Mathematics is a Crucial Subject for 10 students, especially those aiming to pursue science or overall academic performance. Chapter 3: Linear Equations in Two Variables in one Among the key chapters in the NCERT syllabus and stands out as essential. This article covers NCERT Solutions for Maths Chapter 3 Exercise 3.4, with detailed explanations and step-by-step solutions.
For students looking to download NCERT Solutions Class 10 Maths - Chapter 3 Exercise 3.4, they can download the PDF from any reputed study organisation to access step-by-step solutions, explanations of key concepts, solved examples, and practice questions. It will help them hone their skills and prepare themselves well for competitive as well as board examinations.
Before jumping into exercise 3.4, let us understand the basics of linear equations in two variables. Any given linear equation in two variables is of the form ax + by + c = 0, where a, b, and c are real numbers while x, y are variables. The graph of every linear equation in two variables always represents a straight line in the Cartesian plane. The methods to solve these equations include the substitution method, elimination method, and graphical method.
NCERT Solutions Class-10 Maths Chapter 3 Exercise 3.4 explains the elimination method of solving the pair of linear equations in two variables with the help of several examples. In some questions, using the elimination technique is more helpful than the substitution method. Therefore, students are encouraged to use this method unless stated otherwise in the question.
1. Solve the following pair of 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
Sol. (i) Solution by elimination method:
x + y = 5 ...(1)
2x - 3y = 4 ...(2)
Multiplying equation (1) by 3 and equation (2) by 1, and then adding them, we get:
3(x + y) + 1(2x - 3y) = 3 × 5 + 1 × 4
3x + 3y + 2x - 3y = 15 + 4
5x = 19
x = 19/5
Substituting x = 19/5 in equation (1), we get:
19/5 + y = 5
y = 5 - 19/5
y = 6/5
Hence, x = 19/5, y = 6/5.
(i) Solution by substitution method:
x + y = 5 ...(1)
2x - 3y = 4 ...(2)
From equation (1), we get:
y = 5 - x ...(3)
Substituting y from equation (3) in equation (2), we get:
2x - 3(5 - x) = 4
2x - 15 + 3x = 4
5x = 19
x = 19/5
Substituting x = 19/5 in equation (3), we get:
y = 5 - 19/5
y = 6/5
Hence, x = 19/5, y = 6/5.
(ii) Solution by elimination method:
3x + 4y = 10 ...(1)
2x - 2y = 2 ...(2)
Multiplying equation (2) by 2, we get:
4x - 4y = 4 ...(3)
Adding equation (1) and (3), we get:
7x = 14
x = 2
Substituting x = 2 in equation (1), we get:
3 × 2 + 4y = 10
6 + 4y = 10
4y = 4
y = 1
Hence, x = 2, y = 1.
(ii) Solution by substitution method:
3x + 4y = 10 ...(1)
2x - 2y = 2 ...(2)
From equation (2), we get:
y = (2x - 2) / 2 = x - 1 ...(3)
Substituting y = x - 1 in equation (1), we get:
3x + 4(x - 1) = 10
3x + 4x - 4 = 10
7x = 14
x = 2
Substituting x = 2 in equation (3), we get:
y = 2 - 1 = 1
Hence, x = 2, y = 1.
(iii) Solution by elimination method:
3x - 5y = 4 ...(1)
9x = 2y + 7 ...(2)
Multiplying equation (1) by 3, we get:
9x - 15y = 12 ...(3)
Subtracting equation (3) from equation (2), we get:
9x - 9x + 15y = 2y + 7 - 12
13y = -5
y = -5/13
Substituting y = -5/13 in equation (1), we get:
3x = 5 × (-5/13) + 4
3x = -25/13 + 4
3x = (-25 + 52) / 13
3x = 27/13
x = 9/13
Hence, y = -5/13, x = 9/13.
(iii) Solution by substitution method:
3x - 5y = 4 ...(1)
9x = 2y + 7 ...(2)
From equation (1), we get:
x = (4 + 5y) / 3 ...(3)
Substituting x = (4 + 5y) / 3 in equation (2), we get:
9 × (4 + 5y) / 3 = 2y + 7
12 + 15y = 2y + 7
13y = -5
y = -5/13
Substituting y = -5/13 in equation (3), we get:
x = (4 + 5 × (-5/13)) / 3
x = (4 - 25/13) / 3
x = (52 - 25) / 39
x = 27/39 = 9/13
Hence, y = -5/13, x = 9/13.
(iv) Solution by elimination method:
x/2 + 2y/3 = -1 ...(1)
x - y/3 = 3 ...(2)
Multiplying equation (2) by 2, we get:
2x - 2y/3 = 6 ...(3)
Adding equation (1) and (3), we get:
2x + x/2 = -1 + 6
5x/2 = 5
x = 2
Substituting x = 2 in equation (2), we get:
2 - y/3 = 3
-1 = y/3
y = -3
Hence, x = 2, y = -3.
(iv) Solution by substitution method:
x/2 + 2y/3 = -1 ...(1)
x - y/3 = 3 ...(2)
From equation (2), we get:
x = 3 + y/3 ...(3)
Substituting x from equation (3) in equation (1), we get:
(3 + y/3) / 2 + 2y/3 = -1
3/2 + y/6 + 2y/3 = -1
(y + 4y) / 6 = -1 - 3/2
5y / 6 = -5/2
y = -3
Substituting y = -3 in equation (3), we get:
x = 3 + (-3) / 3
x = 3 - 1
x = 2
Hence, 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.
Solutions:
(i) Fraction Problem
Let the fraction be x/y, where y ≠ 0.
Given:
(x+1)/(y-1) = 1 ...(1)
x/(y+1) = 1/2 ...(2)
From (1), x + 1 = y - 1, which simplifies to x - y = -2.
From (2), 2x = y + 1, which simplifies to 2x - y = 1.
Multiplying equation (1) by 2 gives 2x - 2y = -4.
Subtracting the equation 2x - y = 1 from 2x - 2y = -4, we get:
-y = -5, so y = 5.
Substituting y = 5 into 2x - y = 1, we get:
2x - 5 = 1, so 2x = 6, and x = 3.
Therefore, the fraction is 3/5.
(ii) Age Problem
Let Nuri's present age be x years and Sonu's present age be y years.
Five years ago:
x - 5 = 3(y - 5)
x - 5 = 3y - 15
x - 3y = -10 ...(1)
Ten years later:
x + 10 = 2(y + 10)
x + 10 = 2y + 20
x - 2y = 10 ...(2)
Subtracting equation (2) from equation (1):
-y = -20, so y = 20.
Substituting y = 20 into equation (2):
x - 2(20) = 10
x - 40 = 10
x = 50.
Therefore, Nuri's present age is 50 years and Sonu's present age is 20 years.
(iii) Two-Digit Number Problem
Let the unit digit be x and the ten's digit be y.
The original number is 10y + x.
Given:
x + y = 9 ...(1)
9(10y + x) = 2(10x + y)
90y + 9x = 20x + 2y
88y = 11x
x = 8y ...(2)
Substituting (2) into (1):
8y + y = 9
9y = 9
y = 1
Substituting y = 1 into x = 8y, we get x = 8.
Therefore, the number is 10(1) + 8 = 18.
(iv) Bank Notes Problem
Let the number of ₹ 50 notes be x and the number of ₹ 100 notes be y.
Given:
x + y = 25 ...(1)
50x + 100y = 2000, which simplifies to x + 2y = 40 ...(2)
Subtracting equation (1) from equation (2):
y = 15.
Substituting y = 15 into equation (1):
x + 15 = 25
x = 10.
Therefore, Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.
(v) Lending Library Problem
Let the fixed charge be ₹ x and the charge for each extra day be ₹ y.
Given:
x + 4y = 27 ...(1) (Saritha's case: 7 days = 3 fixed + 4 extra)
x + 2y = 21 ...(2) (Susy's case: 5 days = 3 fixed + 2 extra)
Subtracting equation (2) from equation (1):
2y = 6
y = 3.
Substituting y = 3 into equation (1):
x + 4(3) = 27
x + 12 = 27
x = 15.
Therefore, the fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.
(Session 2025 - 26)