = 36 − 20 = 16. So the garden is 20 m long and 16 m wide (check: 20 + 16 = 36 and 20 − 16 = 4). ✓

6. 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

SOLUTION Here a1 = 2, b1 = 3, c1 = −8. (Many correct answers are possible; one set is given.) (i) Intersecting: need a1/a2 ≠ b1/b2. Example: 3x − 2y − 5 = 0 (2/3 ≠ 3/−2). (ii) Parallel: need a1/a2 = b1/b2 ≠ c1/c2. Example: 4x + 6y − 9 = 0 (2/4 = 3/6 = 1/2, but −8/−9 ≠ 1/2). (iii) Coincident: need a1/a2 = b1/b2 = c1/c2. Example: 4x + 6y − 16 = 0 (just 2 × the given equation).

7. Draw the graphs of the equations x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

SOLUTION For x − y + 1 = 0 (y = x + 1): points (0, 1) and (−1, 0). For 3x + 2y − 12 = 0 (y = (12 − 3x)/2): points (0, 6) and (4, 0). The two lines meet where x + 1 = (12 − 3x)/2 → 2x + 2 = 12 − 3x → 5x = 10 → x = 2, y = 3, i.e. at (2, 3). Each line cuts the x-axis: x − y + 1 = 0 at (−1, 0); 3x + 2y − 12 = 0 at (4, 0). So the triangle has vertices (−1, 0), (4, 0) and (2, 3). Its base on the x-axis is 5 units and height is 3 units, so its area is ½ × 5 × 3 = 7.5 sq units. Shade the region enclosed by the two lines and the x-axis.

NCERT Solutions — Exercise 3.2

1. Solve the following pair of linear equations by the substitution method. (i) x + y = 14 ;  x − y = 4 (ii) s − t = 3 ;  (s/3) + (t/2) = 6 (iii) 3x − y = 3 ;  9x − 3y = 9 (iv) 0.2x + 0.3y = 1.3 ;  0.4x + 0.5y = 2.3 (v) √2 x + √3 y = 0 ;  √3 x − √8 y = 0 (vi) (3x/2) − (5y/3) = −2 ;  (x/3) + (y/2) = 13/6

SOLUTION (i) From x − y = 4, x = 4 + y. Substitute in x + y = 14: (4 + y) + y = 14 → 2y = 10 → y = 5. Then x = 4 + 5 = 9. So x = 9, y = 5. (ii) From s − t = 3, s = 3 + t. Substitute in (s/3) + (t/2) = 6 → multiply by 6: 2s + 3t = 36 → 2(3 + t) + 3t = 36 → 6 + 5t = 36 → t = 6. Then s = 3 + 6 = 9. So s = 9, t = 6. (iii) From 3x − y = 3, y = 3x − 3. Substitute in 9x − 3y = 9: 9x − 3(3x − 3) = 9 → 9x − 9x + 9 = 9 → 9 = 9 (always true). So the pair has infinitely many solutions (the equations are equivalent), of the form x = t, y = 3t − 3. (iv) Multiply both by 10: 2x + 3y = 13 and 4x + 5y = 23. From the first, x = (13 − 3y)/2. Substitute: 4(13 − 3y)/2 + 5y = 23 → 2(13 − 3y) + 5y = 23 → 26 − 6y + 5y = 23 → −y = −3 → y = 3. Then x = (13 − 9)/2 = 2. So x = 2, y = 3. (v) From √2 x + √3 y = 0, x = −(√3/√2) y. Substitute in √3 x − √8 y = 0: √3(−√3/√2) y − √8 y = 0 → (−3/√2) y − 2√2 y = 0 → y(−3/√2 − 2√2) = 0. The bracket is non-zero, so y = 0, and then x = 0. So x = 0, y = 0. (vi) Multiply the first by 6: 9x − 10y = −12. Multiply the second by 6: 2x + 3y = 13, giving x = (13 − 3y)/2. Substitute: 9(13 − 3y)/2 − 10y = −12 → multiply by 2: 9(13 − 3y) − 20y = −24 → 117 − 27y − 20y = −24 → −47y = −141 → y = 3. Then x = (13 − 9)/2 = 2. So x = 2, y = 3.

2. Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

SOLUTION From 2x + 3y = 11, x = (11 − 3y)/2. Substitute in 2x − 4y = −24: (11 − 3y) − 4y = −24 → 11 − 7y = −24 → −7y = −35 → y = 5. Then x = (11 − 15)/2 = −2. So x = −2, y = 5. Put x = −2, y = 5 in y = mx + 3: 5 = m(−2) + 3 → 2 = −2m → m = −1.

3. Form the pair of linear equations for the following problems and find their solution by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. Find them. (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. (iii) 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. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? (v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

SOLUTION (i) Let the numbers be x and y (x > y). x − y = 26 and x = 3y. Substitute: 3y − y = 26 → 2y = 26 → y = 13, x = 39. So the numbers are 39 and 13. (ii) Let the angles be x (larger) and y (smaller). Supplementary: x + y = 180. Also x = y + 18. Substitute: (y + 18) + y = 180 → 2y = 162 → y = 81, x = 99. So the angles are 99° and 81°. (iii) Let one bat = ₹ x and one ball = ₹ y. 7x + 6y = 3800 and 3x + 5y = 1750, so x = (1750 − 5y)/3. Substitute: 7(1750 − 5y)/3 + 6y = 3800 → 7(1750 − 5y) + 18y = 11400 → 12250 − 35y + 18y = 11400 → −17y = −850 → y = 50. Then x = (1750 − 250)/3 = 500. So bat = ₹ 500, ball = ₹ 50. (iv) Let fixed charge = ₹ x and per-km charge = ₹ y. x + 10y = 105 and x + 15y = 155, so x = 105 − 10y. Substitute: (105 − 10y) + 15y = 155 → 5y = 50 → y = 10, x = 5. So fixed charge = ₹ 5, charge per km = ₹ 10. For 25 km: 5 + 25(10) = ₹ 255. (v) Let the fraction be x/y. (x + 2)/(y + 2) = 9/11 → 11x + 22 = 9y + 18 → 11x − 9y = −4. (x + 3)/(y + 3) = 5/6 → 6x + 18 = 5y + 15 → 6x − 5y = −3, so x = (5y − 3)/6. Substitute: 11(5y − 3)/6 − 9y = −4 → 11(5y − 3) − 54y = −24 → 55y − 33 − 54y = −24 → y = 9. Then x = (45 − 3)/6 = 7. So the fraction is 7/9. (vi) Let Jacob’s present age = x and son’s = y. Five years hence: x + 5 = 3(y + 5) → x − 3y = 10. Five years ago: x − 5 = 7(y − 5) → x − 7y = −30. From the first, x = 10 + 3y. Substitute: (10 + 3y) − 7y = −30 → −4y = −40 → y = 10, x = 40. So Jacob is 40 years and his son is 10 years old.

NCERT Solutions — Exercise 3.3

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) (x/2) + (2y/3) = −1 and x − (y/3) = 3

SOLUTION (i) Elimination: multiply x + y = 5 by 3: 3x + 3y = 15. Add to 2x − 3y = 4: 5x = 19 → x = 19/5. Then y = 5 − 19/5 = 6/5. Substitution check: y = 5 − x in 2x − 3(5 − x) = 4 → 5x − 15 = 4 → x = 19/5, y = 6/5. So x = 19/5, y = 6/5. (ii) Elimination: multiply 2x − 2y = 2 by 2: 4x − 4y = 4. Multiply 3x + 4y = 10 by 1 and add: 7x = 14 → x = 2. Then 2(2) − 2y = 2 → y = 1. Substitution check: from 2x − 2y = 2, x = y + 1; 3(y + 1) + 4y = 10 → 7y = 7 → y = 1, x = 2. So x = 2, y = 1. (iii) Rewrite: 3x − 5y = 4 and 9x − 2y = 7. Elimination: multiply the first by 3: 9x − 15y = 12. Subtract from 9x − 2y = 7: 13y = −5 → y = −5/13. Then 3x − 5(−5/13) = 4 → 3x = 4 − 25/13 = 27/13 → x = 9/13. Substitution check: x = (4 + 5y)/3 in 9x − 2y = 7 gives 3(4 + 5y) − 2y = 7 → 13y = −5. So x = 9/13, y = −5/13. (iv) Multiply the first by 6: 3x + 4y = −6. Multiply the second by 3: 3x − y = 9. Elimination: subtract: (3x + 4y) − (3x − y) = −6 − 9 → 5y = −15 → y = −3. Then 3x − (−3) = 9 → 3x = 6 → x = 2. Substitution check: x = (9 + y)/3 in 3x + 4y = −6 gives 9 + y + 4y = −6 → 5y = −15. So x = 2, y = −3.

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 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.

SOLUTION (i) Let the fraction be x/y. (x + 1)/(y − 1) = 1 → x − y = −2. Also x/(y + 1) = 1/2 → 2x − y = 1. Eliminate y by subtracting: (2x − y) − (x − y) = 1 − (−2) → x = 3. Then 3 − y = −2 → y = 5. So the fraction is 3/5. (ii) Let Nuri = x, Sonu = y. Five years ago: x − 5 = 3(y − 5) → x − 3y = −10. Ten years later: x + 10 = 2(y + 10) → x − 2y = 10. Eliminate x by subtracting: (x − 2y) − (x − 3y) = 10 − (−10) → y = 20. Then x − 60 = −10 → x = 50. So Nuri is 50 years and Sonu is 20 years old. (iii) Let the digits be x (ten’s) and y (unit’s); the number is 10x + y. Sum: x + y = 9. Nine times the number = twice the reversed number: 9(10x + y) = 2(10y + x) → 90x + 9y = 20y + 2x → 88x − 11y = 0 → 8x − y = 0. Add to x + y = 9: 9x = 9 → x = 1, y = 8. So the number is 18. (iv) Let ₹ 50 notes = x and ₹ 100 notes = y. x + y = 25 and 50x + 100y = 2000 → x + 2y = 40. Subtract: (x + 2y) − (x + y) = 40 − 25 → y = 15. Then x = 25 − 15 = 10. So Meena got 10 notes of ₹ 50 and 15 notes of ₹ 100. (v) Let fixed charge (first 3 days) = ₹ x and extra charge per day = ₹ y. Saritha (7 days = 3 + 4 extra): x + 4y = 27. Susy (5 days = 3 + 2 extra): x + 2y = 21. Subtract: 2y = 6 → y = 3. Then x + 6 = 21 → x = 15. So the fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.

Common Mistakes to Avoid

Watch out for these

  • Comparing ratios without writing both equations in the form ax + by + c = 0 first — move every term to one side so the signs of a, b, c are correct.
  • Forgetting that the test for parallel lines is a1/a2 = b1/b2 ≠ c1/c2, while coincident lines need all three ratios equal.
  • Reading a graphical solution carelessly — always verify the point of intersection in both original equations.
  • In word problems, mixing up which quantity is larger (e.g. supplementary angles, two numbers) — define variables clearly before forming equations.
  • In the substitution method, substituting back into the same equation you rearranged instead of the other one.
  • When eliminating, multiplying only one equation — multiply both by suitable constants so one variable’s coefficients become numerically equal.
  • Treating “0 = 0” as no solution — it means infinitely many solutions; only a false statement like “0 = 9” means no solution.

Practice MCQs & Assertion–Reason

1. The pair of equations 2x + 3y − 9 = 0 and 4x + 6y − 18 = 0 has:

(a) a unique solution    (b) no solution    (c) infinitely many solutions    (d) exactly two solutions

2. For what value is the pair a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 inconsistent?

(a) a1/a2 ≠ b1/b2    (b) a1/a2 = b1/b2 = c1/c2    (c) a1/a2 = b1/b2 ≠ c1/c2    (d) a1 = a2

3. The lines x + y = 5 and 2x + 2y = 10 are:

(a) intersecting    (b) parallel    (c) coincident    (d) perpendicular

4. Solving x + y = 14 and x − y = 4 gives:

(a) x = 9, y = 5    (b) x = 5, y = 9    (c) x = 7, y = 7    (d) x = 10, y = 4

5. The difference of two numbers is 26 and the larger is three times the smaller. The numbers are:

(a) 39 and 13    (b) 36 and 10    (c) 30 and 4    (d) 52 and 26

6. If a pair of linear equations is consistent with a unique solution, the lines are:

(a) parallel    (b) coincident    (c) intersecting at one point    (d) the same line

7. While solving by elimination, if you obtain a false statement such as 0 = 9, the pair has:

(a) a unique solution    (b) two solutions    (c) infinitely many solutions    (d) no solution

8. The pair 3x − y = 3 and 9x − 3y = 9 has:

(a) no solution    (b) a unique solution    (c) infinitely many solutions    (d) x = 0 only

9. The larger of two supplementary angles exceeds the smaller by 18°. The angles are:

(a) 99° and 81°    (b) 100° and 82°    (c) 90° and 72°    (d) 108° and 90°

10. Solving 2x + 3y = 11 and 2x − 4y = −24 gives:

(a) x = 2, y = 5    (b) x = −2, y = 5    (c) x = 5, y = −2    (d) x = −2, y = −5

Answer key: 1-(c), 2-(c), 3-(c), 4-(a), 5-(a), 6-(c), 7-(d), 8-(c), 9-(a), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The pair x − y = 8 and 3x − 3y = 16 has no solution.

Reason: For these equations a1/a2 = b1/b2 ≠ c1/c2, so the lines are parallel.

A-R 2. Assertion: A dependent pair of linear equations is always consistent.

Reason: A dependent pair represents coincident lines, which share infinitely many common points.

A-R 3. Assertion: The pair 2x − 3y = 8 and 4x − 6y = 9 is consistent.

Reason: For these equations a1/a2 = b1/b2 = c1/c2.

A-R 4. Assertion: If a1/a2 ≠ b1/b2, the pair of linear equations has a unique solution.

Reason: When a1/a2 ≠ b1/b2, the two lines intersect at exactly one point.

A-R 5. Assertion: A linear equation in two variables has infinitely many solutions on its own.

Reason: Every point lying on the straight line that represents the equation is a solution of it.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • A pair of linear equations in two variables can be solved by the graphical method or by algebraic methods (substitution, elimination).
  • Graphically: intersecting lines → unique solution (consistent); coincident lines → infinitely many solutions (dependent, consistent); parallel lines → no solution (inconsistent).
  • Unique solution: a1/a2 ≠ b1/b2.
  • Infinitely many solutions: a1/a2 = b1/b2 = c1/c2.
  • No solution: a1/a2 = b1/b2 ≠ c1/c2.
  • Substitution: express one variable from one equation and put it into the other.
  • Elimination: make one variable’s coefficients equal, then add or subtract to remove it.
  • In algebra, a true statement with no variable (0 = 0) means infinitely many solutions; a false one (0 = 9) means no solution.

How to score full marks in this chapter

First rewrite both equations in the standard form ax + by + c = 0 with correct signs, then state which method you are using. For ratio questions, write a1/a2, b1/b2 and c1/c2 explicitly and name the case (intersecting/parallel/coincident). In word problems, clearly define your variables, form two equations, solve neatly, and always end with a one-line verification of your answer in both equations — this guards against sign errors and earns the final accuracy mark.

Frequently Asked Questions

What is Class 10 Maths Chapter 3 about?

Chapter 3, Pair of Linear Equations in Two Variables, deals with solving two linear equations together using the graphical method, the substitution method and the elimination method, and with deciding when a pair has a unique solution, infinitely many solutions or no solution.

How many exercises are there in Class 10 Maths Chapter 3?

There are three exercises — Exercise 3.1 (graphical method and comparison of ratios), Exercise 3.2 (substitution method) and Exercise 3.3 (elimination method) — all solved step by step on this page.

What is the difference between consistent and inconsistent equations?

A consistent pair has at least one solution (the lines intersect or coincide), while an inconsistent pair has no solution because the lines are parallel. You can check this from the ratios of the coefficients a1/a2, b1/b2 and c1/c2.

Are these Class 10 Maths Chapter 3 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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