Pair of Linear Equations in Two Variables Exercise 3.3, Class 10 Maths Chapter 3 PDF

Pair of Linear Equations in Two Variables Exercise 3.3: Exercise 3.3 of Class 10 Maths Chapter 3 focuses on solving pairs of linear equations in two variables using the elimination method, a key topic in the CBSE syllabus. This exercise provides clear, step-by-step explanations to help students systematically eliminate one variable and find the other.

 Practising these problems enhances problem-solving skills, builds confidence, and ensures accuracy. The pair of linear equations in two variables, Class 10 NCERT solutions, exercise 3.3, is perfect for revision and exam preparation.

Class 10 Chapter 3 Exercise 3.3 (Substitution Method in Pair of Linear Equations)

Exercise 3.3 of Class 10 Maths Chapter 3 focuses on solving a pair of linear equations in two variables using the Elimination Method. In this method, we eliminate one variable by adding or subtracting the two equations after making the coefficients of one variable equal (or opposite). Once one variable is eliminated, the remaining variable can be solved easily. After that, we substitute the value back to find the second variable.

Pair of Linear Equations in Two Variables class 10 NCERT Solutions Exercise 3.3

Below is the NCERT solutions Class 10 Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3, helping students practice the questions step by step. Here you can check the solutions of 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

Solutions:

(i) x + y = 5 and 2x – 3y = 4

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get 2x + 2y = 10 ……………………………(iii)

When equation (ii) is subtracted from (iii), we get 5y = 6 y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get x=5−6/5 = 19/5

∴ x = 19/5 , y = 6/5

By the method of substitution.

From equation (i), we get x = 5 – y………………………………….. (v)

When the value is put in equation (ii), we get 2(5 – y) – 3y = 4 -5y = -6 y = 6/5

When the values are substituted in equation (v), we get x =5− 6/5 = 19/5

∴ x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

By the method of elimination.

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When equations (i) and (ii) are multiplied by 2, we get 4x – 4y = 4 ………………………..(iii)

When equations (i) and (iii) are added, we get 7x = 14 x = 2 ……………………………….(iv)

Substituting equation (iv) in (i), we get 6 + 4y = 10 4y = 4 y = 1

Hence, x = 2 and y = 1

By the method of substitution

From equation (ii), we get x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i), we get 3(1 + y) + 4y = 10 7y = 7 y = 1.

When y = 1 is substituted in equation (v), we get A = 1 + 1 = 2

Therefore, A = 2 and B = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By the method of elimination.

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7 9x – 2y – 7 = 0 ………………………………… (ii)

When the equation (i) is multiplied by 3, we get 9x – 15y – 12 = 0 ……………………………… (iii)

When equation (iii) is subtracted from equation (ii), we get 13y = -5 y = -5/13 …………………………………………. (iv)

When equation (iv) is substituted in equation (i), we get 3x +25/13 −4=0 3x = 27/13 x =9/13

∴ x = 9/13 and y = -5/13

By the method of substitution.

From equation (i), we get x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii), we get 9(5y+4)/3 −2y-7=0 13y = -5 y = -5/13.

Substituting this value in equation (v), we get x = (5(-5/13)+4)/3 x = 9/13

∴ x = 9/13 , y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

By the method of elimination.

3x + 4y = -6 …………………………. (i)

x-y/3 = 3 3x – y = 9 ……………………………. (ii)

When equation (ii) is subtracted from equation (i), we get 5y = -15y = -3 ………………………………….(iii)

When equation (iii) is substituted in (i), we get 3x – 12 = -6 3x = 6 x = 2

Hence, x = 2 , y = -3

By the method of substitution.

From equation (ii), we get x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i), we get 3(y+9)/3 +4y =−6 5y = -15 y = -3.

When y = -3 is substituted in equation (v), we get x = (-3+9)/3 = 2

Therefore, x = 2 and 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 if we only add 1 to the denominator. What is the fraction?

Solution:

Let the fraction be a/b According to the given information, (a+1)/(b-1) = 1 => a – b = -2 ………………………………..(i)

a/(b+1) = 1/2 => 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii), we get a = 3 ………………………………………………….. (iii)

When a = 3 is substituted in equation (i), we get 3 – b = -2 -b = -5 b = 5

Hence, the fraction is 3/5.

(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?

Solution:

Let us assume the present age of Nuri is x. And the present age of Sonu is y.

According to the given condition, we can write as x – 5 = 3(y – 5) x – 3y = -10…………………………………..(1)

Now, x + 10 = 2(y +10) x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2 to get y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get x – 3.20 = -10 x – 60 = -10 x = 50.

Therefore, Age of Nuri is 50 years, and the age of Sonu is 20 years

(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.

Solution:

Let the unit digit and tens digit of a number be x and y, respectively. Then, number (n) = 10B + A N after reversing the order of the digits = 10A + B

According to the given information, A + B = 9…………………….(i)

9(10B + A) = 2(10A + B) 88 B – 11 A = 0 -A + 8B = 0 ………………………………………………………….. (ii)

Adding the equations (i) and (ii), we get 9B = 9 B = 1……………………………………………………………………….(3)

Substituting this value of B in equation (i), we get A= 8

Hence, the number (N) is 10B + A = 10 x 1 +8 = 18

(iv) Meena went to a bank to withdraw Rs.2,000. She asked the cashier to give her Rs.50, and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.

Solution:

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B.

According to the given information, A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied by (ii), we get 50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting equation (iii) from equation (ii), we get 50B = 750 B = 15. Substituting in equation (i), we get A = 10

Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.

(v) A 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 a book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs.B.

According to the information given, A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i), we get 2B = 6 B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i), we get A + 12 = 27 A = 15

Hence, the fixed charge is Rs.15.

And the charge per day is Rs.3.

Exercise 3.3 of Pair of Linear Equations in Two Variables of Class 10 PDF

Exercise 3.3 of Class 10 Maths Chapter 3, Pair of Linear Equations in Two Variables Exercise 3.3, focuses on solving systems of equations using the elimination method.

These step-by-step solutions from pair of linear equations in two variables 3.3 class 10 make it easier to understand, practice, and build confidence for exams. A detailed PDF is provided below for easy learning and revision.

NCERT solutions Class 10 Maths PDF

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