Pair of Linear Equations in Two Variables Class 10 NCERT Solutions Exercise 3.2 PDF

Pair of Linear Equations in Two Variables Class 10 NCERT Solutions Exercise 3.2: Exercise 3.2 in Class 10 Chapter 3 is about solving pairs of linear equations using the substitution method, not elimination.

It teaches students how to solve one equation for a variable and substitute it into the other to find both values. Practising pair of linear equations in two variables exercise 3.2 and pair of linear equations in two variables 3.2 class 10 helps strengthen understanding and accuracy.

Pair of Linear Equations in Two Variables Exercise 3.2 Questions and Solutions 

Below is the NCERT solutions Class 10 Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3-

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.

Pair of linear equations in Two Variables Class 10 NCERT Solutions Exercise 3.2 PDF

Chapter 3 of Class 10 Maths, Pair of Linear Equations in Two Variables, is an important part of the CBSE Class 10 Maths syllabus. Exercise 3.3 focuses on solving linear equations using the elimination method, where one variable is systematically removed to find the other.

These step-by-step solutions help students practice effectively and strengthen problem-solving skills. For a complete understanding, you can also refer to class 10 maths pair of linear equations in two variables 3.2 pdf. The link is given below to access Exercise 3.2 of pair of linear equations in two variables of class 10 PDF

Pair of Linear Equations in Two Variables Exercise 3.2 PDF

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