Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2 NCERT Solutions

Exercise 3.2 of Pair of Linear Equations in Two Variables Class 10 Maths focuses on solving pairs of linear equations using the substitution method. This approach involves rewriting one equation to express a variable in terms of another and then substituting it into the second equation to simplify the problem, as per the CBSE 10th Maths syllabus. 

These NCERT solutions are explained in a concise manner to help you follow each step logically. Practising this method improves algebraic manipulation skills and builds confidence in solving equations accurately, which is important for Class 10 exams.

NCERT Solutions for Class 10 Maths Chapter 3 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)

Solutions:

(i) Given, x + y = 14 and x – y = 4 are the two equations. From the 1 st equation, we get x = 14 – y.

Now, substitute the value of x to the second equation to get, (14 – y) – y = 4 14 – 2y = 4 2y = 10 Or y = 5

By the value of y, we can now find the exact value of x. ∵ x = 14 – y 

∴ x = 14 – 5 Or x = 9 

Hence, x = 9 and y = 5

(ii) Given, s – t = 3 and (s/3) + (t/2) = 6 are the two equations.

From the 1 st equation, we get, s = 3 + t ________________(1)

Now, substitute the value of s to the second equation to get, 

(3+t)/3 + (t/2) = 6
⇒ (2(3+t) + 3t )/6 = 6
⇒ (6+2t+3t)/6 = 6
⇒ (6+5t) = 36
⇒5t = 30
⇒t = 6

Now, substitute the value of t to the equation (1) s = 3 + 6 = 9.

Therefore, s = 9 and t = 6

(iii) Given, 3x – y = 3 and 9x – 3y = 9 are the two equations.

From the 1 st equation, we get, x = (3+y)/3

Now, substitute the value of x to the second equation to get, 

9(3+y)/3 – 3y = 9
⇒9 +3y -3y = 9
⇒ 9 = 9

Therefore, y has infinite values, and since x = (3+y) /3, x also has infinite values.

(iv) Given, 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 are the two equations. From the 1 st equation,

we get, x = (1.3- 0.3y)/0.2 _________________(1)

Now, substitute the value of x to the second equation to get, 

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3
⇒2(1.3 – 0.3y) + 0.5y = 2.3
⇒ 2.6 – 0.6y + 0.5y = 2.3
⇒ 2.6 – 0.1 y = 2.3
⇒ 0.1 y = 0.3
⇒ y = 3

Now, substitute the value of y in equation (1), and we get x = (1.3-0.3(3))/0.2
= (1.3-0.9)/0.2
= 0.4/0.2 = 2.

Therefore, x = 2 and y = 3

(v) Given, √2 x + √3 y = 0 and √3 x – √8 y = 0 are the two equations.

From the 1 st equation, we get, x = – (√3/√2)y __________________(1)

Putting the value of x in the given second equation to get, 

√3(-√3/√2)y – √8y = 0
⇒ (-3/√2)y- √8 y = 0
⇒ y = 0

Now, substitute the value of y in equation (1), and we get x = 0. Therefore, x = 0 and y = 0

(vi) Given, (3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations. From 1 st equation, we get,
(3/2)x = -2 + (5y/3)
⇒ x = 2(-6+5y)/9
= (-12+10y)/9 ………………………(1)

Putting the value of x in the second equation, we get, ((-12+10y)/9)/3 + y/2 = 13/6 ⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6

Now, substitute the value of y in equation (1), and we get,
(3x/2) – 5(3)/3 = -2
⇒ (3x/2) – 5 = -2
⇒ x = 2.
Therefore, x = 2 and y = 3 

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

Solution:

2x + 3y = 11…………………………..(I)

2x – 4y = -24………………………… (II)

From equation (II), we get x = (11-3y)/2 ………………….(III) 

Substituting the value of x to equation (II), we get
2(11-3y)/2 – 4y
= 24 11 – 3y – 4y
= -24 -7y = -35 y = 5…………………………………….. (IV)

Putting the value of y in equation (III), we get x
= (11-3×5)/2
= -4/2
= -2. 

Hence, x = -2, y = 5. Also, y = mx + 3 5 = -2m +3 -2m = 2 m = -1

Therefore, the value of m is -1. 

3. Form the pair of linear equations for the following problems and find their solution by the substitution method.

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

Solution:

Let the two numbers be x and y, respectively, such that y > x.

According to the question, y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of (1) to (2), we get 3x – x = 26 x = 13 ……………. (3)

Substituting (3) in (1), we get y = 39

Hence, the numbers are 13 and 39. 

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

Solution:

Let the larger angle be x o and the smaller angle be y o. We know that the sum of two supplementary pairs of angles is always 180 o.

According to the question, x + y = 180 o ……………. (1)

x – y = 18 o ……………..(2) From (1), we get x = 180 o – y …………. (3)

Substituting (3) in (2), we get 180 o – y – y =18 o 162 o = 2y y = 81 o ………….. (4)

Using the value of y in (3), we get x = 180 o – 81 o = 99 o

Hence, the angles are 99 o and 81 o.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.

Solution:

Let the cost of a bat be x and the cost of a ball be y. According to the question, 7x + 6y = 3800 ………………. (I)

3x + 5y = 1750 ………………. (II)

From (I), we get y = (3800-7x)/6………………..(III)

Substituting (III) to (II), we get,
3x+5(3800-7x)/6 =1750
⇒3x+ 9500/3 – 35x/6 = 1750
⇒3x- 35x/6 = 1750 – 9500/3
⇒(18x-35x)/6 = (5250 – 9500)/3
⇒-17x/6 = -4250/3
⇒-17x = -8500 x = 500 ……………………….. (IV)

Substituting the value of x to (III), we get y = (3800-7 ×500)/6 = 300/6 = 50

Hence, the cost of a bat is Rs 500, and the cost of a ball is Rs 50. 

(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 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?

Solution:

Let the fixed charge be Rs x and the per km charge be Rs y.

According to the question, x + 10y = 105 …………….. (1)

 x + 15y = 155 …………….. (2) From (1), we get x = 105 – 10y ………………. (3)

Substituting the value of x to (2), we get
105 – 10y + 15y
= 155 5y
= 50 y = 10 …………….. (4)

Putting the value of y in (3), we get x = 105 – 10 × 10 = 5

Hence, the fixed charge is Rs 5 and the per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255 

(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.

Solution:

Let the fraction be x/y.

According to the question,
(x+2) /(y+2)
= 9/11 11x + 22
= 9y + 18 11x – 9y
= -4 …………….. (1)

(x+3) /(y+3)
= 5/6 6x + 18
= 5y +15 6x – 5y = -3 ………………. (2)

From (1), we get x = (-4+9y)/11 …………….. (3)

Substituting the value of x to (2), we get
6(-4+9y)/11 -5y
= -3 -24 + 54y – 55y
= -33 -y
= -9 y = 9 ………………… (4)

Substituting the value of y to (3), we get x = (-4+9×9 )/11 = 7

Hence, the fraction is 7/9. 

(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?

Solutions:

Let the ages of Jacob and his son be x and y, respectively.

According to the question, (x + 5) = 3(y + 5) x – 3y = 10 …………………………………….. (1)

(x – 5) = 7(y – 5) x – 7y = -30 ………………………………………. (2)

From (1), we get x = 3y + 10 ……………………. (3)

Substituting the value of x to (2), we get
3y + 10 – 7y
= -30 -4y
= -40 y = 10 ………………… (4)

Substituting the value of y to (3), we get x = 3 x 10 + 10 = 40

Hence, the present age of Jacob and his son is 40 years and 10 years, respectively.

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

• Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

• Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

• Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

• Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

• Solve Previous Year Papers:

Practising CBSE Class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.

• Attempt Sample Papers:

Solving PW Class 10 Maths sample papers improves time management and gives you exam-like practice.

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