Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables
Chapter 3 covers methods to solve a pair of linear equations in two variables - graphical, substitution, elimination, and cross-multiplication. Students learn to identify consistent, inconsistent, and dependent systems by comparing ratios of coefficients. This chapter is the highest-scoring topic in Algebra, carrying 8-10 marks in the CBSE Class 10 board exam 2027.
Syllabus Update 2026-27
For CBSE 2026-27, this chapter has 7 exercises covering graphical method, substitution, elimination, cross-multiplication, and word problems. All exercises are retained in the current syllabus.
Table of Contents
Exercise 3.1 - Graphical Representation of Linear Equations
3 questions
Exercise 3.2 - Consistency of Linear Systems (Ratio Method)
4 questions
Exercise 3.3 - Substitution Method
3 questions
Exercise 3.4 - Elimination Method
3 questions
Exercise 3.5 - Cross-Multiplication Method
4 questions
Exercise 3.6 - Equations Reducible to Linear Form
3 questions
Exercise 3.7 - Word Problems (Optional but Important)
5 questions
Previous Year Questions
15 questions (CBSE 2019-2026)
MCQ Practice
12 questions
3.1Exercise 3.1 - Graphical Representation of Linear Equations
Q1Aftab tells his daughter, '7 years ago, I was seven times as old as you were then. Also, 3 years from now, I shall be three times as old as you will be.' Represent this situation algebraically and graphically.
Let Aftab's present age = x years and daughter's present age = y years.
7 years ago: Aftab was (x - 7), daughter was (y - 7).
Condition: x - 7 = 7(y - 7)
x - 7 = 7y - 49
x - 7y = -42 ... (1)
3 years from now: Aftab will be (x + 3), daughter will be (y + 3).
Condition: x + 3 = 3(y + 3)
x + 3 = 3y + 9
x - 3y = 6 ... (2)
For graphical representation:
From (1): x = 7y - 42
When y = 5, x = -7; when y = 6, x = 0; when y = 7, x = 7
From (2): x = 3y + 6
When y = 0, x = 6; when y = 3, x = 15; when y = -2, x = 0
Plotting these points and drawing the lines, the two lines intersect at the point (42, 12).
So Aftab's present age = 42 years and daughter's present age = 12 years.
Q2The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Let cost of one bat = Rs x and cost of one ball = Rs y.
Condition 1: 3x + 6y = 3900
Simplifying: x + 2y = 1300 ... (1)
Condition 2: x + 3y = 1300 ... (2)
For graphical representation:
From (1): x = 1300 - 2y
When y = 0, x = 1300; when y = 200, x = 900; when y = 500, x = 300
From (2): x = 1300 - 3y
When y = 0, x = 1300; when y = 200, x = 700; when y = 300, x = 400
Plotting these points, the two lines intersect at (1300, 0).
So cost of one bat = Rs 1300 and cost of one ball = Rs 0.
Note: This means the balls were given free. The system has a unique solution as the lines intersect at exactly one point.
Q3The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.
Let cost of 1 kg apples = Rs x and cost of 1 kg grapes = Rs y.
Condition 1: 2x + y = 160 ... (1)
Condition 2: 4x + 2y = 300, i.e., 2x + y = 150 ... (2)
For graphical representation:
From (1): y = 160 - 2x
When x = 0, y = 160; when x = 40, y = 80; when x = 80, y = 0
From (2): y = 150 - 2x
When x = 0, y = 150; when x = 40, y = 70; when x = 75, y = 0
Plotting these points, the two lines are parallel (same slope but different y-intercepts).
This means the system is inconsistent - no solution exists. The prices have changed in a way that no single pair (x, y) satisfies both conditions simultaneously.
3.2Exercise 3.2 - Consistency of Linear Systems (Ratio Method)
Q1Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the graphical method: (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.
Let number of boys = x and number of girls = y.
Total students: x + y = 10 ... (1)
Girls are 4 more than boys: y - x = 4, i.e., -x + y = 4 ... (2)
From (1): y = 10 - x
When x = 0, y = 10; when x = 4, y = 6; when x = 10, y = 0
From (2): y = x + 4
When x = 0, y = 4; when x = 3, y = 7; when x = 6, y = 10
The lines intersect at (3, 7).
Number of boys = 3, Number of girls = 7.
Q2On 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 and 7x + 6y - 9 = 0 (ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
(i) 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0
a1/a2 = 5/7, b1/b2 = -4/6 = -2/3
Since a1/a2 != b1/b2 (5/7 != -2/3), the lines intersect at a point (unique solution).
(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
a1/a2 = 9/18 = 1/2, b1/b2 = 3/6 = 1/2, c1/c2 = 12/24 = 1/2
Since a1/a2 = b1/b2 = c1/c2 (1/2 = 1/2 = 1/2), the lines are coincident (infinitely many solutions).
Q4Which 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
(i) x + y = 5 and 2x + 2y = 10
a1/a2 = 1/2, b1/b2 = 1/2, c1/c2 = 5/10 = 1/2
Since a1/a2 = b1/b2 = c1/c2, lines are coincident.
Consistent (dependent) - infinitely many solutions.
Every point on the line x + y = 5 is a solution.
(ii) x - y = 8 and 3x - 3y = 16
a1/a2 = 1/3, b1/b2 = -1/-3 = 1/3, c1/c2 = 8/16 = 1/2
Since a1/a2 = b1/b2 != c1/c2 (1/3 = 1/3 != 1/2), lines are parallel.
Inconsistent - no solution.
(iii) 2x + y - 6 = 0 and 4x - 2y - 4 = 0
a1/a2 = 2/4 = 1/2, b1/b2 = 1/(-2) = -1/2
Since a1/a2 != b1/b2 (1/2 != -1/2), lines intersect.
Consistent - unique solution.
From equation (1): y = 6 - 2x
Substituting in (2): 4x - 2(6 - 2x) - 4 = 0 -> 4x - 12 + 4x - 4 = 0 -> 8x = 16 -> x = 2
y = 6 - 4 = 2
Solution: x = 2, y = 2.
Q6Given 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
Given: 2x + 3y - 8 = 0 (a1=2, b1=3, c1=-8)
(i) Intersecting lines: We need a1/a2 != b1/b2.
Choose: 4x + 3y - 8 = 0 (a1/a2 = 2/4 = 1/2, b1/b2 = 3/3 = 1, and 1/2 != 1)
(ii) Parallel lines: We need a1/a2 = b1/b2 != c1/c2.
Choose: 4x + 6y - 10 = 0 (a1/a2 = 2/4 = 1/2, b1/b2 = 3/6 = 1/2, c1/c2 = -8/-10 = 4/5, and 1/2 != 4/5)
(iii) Coincident lines: We need a1/a2 = b1/b2 = c1/c2.
Choose: 4x + 6y - 16 = 0 (a1/a2 = 1/2, b1/b2 = 1/2, c1/c2 = -8/-16 = 1/2, all equal)
3.3Exercise 3.3 - Substitution Method
Q1Solve 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
(i) x + y = 14 ... (1), x - y = 4 ... (2)
From (2): x = y + 4
Substituting in (1): (y + 4) + y = 14
2y + 4 = 14
2y = 10
y = 5
x = 5 + 4 = 9
x = 9, y = 5
(ii) s - t = 3 ... (1), s/3 + t/2 = 6 ... (2)
From (1): s = t + 3
Substituting in (2): (t + 3)/3 + t/2 = 6
Multiply by 6: 2(t + 3) + 3t = 36
2t + 6 + 3t = 36
5t = 30
t = 6
s = 6 + 3 = 9
s = 9, t = 6
(iii) 3x - y = 3 ... (1), 9x - 3y = 9 ... (2)
From (1): y = 3x - 3
Substituting in (2): 9x - 3(3x - 3) = 9
9x - 9x + 9 = 9
9 = 9 (always true)
This means the equations are dependent (coincident lines).
Infinitely many solutions of the form y = 3x - 3.
Q2Solve 2x + 3y = 11 and 2x - 4y = -24 and hence find the value of 'm' for which y = mx + 3.
2x + 3y = 11 ... (1)
2x - 4y = -24 ... (2)
From (1): 2x = 11 - 3y
Substituting in (2): (11 - 3y) - 4y = -24
11 - 7y = -24
-7y = -35
y = 5
From (1): 2x + 15 = 11 -> 2x = -4 -> x = -2
x = -2, y = 5
Now, y = mx + 3:
5 = m(-2) + 3
5 - 3 = -2m
2 = -2m
m = -1
Q3Form the pair of linear equations for the following problem and find their solution by substitution method: The difference between two numbers is 26 and one number is three times the other. Find them.
Let the two numbers be x and y (x > y).
Condition 1: x - y = 26 ... (1)
Condition 2: x = 3y ... (2)
Substituting (2) in (1):
3y - y = 26
2y = 26
y = 13
From (2): x = 3(13) = 39
The two numbers are 39 and 13.
3.4Exercise 3.4 - Elimination Method
Q1Solve 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
(i) x + y = 5 ... (1), 2x - 3y = 4 ... (2)
Elimination method:
Multiply (1) by 3: 3x + 3y = 15 ... (3)
Add (2) and (3): 2x - 3y + 3x + 3y = 4 + 15
5x = 19 -> x = 19/5
From (1): y = 5 - 19/5 = 6/5
x = 19/5, y = 6/5
(ii) 3x + 4y = 10 ... (1), 2x - 2y = 2 ... (2)
Elimination method:
Multiply (2) by 2: 4x - 4y = 4 ... (3)
Add (1) and (3): 7x = 14 -> x = 2
From (2): 4 - 2y = 2 -> 2y = 2 -> y = 1
x = 2, y = 1
(iii) 3x - 5y = 4 ... (1), 9x - 2y = 7 ... (2)
Elimination method:
Multiply (1) by 3: 9x - 15y = 12 ... (3)
Subtract (2) from (3): -15y + 2y = 12 - 7 -> -13y = 5 -> y = -5/13
From (1): 3x - 5(-5/13) = 4 -> 3x + 25/13 = 4 -> 3x = 27/13 -> x = 9/13
x = 9/13, y = -5/13
Q2Form 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?
(i) Let numerator = x, denominator = y.
(x + 1)/(y - 1) = 1 -> x + 1 = y - 1 -> x - y = -2 ... (1)
x/(y + 1) = 1/2 -> 2x = y + 1 -> 2x - y = 1 ... (2)
Subtract (1) from (2): 2x - y - x + y = 1 + 2
x = 3
From (1): 3 - y = -2 -> y = 5
The fraction is 3/5.
(ii) Let Nuri's present age = x, Sonu's present age = y.
5 years ago: x - 5 = 3(y - 5) -> x - 3y = -10 ... (1)
10 years later: x + 10 = 2(y + 10) -> x - 2y = 10 ... (2)
Subtract (1) from (2): -2y + 3y = 10 + 10
y = 20
From (2): x - 40 = 10 -> x = 50
Nuri's age = 50 years, Sonu's age = 20 years.
Q3Form the pair of linear equations in the following problem, and find their solution by the elimination method: The sum of a two-digit number and the number formed by interchanging its digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the digits in the first number. Find the first number.
Let tens digit = x, units digit = y. Number = 10x + y.\n\nReversed number = 10y + x.\nCondition 1: (10x + y) + (10y + x) = 110 -> 11x + 11y = 110 -> x + y = 10 ... (1)\nCondition 2: (10x + y) - 10 = 5(x + y) + 4\n10x + y - 10 = 5x + 5y + 4\n5x - 4y = 14 ... (2)\n\nMultiply (1) by 4: 4x + 4y = 40 ... (3)\nAdd (2) and (3): 9x = 54 -> x = 6\nFrom (1): y = 4\n\nThe number is 64.\n\nVerification: 64 + 46 = 110 and 64 - 10 = 54 = 5(10) + 4.
3.5Exercise 3.5 - Cross-Multiplication Method
Q1Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? (i) x - 3y - 3 = 0 and 3x - 9y - 2 = 0 (ii) 2x + y = 5 and 3x + 2y = 8
(i) x - 3y - 3 = 0 and 3x - 9y - 2 = 0
a1/a2 = 1/3, b1/b2 = -3/-9 = 1/3, c1/c2 = -3/-2 = 3/2
Since a1/a2 = b1/b2 != c1/c2 (1/3 = 1/3 != 3/2)
No solution (parallel lines, inconsistent).
(ii) 2x + y = 5 and 3x + 2y = 8
a1/a2 = 2/3, b1/b2 = 1/2
Since a1/a2 != b1/b2 (2/3 != 1/2)
Unique solution (intersecting lines, consistent).
Q2Solve the following pair of linear equations by cross-multiplication method: (i) x + 2y + 1 = 0 and 2x - 3y - 12 = 0
x + 2y + 1 = 0 ... (1)
2x - 3y - 12 = 0 ... (2)
Using cross-multiplication:
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
x/[2(-12) - (-3)(1)] = y/[1(2) - (-12)(1)] = 1/[1(-3) - 2(2)]
x/(-24 + 3) = y/(2 + 12) = 1/(-3 - 4)
x/(-21) = y/14 = 1/(-7)
x = -21/(-7) = 3
y = 14/(-7) = -2
x = 3, y = -2
Q3Form the pair of linear equations in the following problem and find their solution (if they exist) by any algebraic method: A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Let fixed charge = Rs x and cost of food per day = Rs y.
Student A (20 days): x + 20y = 1000 ... (1)
Student B (26 days): x + 26y = 1180 ... (2)
Subtracting (1) from (2):
6y = 180
y = 30
From (1): x + 600 = 1000
x = 400
Fixed charges = Rs 400, Cost of food per day = Rs 30.
Q4A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
Let the fraction be x/y.
Condition 1: (x - 1)/y = 1/3 -> 3(x - 1) = y -> 3x - y = 3 ... (1)
Condition 2: x/(y + 8) = 1/4 -> 4x = y + 8 -> 4x - y = 8 ... (2)
Subtracting (1) from (2):
4x - y - 3x + y = 8 - 3
x = 5
From (1): 15 - y = 3 -> y = 12
The fraction is 5/12.
3.6Exercise 3.6 - Equations Reducible to Linear Form
Q1Solve the following pairs of equations by reducing them to a pair of linear equations: (i) 1/(2x) + 1/(3y) = 2 and 1/(3x) + 1/(2y) = 13/6 (ii) 2/sqrt(x) + 3/sqrt(y) = 2 and 4/sqrt(x) - 9/sqrt(y) = -1
(i) Let 1/x = u and 1/y = v.
u/2 + v/3 = 2 -> 3u + 2v = 12 ... (1)
u/3 + v/2 = 13/6 -> 2u + 3v = 13 ... (2)
Multiply (1) by 3 and (2) by 2:
9u + 6v = 36 ... (3)
4u + 6v = 26 ... (4)
Subtract (4) from (3): 5u = 10 -> u = 2
From (1): 6 + 2v = 12 -> v = 3
So 1/x = 2 -> x = 1/2 and 1/y = 3 -> y = 1/3
x = 1/2, y = 1/3
(ii) Let 1/sqrt(x) = u and 1/sqrt(y) = v.
2u + 3v = 2 ... (1)
4u - 9v = -1 ... (2)
Multiply (1) by 3: 6u + 9v = 6 ... (3)
Add (2) and (3): 10u = 5 -> u = 1/2
From (1): 1 + 3v = 2 -> v = 1/3
1/sqrt(x) = 1/2 -> sqrt(x) = 2 -> x = 4
1/sqrt(y) = 1/3 -> sqrt(y) = 3 -> y = 9
Q2Formulate the following problem as a pair of equations, and hence find their solution: Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Let speed of rowing in still water = x km/h and speed of current = y km/h.
Downstream speed = x + y, Upstream speed = x - y.
Downstream: Distance/Speed = Time -> 20/(x + y) = 2 -> x + y = 10 ... (1)
Upstream: 4/(x - y) = 2 -> x - y = 2 ... (2)
Adding (1) and (2): 2x = 12 -> x = 6
From (1): y = 10 - 6 = 4
Speed in still water = 6 km/h, Speed of current = 4 km/h.
Q3Solve the following pair of equations: 6x + 3y = 6xy and 2x + 4y = 5xy.
Dividing both equations by xy (x, y not zero):\n6/y + 3/x = 6 and 2/y + 4/x = 5.\n\nLet 1/x = u and 1/y = v.\n3u + 6v = 6 -> u + 2v = 2 ... (1)\n4u + 2v = 5 ... (2)\n\nSubtract (1) from (2): 3u = 3 -> u = 1\nFrom (1): 1 + 2v = 2 -> v = 1/2\n\n1/x = 1 -> x = 1\n1/y = 1/2 -> y = 2\n\nx = 1, y = 2.
3.7Exercise 3.7 - Word Problems (Optional but Important)
Q1The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Let Ani's age = x years, Biju's age = y years.
Case 1: x - y = 3 ... (1)
Dharam's age = 2x, Cathy's age = y/2.
2x - y/2 = 30 -> 4x - y = 60 ... (2)
From (1): x = y + 3. Substituting in (2):
4(y + 3) - y = 60
4y + 12 - y = 60
3y = 48 -> y = 16
x = 19
Check Case 2: y - x = 3 -> y = x + 3
4x - (x + 3) = 60 -> 3x = 63 -> x = 21, y = 24
Case 1: Ani = 19 years, Biju = 16 years.
Case 2: Ani = 21 years, Biju = 24 years.
Q3Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Let number of correct answers = x and wrong answers = y.
Condition 1: 3x - y = 40 ... (1)
Condition 2: 4x - 2y = 50 -> 2x - y = 25 ... (2)
Subtract (2) from (1): x = 15
From (2): 30 - y = 25 -> y = 5
Total questions = x + y = 15 + 5 = 20 questions.
Verification: 3(15) - 5 = 40 ✓ and 4(15) - 2(5) = 50 ✓
Q5A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Let fixed charge (first 3 days) = Rs x and charge per extra day = Rs y.
Saritha (7 days = 3 fixed + 4 extra): x + 4y = 27 ... (1)
Susy (5 days = 3 fixed + 2 extra): x + 2y = 21 ... (2)
Subtracting (2) from (1): 2y = 6 -> y = 3
From (2): x + 6 = 21 -> x = 15
Fixed charge = Rs 15, Charge per extra day = Rs 3.
Q7The 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 Rs 105 and for a journey of 15 km, the charge paid is Rs 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?
Let fixed charge = Rs x and charge per km = Rs y.
10 km journey: x + 10y = 105 ... (1)
15 km journey: x + 15y = 155 ... (2)
Subtracting (1) from (2): 5y = 50 -> y = 10
From (1): x + 100 = 105 -> x = 5
Fixed charge = Rs 5, Charge per km = Rs 10.
For 25 km: Charge = 5 + 25(10) = 5 + 250 = Rs 255.
Q8A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Let speed = x km/h and time = y hours.
Distance = xy.
If 10 km/h faster, 2 hours less: (x + 10)(y - 2) = xy
xy - 2x + 10y - 20 = xy
-2x + 10y = 20 -> -x + 5y = 10 ... (1)
If 10 km/h slower, 3 hours more: (x - 10)(y + 3) = xy
xy + 3x - 10y - 30 = xy
3x - 10y = 30 ... (2)
From (1): x = 5y - 10. Substituting in (2):
3(5y - 10) - 10y = 30
15y - 30 - 10y = 30
5y = 60 -> y = 12
x = 60 - 10 = 50
Distance = xy = 50 x 12 = 600 km.
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Find the value of k for which the pair of equations kx + 3y = k - 3 and 12x + ky = k has no solution.
For no solution: a1/a2 = b1/b2 != c1/c2.
k/12 = 3/k -> k^2 = 36 -> k = 6 or k = -6.
Check c1/c2: For k = 6: (k-3)/k = 3/6 = 1/2 and a1/a2 = 6/12 = 1/2. Since a1/a2 = c1/c2, this gives infinite solutions, not no solution.
For k = -6: a1/a2 = -6/12 = -1/2, c1/c2 = (-6-3)/(-6) = -9/-6 = 3/2. Since -1/2 != 3/2, this satisfies no solution.
k = -6
Solve the pair of linear equations 2x + 3y = 11 and 2x - 4y = -24 by elimination method.
2x + 3y = 11 ... (1) and 2x - 4y = -24 ... (2). Subtracting (2) from (1): 3y - (-4y) = 11 - (-24), so 7y = 35, giving y = 5. Substituting in (1): 2x + 15 = 11, 2x = -4, x = -2. Solution: x = -2, y = 5. Verification: 2(-2) + 3(5) = -4 + 15 = 11 and 2(-2) - 4(5) = -4 - 20 = -24. Both equations are satisfied.
A father's age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.
Let father's age = x, sum of children's ages = y. Given: x = 3y ... (1). After 5 years: x + 5 = 2(y + 10) -> x + 5 = 2y + 20 -> x - 2y = 15 ... (2). Substituting (1) in (2): 3y - 2y = 15 -> y = 15. From (1): x = 45. Father's present age = 45 years.
Solve for x and y: x/a + y/b = 2 and ax - by = a^2 - b^2.
From equation 1: bx + ay = 2ab ... (1). Equation 2: ax - by = a^2 - b^2 ... (2). Multiply (1) by b and (2) by a: b^2x + aby = 2ab^2 ... (3) and a^2x - aby = a^3 - ab^2 ... (4). Adding (3) and (4): (a^2 + b^2)x = a^3 - ab^2 + 2ab^2 = a^3 + ab^2 = a(a^2 + b^2). So x = a. From (1): ab + ay = 2ab -> ay = ab -> y = b.
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 tens digit = x, units digit = y. Number = 10x + y. Reversed number = 10y + x. Condition 1: x + y = 9 ... (1). Condition 2: 9(10x + y) = 2(10y + x) -> 90x + 9y = 20y + 2x -> 88x - 11y = 0 -> 8x - y = 0 -> y = 8x ... (2). Substituting in (1): x + 8x = 9 -> 9x = 9 -> x = 1, y = 8. The number is 18. Verification: 9 x 18 = 162 and 2 x 81 = 162.
For what value of k, do the equations 3x - y + 8 = 0 and 6x - ky = -16 represent coincident lines?
For coincident lines: a1/a2 = b1/b2 = c1/c2. Writing in standard form: 3x - y + 8 = 0 and 6x - ky + 16 = 0. So 3/6 = (-1)/(-k) = 8/16. From 3/6 = 1/k: k = 2. Verification: 1/2 = 1/2 = 1/2. k = 2.
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Let speed in still water = x km/h, speed of stream = y km/h. Upstream speed = (x-y), downstream speed = (x+y). Let 1/(x-y) = u, 1/(x+y) = v. Equation 1: 30u + 44v = 10 ... (1). Equation 2: 40u + 55v = 13 ... (2). Multiply (1) by 4 and (2) by 3: 120u + 176v = 40, 120u + 165v = 39. Subtracting: 11v = 1, v = 1/11. From (1): 30u + 4 = 10, u = 1/5. So x - y = 5 and x + y = 11. Solving: x = 8, y = 3. Speed in still water = 8 km/h, Speed of stream = 3 km/h.
Find the values of a and b for which the following pair of linear equations has infinitely many solutions: 2x + 3y = 7 and (a - b)x + (a + b)y = 3a + b - 2.
For infinitely many solutions: a1/a2 = b1/b2 = c1/c2. So 2/(a-b) = 3/(a+b) = 7/(3a+b-2). From first two ratios: 2(a+b) = 3(a-b) -> 2a + 2b = 3a - 3b -> a = 5b ... (i). From first and third: 2(3a+b-2) = 7(a-b) -> 6a + 2b - 4 = 7a - 7b -> a = 9b - 4 ... (ii). From (i) and (ii): 5b = 9b - 4 -> 4b = 4 -> b = 1, a = 5. a = 5, b = 1.
Solve the following pair of equations: 1/(3x+y) + 1/(3x-y) = 3/4 and 1/(2(3x+y)) - 1/(2(3x-y)) = -1/8.
Let 1/(3x+y) = u and 1/(3x-y) = v. Then: u + v = 3/4 ... (1) and u/2 - v/2 = -1/8 -> u - v = -1/4 ... (2). Adding (1) and (2): 2u = 1/2 -> u = 1/4. From (1): v = 3/4 - 1/4 = 1/2. So 3x + y = 4 ... (3) and 3x - y = 2 ... (4). Adding: 6x = 6, x = 1. From (3): 3 + y = 4, y = 1. x = 1, y = 1.
The sum of two numbers is 8 and the sum of their reciprocals is 8/15. Find the numbers.
Let the numbers be x and y. x + y = 8 ... (1). 1/x + 1/y = 8/15 -> (x+y)/(xy) = 8/15 -> 8/(xy) = 8/15 -> xy = 15 ... (2). From (1): y = 8 - x. Substituting in (2): x(8-x) = 15 -> 8x - x^2 = 15 -> x^2 - 8x + 15 = 0 -> (x-3)(x-5) = 0. So x = 3 or x = 5. The two numbers are 3 and 5.
Find the value of p for which the pair of linear equations px + 3y = 7 and 2x + 6y = 14 has infinitely many solutions.
For infinitely many solutions: a1/a2 = b1/b2 = c1/c2. So p/2 = 3/6 = 7/14. Since 3/6 = 1/2 and 7/14 = 1/2, we need p/2 = 1/2. Therefore p = 1. Verification: 1/2 = 1/2 = 1/2, all ratios are equal confirming infinite solutions.
Solve for x and y: (x+1)/2 + (y-1)/3 = 8 and (x-1)/3 + (y+1)/2 = 9.
Simplifying equation 1: 3(x+1) + 2(y-1) = 48 -> 3x + 3 + 2y - 2 = 48 -> 3x + 2y = 47 ... (1). Simplifying equation 2: 2(x-1) + 3(y+1) = 54 -> 2x - 2 + 3y + 3 = 54 -> 2x + 3y = 53 ... (2). Multiply (1) by 3 and (2) by 2: 9x + 6y = 141 and 4x + 6y = 106. Subtracting: 5x = 35, x = 7. From (1): 21 + 2y = 47, y = 13. x = 7, y = 13.
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Let speed of car from A = x km/h and speed of car from B = y km/h (x > y). Same direction: Relative speed = (x - y). They meet when distance = 100 km. 5(x - y) = 100 -> x - y = 20 ... (1). Opposite direction: Relative speed = (x + y). 1(x + y) = 100 -> x + y = 100 ... (2). Adding (1) and (2): 2x = 120 -> x = 60. From (2): y = 40. Speed of car from A = 60 km/h, Speed of car from B = 40 km/h.
For which value(s) of k will the pair of equations kx + 3y = k - 3 and 12x + ky = k have no solution?
For no solution: a1/a2 = b1/b2 != c1/c2. k/12 = 3/k gives k^2 = 36, so k = 6 or k = -6. For k = 6: c1/c2 = 3/6 = 1/2 = a1/a2, so this gives infinite solutions. For k = -6: a1/a2 = -1/2, c1/c2 = (-9)/(-6) = 3/2, and -1/2 != 3/2. k = -6.
A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is subtracted from the denominator. Find the fraction.
Let fraction = x/y. Condition 1: (x-2)/y = 1/3 -> 3x - 6 = y -> 3x - y = 6 ... (1). Condition 2: x/(y-1) = 1/2 -> 2x = y - 1 -> 2x - y = -1 ... (2). Subtracting (2) from (1): x = 7. From (2): 14 - y = -1 -> y = 15. The fraction is 7/15. Verification: (7-2)/15 = 5/15 = 1/3 and 7/(15-1) = 7/14 = 1/2.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The pair of linear equations x + 2y = 5 and 3x + 6y = 15 has infinitely many solutions.
Reason (R): If a1/a2 = b1/b2 = c1/c2, the pair of linear equations is consistent and dependent.
Answer: (A)
Assertion: 1/3 = 2/6 = 5/15 (all equal to 1/3), so infinitely many solutions - TRUE. Reason: This is the correct condition for coincident lines (dependent system) - TRUE. Reason explains assertion.
AR2
Assertion (A): The pair of linear equations 2x + 3y = 9 and 4x + 6y = 18 represents coincident lines.
Reason (R): If two lines are coincident, the system has a unique solution.
Answer: (C)
Assertion is TRUE: 2/4 = 3/6 = 9/18 = 1/2, so lines are coincident. Reason is FALSE: coincident lines have infinitely many solutions, not a unique solution.
AR3
Assertion (A): The pair of equations x + y = 3 and 2x + 2y = 8 has no solution.
Reason (R): When a1/a2 = b1/b2 but not equal to c1/c2, the system is inconsistent (parallel lines).
Answer: (A)
Assertion: a1/a2 = 1/2, b1/b2 = 1/2, c1/c2 = 3/8. Since 1/2 = 1/2 != 3/8, lines are parallel - TRUE. Reason: This is the correct condition for no solution - TRUE. Reason explains the assertion.
AR4
Assertion (A): The graphical representation of the equations x + 2y = 4 and 2x + 4y = 12 gives two parallel lines.
Reason (R): The condition for parallel lines is a1/a2 = b1/b2 = c1/c2.
Answer: (C)
Assertion is TRUE: a1/a2 = 1/2, b1/b2 = 2/4 = 1/2, c1/c2 = 4/12 = 1/3. Since 1/2 = 1/2 != 1/3, lines are parallel. Reason is FALSE: a1/a2 = b1/b2 = c1/c2 is the condition for coincident lines, not parallel lines.
AR5
Assertion (A): If a pair of linear equations has a unique solution, then the lines representing them are intersecting.
Reason (R): For a unique solution, the condition is a1/a2 != b1/b2.
Answer: (A)
Both are TRUE. When a1/a2 != b1/b2, the lines have different slopes, so they must intersect at exactly one point giving a unique solution. The reason correctly explains the assertion.
AR6
Assertion (A): The pair of equations 3x + 4y = 6 and 6x + 8y = 18 is consistent.
Reason (R): The ratios a1/a2 = b1/b2 != c1/c2 indicate a consistent system.
Answer: (D)
Assertion is FALSE: 3/6 = 4/8 = 1/2 but 6/18 = 1/3. Since a1/a2 = b1/b2 != c1/c2, the system is inconsistent (parallel lines, no solution). Reason is also FALSE: this ratio condition indicates an inconsistent system, not a consistent one.
Case Study Questions
Ravi lives in a city where taxi fare consists of a fixed charge plus a per-km rate. He observed the following: For a 10 km ride, the total fare was Rs 75. For a 15 km ride, the total fare was Rs 110. Let the fixed charge be Rs x and the per-km charge be Rs y.
(i) Write the pair of linear equations representing the above situation. (1 mark)
For 10 km ride: x + 10y = 75 ... (1)
For 15 km ride: x + 15y = 110 ... (2)
The pair of linear equations is x + 10y = 75 and x + 15y = 110.
(ii) Find the fixed charge and the per-km charge. (2 marks)
Subtracting (1) from (2): 5y = 35 -> y = 7.
From (1): x + 70 = 75 -> x = 5.
Fixed charge = Rs 5, Per-km charge = Rs 7.
(iii)(a) What will be the fare for a 25 km ride? (1 mark)
Fare = x + 25y = 5 + 25(7) = 5 + 175 = Rs 180.
(iii)(b) If the fare paid is Rs 194, what is the distance travelled? (1 mark)
5 + d(7) = 194 -> 7d = 189 -> d = 27.
Distance travelled = 27 km.
In a family, the father's age is 3 years more than 3 times the son's age. After 3 years, the father's age will be 10 years more than twice the son's age. Let the father's present age be x years and the son's present age be y years.
(i) Write the pair of linear equations for the given situation. (1 mark)
Father's present age: x = 3y + 3 -> x - 3y = 3 ... (1)
After 3 years: (x + 3) = 2(y + 3) + 10 -> x + 3 = 2y + 16 -> x - 2y = 13 ... (2)
The equations are x - 3y = 3 and x - 2y = 13.
(ii) Find the present ages of father and son. (2 marks)
Subtracting (1) from (2): -2y + 3y = 13 - 3 -> y = 10.
From (1): x - 30 = 3 -> x = 33.
Father's present age = 33 years, Son's present age = 10 years.
(iii)(a) What is the ratio of father's age to son's age after 3 years? (1 mark)
After 3 years: Father = 36, Son = 13.
Ratio = 36:13.
Ratio = 36:13.
(iii)(b) After how many years will the father be twice as old as the son? (1 mark)
Let after t years: 33 + t = 2(10 + t)
33 + t = 20 + 2t
13 = t
After 13 years, the father will be twice as old as the son.
MCQ Practice
MCQ1. The pair of equations x + 2y - 5 = 0 and -4x - 8y + 20 = 0 has:
(A) Unique solution
(B) Exactly two solutions
(C) Infinitely many solutions
(D) No solution
Answer: (C) - a1/a2 = 1/(-4) = -1/4, b1/b2 = 2/(-8) = -1/4, c1/c2 = -5/20 = -1/4. All ratios equal, so infinitely many solutions (coincident lines).
MCQ2. If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is:
(A) -5/4
(B) 2/5
(C) 15/4
(D) 3/2
Answer: (C) - For parallel lines: a1/a2 = b1/b2 != c1/c2. 3/2 = 2k/5 -> k = 15/4. Check c1/c2: -2/1 = -2. Since 3/2 != -2, confirmed parallel.
MCQ3. The value of c for which the pair of equations cx - y = 2 and 6x - 2y = 3 will have infinitely many solutions is:
(A) 3
(B) -3
(C) -12
(D) No value
Answer: (D) - For infinite solutions: a1/a2 = b1/b2 = c1/c2. b1/b2 = -1/(-2) = 1/2 and c1/c2 = 2/3. Since 1/2 != 2/3, no value of c gives infinite solutions.
MCQ4. If a pair of linear equations is consistent, then the lines will be:
(A) Parallel
(B) Always coincident
(C) Intersecting or coincident
(D) Always intersecting
Answer: (C) - A consistent system has at least one solution. This happens when lines either intersect (unique solution) or coincide (infinitely many solutions).
MCQ5. The pair of equations y = 0 and y = -7 has:
(A) One solution
(B) Two solutions
(C) Infinitely many solutions
(D) No solution
Answer: (D) - y = 0 is the x-axis and y = -7 is a horizontal line parallel to it. Parallel lines have no point of intersection, hence no solution.
MCQ6. One equation of a pair of dependent linear equations is -5x + 7y = 2. The second equation can be:
(A) 10x + 14y + 4 = 0
(B) -10x - 14y + 4 = 0
(C) -10x + 14y + 4 = 0
(D) 10x - 14y = -4
Answer: (D) - Dependent means coincident: same line. Multiplying -5x + 7y = 2 by -2 gives 10x - 14y = -4, which is option (D).
MCQ7. A pair of linear equations which has a unique solution x = 2, y = -3 is:
(A) x + y = -1 and 2x - 3y = -5
(B) 2x + 5y = -11 and 4x + 10y = -22
(C) 2x - y = 1 and 3x + 2y = 0
(D) x - 4y - 14 = 0 and 5x - y - 13 = 0
Answer: (D) - Substitute x=2, y=-3 in each option. Option D: 2-4(-3)-14 = 2+12-14 = 0 and 5(2)-(-3)-13 = 10+3-13 = 0. Both satisfied. Also a1/a2 = 1/5 != b1/b2 = -4/-1 = 4, so unique solution.
MCQ8. If x = a, y = b is the solution of the equations x - y = 2 and x + y = 4, then the values of a and b are:
(A) 3 and 5
(B) 5 and 3
(C) 3 and 1
(D) -1 and -3
Answer: (C) - Adding: 2x = 6 -> x = 3. From x - y = 2: y = 1. So a = 3, b = 1.
MCQ9. The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages of the son and the father are:
(A) 4 and 24
(B) 5 and 30
(C) 6 and 36
(D) 3 and 18
Answer: (C) - Let son = y, father = x. x = 6y ... (1). x + 4 = 4(y + 4) -> x = 4y + 12 ... (2). From (1) and (2): 6y = 4y + 12 -> y = 6, x = 36.
MCQ10. If the system of equations 2x + 3y = 7 and 2ax + (a+b)y = 28 has infinitely many solutions, then:
(A) a = 2b
(B) b = 2a
(C) a + 2b = 0
(D) 2a + b = 0
Answer: (B) - For infinite solutions: 2/(2a) = 3/(a+b) = 7/28. From 7/28 = 1/4, so 2/(2a) = 1/4 -> a = 4. And 3/(a+b) = 1/4 -> a + b = 12 -> b = 8. Check: b = 2a -> 8 = 2(4). True.
MCQ11. Graphically, the pair of equations 6x - 3y + 10 = 0 and 2x - y + 9 = 0 represents two lines which are:
(A) Intersecting at exactly one point
(B) Intersecting at exactly two points
(C) Coincident
(D) Parallel
Answer: (D) - a1/a2 = 6/2 = 3, b1/b2 = -3/(-1) = 3, c1/c2 = 10/9. Since a1/a2 = b1/b2 != c1/c2 (3 = 3 != 10/9), the lines are parallel.
MCQ12. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. The fraction is:
(A) 4/8
(B) 5/7
(C) 6/6
(D) 3/9
Answer: (B) - Let fraction = x/y. x + y = 12 ... (1). x/(y+3) = 1/2 -> 2x = y + 3 -> 2x - y = 3 ... (2). Adding (1) and (2): 3x = 15 -> x = 5, y = 7. Fraction = 5/7.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Write the condition for which the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution.
The pair has a unique solution when a1/a2 != b1/b2. This means the lines intersect at exactly one point.
Is the pair of equations x + 2y - 4 = 0 and 2x + 4y - 12 = 0 consistent or inconsistent?
a1/a2 = 1/2, b1/b2 = 2/4 = 1/2, c1/c2 = -4/-12 = 1/3. Since a1/a2 = b1/b2 != c1/c2, the pair is **inconsistent** (no solution).
For what value of k does the pair kx - y = 2 and 6x - 2y = 3 have a unique solution?
For unique solution: a1/a2 != b1/b2 -> k/6 != (-1)/(-2) -> k/6 != 1/2 -> **k != 3**. So any value of k except 3.
If the pair of equations 2x + 3y = 5 and 4x + ky = 10 has infinitely many solutions, find k.
For infinite solutions: 2/4 = 3/k = 5/10. From 2/4 = 3/k: k = 6. Verify: 5/10 = 1/2 = 2/4 = 3/6. **k = 6.**
What does it mean graphically if a pair of linear equations has no solution?
If a pair of linear equations has no solution, the lines representing them are **parallel** - they never intersect.
2-3 Mark Questions
Solve by elimination method: 3x + 4y = 10 and 2x - 2y = 2.
Multiply equation 2 by 2: 4x - 4y = 4 ... (3). Add equation 1 and (3): 7x = 14, x = 2. From equation 2: 4 - 2y = 2, y = 1. x = 2, y = 1.
Find the values of a and b for which the pair 2x + 3y = 7 and (a-b)x + (a+b)y = 3a + b - 2 has infinitely many solutions.
2/(a-b) = 3/(a+b) = 7/(3a+b-2). From first two: 2(a+b) = 3(a-b) -> a = 5b. From first and third: 2(3a+b-2) = 7(a-b) -> a = 9b - 4. Solving: 5b = 9b - 4, b = 1, a = 5. a = 5, b = 1.
The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits differ by 2, find the number.
Let tens digit = x, units = y. (10x+y) + (10y+x) = 66 -> 11(x+y) = 66 -> x + y = 6 ... (1). x - y = 2 ... (2) (or y - x = 2). Adding: 2x = 8, x = 4, y = 2. Numbers: 42 or 24.
Solve by cross-multiplication: 2x + 3y = 46 and 3x + 5y = 74.
Using cross-multiplication: x/(3x(-74)-5x(-46)) = y/((-46)x3-(-74)x2) = 1/(2x5-3x3). x/(-222+230) = y/(-138+148) = 1/(10-9). x/8 = y/10 = 1/1. x = 8, y = 10.
3-4 Mark Questions
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Find the speed of the boat in still water and the speed of the stream.
Let 1/(x-y) = u, 1/(x+y) = v. 30u + 44v = 10, 40u + 55v = 13. Multiply eq1 by 4 and eq2 by 3: 120u + 176v = 40, 120u + 165v = 39. Subtracting: 11v = 1, v = 1/11. Then 30u + 4 = 10, u = 1/5. x - y = 5, x + y = 11. x = 8 km/h (still water), y = 3 km/h (stream).
The taxi charges in a city consist of a fixed charge together with the charge for distance covered. For 10 km, the charge is Rs 105 and for 15 km, the charge is Rs 155. Find the fixed charge and per-km charge. How much would a person have to pay for 25 km?
Let fixed = x, per-km = y. x + 10y = 105, x + 15y = 155. Subtracting: 5y = 50, y = 10. x = 5. For 25 km: 5 + 250 = Rs 255. Fixed = Rs 5, per km = Rs 10.
Solve: 2/x + 3/y = 13 and 5/x - 4/y = -2, where x != 0 and y != 0.
Let 1/x = u, 1/y = v. 2u + 3v = 13 ... (1), 5u - 4v = -2 ... (2). Multiply (1) by 4 and (2) by 3: 8u + 12v = 52, 15u - 12v = -6. Adding: 23u = 46, u = 2. From (1): 4 + 3v = 13, v = 3. x = 1/2, y = 1/3.
Key Concepts
Important Formulas
Study Plan - Pair of Linear Equations in Two Variables
Graphical Method and Ratio Conditions
Read NCERT theory on linear equations in two variables. Understand graphical representation. Solve Exercise 3.1 and Exercise 3.2. Memorise the three ratio conditions for unique/no/infinite solutions.
Substitution and Elimination Methods
Learn substitution method step-by-step. Solve Exercise 3.3 completely. Learn elimination method. Solve Exercise 3.4 completely. Compare both methods on the same problem.
Cross-Multiplication and Reducible Equations
Learn the cross-multiplication formula. Solve Exercise 3.5. Practice reducing equations to linear form (1/x, 1/y substitution). Solve Exercise 3.6.
Word Problems Mastery
Solve Exercise 3.7 (all 8 questions). Focus on age problems, speed-distance, fraction problems, and geometry problems. Practice converting word problems to equations.
Previous Year Questions and Practice
Solve all 15 PYQs from this page. Focus on finding k for no/infinite solutions and word problems. Practice MCQs and assertion-reason questions under timed conditions.
Full Revision and Mock Test
Attempt all important questions (mark-wise) as a timed test. Review common mistakes. Solve case study questions. Write down all formulas and ratio conditions from memory.
Common Mistakes to Avoid
- ✗Confusing the ratio conditions: a1/a2 = b1/b2 != c1/c2 means NO solution (parallel), NOT infinite solutions. Students often mix up parallel and coincident conditions.
- ✗Not converting equations to standard form (a1x + b1y + c1 = 0) before comparing ratios. The signs of c1 and c2 matter - always rewrite as ax + by + c = 0.
- ✗In word problems on fractions, forgetting that the denominator cannot be zero and setting up equations incorrectly by confusing numerator/denominator roles.
- ✗In speed-distance problems, forgetting to use 1/(x+y) and 1/(x-y) as substitution variables, and instead trying to solve non-linear equations directly.
- ✗In elimination method, forgetting to multiply BOTH sides of the equation (including the constant term) when making coefficients equal.
- ✗In cross-multiplication, making sign errors with the formula. Remember the pattern: b1c2-b2c1, c1a2-c2a1, a1b2-a2b1 (cyclic order b-c, c-a, a-b).
Tips for Board Exam
- ✓For MCQs on ratio conditions, always write both equations in standard form ax + by + c = 0 first. Then compare a1/a2, b1/b2, c1/c2 systematically - this takes under 30 seconds.
- ✓For word problems, spend 1-2 minutes clearly defining variables before writing equations. Label your variables (let x = father's age, y = son's age). This prevents confusion in longer problems.
- ✓When asked to 'find the value of k for no solution/infinite solutions', use the ratio conditions directly. For no solution: equate a1/a2 = b1/b2 to find k, then verify c1/c2 is different. For infinite solutions: equate all three ratios.
- ✓Elimination method is usually the fastest for board exams. Use it as your default unless the question specifically asks for substitution or cross-multiplication.
- ✓In speed-distance-time word problems, always set up the table: columns for Upstream/Downstream, rows for Speed/Distance/Time. This visual organization prevents errors and earns method marks.
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Start Pair of Linear Equations in Two Variables - Choose Your ModeAbout Pair of Linear Equations in Two Variables - Class 10 Mathematics
Chapter 3 Pair of Linear Equations in Two Variables is one of the most important chapters in CBSE Class 10 Mathematics. It belongs to the Algebra unit, which carries 20 marks in total, and this chapter alone accounts for 8-10 marks in the board exam. The chapter teaches students four different methods to solve a system of two linear equations and how to determine whether a system has a unique solution, no solution, or infinitely many solutions.
What Are Linear Equations in Two Variables?
A linear equation in two variables is an equation of the form ax + by + c = 0, where a, b, and c are real numbers and both a and b are not zero simultaneously. The graph of such an equation is always a straight line. When we have two such equations involving the same two variables (x and y), they form a pair of linear equations. The solution of this pair is a set of values of x and y that satisfies both equations simultaneously. Geometrically, this solution is the point where the two lines intersect.
Methods of Solving
The NCERT textbook presents four algebraic methods for solving a pair of linear equations:
1. Graphical Method (Exercise 3.1): Plot both equations as straight lines on a graph. The point of intersection gives the solution. This method is useful for understanding the concept but is less practical for exact answers, especially when solutions involve fractions or decimals.
2. Substitution Method (Exercise 3.3): Express one variable in terms of the other from one equation, then substitute this expression into the second equation. This reduces the system to a single equation in one variable. Best used when one equation already has a variable with coefficient 1 (like x = ... or y = ...).
3. Elimination Method (Exercise 3.4): Multiply the equations by suitable numbers to make the coefficients of one variable equal (or equal in magnitude but opposite in sign). Then add or subtract the equations to eliminate that variable. This is the most commonly used method in board exams because it works efficiently for all types of equations.
4. Cross-Multiplication Method (Exercise 3.5): For equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, use the formula: x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1). This method is direct but requires careful handling of signs.
Graphical Interpretation
The behaviour of a pair of linear equations can be understood graphically:
- Intersecting lines: The system has exactly one solution (the point of intersection). The system is called consistent.
- Parallel lines: The system has no solution. The lines never meet. The system is called inconsistent.
- Coincident lines: The system has infinitely many solutions. Both equations represent the same line. The system is called consistent and dependent.
Conditions for Consistency
The most important concept in this chapter is the ratio test for determining the nature of solutions without actually solving the equations. For two equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0:
- Unique solution (intersecting lines): a1/a2 is not equal to b1/b2. The lines have different slopes, so they must intersect.
- No solution (parallel lines): a1/a2 = b1/b2 but not equal to c1/c2. The lines have the same slope but different y-intercepts.
- Infinitely many solutions (coincident lines): a1/a2 = b1/b2 = c1/c2. One equation is a scalar multiple of the other.
This ratio test is the most frequently asked concept in MCQs and 1-2 mark questions in the board exam. Students should be able to apply it quickly and accurately.
Cross-Multiplication Method
The cross-multiplication method provides a direct formula for finding the solution. Given a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, the solution is:
x = (b1c2 - b2c1)/(a1b2 - a2b1) and y = (c1a2 - c2a1)/(a1b2 - a2b1)
This works when a1b2 - a2b1 is not equal to 0 (i.e., the system has a unique solution). The method is efficient but students often make sign errors. A useful memory trick is the cyclic pattern: b-c, c-a, a-b.
Word Problems Strategy
Word problems from this chapter are the most commonly asked 3-5 mark questions in CBSE board exams. The key categories are:
Age Problems: Define present ages as variables. Use conditions like 'x years ago' (subtract x from ages) and 'x years from now' (add x to ages). Always set up two equations from two conditions.
Speed-Distance-Time Problems: Use Distance = Speed x Time. For upstream/downstream problems, define speed in still water as u and speed of current as v. Downstream speed = u + v, Upstream speed = u - v. Often requires substitution like 1/(u+v) = p and 1/(u-v) = q to convert to linear equations.
Fraction Problems: Let numerator = x, denominator = y. Carefully translate conditions like 'fraction becomes 1/3 when 1 is subtracted from numerator' into equations: (x-1)/y = 1/3.
Number Problems: For two-digit numbers, use 10x + y (where x is tens digit, y is units digit). The reversed number is 10y + x. Sum of digits = x + y.
Geometry Problems: Involve perimeter, area, or angle conditions. For rectangles: 2(l + b) = perimeter. For triangles: sum of angles = 180 degrees.
How This Chapter Is Tested in Board Exams
In the CBSE Class 10 board exam, this chapter typically contributes:
- 1-2 MCQs (1-2 marks): Usually testing ratio conditions (unique/no/infinite solutions), finding k for a given condition, or identifying the type of system.
- 1 Short Answer (2-3 marks): Solving a pair of equations using a specified method, or finding values of parameters for no solution/infinite solutions.
- 1 Long Answer/Word Problem (4-5 marks): A real-life word problem involving age, speed-distance, fractions, or cost. This is the highest-scoring question from this chapter.
- Case Study (4 marks): Since 2023, CBSE has included case study questions that present a real-world scenario and ask students to form and solve linear equations.
Connection to Other Chapters
This chapter connects closely with Chapter 4 (Quadratic Equations) - some word problems can lead to quadratic equations after substitution. It also links to Chapter 7 (Coordinate Geometry) since every linear equation represents a line on the coordinate plane, and Chapter 8 (Trigonometry) in problems involving angles of triangles. Understanding linear equations is foundational for higher mathematics, including matrices, linear algebra, and optimization problems that students will encounter in Classes 11 and 12.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 3 for CBSE 2026-27?
Chapter 3 Pair of Linear Equations in Two Variables has 7 exercises in the NCERT textbook for CBSE 2026-27: Exercise 3.1 (graphical method, 3 questions), Exercise 3.2 (ratio comparison, 7 questions), Exercise 3.3 (substitution, 3 questions), Exercise 3.4 (elimination, 2 questions), Exercise 3.5 (cross-multiplication, 4 questions), Exercise 3.6 (reducible equations, 2 questions), and Exercise 3.7 (word problems, 8 questions). All exercises are retained in the current syllabus.
What are the conditions for consistent and inconsistent linear equations?
For a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: (1) If a1/a2 is not equal to b1/b2, the system is consistent with a unique solution (intersecting lines). (2) If a1/a2 = b1/b2 = c1/c2, the system is consistent with infinitely many solutions (coincident lines). (3) If a1/a2 = b1/b2 but not equal to c1/c2, the system is inconsistent with no solution (parallel lines).
How many marks does Chapter 3 Pair of Linear Equations carry in CBSE Class 10 board exam?
Chapter 3 is part of the Algebra unit which carries 20 marks total. This chapter alone carries 8-10 marks, making it the highest-scoring topic in Algebra. Typically: 1-2 MCQs (1-2 marks) + 1 short answer on solving equations or finding k values (2-3 marks) + 1 word problem (4-5 marks).
Which method is best for solving linear equations in CBSE board exam?
The elimination method is generally the fastest and most reliable for CBSE board exams. It works well for all types of equations and is less error-prone than cross-multiplication. However, use substitution when one variable already has coefficient 1 (like x = 3y + 5), and use cross-multiplication when specifically asked. For MCQs, the ratio comparison method (a1/a2, b1/b2, c1/c2) is the quickest approach.
How to find the value of k for which a pair of linear equations has no solution?
For no solution (parallel lines), the condition is a1/a2 = b1/b2 but not equal to c1/c2. Step 1: Set a1/a2 = b1/b2 and solve for k. Step 2: Check that c1/c2 is not equal to a1/a2 for the obtained value of k. If it is equal, that value gives infinite solutions, not no solution. Example: For kx + 3y = k-3 and 12x + ky = k, set k/12 = 3/k to get k = 6 or -6, then verify which value satisfies the no-solution condition.
What types of word problems come from Chapter 3 in CBSE board exam?
Five main types of word problems appear: (1) Age problems - relating ages of family members at different times. (2) Speed-distance-time problems - boats upstream/downstream, cars travelling in same/opposite directions. (3) Fraction problems - finding numerator and denominator given conditions. (4) Number problems - two-digit numbers, sum of digits, reversing digits. (5) Cost/charges problems - fixed charge plus variable rate (taxi fare, hostel charges).
Is Chapter 3 Pair of Linear Equations deleted from CBSE 2026-27 syllabus?
No, Chapter 3 is NOT deleted from the CBSE 2026-27 syllabus. All 7 exercises are retained. Exercise 3.7 (word problems) is marked as optional in the textbook but word problems from this chapter are regularly asked in board exams. Students should prepare all exercises thoroughly as this chapter carries 8-10 marks.
What is the cross-multiplication method for solving linear equations?
For equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, the cross-multiplication formula gives: x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1). This works when a1b2 - a2b1 is not zero (unique solution exists). The trick to remember the pattern is the cyclic order: b-c, c-a, a-b. Each fraction's denominator involves cross-multiplication of coefficients in a specific cyclic pattern.